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Solving ICSE Class 10 Chemistry Previous Year Question Papers ICSE Class 10 Chemistry Question Paper 2018  is the best way to boost your preparation for the board exams.

ICSE Class 10 Chemistry Question Paper 2018 Solved

Time: 2 hours
Maximum Marks: 80

General Instructions:

• Answers to this Paper must be written on the paper provided separately.
• You will not be allowed to write during the first 15 minutes.
• This time is to be spent in reading the Question Paper.
• The time given at the head of this paper is the time allowed for writing the answers.
• Section I is compulsory. Attempt any four questions from Section II.
• The intended marks for questions or parts of questions are given in brackets [ ].

Section – I (40 MARKS)
(Attempt all questions from this Section)

Question 1.
(a) Choose the correct answer from the options given below: [5]
(i) The salt solution which does not react with ammonium hydroxide is :
(A) Calcium Nitrate
(B) Zinc Nitrate
(C) Lead Nitrate
(D) Copper Nitrate
Answer:
(A) Calcium Nitrate

(ii) The organic compound which undergoes substitution reaction is :
(A) C₂H₂
(B) C₂H₄
(C) C₁₀H₁₈
(D) C₂H₆
Answer:
(D) C₂H₆

(iii) The electrolysis of acidified water is an example of:
(A) Reduction
(B) Oxidation
(C) Redox reaction
(D) Synthesis
Answer:
(C) Redox reaction

(iv) The TUPAC name of dimethyl ether is :
(A) Ethoxy methane
(B) Methoxy methane
(C) Methoxy ethane
(D) Ethoxy ethane
Answer:
(B) Methoxy methane

(v) The catalyst used in the Contact Process is :
(A) Copper
(B) Iron
(C) Vanadium pentoxide
(D) Manganese dioxide
Answer:
(C) Vanadium pentoxide

(b) Give one word or a phrase for the following statements: [5]
(i) The energy released when an electron is added to a neutral gaseous isolated atom to form a negatively charged ion.
(ii) Process of formation of ions from molecules which are not in ionic state.
(iii) The tendency ofan element to form chains of identical atoms.
(iv) The property by which certain hydrated salts, when left exposed to atmosphere, lose their water of crystallization and crumble into powder.
(v) The process by which sulphide ore is concentrated.
Answer:
(i) Electron affinity
(ii) Ionisation
(iii) Catenation
(iv) Efflorescence
(v) Froth floatation process.

(c) Write, a balanced chemical equation for each of the following: [5]
(i) Action of concentrated sulphuric acid on carbon.
Answer:
C + 2H₂SO₄ → 2H₂O + 2SO₂↑ + CO₂ ↑

(ii) Reaction of sodium hydroxide solution with iron (III) chloride solution.
Answer:
FeCl₃ + 3NaOH → Fe(OH)₃ + 3 NaCl

(iii) Action of heat on aluminium hydroxide.
Answer:

(iv) Reaction of zinc with potassium hydroxide solution.
Answer:
Zn + 2KOH → K₂ZnO₂ + H₂O

(v) Action of dilute hydrochloric acid on magnesium sulphite.
Answer:
MgSO₃ +2HCl → MgCl₂ + H₂O + SO₂↑

(d) (i) Give the IUPAC name for each of the following: [5]

Answer:
(1) Methanal
(2) Propanol
(3) But-2-ene

(ii) Write the structural formula of the two isomers of butane.
Answer:

(e) State one relevant observation for each of the following: [5]
(i) Lead nitrate solution is treated with sodium hydroxide solution drop wise till it is in excess.
Answer:
Pb(NO₃)₂ + 2 NaOH → Pb(OH)₂ ↓ (white precipitate) + 2NaNO₃
Pb(OH)₂ + 2NaOH → Na₂PbO₂ (soluble) + 2H₂O
when drop by drop of sodium hydroxide is added, chalky white precipitates appear. These precipitates dissolve in excess of sodium hydroxide forming sodium plumbite.

(ii) At the anode, when molten lead bromide is electrolyzed using graphite electrodes.
Answer:

At Anode
2Br^– – 2e → 2Br
Br + Br → Br₂ ↑
At anode reddish vapours of bromine escape in air from lead bromide.

(iii) Lead nitrate solution is mixed with dilute hydrochloric acid and heated.
Answer:

White precipitates of lead chloride are formed which are soluble in hot water and insoluble in cold water.

(iv) Anhydrous calcium chloride is exposed to air for sometime.
Answer:
Anhydrous calcium chloride absorbs moisture becomes moist and loose its crystalline form showing its deliquescent nature.

(v) Barium chloride solution is slowly added to sodium sulphate solution.
Answer:
BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl

(f) Give a reason for each of the following: [5]
(i) Ionic compounds have a high melting point.
Answer:
Ionic compounds have ions held strongly by electrostatic forces of attraction. These strong forces need more energy to be broken apart.Hence, they have high melting point.

(ii) Inert gases do not form ions.
Answer:
Inert gases do not form ions because they have completely filled octet. They are extremely stable. Hence, they neither loose, nor gain electrons.

(iii) Ionisation potential increases across a period, from left to right.
Answer:
As we move from left to right along a period, the atomic size decreases due to the increase in nuclear charge thus more energy is required to remove the electron, hence ionisation potential increase.

(iv) Alkali metals are good reducing agents.
Answer:
Alkali metals have free electrons. They can easily loose electrons to form positive ions. The loss of electron is known as oxidation and the substance/element that lose electrons is said to be reducing agent.

Example:

(v) Conductivity of dilute hydrochloric acid is greater than that of acetic acid.
Answer:
Hydrochloric acid is a strong acid. It splits to give more hydrogen ions as compared to acetic acid. Hence conductivity of dilute hydrochloric acid is more than that of acetic acid.

(g) Name the gas that is produced in each of the following cases : [5]
(i) Sulphur is oxidized by concentrated nitric acid.
(ii) Action of dilute hydrochloric acid on sodium sulphide.
(iii) Action of cold and dilute nitric acid on copper.
(iv) At the anode during the electrolysis of acidified water.
(v) Reaction of ethanol and sodium.
Answer:
(i) Nitrogen dioxide
(ii) hydrogen sulphide
(iii) nitrogen monoxide
(iv) oxygen
(v) hydrogen

(h) Fill up the blanks with the correct choice given in brackets. [5]
(i) Ionic or electrovalent compounds do not conduct electricity in their ______________ state, (fused/solid)
(ii) Electrolysis of aqueous sodium chloride solution will form ______________ at the cathode. (hydrogen gas/sodium metal)
(iii) Dry hydrogen chloride gas can be collected by displacement of air.
(downward/upward)
(iv) The most common ore of iron is .
(calamine / haematite) (v) The salt prepared by the method ofdirect combination is .
(iron (II) chloride / iron (III) chloride)
Answer.
(i) fused
(ii) hydrogen gas
(iii) upward
(iv) haematite
(v) Iron (III) chloride

Section – II (40 Marks)
(Attempt any four questions from this Section)

Question 2.
(a) (i) What do you understand by a lone pair of electrons? [3]
(ii) Draw the electron dot diagram of Hydronium ion. (H = 1, O = 8)
Answer:
(i) The unshared pair of electron that does not normally take part in a chemical reaction is known as lone pair.

(ii) Hydronium ion Formation of proton :

(b) In Period 3 of the Periodic Table, element B is placed to the left of element A. [3]
On the basis of this information, choose the correct word from the brackets to complete the following statements :
(i) The element B would have (lower/higher) metallic character than A.
(ii) The element A would probably have (lesser/higher) electron affinity than B.
(iii) The element A would have (greater/smaller) atomic size than B.
Answer:
(i) Higher
(ii) higher
(iii) smaller

(c) Copy and complete the following table which refers to the conversion of ions to neutral particles. [4]

Answer:

Conversion	Ionic equation	Oxidation/Reduction
Chloride ion to chlorine molecule	(i) Cl– – 1 e– → Cl Cl + Cl → Cl2	(ii) Oxidation
Lead (ii) ion to lead	(ii) pb2+ + 2e– → Pb	(iv) Reduction

Question 3.
(a) (i) Write the balanced chemical equation to prepare ammonia gas in the laboratory by using an alkali. [3]
(ii) State why concentrated sulphuric acid is not used for drying ammonia gas.
(iii) Why is ammonia gas not collected over water ?
Answer:
(i)

(ii) Concentrated sulphuric acid is not used for drying ammonia gas because it reacts with ammonia.
2NH₃ + H₂SO₄ → (NH₄)₂SO₄

(iii) Ammonia is highly soluble gas one volume of water can dissolve 702 volumes of ammonia at 20°C and at 1 atmospheric pressure. Hence it is not collected over water.

(b) (i) Name the acid used for the preparation of hydrogen chloride gas in the laboratory. Why is this particular acid preferred to other acids ? [3]
(ii) Write the balanced chemical equation for the laboratory preparation of hydrogen chloride gas.
Answer:
(i) Sulphuric acid
[It is preferred to other acids because it is non-volatile acid]

(ii)

(c) For the preparation of hydrochloric acid in the laboratory: [2]
(i) Why is direct absorption of hydrogen chloride gas in water not feasible?
(ii) What arrangement is done to dissolve hydrogen chloride gas in water?
Answer:
(i) Direct absorption of hydrogen chloride gas in water is not feasible as it leads to back suction.
(ii) “Inverted funnel arrangement” is done to dissolve hydrogen chloride gas in water.

(d) For the electrorefining of copper [2]
(i) What is the cathode made up of ?
(ii) Write the reaction that takes place at the anode.
Answer:
(i) Cathode is made-up of thin sheets of pure copper connected in parallel.
(ii) Cu – 2e^– → Cu²⁺
Copper anode undergoes oxidation forming Cu²⁺ ions which pass into the solution.

Question 4.
(a) The percentage composition of a gas is: [2]
Nitrogen 82.35%, Hydrogen 17.64%
Find the empirical formula of the gas. [N = 14, H = 1]
Answer:

(b) Aluminium carbide reacts with water according to the following equation: [4]
Al₄C₃ + 12H₂O → 4Al(OH)₃ + 3CH₄
(i) What mass of aluminum hydroxide is formed from 12 g of aluminum carbide ?
(ii) What volume of methane at s.t.p. is obtained from 12 g of aluminum carbide ?
[Relative molecular weight of Al₄Cl₃ = 144; Al(OH)₃ = 78]
Answer:

144 g of Al₄Cl₃ produce 312 g of Al(OH)₃
1 g of Al₄C₃ produce \(\frac{312}{144}\)g of Al(OH)₃
12 g of Al₄C₃ produce \(\frac{312}{144}\) × 12 = 26 g of Al(OH)₃

(ii)

144 g of Al₄C₃ will produce 67.2 l of methane
1 g of Al₄C₃ will produce \(\frac{67.2}{144}\)l of methane
12g of Al₄C₃ will produce \(\frac{67.2}{144}\) × 12 = 5.6 g

(c) (i) If 150 cc of gas A contains X molecules, how many molecules of gas B will be present in 75 cc ofB ? [2]
The gases A and B all under the same conditions of temperature and pressure.
(ii) Name the law on which the above problem is based.
Answer:
(i) 150 ccofgas A contain X molecules
150 cc of gas B will also contain X molecules
75 cc of gas B will contain \(\frac{X}{2}\) molecules under the same conditions of temperature and pressure.

(ii) Avogadro’s law

(d) Name the main component of the following alloys : [2]
(i) Brass
(ii) Duralumin
Answer:
(i) Brass → Copper, zinc and tin
(ii) Duralumin → Aluminium, copper, magnesium and manganese

Question 5.
(a) Complete the following table which relates to the homologous series of hydrocarbons. [6]

Answer:
(A) Alkynes
(B) triple bond
(C) ethyne
(D) Alkane
(E) Single bond
(F) Methane

(b) (i) Name the most common ore of the metal aluminum from which the metal is extracted. Write the chemical formula of the ore. [4]
(ii) Name the process by which impure ore of aluminum gets purified by using concentrated solution of an alkali.
(iii) Write the equation for the formation of aluminum at the cathode during the electrolysis of alumina.
Answer:
(i) Bauxite (Al₂O₃.2H₂O)
(ii) Bayer’s process
(iii)

Cathode
Al³⁺ + 3e^– → Al

Question 6.
(a) A compound X (having vinegar like smell) when treated with ethanol in the presence of the acid Z, gives a compound Y which has a fruity smell. [4]
The reaction is:

(i) Identify Y and Z.
(ii) Write the structural formula of X.
(iii) Name the above reaction.
Answer:

(i) Y is ethyl ethanoate and Z is concentrated sulphuric acid.

(ii)

(iii) Esterification reaction

(b) Ethane burns in oxygen to form CO₂ and H₂O according to the equation: [4]
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
If 1250 cc of oxygen is burnt with 300 cc of ethane. Calculate:
(i) the volume of CO₂ formed.
(ii) the volume of unused O₂.
Answer:

(i)

2 vol. of ethane forms 4 volumes of carbon dioxide
1 vol. of ethane forms \(\frac{4}{2}\) volumes of carbon dioxide.
300cc will form \(\frac{4}{2}\) × 300 = 600 cc of CO₂.

(ii) 2 vol. of ethane requires 7 volumes of oxygen
2 vol. of ethane requires \(\frac{7}{2}\) volumes of oxygen
300 cc of ethane requires \(\frac{7}{2}\) × 300 = 1050 cc
Total volume of Oxygen= 1250 cc
Volume of oxygen used= 1050 cc
Unused oxygen= (1250 – 1050) cc = 200 cc

(c) Three solutions P, Q and R have pH value of 3.5, 5.2 and 12.2 respectively. [2]
Which one of these is a :
(i) Weak acid ?
(ii) Strong alkali ?
Answer:
P = 3.5
Q = 5.2
R = 12.2

(i) Weak acid = Q
(ii) Strong alkali = R

Question 7.
(a) Give a chemical test to distinguish between the following pairs of chemicals : [4]
(i) Lead nitrate solution and Zinc nitrate solution.
(ii) Sodium chloride solution and Sodium nitrate solution.
Answer:
(i) Add ammonium hydroxide to the solutions of lead nitrate and zinc nitrate dropwise and then in excess.

Lead Nitrate	Zinc Nitrate
(i) On adding ammonium hydroxide chalky white precipitates of lead hydroxide are formed. Pb(NO3)2 + 2NH4OH → Pb(OH)2 + 2NH4NO3	(i) On adding ammonium hydroxide gelatinous white precipitates of zinc hydroxide are formed. Zn(NO3)2 + 2NH4OH → Zn(OH)2 + 2NH4NO3
(ii)On adding excess of ammonium hydroxide, the precipitates do not dissolve.	(ii) On adding excess of ammonium hydroxide, the precipitates dissolve forming a soluble complex. Zn(OH) 2 + 2NH4NO23 + 2NH4OH → [Zn(NH3) 4] (NO3)2 + 4H2O

(ii) Add silver nitrate solution to both sodium chloride and sodium nitrate solution and observe.

Sodium	Chloride
(i) Add silver nitrate solution to sodium chloride NaCl + AgNO3 → AgCl ↓ + NaNO3 You will observe curdy white precipitates of silver chloride.	(i) Add silver nitrate solution to sodium nitrate. No reaction will take place. NaNO3 + AgNO3 → No reaction

(b) Write a balanced equation for the preparation of each of the following salts : [2]
(i) Copper sulphate from Copper carbonate
(ii) Zinc carbonate from Zinc sulphate.
Answer:
(i) CuCO₃ + H₂SO₄ → CuSO₄ + H₂O + CO₂↑
(ii) ZnSO₄ + Na₂CO₃ → ZnCO₃ + Na₂SO₄

(c) (i) What is the type of salt formed when the reactants are heated at a suitable temperature for the preparation of Nitric acid ? [2]
(ii) State why for the preparation of Nitric acid, the complete apparatus is made up of glass.
(d) Which property of sulphuric acid is shown by the reaction of concentrated sulphuric acid with: [2]
(i) Ethanol ?
(ii) Carbon ?
Answer:
(i) Dehydrating agent

(ii) Oxidising agent
C + 2H₂SO₄ → CO₂ + 2H₂O + 2SO₂

Solving ICSE Class 10 Maths Previous Year Question Papers ICSE Class 10 Maths Question Paper 2019 is the best way to boost your preparation for the board exams.

ICSE Class 10 Maths Question Paper 2019 Solved

Section – A (40 MARKS)
(Attempt all questions from this Section)

Question 1.
(a) Solve the following inequation and write down the solution set: [3]
11x – 4 < 15x + 4 ≤ 13x + 14, x ∈ W
Represent the solution on a real number line.
Solution:
11x – 4 ≤ 15x + 4 ;
15x + 4 ≤ 13x + 4, x ∈ W
-4 – 4 < 15x – 11x;
15x – 13x ≤ 4 – 4
– 8 < 4x ;
2x <0
– 2 < x
x ≤ 0
Thus, x = 0, x ∈ W

(b) A man invests ₹ 4500 in shares of a company which is paying 7.5% dividend. [3]
If ₹ 100 shares are available at a discount of 10%. Find :
(i) Number of shares he purchases.
(ii) His annual income.
Solution:
Total investment = ₹ 4500
Face value of a share = ₹ 100
Discount = 10%
Market value of a share = ₹ (100 – 10)
= ₹ 90
Now, Number of shares purchased
= \(\frac{4500}{90}\)
= 60
Annual income = ₹ 7.5%
of 50 × 100= ₹ \(\frac{7.5}{100}\) × 5000
= ₹ 375

(c) In a class of 40 students, marks obtained by the students in a class test (out of 10) are given below: [4]
Calculate the following for the given distribution :

Marks	Number of Students
1	1
2	2
3	3
4	3
5	6
6	10
7	5
8	4
9	3
10	3

(i) Median
(ii) Mode
Solution:

Marks	Number of Students	Cumulative
1	1	1
2	2	3
3	3	6
4	3	9
5	6	15
6	10	25
7	5	30
8	4	34
9	3	37
10	3	40
Total	40

Here, N = 40 and \(\frac{\mathrm{N}}{\mathrm{2}}\) = \(\frac{40}{2}\) = 20
Marks corresponding to cumulative frequency 20 is 6.
Thus, the required median is 6.
Clearly, 6 occurs 10 times which is maximum.
Hence, mode is 6.

Question 2.
(a) Using the factor theorem, show that (x – 2) is a factor of x³ + x² – 4x – 4. [3]
Hence, factorise the polynomial completely.
Solution:
Given polynomial is p(x) = x³ + x² – 4x – 4
x – 2 is its factor, if p(2) = 0
∴ p(2) = (2)³ + (2)² – 4(2) – 4
= 8 + 4 – 8 – 4
= 0
Thus, x – 2 is a factor of p(x).
Now, x³ + x² – 4x – 4 = x²(x + 1) – 4(x + 1)
=(x + 1)(x² – 4)
=(x + 1)(x + 2)(x – 2)
Hence, the required factors are (x + 1), (x + 2) and (x – 2).

(b) Prove that: (cosec θ – sin θ) (sec θ – cos θ)(tan θ + cot θ) = 1 [3]
Solution:
L.H.S. = (cosec θ – sin θ)(sec θ – cos θ)(tan θ + cot θ)

= 1 = R.H.S.

(c) In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find the : [4]
(i) first term
(ii) common difference
(iii) sum of the first 20 terms.
Solution:
Here, a₄ = 8
⇒ a + 3d = 8 ……………… (i)
a₆ = 14
⇒ a + 5d = 14 …………….. (ii)
Subtracting (i) from (ii), we have
2d = 6
⇒ d = 3
From (i), we have
a + 3(3) = 8
⇒ 0 = 8 – 9 = -1
Now, S_n = \(\frac{\mathrm{N}}{\mathrm{2}}\) [2a + (n – 1)d]
S₂₀ = \(\frac{20}{2}\) [2(-1) + (20 – 1)(3)]
= 10 [-2 + 57]
= 10 [55]
= 550
Hence, first term is -1, common difference is 3 and sum of the first 20 terms is 550.

Question 3.
(a) Simplify : sin A \(\left[\begin{array}{ll}
\sin A & -\cos A \\
\cos A & \sin A
\end{array}\right]\) + cos A \(\left[\begin{array}{ll}
\cos A & \sin A \\
-\sin A & \cos A
\end{array}\right]\) [3]
Solution:

(b) M and N are two points on the X axis and Y axis respectively. P(3, 2) divides the line segment MN in the ratio 2 : 3. [3]
Find: (i) the coordinates of M and N
(ii) slope of the line MN.
Solution:
Let the coordinates of M and N be (x, 0) and (0, y)

P(3, 2) = P \(\left(\frac{2 \times 0+3 x}{2+3}, \frac{2 y+3 \times 0}{2+3}\right)\)
= P \(\left(\frac{3 x}{5}, \frac{2 y}{5}\right)\)
∴ 3 = \(\frac{3 x}{5}\) ⇒ x = 5 and 2 = \(\frac{2 y}{5}\) ⇒ y = 5
Thus, the coordinates of M and N are M(5, 0) and N(0, 5).
Slope of the line MN = \(\frac{y_2-y_1}{x_2-x_1}=\frac{y-0}{0-x}\)
= – \(\frac{y}{x}\) = – \(\frac{5}{5}\) – 1
Hence, the slope of the line MN is -1.

(c) A solid metallic sphere of radius 6 cm is melted and made into a solid cylinder of height 32 cm. Find the: [4]
(i) radius of the cylinder .
(ii) curved surface area of the cylinder Take π = 3.1
Solution:
Radius of metallic sphere (R) = 6 cm
Height of cylinder (h) = 32 cm
∴ Volume of cylinder= Volume of metallic sphere
πr²h = \(\frac{4}{3}\) πR³
⇒ r²(32) = \(\frac{4}{3}\) (6)³
⇒ r² = \(\frac{4}{3} \times \frac{6 \times 6 \times 6}{32}\) = 9
∴ r = 3 cm
Curved surface area of the cylinder = 2πrh
= 2 × 3.1 × 3 × 32 = 595.2 cm²

Question 4.
(a) The following numbers, K + 3, K + 2, 3K – 7 and 2K – 3 are in proportion. Find K. [3]
Solution:
Here, \(\frac{\mathrm{K}+3}{\mathrm{~K}+2}=\frac{3 \mathrm{~K}-7}{2 \mathrm{~K}-3}\)
⇒ (K + 3) (2K – 3) = (K + 2) (3K – 7)
⇒ 2K² – 3K + 6K – 9 = 3K² – 7K + 6K – 14
⇒ K² – 4K – 5 = 0
⇒ (K – 5) (K + 1) = 0
⇒ K = 5 or K = -1

(b) Solve for x the quadratic equation x² – 4x – 8 = 0 [3]
Give your answer correct to three significant figures.
Solution:
Given quadratic equation is x² – 4x – 8 = 0
By using quadratic formula, we have
∴ x = \(\frac{-(-4) \pm \sqrt{(-4)^2-4(1)(-8)}}{2(1)}\)
= \(\frac{4 \pm \sqrt{16+32}}{2}\)
= \(\frac{4 \pm \sqrt{48}}{2}=\frac{4 \pm 4 \sqrt{3}}{2}\) = 2 ± 2\(\sqrt{3}\)
= 2(1 ± \(\sqrt{3}\)) = 2(1 ± 1.73205)
= 2(2.73205) or 2(-0.73205)
= 5.46410 or -1.4641
= 5.46 or -1.46

(c) Use ruler and compass only for answering this question. [4]
Draw a circle of radius 4 cm. Mark the centre as O. Mark a point P outside the circle at a distance of 7 cm from the centre. Construct two tangents to the circle from the external point P.
Measure and write down the length of any one tangent.
Solution:
Steps of Construction :

1. Draw a circle of radius 4 cm and centre O.
2. Draw a radius and produce it to P, such that OP = 7 cm.
3. Bisect OP at M.
4. With M as centre and MP as radius, draw a circle to intersect the given circle at Q and R.
5. Join PQ and PR.
   PQ and PR are the required tangents and length of the tangents is 5.74 cm.

Section – B (40 MARKS)
(Attempt any four questions)

Question 5.
(a) There are 25 discs numbered 1 to 25. They are put in a closed box and shaken thoroughly. A disc is drawn at random from the box. [3]
Find the probability that the number on the disc is :
(i) an odd number
(ii) divisible by 2 and 3 both
(iii) a number less than 16.
Solution:
Sample space = 25 discs numbered from 1 to 25.
(i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25
Probability (an odd number) = \(\frac{13}{24}\)

(ii) Numbers divisible by 2 and 3 both are 6, 12, 18, 24
Probability (divisible by 2 and 3 both) = \(\frac{4}{25}\)

(iii) Numbers less than 16 are 1 to 15
Probability (a no. less than 16) = \(\frac{15}{25}\) or \(\frac{3}{5}\)

(b) Rekha opened a recurring deposit account for 20 months. The rate of interest is 9% per annum and Rekha receives ₹ 441 as interest at the time of maturity.
Find the amount Rekha deposited each month. [3]
Solution:
Here, n = 20, R = 9% p.a., Interest = ₹ 441
Let the monthly deposit be of ₹ x.
∴ x\(\left(\frac{20 \times 21}{2 \times 12}\right)\) × \(\frac{9}{100}\) = 441
x = \(\frac{441 \times 100 \times 24}{20 \times 21 \times 9}\)
= 280
Hence, the monthly deposit is ₹ 280.

(c) Use a graph sheet for this question. [4]
Take 1 cm = 1 unit along both x and y axis.
(i) Plot the following points :
A(0, 5), B(3, 0), C(1, 0) and D(1, -5)
(ii) Reflect the points B, C and D on the y-axis and name them as B’, C’, D’ respectively.
(iii) Write down the coordinates of B’, C’ and D’.
(iv) Join the points A, B, C, D, D’, C’, B’, A in order and give a name to the closed figure ABCDD’C’B’.
Solution:
(i) and (ii)

(iii) B’ (-3, 0), C'(-1, 0) and D'(-1, -5)
(iv) Arrow Head.

Question 6.
(a) In the given figure, ∠PQR = ∠PST = 90°, PQ = 5 cm and PS = 2 cm. [3]
(i) Prove that ∆PQR ~ ∆PST.
(ii) Find Area of ∆PQR : Area of quadrilateral SRQT.

Solution:
In rt. ∠ed ∆PQR and ∆PST
∠P = ∠P [common]
∠Q = ∠S [each = 90°]
∴ ∆PQR ~ ∆PST [by AA similarity rule]

(b) The first and last term of a Geometrical Progression (G.P.) are 3 and 96 respectively. If the common ratio is 2, find: [3]
(i) ‘n’ the number of terms of the G.P.
(ii) Sum of the n terms.
Solution:
Given that, a = 3 and a_n = 96, r = 2
∴ ar^{n – 1} = 96
⇒ 3(2)^{n – 1} = 96
⇒ (2)^{n – 1} = 32 = 25
⇒ n – 1= 5
⇒ n = 6
Now, S_n = \(\frac{a\left(r^n-1\right)}{r-1}\), r > 1
= \(\frac{3\left(2^n-1\right)}{2-1}\)
= 3(2ⁿ – 1)

(c) A hemispherical and conical hole is scooped out of a solid wooden cylinder. [4]
Find the volume of the remaining solid where the measurements are as follows :
The height of the solid cylinder is 7 cm, radius of each of hemisphere, cone and cylinder is 3 cm. Height of cone is 3 cm.
Give your answer correct to the nearest whole number. Take π = \(\frac{22}{7}\)

Solution:
Given that:
Radius of each of hemisphere, cone and cylinder (r) = 3 cm
Height of cylinder = 7 cm
Height of cone = 3 cm
Volume of remaining solid
= Vol. of cylinder – Vol. of cone – Vol. of hemisphere
= πr²h \(\frac{1}{3}\) πr²h – \(\frac{2}{3}\) πr³
= πr² [h – \(\frac{\mathrm{h}}{\mathrm{3}}\) – \(\frac{\mathrm{2r}}{\mathrm{3}}\)]
= \(\frac{22}{7}\) × 3 × 3 [7 – \(\frac{3}{2}\) – \(\frac{2}{3}\) × 3]
= \(\frac{198}{7}\) × 4
= 113.14 cm³

Question 7.
(a) In the given figure, AC is a tangent to the circle with centre O.
If ∠ADB = 55°, find x and y. Give reasons for your answers. [3]

Solution:
We know that angle between the radius and the tangent at the point of contact is right angle.
∴ ∠A = 90°
Also, in ∆OBE, OB = OE = radius (r)
∴ ∠B = ∠OEB ………….. (i)
In ∆ABD,
∠A + ∠B + ∠ADB = 180°
90° + ∠B + 55° = 180°
∠B = 180° – 90° – 55°
= 35°
Thus, ∠B = ∠OEB = 35°
∠OEB = ∠DEC = 35° [vertically opp. ∠s]
∠EDC + ∠ADE = 180°
∠EDC + 55° = 180°
⇒ ∠EDC = 180° – 55°
= 125°
In ∆EDC,
∠DEC + ∠EDC + x = 180°
35° +125° + x = 180°
⇒ x = 180° – 35° – 125°
= 20°
In ∆AOC, ∠A + y° + x° = 180°
⇒ 90° + y° + 20° = 180°
⇒ y° = 180° – 90° – 20°
= 70°
Hence, x – 20° and y – 70°.

(b) The model of a building is constructed with the scale factor 1: 30.   [3]
(i) If the height of the model is 80 cm, find the actual height of the building in metres.
(ii) If the actual volume of a tank at the top of the building is 27m³, find the volume of the tank on the top of the model.
Solution:
Here, scale factor (k) = \(\frac{1}{30}\)
(i) Height of the model = k (Actual height of the building)
⇒ 80 cm = \(\frac{1}{30}\) (Actual height of the building)
⇒ Actual height of the building = 30 × 80 = 2400 cm

(ii) Volume of the tank at the top of the model
= k³ (Actual volume of the tank)
⇒ Volume of the tank at the top of the model
= \(\left(\frac{1}{30}\right)^3\) × 27 m³ = 0.001 m³

(c) Given \(\left[\begin{array}{ll}
4 & 2 \\
-1 & 1
\end{array}\right]\) M = 6I, where M is a matrix and I is unit matrix of order 2 × 2.
(i) State the order of matrix M.
(ii) Find the matrix M. [4]
Solution:

= 4 – 6
= -2
Hence, M = \(\left[\begin{array}{rr}
1 & -2 \\
1 & 4
\end{array}\right]\)

Question 8.
(a) The sum of the first three terms of an Arithmetic Progression (A.P.) is 42 and the product of the first and third term is 52. Find the first term and the common difference. [3]
Solution:
Let the first three terms of an A.P. be a – d, a and a + d.
According to the statement, we have
a – d + a + a + d = 42
3 a = 42
a = 14
Now, (a – d) (a + d) = 52
a² – d² = 52
14² – d² = 52
⇒ d² = 196 – 52 = 144
⇒ d = ± 12
Hence, the first term is 14 and common difference is +12.

(b) The vertices of a ∆ABC are A (3, 8), B(-1, 2) and C(6, -6). Find : [3]
(i) Slope of BC.
(ii) Equation of a line perpendicular to BC and passing through A.
Solution:
Vertices of a ∆ABC are A(3, 8), B(-1, 2) and C(6, – 6)
Slope of BC = \(\frac{-6-2}{6+1}=\frac{-8}{7}\)
Slope of the line perpendicular to BC = \(\frac{7}{8}\)
Now, equation of the line perpendicular to BC and passing through A is
y – 8 = \(\frac{7}{8}\) (x – 3)
8y – 64 = 7x – 21
⇒ 7x – 8y + 43 = 0

(c) Using ruler and a compass only construct a semi-circle with diameter BC = 7 cm. Locate a point A on the circumference of the semicircle such that A is equidistant from B and C. Complete the cyclic quadrilateral ABCD, such that D is equidistant from AB and BC. Measure ∠ADC and write it down. [4]
Solution:
Steps of Construction :

1. Draw a line segment BC = 7 cm.
2. Draw its perpendicular bisector l and let it intersect BC in M.
3. With M as centre and radius equal to BM or CM, draw a semi-circle and let the semi-circle intersect the perpendicular bisector of line segment BC in A. Join BA.
   
4. Draw the angle bisector of ∠ABC and let it intersect the semi-circle in D.
5. Join AD and CD.
   Hence, ∠ADC = 135°

Question 9.
(a) The data on the number of patients attending a hospital in a month are given below. [3]
Find the average (mean) number of patients attending the hospital in a month by using the shortcut method. Take the assumed mean as 45. Give your answer correct to 2 decimal places.

Number of Patients	Number of Days
10 – 20	5
20 – 30	2
30 – 40	7
40 – 50	9
50 – 60	2
60 – 70	5

Solution:

Mean (\(\overline{\mathrm{X}}\)) = a + \(\frac{\sum f_i d_i}{\sum f_i}\)
Mean (\(\overline{\mathrm{X}}\)) = 45 + \(\frac{-140}{30}\)
= 45 – 4.67
= 40.33

(b) Using properties of proportion solve for x, given [3]
\(\frac{\sqrt{5 x}+\sqrt{2 x-6}}{\sqrt{5 x}-\sqrt{2 x-6}}\) = 4
Solution:
\(\frac{\sqrt{5 x}+\sqrt{2 x-6}}{\sqrt{5 x}-\sqrt{2 x-6}}=\frac{4}{1}\)
Applying componendo and dividendo, we have
\(\frac{\sqrt{5 x}+\sqrt{2 x-6}+\sqrt{5 x}-\sqrt{2 x-6}}{\sqrt{5 x}+\sqrt{2 x-6}-\sqrt{5 x}+\sqrt{2 x-6}}=\frac{4+1}{4-1}\)
\(\frac{2 \sqrt{5 x}}{2 \sqrt{2 x-6}}=\frac{5}{3}\)
\(\frac{\sqrt{5 x}}{\sqrt{2 x-6}}=\frac{5}{3}\)
Squaring both sides, we obtain
\(\frac{5 x}{2 x-6}=\frac{25}{9}\)
45x = 50x – 150
50x – 45x = 150
5x = 150
x = 30

(c) Sachin invests ₹ 8500 in 10%, ₹ 100 shares at ₹ 170. He sells the shares when the price of each share rises by ₹ 30. He invests the proceeds in 12% ₹ 100 shares at ₹ 125. Find: [4]
(i) the sale proceeds.
(ii) the number of ₹ 125 shares he buys.
(iii) the change in his annual income.
Solution:
Total investment = ₹ 8500
Market value of each share = ₹ 170
Number of shares purchased = \(\frac{8500}{170}\) = 50
Dividend received = ₹ \(\frac{10}{100}\) × 50 × 100 = ₹ 500
Now, market value of each share
= ₹ (170 + 30) = ₹ 200
Amount received on selling = ₹ (50 × 200)
= ₹ 10000
Market value of new shares = ₹ 125 each
Number of shares purchased = \(\frac{10000}{125}\) = 80
Dividend received = ₹ \(\frac{12}{100}\) × 80 × 100
= ₹ 960
Change in income = ₹ (960 – 500)
= ₹ 460

Question 10.
(a) Use graph paper for this question. [6]
The marks obtained by 120 students in an English test are given below:

Marks	No. of Students
0-10	5
10-20	9
20-30	16
30-40	22
40-50	26
50-60	18
60-70	11
70-80	6
80-90	4
90-100	3

Draw the ogive and hence, estimate :
(i) the median marks.
(ii) the number of students who did not pass the test if the pass percentage was 50.
(iii) the upper quartile marks.
Solution:

Marks	No, of Students	c.f. Less Than Type
0-10	5	Less than 10	5
10-20	9	Less than 20	14
20-30	16	Less than 30	30
30-40	22	Less than 40	52
40-50	26	Less than 50	78
50-60	18	Less than 60	96
60-70	11	Less than 70	107
70 – 80	6	Less than 80	113
80-90	4	Less than 90	117
90 – 100	3	Less than 100	120
Total	120

Plot the points (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117), (100, 120).
On the graph paper by taking upper limits on x-axis and number of students ony-axis. Join them free hand to get smooth curve.

Here, N = 120
\(\frac{\mathrm{N}}{\mathrm{2}}\) = \(\frac{120}{2}\) = 60
Median marks = 42 marks
Number of students who did not pass = 78 students
Upper quartile marks = 57 marks

(b) A man observes the angle of elevation of the top of the tower to be 45°. He walks towards it in a horizontal line through its base. On covering 20 m the angle of elevation changes to 60°. Find the height of the tower correct to 2 significant figures. [4]
Solution:
Let AB be the tower of height h m. P and Q are the two observing points, such that
∠APB = 45°, ∠AQB = 60°, PQ = 20 m
In rt. ∠ed ∆QBA,
\(\frac{\mathrm{AB}}{\mathrm{QB}}\) = tan 60°

= \(\frac{20(3+\sqrt{3})}{2}\)
= 10(3 + 1.73) = 47.3
Hence, the height of the tower is 47.3 m.

Question 11.
(a) Using the Remainder Theorem find the remainders obtained when x³ + (kx + 8)x + k is divided by x + 1 and x – 2. [3]
Hence, find k if the sum of the two remainders is 1.
Solution:
Given polynomial is
p(x) = x³ +(kx + 8)x + k
g(x) = x + 1
∴ R₁ = P(-1)
= (-1)³ + {k(-1)} + 8} (-1) + k
= -1 + k – 8 + k
= 2k – 9
h(x) = x – 2
∴ R₂ = P(2)
= (2)³ + (2k + 8)² + k
= 8 + 4k + 16 + k
= 5k + 24
Now, R₁ + R₂ = 1
⇒ 2k – 9 + 5k + 24 = 1
⇒ 7k = 1 + 9 – 24
⇒ 7k = – 14
⇒ k = – 2

(b) The product of two consecutive natural numbers which are multiples of 3 is equal to 810. Find the two numbers. [3]
Solution:
Let the two consecutive natural numbers which are multiples of 3 be 3x and 3(x + 1).
Now, 3x(3x + 3) = 810
⇒ x² + x = 90
⇒ x² + x – 90 = 0
⇒ (x + 10) (x – 9) = 0
⇒ x = 9 or
x = – 10
Rejecting negative value of x, because numbers are natural. We have x = 9.
Hence, the required numbers are 27 and 30.

(c) In the given figure, ABCDE is a pentagon inscribed in a circle such that AC is a diameter and side BC||AE. If ∠BAC = 50°, find giving reasons: [4]
(i) ∠ACB
(ii) ∠EDC
(iii) ∠BEC

Hence, prove that BE is also a diameter.
Solution:
Since AC is a diameter and angle in a semi-circle is right
∴ ∠B = 90°
and ∠ACB = 40°
Also, BC || AE
∴ ∠EAC = ∠ACB
= 40° [alt. int. ∠s]

In cyclic quadrilateral ACDE
∠EAC + ∠EDC = 180°
40° + ∠EDC = 180°
⇒ ∠EDC = 180° – 40° = 140°
∠BEC = ∠BAC
= 50°
[∠s in the same segment]
Also, ∠EAC = ∠EBC
= 40°
[∠s in the same segment]
∴ ∠ABE = ∠ABC – ∠EBC
= 90° – 40°
= 50°
Again, ∠ABE = ∠ACE = 50°
[∠s in the same segment]
Now, ∠ACE + ∠ACB
= 50° + 40°
= 90°
⇒ ∠BCE = 90°
Hence, BE is a diameter, because angle is a semi-circle is right angle.

Solving ICSE Class 10 Maths Previous Year Question Papers ICSE Class 10 Maths Question Paper 2021 Semester 1 is the best way to boost your preparation for the board exams.

ICSE Class 10 Maths Question Paper 2021 Solved Semester 1

Maximum Marks: 40
Time Allowed: One and a half hours

• You will not be allowed to write during the first 10 minutes.
• This time is to be spent in reading the question paper.
• ALL QUESTIONS ARE COMPULSORY.
• The marks intended for questions are given in brackets [ ]

Select the correct option

Question 1.
If (x + 2) is a factor of polynomial x³ – kx² – 5x + 6, then the value ofk is : [1]
(a) 1
(b) 2
(c) 3
(d) -2
Solution:
(b) 2
[Here, (- 2)³ – k(- 2)² – 5(- 2) + 6 = 0
-8 – 4k + 10 + 6 = 0
4k = 8
k = 2]

Question 2.
The solution set of the inequation x – 3 ≥ – 5, x ∈ R is : [1]
(a) {x : x > – 2, x ∈ R}
(b) {x : x ≤ – 2, x ∈ R}
(c) {x : x ≥ -2, x ∈ R}
(d) {-2, -1, 0, 1, 2}
Solution:
(c) {x : x ≥ – 2, x ∈ R}
[Here, x – 3 ≥ -5, x ∈ R
x ≥ – 5 + 3
x ≥ -2]

Question 3.
The product AB of two matrices A and B is possible if : [1]
(a) A and B have the same number of rows.
(b) the number of columns of A is equal to the number of rows of B.
(c) the number of rows of A is equal to the number of columns of B.
(d) A and B have the same number of columns.
Solution:
(b) the number of columns of A is equal to the number of rows of B.

Question 4.
If 70, 75, 80, 85 are the first four terms of an Arithmetic Progression, then the 10^th term is : [1]
(a) 35
(b) 25
(c) 115
(d) 105
Solution:
(c) 115
[Here, a = 70, d = 75 – 70 = 5 and n = 10
∴ a₁₀ = a + 9 d
= 70 + 9 × 5
= 70 + 45
= 115]

Question 5.
The selling price of a shirt excluding GST is ₹ 800. If the rate of GST is 12%, then the total price of the shirt is : [1]
(a) ₹ 704
(b) ₹ 96
(c) ₹ 896
(d) ₹ 1848
Solution:
(c) ₹ 896
[Total price of the shirt = ₹ 800 + ₹ \(\frac{12}{100}\) × 800 = ₹ 896]

Question 6.
Which of the following quadratic equations has 2 and 3 as its roots ? [1]
(a) x² – 5x + 6 = 0
(b) x² + 5x + 6 = 0
(c) x² – 5x – 6 = 0
(d) x² + 5x – 6 = 0
Solution:
(a) x² – 5x + 6 = 0
[Required quadratic equation
= (x – 2) (x – 3)
= x² – 5x + 6]

Question 7.
If x, 5.4, 5, 9 are in proportion, then x is : [1]
(a) 3
(b) 9.72
(c) 25
(d) \(\frac{25}{3}\)
Solution:
(a) 3
[Here, x : 5.4 = 5 : 9
⇒ x = \(\frac{5 \times 5.4}{9}\) = 3]

Question 8.
Mohit opened a Recurring deposit account in a bank for 2 years. He deposits ₹ 1000 every month and receives ₹ 25500 on maturity. The interest he earned in 2 years is: [1]
(a) ₹ 13500
(b) ₹ 3000
(c) ₹ 24000
(d) ₹ 1500
Solution:
(d) ₹1500 ;
[Here, Interest earned in 2 years
= ₹ 25500 – ₹ 2 × 12 × 1000
= ₹ 25500 – ₹ 24000
= ₹ 1500]

Question 9.
In the given figure AB = 24 cm, AC = 18 cm, DE = 12 cm, DF = 9 cm and ∠BAC = ∠EDF. Then ∆ABC ~ ∆DEF by the condition : [1]

(a) AAA
(b) SAS
(c) SSS
(d) AAS
Solution:
(b) SAS
[Here, ∠BAC = ∠EDF, \(\frac{\mathrm{AB}}{\mathrm{DE}}\) = \(\frac{24}{12}\) = 2, \(\frac{\mathrm{AC}}{\mathrm{DF}}\) = \(\frac{18}{9}\) = 2]

Question 10.
If A = \(\left[\begin{array}{ll}
5 & 10 \\
3 & -4
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), then AI is equal to : [1]
(a) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
(b) \(\left[\begin{array}{rr}
5 & 10 \\
-3 & 4
\end{array}\right]\)
(c) \(\left[\begin{array}{rr}
5 & 10 \\
3 & -4
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
15 & 15 \\
-1 & -1
\end{array}\right]\)
Solution:
(c) \(\left[\begin{array}{rr}
5 & 10 \\
3 & -4
\end{array}\right]\)

Question 11.
The polynomial x³ – 2x² + ax + 12 when divided by (x + 1) leaves a remainder 20, then ‘a’ is equal to : [1]
(a) -31
(b) 9
(c) 11
(d) -11
Solution:
(b) 9
[Here, (- 1)³ – 2(- 1)² + a(- 1) + 12 = 20
-1 – 2 – a + 12 = 0 ⇒ a = 9]

Question 12.
In an Arithmetic Progression (A.P.) if, first term is 5, common difference is -3 and the n^th term is 7, then n is equal to . [1]
(a) 5
(b) 17
(c) -13
(d) 7
Solution:
(a) 5
[Here, a_n = – 7, a = 5, d = – 3
∴ a + (n – 1)d = -7
⇒ 5 + (n – 1) (-3) = – 7
⇒ n – 1 = \(\frac{-7-5}{-3}\) = 4
⇒ n = 1 + 4 = 5]

Question 13.
In the given figure PQ is parallel to TR, then by using condition of similarity: [1]

Solution:
(b) \(\frac{P Q}{R T}=\frac{O P}{O R}=\frac{O Q}{O T}\)
[Here, ∆PQO ~ ∆RTO by AA similarity .
axiom \(\frac{P Q}{R T}=\frac{O P}{O R}=\frac{O Q}{O T}\)].

Question 14.
If a, b, c and d are proportional, then \(\frac{a+b}{a-b}\) is equal to : [1]
(a) \(\frac{c}{d}\)
(b) \(\frac{d}{c}\)
(c) \(\frac{c-d}{c+d}\)
(d) \(\frac{c+d}{c-d}\)
Solution:
(d) \(\frac{c+d}{c-d}\)
[Here, a, b, c and d are in proportion
∴\(\frac{a}{b}\) = \(\frac{c}{d}\)
By using componendo and dividendo, we have
\(\frac{a+b}{a-b}\) = \(\frac{c+d}{c-d}\)]

Question 15.
The first four terms, of an Arithmetic Progression (A.P.), whose first term is 4 and common difference is-6 are: [1]
(a) 4, -10, -16, -22
(b) 4, 10, 16, 22
(c) 4, – 2, -8, -14
(d) 4, 2, 8, 14
Solution:
(c) 4, -2, -8, -14
[Here, a = 4 and d = – 6
∴ a₁ = 4, a₂ = 4 – 6 = – 2, a₃ = – 2 – 6 = – 8 and a₄ = – 8 – 6 = – 14]

Question 16.
One of the roots of the quadratic equation x² – 8x + 5 = 0 is 7.3166. The root of the equation correct to 4 significant figures is: [1]
(a) 7.3166
(b) 7.317
(c) 7.316
(d) 7.32
Solution:
(c) 7.316

Question 17.
(x + 2) and (x + 3) are two factors of the polynomial x³ + 6x² + 11x + 6. If this polynomial is completely factorised the result is: [2]
(a) (x – 2) (x + 3) (x + 1)
(b) (x + 2) (x – 3) (x – 1)
(c) (x + 2) (x + 3) (x – 1)
(d) (x + 2) (x + 3) (x + 1)
Solution:
(d) (x + 2)(x + 3)(x + 1)
[Here, (x + 2)(x + 3) = x² + 5x + 6

Question 18.
The sum of the first 20 terms of the Arithmetic Progression 2, 4, 6, 8, … is :
(a) 400
(b) 840
(c) 420
(d) 800
Solution:
(c) 420
[Here, a = 2, d = 4 – 2 = 2, n = 20
∴ S₂₀ = \(\frac{20}{2}\)[2 × 2 + (20 – 1)²]
= 10[4 + 38]
= 420]

Question 19.
The solution set on the number line of the linear inequation : 2y – 6 < y + 2 ≤ 2y, y ∈ N is:  [2]

Solution:

[Here, 2y – 6 < y + 2 ; y + 2 ≤ 2y, y ∈ N y < 8 ; 2 ≤ y ⇒ 2 ≤ y < 8; y ∈ N]

Question 20.
If x, y, z are in continued proportion, then (y² + z²) : (x² + y²) is equal to: [2]
(a) z : x
(b) x : z
(c) zx
(d) (y + z) : (x+y)
Solution:
(a) z : x
[Given that x, y, z are in continued proportion
∴ \(\frac{\mathrm{x}}{\mathrm{y}}\) = \(\frac{\mathrm{y}}{\mathrm{z}}\) ⇒ y² = xz
Now \(\frac{y^2+z^2}{x^2+y^2}=\frac{x z+z^2}{x^2+x z}=\frac{z(x+z)}{x(x+z)}=\frac{z}{x}\) or z : x]

Question 21.
The marked price of an article is ₹ 5000. The shopkeeper gives a discount of 10%. If the rate of GST is 12%, then the amount paid by the customer including GST is: [2]
(a) ₹ 5040
(b) ₹ 6100
(c) ₹ 6272
(d) ₹ 6160
Solution:
Solution :
(a) ₹ 5040
[Amount paid by the customer
= (5000 – 10% of 5000) + 12% of (5000 – 10% of 5000)
= 4500 + 540 = ₹ 5040]

Question 22.
If A = \(\left[\begin{array}{ll}
3 & 5 \\
I & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
2 & 4 \\
0 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{rr}
I & -I \\
2 & I
\end{array}\right]\), then 5A – BC is equal to :

Solution:

Question 23.
In the given figure ABCD is a trapezium in which DC is parallel to AB.
AB = 16 cm and DC = 8 cm, OD = 5 cm, OB = (y + 3) cm, OA = 11 cm and OC = (x – 1) cm.
Using the given information answer the following questions.

(i) From the given figure name the pair of similar triangles: [1]
(a) ∆OAB, ∆OBC
(b) ∆COD, ∆AOB
(c) ∆ADB, ∆ACB
(d) ∆COD, ∆COB
Solution:
(b) ∆COD, ∆AOB
[∵ ∆COD ~ ∆AOB by AA similarity axiom]

(ii) The corresponding proportional sides with respect to the pair of similar triangles obtained in (i) : [ 1 ]
(a) \(\frac{C D}{A B}=\frac{O C}{O A}=\frac{O D}{O B}\)
(b) \(\frac{A D}{B C}=\frac{O C}{O A}=\frac{O D}{O B}\)
(c) \(\frac{A D}{B C}=\frac{B D}{A C}=\frac{A B}{D C}\)
(d) \(\frac{O D}{O B}=\frac{C D}{C B}=\frac{O C}{O A}\)
Solution:
(a) \(\frac{C D}{A B}=\frac{O C}{O A}=\frac{O D}{O B}\)

(iii) The ratio of the sides of the pair of similar triangles is: [1]
(a) 1 : 3
(b) 1 : 2
(c) 2 : 3
(d) 3 : 1
Solution:
(b) 1 : 2
[Here, \(\frac{\mathrm{CD}}{\mathrm{AB}}=\frac{\mathrm{OC}}{\mathrm{OA}}=\frac{\mathrm{OD}}{\mathrm{OB}}=\frac{8 \mathrm{~cm}}{16 \mathrm{~cm}}=\frac{1}{2}\)]

(iv) Using the ratio of sides of the pair of similar triangles the values of x and y are respectively: [1]
(a) x = 4.6, y = 7
(b) x = 7, y = 7
(c) x = 6.5, y = 7
(d) x = 6.5, y = 2
Solution:
(c) x = 6.5, y = 7
[Now, \(\frac{x-1}{11}=\frac{5}{y+3}=\frac{1}{2}\)
2x – 2 = 11 and y + 3 = 10
x = \(\frac{13}{2}\) = 6.5 and y = 7]

Question 24.
7vvo cars X and Y use 1 litre of diesel to travel x km and (x + 3) km respectively. If both the cars covered a distance of 72 km, then:

(i) the number of litres of diesel used by car X is : [ 1 ]
(a) \(\frac{72}{x-3}\) litres
(b) \(\frac{72}{x+3}\) litres
(c) \(\frac{72}{x}\) litres
(d) \(\frac{12}{x}\) litres
Solution:
(c) \(\frac{72}{x}\) litres

(ii) the number of litres of diesel used by car Y is : [ 1 ]
(a) \(\frac{72}{x-3}\) litres
(b) \(\frac{72}{x+3}\) litres
(c) \(\frac{72}{x}\) litres
(d) \(\frac{12}{x+3}\) litres
Solution:
(b) \(\frac{72}{x+3}\) litres

(iii) If car X used 4 litres of diesel more than car Y in the journey. then: [1]

Solution:
(c) \(\frac{72}{x}-\frac{72}{x+3}\) = 4

(iv) The amount of diesel used by the car X is : [1]
(a) 6 litres
(b) 12 litres
(c) 18 litres
(d) 24 litres
Solution:
(b) 12 litres
[Here, 72 \(\left(\frac{x+3-x}{x^2+3 x}\right)\) = 4
⇒ x² + 3x – 54 = 0
⇒ (x – 6) (x + 9) = 0
⇒ x = 6 as x ≠ -9]
Required number of litres = \(\frac{72}{6}\) = 12]

Question 25.
Joseph has a recurring deposit account in a bank for two years at the rate of 8% per annum simple interest.

(i) If at the time of maturity Joseph receives ₹ 2000 as interest, then the monthly instalment is: [1]
(a) ₹ 1200
(b) ₹ 600
(c) ₹ 1000
(d) ₹ 1600
Solution:
(c) ₹ 1000
[Here, P × \(\frac{n(n+1)}{2 \times 12} \times \frac{r}{100}\) = 1
⇒ P × \(\frac{24 \times 25}{24} \times \frac{8}{100}\) = 2000
⇒ P = 2000 ÷ 2 = 1000]

(ii) The total amount deposited in the bank: [1]
(a) ₹ 25000
(b) ₹ 24000
(c) ₹ 26000
(d) ₹ 23000
Solution:
(b) ₹ 24000
[Amount deposited = ₹ 1000 × 2 × 12
= ₹ 24000]

(iii) The amount Joseph receives on maturity is : [ 1 ]
(a) ₹ 27000
(b) ₹ 25000
(c) ₹ 26000
(d) ₹ 28000
Solution:
(c) ₹ 26000
[Amount on maturity = ₹ 24000 + ₹ 2000 = ₹ 26000]

(iv) If the monthly instalment is ₹ 100 and the rate of interest is 8%, in how many months Joseph will receive ₹ 52 as interest ? [1]
(a) 18
(b) 30
(c) 12
(d) 6
Solution:
(c) 12
[Here, P = ₹ 100, r = 8%, Interest = ₹ 52
Now, 100 × \(\frac{n(n+1)}{2 \times 12}\) \(\frac{8}{100}\) = 52
n(n + 1) = \(\frac{52 \times 2 \times 12}{8}\) = 156 = 12(13)
= 12(12 + 1)
Hence, n = 12

Solving ICSE Class 10 Maths Previous Year Question Papers ICSE Class 10 Maths Question Paper 2022 Semester 2 is the best way to boost your preparation for the board exams.

ICSE Class 10 Maths Question Paper 2022 Solved Semester 2

Time : 1 1/2 hour
Maximum Marks : 40

• Answers to this Paper must be written on the paper provided separately.
• You will not be allowed to write during the first 10 minutes.
• This time is to be spent in reading the question paper.
• The time given at the head of this Paper is the time allowed for writing the answers.
• Attempt all questions from Section A and any three questions from Section B.
• The marks intended for questions are given in brackets [ ].
• Mathematical tables are provided.

Section – A
(Attempt all questions)

Question 1.
Choose the correct answers to the questions from the given options. (Do not copy the question. Write the correct answer only.) [10]

(i) The probability of getting a number divisible by 3 in throwing a dice is:
(a) \(\frac{1}{6}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{2}{3}\)
Answer:
(b) \(\frac{1}{3}\)
[Here, S = {1, 2, 3, 4, 5} and E = {3, 6}]
Required probability = \(\frac{2}{6}\) = \(\frac{1}{3}\)

(ii) The volume of a conical tent is 462 m³ and the area of (he base is 154 m². The height of the cone is:
(a) 15 m
(b) 12 m
(c) 9 m
(d) 24 m
Answer:
(c) 9 m
[Here, \(\frac{1}{3}\) πr²h = 462
\(\frac{1}{3}\) × 154 × h = 462 ⇒ h = \(\frac{462}{154}\) × 3 = 9 m]

(iii) The median class for the given distribution is :

Class Interval	Frequency
0 – 10	2
10 – 20	4
20 – 30	3
30 – 40	5

(a) 0-10
(b)10-20
(c) 20 – 30
(d) 30-40
Answer:
(c) 20 – 30
[Here, N = 2 + 4 + 3 + 5 = 14 and \(\frac{\mathrm{N}}{\mathrm{2}}\) = \(\frac{14}{2}\) = 7
Class interval corresponding to \(\frac{\mathrm{N}}{\mathrm{2}}\) is the median class = 20 – 30]

(iv) If two lines are perpendicular to one another then the relation between their slopes m₁ and m₂ is :
(a) m₁ = m₂
(b) m₁ = \(\frac{1}{m_2}\)
(c) m₁ = -m₂
(d) m₁ × m₂ = -1
Answer:
(d) m₁ × m₂ = -1

(v) A lighthouse is 80 m high. The angle of elevation of its top from a point 80 m away from its foot along the same horizontal line is :
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Answer:
(b) 45°

[Here, tan θ = \(\frac{\mathrm{AB}}{\mathrm{OB}}\) = \(\frac{80}{80}\)
= 1
= tan 45°

(vi) The modal class of a given distribution always corresponds to the :
(a) interval with highest frequency
(b) interval with lowest frequency
(c) the first interval
(d) the last interval
Answer:
(a) interval with highest frequency

(vii) The coordinates of the point P (- 3, 5) on reflecting on the x-axis are :
(a) (3, 5)
(b) (- 3, -5)
(c) (3, – 5)
(d) (-3, 5)
Answer:
(b) (- 3, -5)

(viii) ABCD is a cyclic quadrilateral. If ∠BAD = (2x + 5)° and ∠BCD = (x + 10)° then x is equal to:
(a) 65°
(b) 450
(c) 55°
(d) 50
Answer:
(c) 55°
[Here, ∠A + ∠C = 180°
2x + 5 + x + 10 = 180
⇒ 3x = 180 – 15 = 165
⇒ x = 55

(ix) A(1, 4), B(4, 1) and C(x, 4) are the vertices of ∆ABC. 1f the centroid of the triangle is G(4, 3) then x is equal to
(a) 2
(b) 1
(c) 7
(d) 4
Answer:
(c) 7
[Here, \(\left(\frac{1+4+x}{3}, \frac{4+1+4}{3}\right)\) = (4, 3)
⇒ \(\left(\frac{5+x}{3}, \frac{9}{3}\right)\) = (4, 3)
⇒ 5 + x = 12
⇒ x = 7]

(x) The radius of a roller 100 cm long is 14 cm. The curved surface area of the roller is :
(Take π = \(\frac{22}{7}\))
(a) 13200 cm²
(b) 15400 cm²
(c) 4400 cm²
(d) 8800 cm²
Answer:
(d) 8800 cm²
[Given that, r = 14 cm, h = 100 cm
Curved surface area = 2πrh
= 2 × \(\frac{22}{7}\) × 14 × 100 = 8800 cm²]

Section – B
(Attempt any three questions from this section.)

Question 2.
(i) Prove that: \(\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}\) = 2 sec² θ [2]
Answer:
L.H.S.

Hence Proved.

(ii) Find ‘a’, if A (2a + 2, 3), B (7, 4) and C (2a + 5, 2) are collinear. [2]
Answer:
Here, ar(∆ABC) = 0
\(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
(2a + 2) (4 – 2) + 7(2 – 3) + (2a + 5)(3 – 4) = 0
4a + 4 – 7 – 2a – 5 = 0
2a = 8
⇒ a = 4

(iii) Calculate the mean of the following frequency distribution. [3]

Class Interval	Frequency
5 – 15	2
15 – 25	6
25 – 35	4
35 – 45	8
45 – 55	4

Answer:

Class Interval	Class Marks (xi)	Frequency (fi)	fixi
5 – 15	10	2	20
15 – 25	20	6	120
25 – 35	30	4	120
35 – 45	40	8	320
45 – 55	50	4	200
Total		24	780

Mean = \(\frac{\sum x_i f_i}{\sum f_i}\) = \(\frac{780}{24}\) = 32.5

(iv) In the given figure O is the centre of the circle. PQ and PR are tangents and ZQPR = 70°. Calculate : [3]

(a) ∠QOR
(b) ∠QSR
Answer:
(a) In quad. PROQ, ∠PQO = 90°, ∠PRO = 90°

∠QOR + ∠P = 180°
∠QOR + 70 = 180°
∠QOR = 180° – 70° = 110°

(b) ∠QTR = \(\frac{1}{2}\)∠QOR = \(\frac{1}{2}\) × 110°
= 55°
Since QSRT is a cyclic quadrilateral
∠QSR + ∠QTR = 180°
∠QSR = 180° – ∠QTR
= 180° – 55°
= 125°

Question 3.
(i) A bag contains 5 white, 2 red and 3 black balls. A ball is drawn at random. What is the probability that the ball drawn is a red ball ? [2]
Answer:
We have, 5 white, 2 red and 3 black balls
Total number of balls = 5 + 2 + 3 = 10
Number of balls of red colour = 2
∴ Required probability = \(\frac{2}{10}=\frac{1}{5}\)

(ii) A solid cone of radius 5 cm and height 9 cm is melted and made into small cylinders of radius of 0.5 cm and height 1.5 cm. Find the number of cylinders so formed. |2]
Answer:
Let n be the number of cylinders
∴ n × Volume of cylinder = Volume of cone
n × πR²H = \(\frac{1}{3}\) πr²h
n × 0.5 × 0.5 × 1.5 = \(\frac{1}{3}\) × 5 × 5 × 9
n = \(\frac{5 \times 5 \times 3}{5 \times 5 \times 15}\) × 1000
= 200

(iii) Two lamp posts AB and CD each of height 100 m are on either side of the road. P is a point on the road between the two lamp posts. The angles of elevation of the top of the lamp posts from the point P are 60° and 40°. Find the distances PB and PD. [3]

Answer:
In rt. ∠ed ∆DPC
\(\frac{\mathrm{CD}}{\mathrm{PD}}\) = tan 60° ⇒ \(\frac{\mathrm{100}}{\mathrm{PD}}\) = \(\sqrt{3}\)
⇒ PD = \(\frac{100}{\sqrt{3}}\) m
In rt. ∠ed ∆BPA
\(\frac{\mathrm{AB}}{\mathrm{PB}}\) = tan 40° ⇒ \(\frac{\mathrm{100}}{\mathrm{PB}}\) = 0.8391
⇒ PD = \(\frac{100}{0.8391}\) = 119.18 m

(iv) Marks obtained by 100 students in an examination are given below : [3]

Marks	No. of Students
0 – 10	5
10 – 20	15
20 – 30	20
30 – 40	28
40 – 50	20
50 – 60	12

Draw a histogram for the given data using a graph paper and find the mode.
Take 2 cm = 10 marks along one axis and 2 cm = 10 students atong the other axis.
Answer:

Mode = 35

Question 4.
(i) Find a point P which divides internally the line segment joining the points A(-3, 9) and B(1, -3) in the ratio 1 : 3. [2]
Answer:
P(x, y) = P \(\left(\frac{1-9}{1+3}, \frac{-3+27}{1+3}\right)\)
= P\(\left(\frac{-8}{4}, \frac{24}{4}\right)\) = P(-2, 6)

(ii) A letter of the word ‘SECONDARY’ is selected at random. What Is the probability that the letter selected Is not a vowel ? [2]
Answer:
Here, S = {S, E, C, O, N, D, A, R, Y} = 9
Event (E) = {S, C, N, D, R, Y} = 6
P(E) = \(\frac{6}{9}\) = \(\frac{2}{3}\)

(iii) Use a graph paper for this question. Take 2 cm = I unit along both the axes. [3]
(a) Plot the points A(0, 4), B(2, 2), C(5, 2) and D(4, 0), E(0, 0) is the origin.
(b) Reflect B, C, D on the y-axis and name them as B’, C’ and D’ respectively.
(c) Join the points ABCDD’C’B’ and A in order and give a geometrical name to the closed figure.
Answer:
(a)

(b) B'(-2, 2)
C'(-5, 2)
D’ (-4, 0)

(c) Boat

(iv) A solid wooden cylinder is of radius 6 cm and height 16 cm. Two cones each of radius 2 cm and height 6 cm are drilled out of the cylinder. Find the volume of the remaining solid. Take π = \(\frac{22}{7}\)

Solution:
Volume of the remaining solid
= \(\frac{22}{7}\) × 6 × 6 × 16 – \(\frac{2}{3}\) × \(\frac{22}{7}\) × 2 × 2 × 6
= \(\frac{12672}{7}\) – \(\frac{352}{7}\) = \(\frac{12320}{7}\) = 1760 cm³

Question 5.
(i) Two chords AB and CD of a circle intersect externally at E. If EC = 2 cm, EA = 3 cm and AB = 5 cm, find the length of CD. [2]

Solution:
Here, two chords AB and CD intersect externally at E.
∴ BE × AE = DE × CE
(5 + 3) × 3 = (CD + 2) × 2
12 = CD + 2
⇒ CD = 12 – 2
= 10 cm

(ii) Line AB is perpendicular to CD. Coordinates of B,C and D are respectively (4, 0), (0, -1) and (4, 3).

Find : [2]
(a) Slope of CD
(b) Equation of AB
Solution:
(a) Slope of CD = \(\frac{3+1}{4-0}=\frac{4}{4}\) = 1
(b) Slope of AB =\(\frac{-1}{\text { Slope of } C D}\) = -1
(∵ AB ⊥ CD)
Equation of AB
y – 0 = 1(x – 4)
= -x + 4
x + y – 4 = 0

(iii) Prove that : \(\frac{(1+\sin \theta)^2+(1-\sin \theta)^2}{2 \cos ^2 \theta}\) = sec²θ + tan²θ [3]
Solution:
L.H.S.

= sec²θ + tan²θ
= R.H.S

(iv) The mean of the following distribution is 50. Find the unknown frequency. [3]

Class Interval	Frequency
0 – 20	6
20 – 40	f
40 – 60	8
60 – 80	12
80 – 100	8

Solution:

Class Interval	Class Marks (xi)	Frequency (fi)	fixi
0 – 20	10	6	60
20 – 40	30	f	30 f
40 – 60	50	8	400
60 – 80	70	12	840
80 – 100	90	8	720
Total		34+ f	30f + 2020

Mean = 50 (given)
\(\frac{\sum f_i x_i}{\sum f_i}\) = 50
\(\frac{30 f+2020}{34+f}\) = 50
30f + 2020 = 1700 + 50f ⇒ 20f = 320
⇒ f = 16

Question 6.
(i) Prove that: 1 + \(\frac{\tan ^2 \theta}{1+\sec \theta}\) = sec²θ [2]
Solution:
L.H.S. = 1 + \(\frac{\tan ^2 \theta}{1+\sec \theta}\) = 1 + \(\frac{\sec ^2 \theta-1}{\sec \theta+1}\)
= 1 + \(\frac{(\sec \theta-1)(\sec \theta+1)}{(\sec \theta+1)}\)
= 1 + sec θ – 1 = sec θ = R.H.S.

(ii) In the given figure A, B, C and D are points on the circle with centre O. Given ∠ABC = 62°.
Find: [2]
(a) ∠ADC
(b) ∠CAB

Solution:
(a) ∠ADC = ∠ABC (angles in the same segment AC)
= 62°

(b) Since AOB is a diameter
∴ ∠BCA = 90° (angle in the semi circle)
In ∆ABC
∠ABC + ∠BCA + ∠CAB = 180°
62° + 900 + LCAB = 180°
∠CAB = 180° – 90° – 62° = 28°

(iii) Find the equation of a line parallel to the line 2x + y – 7 = 0 and passing through the intersection of the lines x + y – 4 = 0 and 2x – y = 8. [3]
Solution:
x + y – 4 = 0 …………… (i); 2x – y = 8 …………….. (ii)
Adding (i) and (ii), we have
3x = 12 ⇒x = 4
From (i), we obtain y = 0
∴ Point of intersection of (i) and (ii) is (4, 0)
Slope of line 2x + y – 7 = 0 = – \(\frac{2}{1}\) = -2
Hence, the equation of required line, is :
y – 0 = -2 (x – 4) or 2x + y – 8 = 0.

(iv) Marks obtained by 40 students in an examination are given below:

Marks	No. of Students
10 – 20	3
20 – 30	8
30 – 40	14
40 – 50	9
50 – 60	4
60 – 70	2

Using graph paper draw an ogive and estimate the median marks.
Take 2 cm = 10 marks along one axis and 2 cm = 5 students along the other axis. [3]
Solution:

Marks	No. of Students	c.f	Points
10 – 20	3	3	(20, 3)
20 – 30	8	11	(30, 11)
30 – 40	14	25	(40, 25)
40 – 50	9	34	(50, 34)
50 – 60	4	38	(60, 38)
60 – 70	2	40	(70, 40)

Plot the points shown in the table on the graph and joining these points by a free hand curve, starting from the lower limit of first class interval and ending at the upper limit of last class, we have the required ogive.
Median = \(\left(\frac{\mathrm{N}}{2}\right)^{\text {th }}\) term = \(\frac{40}{2}\) = 20^th term
Draw NP || x-axis and PM ⊥ x-axis
The value of point M on x-axis is the median.
Hence, the median marks = 37 marks.

Solving ICSE Class 10 Physics Previous Year Question Papers ICSE Class 10 Physics Question Paper 2018 is the best way to boost your preparation for the board exams.

ICSE Class 10 Physics Question Paper 2018 Solved

Time: 1½ hours
Maximum Marks: 80

General Instructions:

• Answers to this paper must be written on the paper provided separately.
• You will NOT be allowed to write during the first 15 minutes.
• This time is to be spent in reading the question paper.
• The time given at the head of this paper is the time allowed for writing the answers.
• Section 1 is compulsory. Attempt any four questions from Section II.
• The intended marks for questions or parts of questions are given in brackets [ ].

SECTION – I (40 Marks)
(Attempt all questions from this Section)

Question 1.
(a) (i) State and define the SI unit of power.
(ii) How is the unit horse power related to the SI unit of power? (2)
Answer:
(i) The SI unit of power is watt (W). The power of an agent is said to be one watt if it does one joule of work in one second.
(ii) 1 hp = 746 W

(b) State the energy changes in the following cases while in use : (2)
(i) An electric iron.
(ii) A ceiling fan.
Answer:
(i) Electrical into heat energy
(ii) Electrical into mechanical energy.

(c) The diagram below shows a lever in use : (2)
(i) To which class of levers does it belong?
(ii) Without changing the dimensions of the lever, if the load is shifted towards the fulcrum what happens to the mechanical advantage of the lever?

Answer:
(i) Second class lever.
(ii) It increases.

(d) (i) Why is the ratio of the velocities of light of wavelengths 4000 A and 8000 A in vacuum 1:1?
(ii) Which of the above wavelengths has a higher frequency? (2)
Answer:
(i) Because all wavelengths of light travel with the same velocity in vacuum.
(ii) 4000 Å.

(e) (i) Why is the motion of a body moving with a constant speed around a circular path said to be accelerated?
(ii) Name the unit of physical quantity obtained by the formula 2K/V2 where K: kinetic energy, V: Linear velocity. (2)
Answer:
(i) Because at every point of motion the direction of speed changes i. e., the body possesses velocity which changes with time.
(ii) J s^-2 m^-2

Question 2.
(a) The power of a lens is -5D. (2)
(i) Find its focal length.
(ii) Name the type of lens.
Answer:
(i) f = \(\frac{1}{P}=\frac{1}{-5}\) = -0-2 m = 20 cm
(ii) Concave

(b) State the position of the object in front of a converging lens if: (2)
(i) It produces a real and same size image of the object.
(ii) It is used as a magnifying lens.
Answer:
(i) At 2F
(ii) Between F and 2F

(c) (i) State the relation between the critical angle and the absolute refractive index of a medium.
(ii) Which colour of light has a higher critical angle? Red light or Green light. (2)
Answer:
(i) n = \(\frac{1}{\sin C}\)
(ii) Red

(d) (i) Define scattering. (2)
(ii) The smoke from a fire looks white. Which of the following statements is true?
(1) Molecules of the smoke are bigger than the wavelength of light.
(2) Molecules of the smoke are smaller than the wavelength of light.
Answer:
(i) Scattering is a general physical process where light is forced to deviate from a straight trajectory by one or more paths due to localized non-uniformities in the medium through which it passes.
(ii) Molecules of the smoke are bigger than the wavelength of light.

(e) The following diagram shows a 60°, 30°, 90° glass prism of critical angle 42°. Copy the diagram and complete the path of incident ray AB emerging out of the prism marking the angle of incidence on each surface. (2)

Answer:
The diagram is as shown

Question 3.
(a) Displacement distance graph of two sound waves A and B, travelling in a medium, are as shown in the diagram below. (2)

Study the two sound waves and compare their:
(i) Amplitudes
(ii) Wavelengths
Answer:
(i) A_A/ A_B = 20 /10 = 2 /1 = 2
(ii) λ_A > λ_B

(b) You have three resistors of values 2 Ω, 3 Ω and 5 Ω. How will you join them so that the total resistance is more than 7 Ω? (2)
(i) Draw a diagram for the arrangement.
(ii) Calculate the equivalent resistance.
Answer:
(i) All of these can be connected in series as shown

(ii) R = 2 + 3 + 5 = 10Ω

(c) (i) What do you understand by the term nuclear fusion? (2)
(ii) Nuclear power plants use nuclear fission reaction to produce electricity. What is the advantage of producing electricity by fusion reaction?
Answer:
(i) It is a process in which two or more smaller nuclei combine to form a bigger nucleus.
(ii) There is no pollution in nuclear fusion.

(d) (i) What do you understand by free vibrations of a body? (2)
(ii) Why does the amplitude of a vibrating body continuously decrease during damped vibrations?
Answer:
(i) If a body vibrates with its own frequency it is called a free vibration.
(ii) Energy is used up in overcoming friction.

(e) (i) How is the emf across primary and secondary coils of a transformer related with the number of turns of coil in them? (2)
(ii) On which type of current do transformers work? (2)
Answer:
(i) \(\frac{\mathrm{V}_{\mathrm{P}}}{\mathrm{V}_{\mathrm{S}}}=\frac{\mathrm{N}_{\mathrm{P}}}{\mathrm{N}_{\mathrm{S}}}\)
(ii) Alternating current.

Question 4.
(a) (i) How can a temperature in degree Celsius be converted into SI unit of temperature? (2)
(ii) A liquid X has the maximum specific heat capacity and is used as a coolant in Car radiators. Name the liquid X.
Answer:
(i) 1°C = 273 K
(ii) Water

(b) A solid metal weighing 150 g melts at its melting point of 800°C by providing heat at the rate of 100 W. The time taken for it to completely melt at the same temperature is 4 mm. What is the specific latent heat of fusion of the metal? (2)
Answer:
Using Q = mL,
100 × 4 × 60 = 150 × 10^-3 × L
or L = \(\frac{100 \times 4 \times 60}{150 \times 10^{-3}}\)
= L = 1.6 × 10⁵ J kg^-1

(c) Identify the following wires used in a household circuit: (2)
(i) The wire is also called as the phase wire.
(ii) The wire is connected to the top terminal of a three-pin socket.
Answer:
(i) Live wire
(ii) Earth wire

(d) (i) What are isobars?
(ii) Give one example of isobars.
Answer:
(i) Elements having same mass number but different atomic number.
(ii) ⁴⁰S, ⁴⁰Cl,⁴⁰Ar, ⁴⁰K, and ⁴⁰Ca.

(e) State any two advantages of electromagnets over permanent magnets. (2)
Answer:
(i) Strength can be increased or decreased during working.
(ii) Polarities can be changed during working.

SECTION – II (40 Marks)
(Attempt any four questions from this Section)

Question 5.
(a) (i) Derive a relationship between SI and C.G.S. unit of work. (3)
(ii) A force acts on a body and displaces it by a distance S in a direction at an angle 0 with the direction of force. What should be the value of 0 to get the maximum positive work?
Answer:
(i) 1 J = 1 kg × 1 m² × 1 s^-2
= 10³ g × 10⁴ cm² × 1 s^-2 g cm² s^-2
= 10⁷ erg

(ii) W = FS cos θ
= FS (maximum) if cos θ
= 1
or θ = 0°

(b) A half metre rod is pivoted at the centre with two weights of 20 gf and 12 gf suspended at a perpendicular distance of 6 cm and 10 cm from the pivot respectively as shown below: (3)

(i) Which of the two forces acting on the rigid rod causes clockwise moment?
(ii) Is the rod in equilibrium?
(iii)The direction of 20 kgf force is reversed. What is the magnitude of the resultant moment of the forces on the rod?
Answer:
(i) 12 kgf
(ii) Yes
(iii) 120 + 120 = 240 dyne cm

(c) (i) Draw a diagram to show a block and tackle pulley system having a velocity ratio of 3 marking the direction of load (L), effort (E) and tension (T).
(ii) The pulley system drawn lifts a load of 150 N when an effort of 60 N is applied. Find its mechanical advantage.
(iii) Is the above pulley system an ideal machine or not? (4)
Answer:
(i) The diagram is as shown

(ii) Actual MA number of pulleys in the system = 3. Observed MA = \(\frac{\text { load }}{\text { effort }}=\frac{150}{60}\) = =2.5
(iii) Since, observed MA is lesser than the theoretical MA therefore, the machine is not an ideal machine.

Question 6.
(a) A ray of light XY passes through a right angled isosceles prism as shown below. (3)

(i) What is the angle through which the incident ray deviates and emerges out of the prism?
(ii) Name the instrument where this action of prism is put into use.
(iii) Which prism surface will behave as a mirror?
Answer:
(i) 90°
(ii) Refracting Periscope
(iii) Surface AB

(b) An object AB is placed between O and F₁ on the principal axis of converging lens as shown in the diagram. (3)

Copy the diagram and by using three standard rays starting from point A, obtain an image of the object AB.
Answer:
The figure is as shown.

(c) An object is placed at a distance of 12 cm from a convex lens of focal length 8 cm. Find : (4)
(i) the position of the image
(ii) nature of the image.
Answer:
Given u = -12 cm, f = +8 cm, v = ?, nature = ?
Using the lens formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
we have
\(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{8}+\frac{1}{-12}=\frac{3-2}{24}=\frac{1}{24}\)
Or v = 24 cm,
(ii) Image will be real or inverted.

Question 7.
(a) Draw the diagram of a right angled isosceles prism which is used to make an inverted image erect. (3)
Answer:
The diagram is as shown.

(b)

The diagram above shows a wire stretched over a sonometer. Stems of two vibrating tuning forks A and B are touched to the wooden box of the sonometer. It is observed that the paper rider (a small piece of paper folded at the centre) present on the wire flies off when the stem of vibrating tuning fork B is touched to the wooden box but the paper just vibrates when the stem of vibrating tuning fork A is touched to the wooden box.
(i) Name the phenomenon when the paper rider just vibrates.
(ii) Name the phenomenon when the paper rider flies off.
(iii) Why does the paper rider fly off when the stem of tuning fork B is touched to the box?
Answer:
(i) Forced vibrations
(ii) Resonance
(iii)The wire of the sonometer begins to vibrate with its natural frequency. This is called resonance. At resonance the amplitude of vibration becomes very large.

(c) A person is standing at the sea shore. An observer on the ship, which is anchored in between a vertical cliff and the person on the shore, fires a gun. The person on the shore hears two sounds, 2 seconds and 3 seconds after seeing the smoke of the fired gun. If the speed of sound in the air is 320 m s^-1 then calculate :
(i) the distance between the observer on the ship and the person on the shore.
(ii) the distance between the cliff and the observer on the ship. (4)

Answer:
Let distance between the observer on the ship and person on the shore be x. For the first sound which the person hears after 2 seconds, we have
v = \(\frac{x}{t}\)
⇒ x = vt = 320 x 2 = 640 m
Let distance between the ship and the cliff be x, then in the second case 2x + 640
v = \(\frac{2 x+640}{t}\)
⇒ 2x + 640 = vt = 320 × 3
= 960
Or 2x = 960 – 640
= 320 m
x = \(\frac{320}{2}\) = 160 m

Question 8.
(a) (i) A fuse is rated 8 A. Can it be used with an electrical appliance rated 5 kW, 200 V? Give a reason. (3)
(ii) Name two safety devices which are connected to the live wire of a household electric circuit.
Answer:
(i) The current which will pass through the electrical device will be I = \(\frac{\mathrm{P}}{\mathrm{V}}=\frac{5000}{200}\) = 25 A. The fuse cannot be used as it will blow off when the current exceeds 8 A.

(ii) Fuse, MCB

(b) (i) Find the equivalent resistance between A and B.

(ii) State whether the resistivity of a wire changes with the change in the thickness of the wire,
Answer:
(i) 6 Ω and 3 Ω are in parallel therefore,
\(\frac{6 \times 3}{6+3}=\frac{18}{9}\) = 2Ω
4 Ω and 12 Ω are in parallel therefore,
\(\frac{4 \times 12}{4+12}=\frac{48}{16}\) = 3Ω
The above two are now in series therefore, R = 2 + 3 = 5Ω

(ii) Resistivity is independent of the dimensions of the wire. Therefore, no change.

(c) An electric iron is rated 220 V, 2 kW. (4)
(i) If the iron is used for 2 h daily find the cost of running it for one week if it costs ₹ 4.25 per kWh.
(ii) Why is the fuse absolutely necessary in a power circuit?
Answer:
(i) V = 220 V, P = 2 kW = 2000 W
Energy consumed E
= \(\frac{P \text { (in watt) } \times t \text { (in hour) }}{1000}=\frac{2000 \times 2 \times 7}{1000}\) = 28 kWh
Hence cost = 28 × 4.25 = ₹ 119

(ii) Power circuits draw large amount of current, Electric shock in this circuit is very fatal hence to avoid it fuse is necessary in the power circuit.

Question 9.
(a) (i) Heat supplied to a solid change it into liquid. What is this change in phase called? (3)
(ii) During the phase change does the average kinetic energy of the molecules of the substance increase?
(iii) What is the energy absorbed during the phase change called?
Answer:
(i) Melting
(ii) No
(iii) Latent heat of fusion.

(b) (i) State two differences between “Heat Capacity” and “Specific Heat Capacity”.
(ii) Give a mathematical relation between Heat Capacity and Specific Heat Capacity.
Answer:
(i) The two differences are

Specific heat capacity	Heat capacity
1. Heat required to raise the temperature of a substance by 1 degree Celsius.	1. Heat required to raise the temperature of a unit mass of a substance by 1 degree Celsius.
2. Independent of mass of the substance.	2. Depend upon of mass of the substance.

(ii) The relation is, heat capacity = Specific heat × Mass of substance.

(c) The temperature of 170 g of water at 50°C is lowered to 5°C by adding certain amount of ice to it. Find the mass of ice added. Given: Specific heat capacity of water = 4200 J kg^-1 °C^-1 and Specific latent heat of ice 336000 J kg^-1. (4)
Answer:
Given mass of ice = mg
Mass of water = 170 g
Initial temperature of water = 50°C
Let the final temperature of the mixture = 5°C
The latent heat required to change m g of ice at 0°C to m g of water at 0°C
= m × 336
= 336 m J
Now, heat required to change m g of water at 0°C to 5°C
= m × 4.2 × (5-0) J
= 21 m J
Now, heat lost by 170 g of water to reach 5°C from 50°C
= 170 × 4.2 × (50 – 5)
= 32 130 J
By principle of calorimetry
Heat lost = Heat gained
336 m+ 21 m = 32  130
or 357 m = 32  130
m = 90 g

Question 10.
(a) The diagram shows a coil wound around a U shape soft iron bar AB.

(i) What is the polarity induced at the ends A and B when the switch is pressed?
(ii) Suggest one way to strengthen the magnetic field in the electromagnet.
(iii) What will be the polarities at A & B if the direction of current is reversed in the circuit? (3)
Answer:
(i) A South and B South.
(ii) Increasing the strength of current through the coil.
(iii) The Polarities of the two ends become north.

(b) The ore of Uranium found in nature contains 92U238 and 92U235. Although both the isotopes are fissionable, it is found out experimentally that one of the two isotopes is more easily fissionable.
(i) Name the isotope of Uranium which is easily fissionable.
(ii) Give a reason for your answer.
(iii) Write a nuclear reaction when Uranium 238 emits an alpha particle to form a Thorium (Th) nucleus. (3)
Answer:
(i) ₉₂U²³⁵
(ii) It readily absorbs thermal neutrons and becomes unstable.
(iii) The reaction is
₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴

(c) Radiations given out from a source when subjected to an electric field in a direction perpendicular to their path are shown below in the diagram. The arrows show the path of the radiation A, B and C. Answer the following questions in terms of A, B and C. (4)
(i) Name the radiation B which is unaffected by the electrostatic field.
(ii) Why does the radiation C deflect more than A?
(iii) Which among the three causes the least biological damage externally?
(iv) Name the radiation which is used in carbon dating.

Answer:
(i) Gamma radiation
(ii) Because C is lighter than A.
(iii) Alpha (A)
(iv) Carbon-14

Solving ICSE Class 10 Physics Previous Year Question Papers ICSE Class 10 Physics Question Paper 2017 is the best way to boost your preparation for the board exams.

ICSE Class 10 Physics Question Paper 2017 Solved

Time: 1½ hours
Maximum Marks: 80

General Instructions:

• Answers to this paper must be written on the paper provided separately.
• You will NOT be allowed to write during the first 15 minutes.
• This time is to be spent in reading the question paper.
• The time given at the head of this paper is the time allowed for writing the answers.
• Section I is compulsory. Attempt any four questions from Section II.
• The intended marks for questions or parts of questions are given in brackets [ ].

Section – I (40 Marks)
(Attempt all questions from this Section)

Question 1.
(a) A brass ball is hanging from a stiff cotton thread. Draw a neat labelled diagram showing the forces acting on the brass ball and the cotton thread. (2)
Answer:
The diagram is as shown.

(b) The distance between two bodies is doubled. How is the magnitude of gravitational force between them affected? (2)
Answer:
It becomes one-fourth.
This is because F = \(\frac{G m_1 m_2}{r^2}\)

(c) Why is a jack screw provided with a long arm? (2)
Answer:
As a large mechanical advantage is required by the jack screw used for cutting metal, hence its handle should be long and the blades short. (MA = effort arm / load arm)

(d) If the power of a motor be 100 kW, at what speed can it raise a load of 50,000 N? (2)
Answer:
P = F ν, therefore, ν = P/F = 100000 / 50000 = 2 m s^-1.

(e) Which class of level will always have MA > I and why? (2)
Answer:
Class II lever, because the effort arm is always greater than the load arm.

Question 2.
(a) Define heat capacity and state its SI unit. (2)
Answer:
It is defined as the amount of heat required in raising the temperature of a unit mass of a substance through 1°C. Its SI unit is J kg^-1 K^-1.

(b) Why is the base of a cooking pan generally made thick? (2)
Answer:
It is because thicker pans can have better heat distribution and heat retention due to thermal conductivity and thermal mass.

(c) A solid of mass 50 g at 150°C is placed in 100 g of water at 11°C, when the final temperature recorded is 20°C. Find the specific heat capacity of the solid.
(Specific heat capacity of water = 4.2 J/g°C)
Answer:
Given mass of solid m_s = 50 × 10^-3 kg
Mass of water m_w = 100 × 10^-3 kg
Temperature of solid = 1500°C
Temperature of water = 110°C
Final temperature = 200°C
Specific heat of water = 4.2 J kg^-1 °C^-1
Now,heat lost by the solid = heat gained by water
50 × 10^-3 × c × (1500 – 200)
= 100 x 10^-3 x 4200 x (200 – 110)
Or c = 581 Jkg^-1 °C^-1

(d) How is the refractive index of a material related to :
(i) real and apparent depth?
(ii) velocity of light in vacuum or air and the velocity of light in a given medium? (2)
Answer:
(i) μ = \(\frac{\text { Real depth }}{\text { Apparent depth }}\) and
(ii) μ = \(\frac{c}{v}\)

(e) State the conditions required for total internal reflection of light to take place. (2)
Answer:
(i) The incident ray should travel from the denser to the rarer medium.
(ii) The angle of incidence in the denser medium should be greater than the critical angle for the given pair of media.

Question 3.
(a) Draw a ray diagram to show the refraction of a monochromatic ray through a prism when it suffers minimum deviation. (2)
Answer:
The ray diagram is as shown :

(b) The human ear can detect continuous sounds in the frequency range from 20 Hz to 20,000 Hz.
Assuming that the speed of sound in air is 330 ms^-1 for all frequencies, calculate the wavelengths corresponding to the given extreme frequencies of the audible range. (2)
Answer:
The corresponding wavelengths are
λ_lowest = \(\frac{\mathrm{V}}{v}=\frac{330}{20000}\) = 0.0165 m and
λ_highest = \(\frac{\mathrm{V}}{v}=\frac{330}{20}\) = 16.5 m.

(c) An enemy plane is at a distance of 300 km from a radar. In how much time will the radar be able to detect the plane? Take velocity of radiowaves as 3 × 10⁸ ms^-1. (2)
Answer:
Given S = 300 km = 300000 m,
c = 3 × 10⁸ m s^-1, t = ?
Using \(\frac{2 \mathrm{~S}}{v}=\frac{2 \times 300000}{3 \times 10^8}\) = 2 × 10^-3 s

(d) How is the frequency of a stretched string related to :
(i) its length?
(ii) its tension? (2)
Answer:
(i) Inversely proportional to its length and
(ii) Directly proportional to the square root of tension.

(e) Define specific resistance and state its SI unit. (2)
Answer:
It is the resistance of unit cube of the material. Its SI unit is ohm-m.

Question 4.
(a) An electric bulb of resistance 500 Q draws a current of 0.4 A. Calculate the power of the bulb and the potential difference at its end. (2)
Answer:
Given R = 500 W, I = 0.4 A, V = ?, Using V = IR,
we have V = 0.4 × 500 = 200 V

(b) State two causes of energy loss in a transformer. (2)
Answer:
Copper loss and iron loss.

(c) State two characteristics of a good thermion emitter. (2)
Answer:

1. Should have high melting point and
2. Should have low value of work function.

(d) State two factors upon which the rate of emission of thermions depends. (2)
Answer:

1. The nature of the metal and
2. Temperature of the metal surface.

(e) When does the nucleus of an atom tend to be radioactive?
Answer:
The nucleus tends to be radioactive when it becomes unstable.

Section – II (40 Marks)
(Attempt any four questions from this Section)

Question 5.
(a) A uniform half-metre rule balances horizontally on a knife edge at the 29 cm mark when a weight of 20 gf is suspended from one end.
(i) Draw a diagram of the arrangement.
(ii) What is the weight of the half-metre rule? (3)
Answer:
Let the weight of the half-meter scale be W.

To balance the half-metre scale at 29 cm marie, the 20 gf weight should be suspended from the 50 cm end as shown.
Moments due to W in anticlockwise direction
= W gf × 4 cm
Moments due to 20 gf in clockwise direction
= 20 gf × 21 cm
By the principle of moments we have
W gf × 4 = 20 gf × 21
Or W = \(\frac{20 \times 21}{4}\) = 105 gf

(b) (i) A boy uses a single fixed pulley to lift a load of 50 kgf to some height. Another boy uses a single movable pulley to lift the same load to the same height. Compare the efforts applied by them. Give a reason to support your answer.
(ii) How does uniform circular motion differ from uniform linear motion?
(iii)Name the process used for producing electricity using nuclear energy. (3)
Answer:
(i) A single fixed pulley simply changes the direction of the action of the force. Therefore, for this pulley E = L = T = 50 kgf. Thus, it has a mechanical advantage 1.

For a single movable pulley, we have L = T + T = 2T and E = T = L/2 = 50/2 = 25 kgf, therefore, its mechanical advantage is 2.

(ii) In uniform circular motion, velocity is not constant, whereas in uniform linear motion velocity is constant. (iii) Nuclear fission

(c) A pulley system with VR = 4 is used to lift a load of 175 kgf through a vertical height of 15 m. The effort required is 50 kgf in the downward direction (g = 10 N kg^-1). (4)
(i) Distance moved by the effort.
(ii) Work done by the effort.
(iii) MA of the pulley system.
(iv) Efficiency of the pulley system.
Answer:
Given VR = 4, L = 175 kgf, distance moved by load = d_L = 15 m, E = 50 kgf, d_E = ?, W_E = ?, MA = ?, h = ?
Using the formula VR
= \(\frac{\text { distance moved by effort }}{\text { distance moved by load }}=\frac{d_{\mathrm{E}}}{d_{\mathrm{L}}}\) ,
therefore, we have
d_E = VR x d_L
= 4 × 15 = 60 m
W_E = Effort × d_E
= 50 × 10 × 60 = 30000 J
Now, MA = \(\frac{\text { load }}{\text { effort }}=\frac{175}{50}\) = 3.5
Now, η = \(\frac{\mathrm{MA}}{\mathrm{VR}}=\frac{3.5}{4}\) = 0.875 = 87.5%

Question 6.
(a) (i) How is the transference of heat energy by radiation prevented in a calorimeter?
(ii) You have a choice of three metals, A, B and C, of specific heat capacities 900 Jkg^-1C^-1, 380 Jkg^-1C^-1 and 460 Jkg^-1C^-1 respectively, to make a calorimeter. Which material will you select? Justify your answer? (3)
Answer:
(i) This is done by polishing the inner and outer surface of the copper calorimeter and the space between the copper vessel and the insulating container is filled with some poor conductor like wood wool or glass wool.

(ii) A, as it will gain less energy in a given time.

(b) Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 30°C such that the final temperature is 5°C. (3)
Specific heat capacity of calorimeter 0.4 J/g°C
Specific heat capacity of water 4.2 J/g°C
Latent heat capacity of ice 330 J/g
Answer:
Given m_w = 150 g, m_i = ?, m_C = 50 g, T_i = 0 °C, T_w = 32 °C, T_f = 5 °C
Specific heat capacity of calorimeter = 0.4 J g^-1 °C^-1
Specific heat capacity of water = 4.2 J g^-1 °C^-1
Latent heat capacity of ice 330 J g^-1
Heat lost by 150 g water + 50 g of calorimeter in falling form 32 °C to 5 °C
= 150 × 4.2× 27 + 50 × 0.4 × 27 = 17550 J
Heat needed to melt ice = m_i L = m_i × 330
= 330 m_i
So 330m_i = 17550 or m_i = 17550/330 = 53.2 g

(c) (i) Name the radiations which are absorbed by greenhouse gases in the earth’s atmosphere.
(ii) A radiation X is focused by a particular device on the bulb of a thermometer and mercury in the thermometer shows a rapid increase. Name the radiation X.
(iii)Name two factors on which the heat energy liberated by a body depends.
Answer:
(i) Infrared
(ii) Infrared
(iii) Temperature and surface area.

Question 7.
(a) A lens forms an upright and diminished image of an object when the object is placed at the focal point of the given lens.
(i) Name the lens.
(ii) Draw a ray diagram to show the image formation. (3)
Answer:
(i) Concave lens.
(ii) The diagram is shown.

(b) A ray of light travels from water to air as shown in the diagram given below:

(i) Copy the diagram and complete the path of the ray. Given that the critical angle for water is 48°.
(ii) State the condition so that total internal reflection occurs in the above diagram. (3)
Answer:
(i) The redrawn diagram is as shown :

(ii) The angle of incidence should be greater than the critical angle.

(c) The diagram below shows a point source P inside a water container. Four rays A, B, C, D starting from the source P are shown up to the water surface.

(i) Show in the diagram the path of these rays after striking the water surface. The critical angle for water-air surface is 48°.
(ii) Name the phenomenon which the rays B and D exhibit. (4)
Answer:
(i) The diagram is as shown.

(ii) Ray B— Refraction, Ray D —Total internal reflection.

Question 8.
(a) Name the factor that determines : (3)
(i) Loudness of the sound heard.
(ii) Quality of the note.
(iii) Pitch of the note.
Answer:
(i) Intensity/Amplitude.
(ii) Frequency of source and relative motion between the source of sound and the listener.
(iii) The number and the frequency of the overtones present in a given sound.

(b) (i) What are damped vibrations? (3)
(ii) Give one example of damped vibrations.
(iii) Name the phenomenon that causes a loud sound when the stem of a vibrating tuning fork is kept pressed on the surface of a table.
Answer:
(i) Those vibrations which die out with time.
(ii) Vibrations of a simple pendulum.
(iii) Resonance

(c) (i) A wire of length 80 cm has a frequency of 256 Hz. Calculate that length of a similar wire ‘ under similar tension, which will have a frequency of 1024 Hz. (4)
(ii) A certain sound has a frequency of 256 hertz and a wavelength of 1.3 m.
(1) Calculate the speed with which this sound travels.
(2) What difference would be felt by a listener between the above sound and another sound travelling at the same speed, but of wavelength 2.6 m?
Answer:
(i) Given L₁ = 80 cm, v₁ = 256 Hz, L₂ = ?, v₂ = 1024 Hz
We know that for a vibrating string
Lv = constant. Therefore, we have
L₂ = \(\frac{\mathrm{L}_1 v_1}{v_2}=\frac{80 \times 256}{1024}\) = 20 m

(ii)

1. v = 256 Hz, λ = 1.3 m, V = ?
   Using V = vλ, we have
   V = 256 × 1.3 = 332.8 ms^-1
2. He will hear a dull sound as the frequency of the sound wave of wavelength 2.6 m will be half the frequency (256/2 = 128 Hz) of the sound of wavelength 1.3 m.

Question 9.
(a) (i) Name the colour code of the wire which has connected to the metallic body of an appliance,
(ii) Draw the diagram of a dual control switch when the appliance is switched ‘ON’. (3)
Answer:
(i) Green or yellow.

(ii)

(b) (i) Which particles are responsible for current in conductors?
(ii) To which wire of a cable in a power circuit should the metal case of a geyser be connected?
(iii) To which wire should the fuse be connected? (3)
Answer:
(i) Electrons
(ii) Earth wire
(iii) Live wire

(c) (i) Explain the meaning of the statement ‘current rating of a fuse is 5A’.
(ii) In the transmission of power the voltage of power generated at the generating stations is stepped up from 11 kV to 132 kV before it is transmitted. Why? (4)
Answer:
(i) It means that it can withstand a maximum current of 5 A before it “blows”.
(ii) Due to this the current flowing in the circuit decreases. As a result heat losses I²R decreases.

Question 10.
(a) Answer the. following questions based on a hot cathode ray tube. (3)
(i) Name the charged particles.
(ii) State the approximate voltage used to heat the filament.
(iii) What will happen to the beam when it passes through the electric field?
Answer:
(i) Electrons.
(ii) 5 to 6 V.
(iii) It is deflected by the electric field.

(b) State three factors on which the rate of emission of electrons from a metal surface depends. (3)
Answer:

1. Nature of the metallic surface
2. Temperature of the metal surface and
3. Surface area of the metal surface.

(c) (i) What are the free electrons?
(ii) Why do they not leave the metal surface on their own?
(iii) How can they be made to leave the metal surface? (State any two ways) (4)
Answer:
(i) These are the electrons which are not bound to the nucleus and are free to move in the metal.
(ii) Because it finds itself being pulled by the positive charge left behind as it becomes.
(iii) They can be made to leave the metal surface by

1. Thermionic emission (heating the metal surface)
2. Photoelectric emission (By shining a radiation of sufficient frequency on the metal surface).

Solving ICSE Class 10 Physics Previous Year Question Papers ICSE Class 10 Physics Question Paper 2016 is the best way to boost your preparation for the board exams.

ICSE Class 10 Physics Question Paper 2016 Solved

Time: 1½ hours
Maximum Marks: 80

General Instructions:

• Answers to this Paper must be written on the paper provided separately.
• You will not be allowed to write during the first 15 minutes.
• This time is to be spent in reading the Question Paper,
• The time given at the head of this Paper is the time alloted for writing the answers.
• Attempt all questions from Section I and any four questions from Section II.
• The intended marks of questions or parts of questions are given in brackets [ ].

SECTION – I (40 Marks)
Attempt all questions from this Section

Question 1.
(a) (i) Give an example of a non contact force which is always of attractive nature.
(ii) How does the magnitude of this non contact force on the two bodies depend on the distance of separation between them? [2]
Answer:
(i) Gravitational force
(ii) F α \(\frac{1}{r^2}\)

(b) A boy weighing 40 kgf climbs up a stair of 30 steps each 20 cm high in 4 minutes and a girl weighing 30 kgf does the same in 3 minutes. Compare:
(i) The work done by them.
(ii) The power developed by them. [2]
Answer:
(i) Equal work.
(ii) Power developed by the girl is more as she does the same work as the boy but is lesser time.

(c) With reference to the terms Mechanical Advantage, Velocity Ratio and efficiency of a machine, name and define the term that will not change for machine of a given design. [2]
Answer:
Velocity ratio:
It is defined as the ratio of the velocity of the effort (v_E) to the velocity of the resistance force (Load) (v_L).

(d) Calculate the mass of ice required to lower the temperature of 300 g of water 40°C to water at 0°C. [2]
(Specific latent heat of ice = 336 J/g, Specific heat capacity of water = 4.2J/g°C)
Answer:
Given,
Mass of water = 300 g
Initial temperature = 40°C
Final temperature = 0°C
Let the mass of ice = m g
∴ Heat to be removed to lower the temperature of 300 g of water from 40°C to 0°C
= 300 × 4.2 × (40 – 0) J
= 300 × 4.2 × 40 J = 50400 J.
Heat lost by ice from 40 °C to 0°C of water
= (m × 336) J.
By principle of calorimetry,
m × 336 = 50400
Therefore, m = \(\frac{50400}{336}\) = 150g

(e) What do you understand by the following statements:
(i) The heat capacity of the body is 60JK1.
(ii) The specific heat capacity of lead is 130 Jkg‘1K’1. [2]
Answer:
(i) It means 60 J of heat is required to raise its temperature by 1 K.
(ii) It means 130 J of heat is required to raise the temperature of 1 kg of lead through 1K.

Question 2.
(a) State two factors upon which the heat absorbed by a body depends. [2]
Answer:

1. Nature of material and
2. Temperature of the body.

(b) A boy uses blue colour of light to find the refractive index of glass. He then repeats the experiment using red colour of light. Will the refractive index be the same or different in the two cases? Give a reason to support your answer. [2]
Answer:
No, the refractive index of a medium depends upon the colour of light used. As n = \(\frac{\mathrm{c}}{\mathrm{v}}\) and different colours have different speed in a medium, therefore, n is different.

(c) Copy the diagram given below and complete the path of the light ray till it emerges out of the prism. The critical angle of glass is 42°. In your diagram mark the angles wherever necessary. [2]

Answer:
The diagram is as shown :

(d) State the dependence of angle of deviation:
(i) On the refractive index of the material of the prism.
(ii) On the wavelength of light [2]
Answer:
(i) Directly proportional to the refractive index of the material of the prism.
(ii) Inversely proportional to the wavelength of light used.

(e) The ratio of amplitude of two waves is 3:4. What is the ratio of their:
(i) loudness?
(ii) Frequencies? [2]
Answer:
(i) \(\frac{L_1}{L_2}=\left(\frac{3}{4}\right)^2=\frac{6}{16}\)
(ii) 1 : 1

Question 3.
(a) State two ways by which the frequency of transverse vibrations of a stretch string can be increased. [2]
Answer:

1. By increasing the tension in the string and
2. Decreasing the length of the string.

(b) What is meant by noise pollution? Name one source of sound causing noise pollution. [2]
Answer:
Unpleasant sounds are called noise. Some of the main ‘noise polluters’ are aircraft, road traffic, greatly amplified music.

(c) The V-I graph for a series combination and for a parallel combination of two resistors is shown in the figure below. Which of the two A or B. represents the parallel combination? Give reasons for your answer. [2]

Answer:
Resistance in parallel combination is less than in series combination for the same set of resistors. The slope of the graph gives the value of resistance. Graph A has less slope and has less resistance, therefore, it represents parallel combination.

(d) A music system draws a current of 400 mA when connected to a 12 V battery. [2]
(i) What is the resistance of the music system?
(ii) The music system if left playing for several hours and finally the battery voltage drops to 320 mA and the music system stops playing when the current.
Answer:
Given, I = 400 mA, V = 12 V, R = ?
R = \(\frac{V}{I}=\frac{12}{400 \times 10^{-3}}\) = 30 Ω
V = IR = 320 × 10^-3 × 30
= 9.6 V

(e) Calculate the quantity of heat produced in a 20 Ω resistor carrying 2.5 A current in 5 minutes. [2]
Answer:
Given, R = 20 W, I = 2.5 A, t = 5 min = 300 s
Using H = I²Rt
= (2.5)² × 20 × 300
= 37500 J

Question 4.
(a) State the characteristics required of a good thermion emitter. [2]
Answer:

1. The emitter metal should have high melting point so that it can be operated at a high temperature to achieve appreciable emission of electrons.
2. The value or work function of the emitter metal should be as low as possible.

(b) An element _ZS^A decays to ₈₅R²²² after emitting 2 a particles and 1 P particle. Find the atomic number and atomic mass of the element S. [2]
Answer:
The decay will follow the following sequence

Therefore, we have
Z – 3 = 85 or Z = 85 + 3 = 88
And A – 8 = 222 or A = 222 + 8 = 230
Z = 88, A = 230

(c) A radioactive substance is oxidized. Will there be any change in the nature of its radioactivity? Give a reason for your answer. [2]
Answer:
No, radioactivity is not affected by any physical process.

(d) State the characteristics required in a material to be used as an effective fuse wire. [2]
Answer:
The material should have high resistivity and low melting point.

(e) Which coil of a step up transformer is made thicker and why? [2]
Answer:
Primary, as it has lesser number of turns.

SECTION – II (40 Marks)
(Attempt any four questions from this Section)

Question 5.
(a) A stone of mass ‘m’ is rotated in a circular path with a uniform speed by tying a strong string with the help of your hand. Answer the following questions : (3)
(i) Is the stone moving with a uniform or variable speed ?
(ii) Is the stone moving with a uniform acceleration ? In which direction does the acceleration act?
(iii) What kind of force acts on the hand and state its direction ?
Answer:
(i) Uniform speed.
(ii) Yes, towards the centre of the circular path.
(iii) Centrifugal outwards.

(b) From the diagram given below, answer the question that follow : (3)

(i) What kind of pulleys are A and B ?
(ii) State the purpose of pulley B.
(iii) What effort has to be applied at C just raise the load L = 20 kgf ?
(Neglect the weight of pulley A and friction)
Answer:
(i) A – Single movable, B – Single fixed.
(ii) It changes the direction of force.
(iii) Distance moved by load d_L = x/2.
Distance moved by the effort d_E = x
Load = 20 kg f, Effort = ?
Now, Load × d_L = Effort × d_E
20 × x/2 = Effort × x
or Effort = 10 kg f

(c) (i) An effort is applied on the bigger wheel of a gear having 32 teeth. It is used to turn a wheel of 8 teeth. Where it is used.
(ii) A pulley system has three pulleys. A load of 120 N is overcome by applying an effort of 50 N.
Calculate the Mechanical Advantage and Efficiency of this system. (4)
Answer:
(i) It is used to gain speed.
= \(\frac{\text { Number of teeth in the driver }}{\text { Number of teeth in the driven }}\)
= \(\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{32}{8}\) = 4

(ii) Given, L = 120 N, E = 50 N, MA = ?, η = ?
MA = \(\frac{\mathrm{L}}{\mathrm{E}}=\frac{120}{50}\) = 2.4
Velocity ratio is equal to the number of pulleys, hence
η = \(\frac{\mathrm{MA}}{\mathrm{VR}}=\frac{2.4}{3}\) = 0.8

Question 6.
(a) (i) What is the principle of method of mixtures ? (3)
(ii) What is the other name given to it ?
(iii) Name the law on which the principle is based
Answer:
(i) “Heat lost by the hot body is equal to the heat gained by the cold body”.
(ii) Principle of calorimetry.
(iii) Law of conservation of energy.

(b) Some ice is heated at a constant rate, and its temperature is recorded after every few seconds, till steam is formed at 100°C. Draw a temperature time graph to represent the change. Label the two phase changes in your graph. (3)
Answer:
The figure for phase change is shown below :

(c) A copper vessel of mass 100 g contains 150 g of water at 50°C. How much ice is needed to cool it to 5°C ? (4)
Given : Specific heat capacity of copper = 0.4 Jg^-1 °C^-1
Specific heat capacity of water = 4.2 Jg^-1 °C^-1
Specific latent heat of fusion ice = 336 Jg^-1
Answer:
Given,
Mass of ice = m g
Mass of copper vessel = 100 g
Mass of water = 150 g
Initial Temperature = 50 °C
Final temperature = 5 °C
Now, we have
Heat lost by water due to decrease in temperature from 50°C to 5°C
= 150 × 4.2 × (50 – 5)
= 28350 J
Heat lost by the calorimeter
= 100 × 0.4 × (50 – 5) J
= 1800 J
Heat gained by ice = m × 336 × (5 – 0) J
= 1680m J
Total heat lost by the water + Calorimeter
= (28350 + 1800) J
= 30150 J
By principle of calorimetry,
Heat lost by water and calorimeter = heat gained by the ice
30150 = 1680 m
m = \(\frac{30150}{1680}\) = 17.94 g

Question 7.
(a) (i) Write a relationship between angle of incidence and angle of refractions for a given pair of media. (3)
(ii) When a ray of light enters from one medium to another having different optical densities it bends. Why does this pehnomenon occur ?
(iii) Write one conditions where it does not bend when entering a medium of different optical density.
Answer:
(i) The relation is n = \(\frac{\sin i}{\sin r}\)
(ii) This occurs because light has different velocities in different media. The change in velocity causes the bending of light.
(iii) When it enters perpendicular to the interface separating the two media.

(b) A lens produces a virtual image between the object and the lens. (3)
(i) Name the lens.
(ii) Draw a ray diagram to show the formation of this image.
Answer:
(i) Concave lens.
(ii) The ray diagrams is a shown :

(c) What do you understand by the term ‘Scattering of light’ ? Which colour of white light is scattered the least and why ? (4)
Answer:
Light scattering is the deflection of a ray of light from a straight path, for example by irregularities in the propagation medium, particles, or in the interface between two media.

Red colour is scattered the least because scattering of light depends inversely upon the four power of wavelength. As red colour has the maximum wavelength in the visible region, therefore, it scattered the least.

Question 8.
(a) (i) Name the waves used for echo depth sounding.
(ii) Give one reason for their use for the above purpose.
(iii) Why are the waves mentioned by you not audible to us ? (3)
Answer:
(i) Ultrasonic.
(ii) Easier to transmit through water.
(iii) Because their frequency is greater than 20 kHz the upper limit of our audible range.

(b) (i) What is an echo
(ii) State two conditions for an echo to take place. (3)
Answer:
(i) An echo is the phenomenon of repetition of sound of a source by reflection from an obstacle.
(ii) Presence of a solid reflecting material and the distance between the obstacle and the source of sound should be such that it takes the reflected sound at least 0.1 s to reach the source.

(c) (i) Name the phenomenon involved in tuning a radio set to a particular station.
(ii) Define the phenomenon named by you in part (i) above.
(iii) What do you understand by loudness of sound ?
(iv) In which units is the loudness of sound measured ? (3)
Answer:
(i) Resonance.
(ii) Resonance or acoustic resonance is a special case of forced vibrations. It occurs when the frequency of the applied periodic external force becomes equal to the natural frequency of the vibrating body.
(iii) Loudness is the degree of sensation of sound produced. It depends upon the intensity of sound waves near the ear and response (sensitivity) of the ear for the waves of that frequency.
(iv) The unit of loudness is phon.

Question 9.
(a) (i) Which particles are responsible for current in conductors ?
(ii) To which wire of a cable in a power circuit should the metal case of a geyser be connected ? (3)
Answer:
(i) Electrons
(ii) Earth wire
(iii) Live wire

(b) (i) Name the transformer used in the power transmitting station of a power plant.
(ii) What type of current is transmitted from the power station
(iii) At what voltage is this current available to our household ? (3)
Answer:
(i) Step up transformer
(ii) Alternating current of low value.
(iii) 220 V

(c) A battery of emf 12 V and internal resistance 2 Ω is connected with two resistors A and B of resistance 4 Ω and 6 Ω. respectively joined in series. (4)

Find:
(i) Current in the circuit
(ii) The terminal voltage of the cell.
(iii) The potential difference across 6 Ω Resistor.
(iv) Electrical energy spent per minute in 4 Ω resistor.
Answer:
Total resistance in the circuit
R = 4 + 6 + 2
= 12 Ω
(i) I = V / R = 12 / 12 = 1 A
(ii) V = E – Ir = 12 – 1 × 2 = 10 V
(iii) V = IR = 1 × 6 = 6V
(iv) E = I² R t = (1)² × 4 × 60 = 240 J

Question 10.
(a) Arrange α, β, and γ rays in ascending order with respect to their (3)
(i) Penetrating power.
(ii) Ionising power.
(iii) Biological effect.
Answer:
(i) Alpha, Beta and Gamma.
(ii) Gamma, Beta and Alpha.
(iii) Gamma, Beta and Alpha.

(b) (i) In a cathode ray tube what is the function of anode ?
(ii) State the energy conversion taking place in a cathode ray tube.
(iii) Write one use of cathode ray tube.
Answer:
(i) It accelerates the electrons along the highly evacuated tube and also focus them into a narrow beam.
(ii) Heat energy into kinetic energy of electrons.
(iii) Picture tube in television.

(c) (i) Represent the change in the nucleus of a radioactive element when a P particle is emitted.
(ii) What is the name given to elements with same mass number and different atomic number.
(iii) Under which conditions does the nucleus of an atom tend to radioactive ?
Answer:
(i) The change is as shown
\({ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{P} \rightarrow{ }_{\mathrm{Z}+1}^{\mathrm{A}} \mathrm{D}+{ }_{-1}^0 e\)
(ii) Isobars.
(iii) When the nucleus is unstable.

Solving ICSE Class 10 Physics Previous Year Question Papers ICSE Class 10 Physics Question Paper 2021 Semester 1 is the best way to boost your preparation for the board exams.

ICSE Class 10 Physics Question Paper 2021 Solved Semester 1

Time : 1 hour
Maximum Marks: 40

General Instructions:

• You will not be allowed to write during the first 10 minutes.
• This time is to be spent in reading the question paper.
• ALL QUESTIONS ARE COMPULSORY
• The intended marks for questions or parts of questions are given in brackets [].
• Select the correct option for each of the following questions

Question 1.
The deviation produced by an equi-lateral prism does not depend on: [1]
(a) the angle of incidence.
(b) the size of the prism.
(c) the material of the prism.
(d) the colour of light used.
Answer:
(b) the size of the prism.

Question 2.
The refractive index of a diamond is 2.4. It means that: [1]
(a) the speed of light in vacuum is equal to \(\frac{1}{2.4}\) times the speed of light in diamond
(b) the speed of light in the diamond is 2.4 times the speed of light in a vacuum.
(c) the speed of light in a vacuum is 2.4 times the speed of light in the diamond,
(d) the wavelength of light in diamond is 2.4 times the wavelength of light in vacuum.
Answer:
(c) the speed of light in a vacuum is 2.4 times the speed of light in the diamond.

Question 3.
An object of height 10 cm is placed in front of a concave lens of focal length 20 cm at a distance 25 cm from the lens. Is it possible to capture this image on a screen? Select a correct option from the following: [1]
(a) Yes, as the image formed will be real.
(b) Yes, as the image formed will be erect.
(c) No, as the image formed will be virtual.
(d) No, as the image formed will be inverted.
Answer:
(c) No, as the image formed will be virtual.

Question 4.
A ray of light IM is incident on a glass slab ABCD as shown in the figure below. The emergent ray for this incident ray is: [1]

(a) NQ
(b) NR
(c) NP
(d) NS
Answer:
(a) NQ

Question 5.
The colour of white light which is deviated least by a prism is: [1]
(a) green
(b) yellow
(c) red
(d) violet
Answer:
(c) red

Question 6.
The wave length range of visible light is: [1]
(a) 40 nm to 80 nm
(b) 4000 nm to 8000 nm
(c) 4 nm to 8 nm
(d) 400 nm to 800 nm
Answer:
(d) 400 nm to 800 nm

Question 7.
Observe the diagram which shows the path of an incident ray through an optical plane LL” of a lens. The focal length of the lens is 20 cm.

(i) If an object is placed at a distance of 30 cm in front of this lens, then: [1]
(a) the image will be virtual
(b) the image will be diminished and inverted.
(c) the image will be diminished.
(d) the image will be real and magnified.
Answer:
(d) the image will be real and magnified.

(ii) This type of lens can be used: [1]
(a) to correct hypermetropia.
(b) to correct myopia.
(c) to diverge light.
(d) in the front door peepholes.
Answer:
(a) to correct hypermetropia.

(iii) An object is placed in front of this lens at a distance of 60 cm. Then the image distance from the lens with proper sign convention is: [1]
(a) + 60 cm
(b) + 30 cm
(c) – 30 cm
(d) + 15 cm
Answer:
(b) +30 cm

(iv) An object is placed in front of this lens at a distance of 60 cm. Then the magnification of the image is: [1]
(a) 0.25
(b) 1.25
(c) – 0.5
(d) 1
Answer:
(c) -0.5

Question 8.
The relation between CGS and S.I. unit of moment of force is:
(a) 1 Nm = 10⁵ dyne cm
(b) 1 Nm = 10⁵ dyne
(C) 1 Nm = 10⁵ dyne cm
(d) 1 dyne cm = 10⁷ Nm
Answer:
(c) 1 Nm = 10⁷ dyne cm

Question 9.
A coolie raises a load upwards against the force of gravity then the work done by the load is: [1]
(a) zero.
(b) positive work,
(c) negative work,
(d) none of these.
Answer:
(c) negative work.

Question 10.
The energy change during photo-synthesis in plants is: [1]
(a) heat to chemical.
(b) light to chemical.
(c) chemical to light.
(d) chemical to heat.
Answer:
(b) light to chemical.

Question 11.
The diagram below shows the balanced position of a metre scale. [1]

Which one of the following diagrams shows the correct position of the scale when it is supported at the centre?

Answer:
(a)

Question 12.
A stone tied at the end of a string is whirled by hand in a horizontal circle with uniform speed.
(i) Name the force required for this circular motion: [1]
(a) Centrifugal force.
(b) Centripetal force.
(c) Force of gravity.
(d) Frictional force.
Answer:
(b) Centripetal force.

(ii) What is the direction of the above-mentioned force? [1]
(a) Towards the centre of the circular path.
(b) Away from the centre of the circular path.
(c) Normal to the radius at a point where the body is present on the circular path.
(d) Direction of this force keeps on changing alternately towards and away from the centre.
Answer:
(a) Towards the centre of the circular path.

Question 13.
A body of mass 200 g falls freely from a height of 15 m. [g = 10 ms^-2]
(i) When the body reaches 10 m above the ground, its potential energy will be: [1]
(a) 20000 J
(b) 10 J
(c) 10000 J
(d) 20 J
Answer:
(d) 20 J

(ii) The gain in kinetic energy of the body when it reaches 10 m above the ground is: [1]
(a) 20 J
(b) 10 J
(c) 30 J
(d) 25 J
Answer:
(b) 10 J

(iii) The total mechanical energy it will possess, when it is just about to strike the ground is: [1]
(a) 30000 J
(b) 20000 J
(c) 30 J
(d) 20 J
Answer:
(c) 30 J

(iv) The velocity in ms^-1 with which the body will hit the ground is: [1]
(a) 30
(b) 10
(c) 10√3
(d) 10√2
Answer:
(c) 10\(\sqrt{3}\)

Question 14.
A woman draws water from a well using a fixed pulley. The mass of the bucket and the water together is 10 kg. The force applied by the woman is 200 N. The mechanical advantage is (g = 10 m/s²): [1]
(a) 2
(b) 20
(c) 0.05
(d) 0.5
Answer:
(d) 0.5

Question 15.
A single fixed pulley is used because: [1]
(a) it changes the direction of applied effort conveniently.
(b) it multiplies speed.
(c) it multiplies effort.
(d) its efficiency is 100%.
Answer:
(a) it changes the direction of applied effort conveniently.

Question 16.
In the diagram shown below, the velocity ratio of the arrangement is: [1]

(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(b) 2

Question 17.
Which one of the following is the correct mathematical relation? [1]
(а) Power = Force/Velocity
(b) Power = Force × Acceleration
(c) Power = Force/Acceleration
(d) Power = Force × Velocity
Answer:
(d) Power = Force × Velocity

Question 18.
Select a correct option with respect to echo depth sounding: [1]
(a) infrasonic waves are used
(b) the frequency of the waves used is between 20 Hz and 20,000 Hz.
(c) ultrasonic waves are used.
(d) supersonic waves are used.
Answer:
(c) ultrasonic waves are used.

Question 19.
Which one of the following diagnostic methods use reflection of sound? [1]
(a) CT scan
(b) Electrocardiogram
(c) Echo cardiogram
(d) MRI
Answer:
(c) Echo cardiogram

Question 20.
A boy standing in front of a wall produces two whistles per second. He notices that the sound of his whistling coincides with the echo. The echo is heard only once when whistling is stopped. Calculate the distance between the boy and the wall. (The speed of sound in air = 320 m/s)
(i) The time in which the boy hears the echo is: [1]
(a) 1 s
(b) 0.5 s
(c) 1.5 s
(d) 2 s
Answer:
(b) 0.5 s

(ii) The distance at which the boy is standing from the wall: [1]
(a) 160 m
(b) 240 m
(c) 320 m
(d) 80 m
Answer:
(d) 80 cm

(iii) If the speed of sound is increased by 16 ms^-1 and the boy moves 4 m away from the wall then in how much time will he hear the echo of the first whistle? [1]
(a) 0.525 s
(b) 0.5 s
(c) 0.48 s
(d) 0.3 s
Answer:
(b) 0.5 cm

(iv) In w hich of the following timings of reflection of the whistle, the echo cannot be heard? [1]
(a) 0.05 s
(b) 0.12 s
(c) 0.2 s
(d) 0.11 s
Answer:
(a) 0.05 s

Question 21.
The ratio of velocities of light of wavelength 400 nm and 800 nm in a vacuum is: [1]
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Answer:
(a) 1 : 1

Question 22.
1 joule = ________ erg [1]
(a) 10⁹
(b) 10⁷
(c) 10⁵
(d) 10⁶
Answer:
(b) 10⁷

Question 23.
A light body A and a heavy body B have the same momentum.
(i) Choose a correct statement from the given options. [1]
(a) kinetic energy of body A and body B will be the same.
(b) kinetic energy of body A is greater than kinetic energy of body B.
(c) kinetic energy of body Bis greater than kinetic energy of body A.
(d) unless we know the velocity, we cannot find which body has greater kinetic energy.
Answer:
(b) kinetic energy of body A is greater than kinetic energy of body B.

(ii) If the ratio of kinetic energies of A and B is 5 : 2 then which one of the following gives the mass ratio of the bodies respectively? [1]
(a) 5 : 2
(b) 25 : 4
(c) 2 : 5
(d) 4 : 24
Answer:
(c) 2 : 5

Question 24.
The diagram below shows a ray of light travelling from air into a glass material as shown below. Answer the questions that follow:

(i) The angle of incidence at the surface AB is: [1]
(a) 43°
(b) 47°
(c) 90°
(d) 0°
Answer:
(b) 47°

(ii) Select a correct statement from the following. [1]
(a) The speed of light at the curved surface AD does not change while entering the block.
(b) The ray at the surface AD is not travelling along the radius of the curved part
(c) The ray at the surface AD is travelling along the radius of the curved part.
(d) Light never refracts when it enters a curved surface.
Answer:
(c) The ray at the surface AD is travelling along the radius of the curved part.

(iii) The angle of incidence on the surface BC is: [1]
(a) 43°
(b) 47°
(c) 90°
(d) 0°
Answer:
(a) 43°

(iv) The critical angle of this material of glass: [1]
(a) 47°
(b) 43°
(b) 42°
(c) 45°
Answer:
(b) 43°

Question 25.
The diagram below shows the path of light passing through a right-angled prism of critical angle 42°

(i) The angle C of the prism is: [1]
(a) 45°
(b) 60°
(c) 90°
(d) 30°
Answer:
(d) 30°

(ii) Which one of the following diagrams shows the correct path of this ray till it emerges out of the prism? [1]

Answer:
(b)

Solving ICSE Class 10 Physics Previous Year Question Papers ICSE Class 10 Physics Question Paper 2014 is the best way to boost your preparation for the board exams.

ICSE Class 10 Physics Question Paper 2014 Solved

Time: 2 hours
Maximum Marks: 80

General Instructions:

• Answers to this paper must be written on the paper provided separately.
• You will not be allowed to write during the first 15 minutes.
• This time is to be spent in reading the Question Paper.
• The time given at the head of this paper is the time allowed for writing the answers.
• Section I is compulsory. Attempt any four questions from Section II.
• The intended marks for questions or parts of questions are given in brackets [ ].

SECTION – I (40 Marks)
(Attempt all questions from this section)

Question 1.
(a) A force is applied on (i) a non-rigid body and (ii) a rigid body. How does the effect of the force differ in the above two cases? [2]
Answer:
A force when applied on a non-rigid body can change the dimensions of the body as well as can bring motion in the body while force applied on rigid body cannot change the dimensions as it cannot change the interspacing between the constituent particles but can cause motion in the body.

(b) A metallic ball is hanging by a string from a fixed support. Draw a neat labelled diagram showing the forces acting on the ball and the string. [2]
Answer:

(c) (i) What is the weight of a body placed at the centre of the earth?
(ii) What is the principle of an ideal machine? [2]
Answer:
(i) Weight of a body placed at the centre of earth will be zero.
(ii) Work done on machine = Work done by machine
Or Work input = Work output is the principle of an ideal machine where there is no dissipation of energy and efficiency of machine is 100%.

(d) Is it possible to have an accelerated motion with a constant speed? Explain. [2]
Answer:
Yes, when a body is moving in a circular motion, its direction keeps on changing but speed remains same. The continuous change in direction of motion means that the velocity is not uniform or body moving with variable velocity, so motion is called accelerated.

(e) (i) When does a force do work?
(ii) What is the work done by the moon when it revolves around the earth? [2]
Answer:
(i) When a force is able to cause displacement in the body work is done. But when force and displacement make an angle of 90° with each other, then no work is done.

(ii) Work done is zero when moon revolves around the earth, the force is the centripetal force and displacement at all instants is along the tangent to the circular path i. e., normal to the direction of force.

Question 2.
(a) Calculate the change in the Kinetic energy of a moving body if its velocity is reduced to 1/3rd of the initial velocity. [2]
Answer:
Let a body of mass ‘m’ is moving with ‘u’ m/s. So, initial K.E. = \(\frac{1}{2}\)mu². If velocity is reduced to \(\frac{1}{3}\) rd. Final K.E. = \(\frac{1}{2}\)m \(\left(\frac{u}{3}\right)^2=\frac{1}{2} \frac{m u^2}{9}\)

So K.E. becomes \(\frac{1}{9}\) th of its initial K.E.

(b) State the energy changes in the following devices while in use :
(i) A loud speaker.
(ii) A glowing electric bulb. [2]
Answer:
(i) A loud speaker converts electrical energy into sound energy.
(ii) A glowing electric bulb converts electrical energy into light energy and heat energy.

(c) (i) What is nuclear energy?
(ii) Name the process used for producing electricity using nuclear energy. [2]
Answer:
(i) Nuclear energy is the energy released by nuclear fission and nuclear fusion processes in which mass of products is less than mass of reactants, so loss of mass gets converted into energy by Einstein formula E = Δmc², where E is energy released, Δm is loss in mass and c is speed of light which is 3 ×10⁸ ms^-1.

(ii) Nuclear fission in a nuclear power plant.

(d) State one important advantage and disadvantage each of using nuclear energy for producing electricity. [2]
Answer:
A very small amount of nuclear fuel can produce a tremendous amount of energy. But its disadvantage is that along with energy it produces highly energetic and penetrating harmful nuclear radiations which are very harmful to human body and also causes environmental pollution.

(e) (i) The conversion of part of the energy into an undesirable form is called
(ii) For a given height h, the length 1 of the inclined plane, lesser will be the effort required. [2]
Answer:
(i) Dissipation of energy.
(ii) More.

Question 3.
(a) Draw the diagram given below and clearly show the path taken by the emergent ray. [2]

Answer:

(b) (i) What is consumed using different electrical appliances, for which electricity bills are paid?
(ii) Name a common device that uses electromagnets.
Answer:
(i) Energy.
(ii) Electric bell, motor of fan, grinder etc.

(c) (i) A ray of light passes from water to air. How does the speed of light change?
(ii) Which colour of light travels fastest in any medium except air?
Answer:
(i) Speed of light increases when it goes from water to air.
(ii) Red colour.

(d) Name the factors affecting the critical angle for the pair of media.
Answer:
Critical angle for a given pair of media depends on the refractive index of that pair and it is affected by

1. colour or wavelength of light – critical angle is least for violet light and most for red light
2. temperature – critical angle increases with increase in temperature.

(e) (i) Name a prism required for obtaining a spectrum of Ultraviolet light.
(ii) Name the radiations which can be detected by a thermopile.
Answer:
(i) Quartz prism.
(ii) Infrared radiation.

Question 4.
(a) Why is the colour red used as a sign of danger? [2]
Answer:
Red colour has longest wavelength so scatters least and can penetrate to a longer distance. Thus, can be seen from large distance.

(b) (i) What are mechanical waves?
(ii) Name one property of waves that do no change when the wave passes from one medium to another. [2]
Answer:
(i) Mechanical waves require a material medium to travel. These waves travel in the medium through the vibrations of the medium particles about their mean positions.

(ii) Frequency of wave remains same.

(c) Find the equivalent resistance between points A and B. [2]

Answer:
Resistance 3 Ω, 3 Ω, and 3 Ω are parallel to each other (in CD).

So, \(\frac{1}{\mathrm{R}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{3}{3}\) ; R = 1 Ω
So, resistance from A to E = 1 + 5 = 6
Resistance across EB
\(\frac{1}{\mathrm{R}}=\frac{1}{4}+\frac{1}{6}=\frac{3+2}{12}=\frac{5}{12}\) ;
R = \(\frac{12}{5}\) = 2.4 Ω
∴ Total resistance across AB = 6 + 2.4 = 8.4 Ω

(d) 50 g of metal piece at 27 °C requires 2400 J of heat energy so as to attain a temperature of 327 °C. Calculate the specific heat capacity of the metal. [2]
Answer:
Mass of metal = 50 g
t₁ = 27°C
t₂ = 327°C
∴ Δt = 327 – 27 = 300°C
Q = 2400 J (Heat energy)
C = ?
Q = mCΔt
C = \(\frac{\mathrm{Q}}{m \Delta t}=\frac{2400}{50 \times 300}\)
= 0.16 J/g/°C

(e) An electron emitter must have work function and melting point. [2]
Answer:
Low work function and high melting point.

SECTION – II (40 Marks)
(Attempt any four questions from this Section)

Question 5.
(a) (i) A man having a box on his head, climbs up a slope and another man having an identical box walks the same distance on a levelled road. Who does more work against the force of gravity and why?
(ii) Two forces each of 5N act vertically upwards and downwards respectively on the two ends of a uniform metre rule which is placed at its mid-point as shown in the diagram. Determine the magnitude of the resultant moment of these forces about the midpoint. [4]

Answer:
(i) Work done by a man having a box on his head and climbing a rope is more as moving on the level road does no work (f × d cos 90° = 0).

(ii) Magnitude of resultant moment of forces about the mid-point
= 5 N × \(\frac{50}{100}\) m + 5 N × \(\frac{50}{100}\) m
= 2.5 Nm + 2.5 Nm
= 5 Nm (anticlockwise)

(b) (i) A body is thrown vertically upwards. Its velocity keeps on decreasing. What happens to its kinetic energy as its velocity becomes zero?
(ii) Draw a diagram to show how a single pulley can be used so as to have its ideal M.A. = 2. [3]
Answer:
(i) Its kinetic energy keeps on getting converted into potential energy when the body is thrown vertically upwards as it gains height.

(ii) L = T + T = 2T
E = T
∴ M.A. = \(\frac{2 \mathrm{~T}}{\mathrm{~T}}\) = 2
So it acts as a force multiplier.

(c) Derive a relationship between mechanical advantages, velocity ratio and efficiency of a machine. [3]
Answer:
M.A., V.R. and η
Efficiency η = \(\frac{\text { Work output }}{\text { Work input }}\)
Work output = Load × Load Arm
Work input = Effort × Effort Arm
η = \(\frac{L \times \text { L.A. }}{E \times \text { E.A }}\), But \(\frac{\text { E.A. }}{\text { L.A. }}\) = V.R. and \(\frac{L}{E}\) = M.A.
η = M.A. × \(\frac{1}{\text { V.R. }}\) or M.A. = V.R. × η

Question 6.
(a) (i) Light passes through a rectangular glass slab and through a triangular glass prism. In what way does the direction of the two emergent beams differ and why?
(ii) Ranbir claims to have obtained an image twice the size of the object with a concave lens. Is he correct? Give a reason for your Answer. [4]
Answer:
(i)

Light when passes through a rectangular glass slab suffers refraction at two surfaces Q and R as it goes from rarer to denser and from denser to rarer. The same bending takes place at Q and R of glass prism also. But ray of light while passing through a glass prism always deviates towards its base i.e., towards surface BC only which is not in case of glass slab. In each refraction through glass prism the ray always bends towards the bases of prism.

(ii) No, a concave lens can always form a virtual, erect, diminished image as it has a virtual focus.

(b) A lens forms an erect, magnified and virtual image of an object.
(i) Name the lens.
(ii) Draw a labelled ray diagram to show the image formation. [3]
Answer:
(i) Convex lens.
(ii) When object is between O and/of a convex lens. A’

(c) (i) Define the power of a lens.
(ii) The lens mentioned in 6(b) above is of focal length 25 cm. Calculate the power of the lens. [3]
Answer:
(i) Power of a lens is a measure of deviation produced by it in the path of rays refracted through it and is measured in Dioptres.
P (in D) = \(\frac{1}{f(m)}\) or \(\frac{100}{f(cm)}\)

(ii) Focal length of convex lens is taken positive.
f = +25 cm
P (in D) = 100/25 = +4 D

Question 7.
(a) The adjacent diagram shows three different modes of vibrations P, Q and R of the same string.
(i) Which vibrations will produce a louder sound and why?
(ii) The sound of which string will have maximum shrillness?
(iii) State the ratio of wavelengths of P and R. [4]

Answer:
(i) R, because its amplitude is more.
(ii) P, as frequency is more.
(iii) Ratio of wavelengths of P and R is 1 : 3.

(b) A type of electromagnetic wave has wavelength 50 A.
(i) Name the wave.
(ii) What is the speed of the wave in vacuum?
(iii) State one use of this type of wave. [3]
Answer:
(i) X – rays.
(ii) 3 × 10⁸ ms-¹.
(iii) They can penetrate through human flesh but are stopped by bones, hence used for detection of fracture and for diagnostic purpose.

(c) (i) State one important property of waves used for echo depth sounding.
(ii) A radar sends a signal to an aircraft at a distance of 30 km away and receives it back after 2 × 10^-4 second. What is the speed of the signal? [4]
Answer:
(i) They (ultrasonic waves) can travel undeviated through long distances and can be confined to a narrow beam. They are not absorbed easily.

(ii) Distance = 30 km = 30000 m
Time = 2 × 10^-4 s (for echo to be heard)
Speed = 2 d/t
= \(\) = 30000 × 10⁴
= 3 × 10⁸ m/s

Question 8.
(a) Two resistors of 4 Ω and 6 Ω are connected in parallel to a cell to draw 0.5 A current from the cell.
(i) Draw a labelled circuit diagram showing the above arrangement.
(ii) Calculate the current in each resistor. [3]
Answer:
(i)

(ii) Resistance in parallel
\(\frac{1}{4}+\frac{1}{6}=\frac{1}{R}\) ; \(\frac{3+2}{12}=\frac{5}{12}=\frac{1}{R}\) ; R = 2.4 Ω
So, PD in parallel = Total I × R (only in parallel)
= 0.5 × 2.4 = 1.2 V
I₁ = \(\frac{3+2}{12}=\frac{5}{12}=\frac{1}{R}\) = 0.3 A ;
I₂ = \(\frac{3+2}{12}=\frac{5}{12}=\frac{1}{R}\) = 0.2 A

(b) (i) What is an Ohmic resistor?
(ii) Two copper wires are of the same length, but one is thicker than the other.
(1) Which wire will have more resistance?
(2) Which wire will have more specific resistance? [3]
Answer:
(i) Ohmic resistor obeys Ohm’s law i.e., PD is directly proportional to current keeping resistance constant in that conductor i.e., at constant temperature, e.g., all metallic conductors are Ohmic.

(ii)

1. Resistance is inversely proportional to thickness, so thick wire will have less resistance and thin wire more resistance when lengths are same.
2. Specific resistance does not depend on length and area of wire so it will be same for both as only depends on material and temperature of wire.

(c) (i) Two sets A and B, of the three bulbs each, are glowing in two separate rooms. When
one of the bulbs in set A is fused, the other two bulbs, alos cease to glow. But in set B, when one bulb fuses, the other two bulbs continue to glow. Explain why this phenomenon occurs.
(ii) Why do we prefer arrangement of Set B for house circuiting? [3]
Answer:
(i) In Set A, the bulbs are arranged in series while in Set B, the bulbs are arranged in parallel. In parallel, even if one bulb fuses other paths are working.

(ii)

1. In parallel, all appliances work independently at same voltage so glow of bulb (or any other appliance) is not affected.
2. In parallel, resistance so current does not decrease, so current does not decrease.
3. All appliances work independently and do not depend on other so each has a separate switch

Question 9.
(a) Heat energy is supplied at a constant rate to lOOg of ice at 0 °C. The ice is converted into
water at 0 °C in 2 minutes. How much time will be required to raise the temperature of water from 0 °C to 20 °C? [Given : sp. heat capacity of water = 4.2 J g^-1 °C^-1, sp. latent heat of ice = 336 J g^-1]. [4]
Answer:
Mass of ice = 100 g
L of ice = 336 J/g
mL (of ice) = P × time
100 × 336 = P × 2 × 60
P = 280 W
mCQ (of water) = P × time
As same appliance with same power is used
100 × 4.2 × 20 = P × time
8400 = 280 × time
time = 30 s
So, time required to raise the temperature of water from 0°C to 20°C is 30 s.

(b) Specific heat capacity of substance A is 3.8 J g^-1 K^-1 whereas the Specific heat capacity of substance B is 0.4 J g^-1 K^-1.
(i) Which of the two is a good conductor of heat?
(ii) How is one led to the above conclusion?
(iii) If substances A and B are liquids then which one would be more useful in car radiators? [3]
Answer:
(i) B is a good conductor of heat.

(ii) Specific heat capacity is the amount of heat energy required to raise the temperature of unit mass of a substance by 1 K. So the one which uses less amount of heat energy is a good conductor.

(iii) A would be useful in car radiators as it can extract large amount of heat energy without raising its own temperature much.

(c) (i) State any two measures to minimize the impact of global warming.
(ii) What is the Greenhouse effect? [3]
Answer:
(i)

1. Cutting of trees must be banned and plantation should be increased.
2. Burning of fossil fuels should be for limited purposes only. We should find and use some alternative sources of energy.

(ii) Trapping of infrared radiations which causes global heating is known as greenhouse effect. It can be done by mainly carbon dioxide and methane gases.

Question 10.
(a) (i) Name two factors on which the magnitude of an induced e.m.f. in the secondary coil depends.
(ii) In the following diagram an arrow shows the motion of the coil towards the bar magnet.
(1) State in which direction the current flows, A to B or B to A?
(2) Name the law used to come to the conclusion.

Answer:
(i) Change in magnetic flux : More the change in magnetic flux in a given time, more is the e.m.f. induced.
Time in which the magnetic flux changes: More rapid the flux changes, more is the e.m.f. induced.

(ii)

1. Induced current is always opposite to the cause which produces it, so current will flow from B to A.
2. Lenz’s law and Faraday’s laws.

(b) A nucleus ₁₁Na²⁴ emits a beta particle to change into Magnesium (Mg)
(i) Write the symbolic equation for the process.
(ii) What are numbers 24 and 11 called?
(iii) What is the general name of ₁₂²⁴ Mg with respect to ₁₁²⁴Na? [3]
Answer:
(i) \({ }_{11}^{24} \mathrm{Na}\) → \({ }_{11}^{24} \mathrm{Mg}\) + \({ }_{-1}^0 e\) + (where, \({ }_{-1}^0 e\) is β)

(ii) 24 is mass number, where 11 is atomic number.

(iii) Isobars.

(c) In a cathode ray tube state :
(i) The purpose of covering cathode by thorium and carbon.
(ii) The purpose of the fluorescent screen.
(iii) How is it possible to increase the rate of emission of electrons. [3]
Answer:
(i) As a good electron emitter because thorium and carbon lower work function.

(ii) Electrical signal applied on the deflecting plates change into visual patterns on the screen.

(iii) By increasing temperature and surface area of metal and by using metal with a low work function – rate of emission of electrons can be increased.

Solving ICSE Class 10 Physics Previous Year Question Papers ICSE Class 10 Physics Question Paper 2022 Semester 2 is the best way to boost your preparation for the board exams.

ICSE Class 10 Physics Question Paper 2022 Solved Semester 2

Time : 1½ hour
Maximum Marks: 40

Genral Instructions

• Answers to this Paper must be written on the paper provided separately.
• You will not be allowed to write during the first 10 minutes.
• This time is to be spent in reading the question paper.
• The time given at the head of this Paper is the time allowed for writing the answers.
• Attempt all questions from Section A and any three questions from Section B.
• The marks intended for questions are given in brackets [ ].

Section – A
(Attempt all questions)

Question 1.
Choose the correct answer to the questions from the given option. (Do not copy the question. Write the correct answer only.) [10]
(i) Free vibrations are :
(a) the vibrations under the influence of a periodic force
(b) the vibrations with larger amplitude
(c) the vibrations when the frequency continuously decreases
(d) the vibrations with a constant frequency and constant ampli-tude.
Answer:
(d) the vibrations with a constant frequency and constant amplitude.

(ii) The diagram below shows four sound waves. Which sound has the highest pitch ?

Answer:

(iii) The graph plotted for potential difference (V) against current (I) for ohmic resistors is :
(a) A curve passing through the origin
(b) A straight line not passing through origin
(c) A straight line passing through origin
(d) A circle centred at the origin
Answer:
(c) A straight line passing through origin

(iv) A main switch in the main distribution board is present in :
(a) a live wire
(b) a neutral wire
(c) a live as well as neutral wire
(d) an earth wire
Answer:
(c) a live as well as neutral wire

(v) A conductor AB is kept along north south direction of the earth above a magnetic needle as shown below. When the key K is closed then:

(a) the needle will not show any deflection
(b) the needle will defect towards east
(c) the needle will turn in the opposite direction Le. towards south
(d) the needle will deflect towards west.
Answer:
(d) the needle will deflect towards west.

(vi) A coil wound around a piece of soft iron can become an electromagnet only when :
(a) the circuit is open
(b) a magnetic compass is present in the vicinity
(c) a galvanometer is connected to the circuit
(d) a current flows in the circuit.
Answer:
(d) a current flows in the circuit.

(vii) If water absorbs 4000 joule heat to increase the temperature of 1 kg water through 1°C then the specific heat capacity of water is:
(a) 4 Jkg^-1 °C^-1
(b) 400 Jg^-1 °C^-1
(c) 4 Jg^-1 °C^-1
(d) 4.2 Jg^-1 °C^-1
Answer:
(c) 4 Jg^-1 °C^-1

(viii) Water is used in car radiators because :
(a) it is a good conductor of heat.
(b) it conducts heat faster as compared to the other substances and cools the engine quickly.
(c) its specific heat capacity is very- low.
(d) its specific heat capacity is very high so it can cool the engine without a greater increase in its own temperature.
Answer:
(d) its specific heat capacity is very high so it can cool the engine without a greater increase in its own temperature.

(ix) The heaviest nuclear radiation is :
(a) x-radiation
(b) α-radiation
(c) γ-radiation
(d) β-radiation
Answer:
(b) α-radiation

(x) To study the age of excavated material of archaeological signification w e study the rate of decay of an isotope of:
(a) Uranium
(b) Cobalt
(c) Carbon
(d) Chlorine
Answer:
(c) Carbon

Section B
(Attempt any three questions from this Section)

Question 2.
(i) The diagram below shows a magnetic compass kept closer to a coil AB wound around a hollow cylindrical cardboard. [3]

(a) After studying the circuit and the magnetic compass carefully, state whether the switch S₁ is open or closed.
Answer:
The switch is closed.

(b) How did you arrive at the conclusion in (a) ?
Answer:
As the magnetic compass is deflected.

(c) What is the purpose of placing the magnetic compass in the above setup ?
Answer:
The compass allows us to find whether a magnetic field is generated or not.

(ii) (a) Give an important reason for copper to be used as a material for a calorimeter. [3]
Answer:
Because copper is a good conductor of heat.
OR
Copper has low specific heat capacity and thus it reaches the equilibrium temperature quickly by absorbing a small amount of heat. It is also highly malleable and hence a very thin box of small mass can be made thus decreasing the heat capacity of the box.

(b) Calculate the thermal capacity of 40 g of water.
[Specific heat capacity of water = 4200 Jkg^-1 °C^-1] [4]
Answer:
Thermal capacity of water (Q) = mCT, where T is the change in temperature of water
= 40 × 4.2 × T= 168TJ

(iii)

In the above circuit diagram, calculate : [3]
(a) the external resistance of the circuit
(b) the current l₂
(c) the current I.
Answer:
(a) Resistors 12Ω and 8Ω are connected in parallel.
This combination is connected in series with the 2 Ω resistor.
In parallel, we have R = \(\frac{12 \times 8}{12+8}\) = \(\frac{96}{20}\)
= \(\frac{24}{5}\) = 4.8Ω
Net resistance = 2 + 4.8 = 6.8Ω

(b) Now, I₁R₁ = I₂R₂ or I₂ = \(\frac{\mathrm{I}_1 \mathrm{R}_1}{\mathrm{R}_2}\)
= \(\frac{0.4 \times 12}{8}\) = 0.6A

(c) I = I₁ + I₂ = 0.4 + 0.6 = 1A

Question 3.
(i) Three wires with proper colour coding are connected to the three terminals of a three- pin socket. Match the colour of the wire with the proper terminals A, B and C of the socket. [3]
(a) Brown
(b) Green
(c) Light blue

Answer:
(i) A – Green,
B – Brown,
C – Light blue

(ii) (a) Why does it become colder after a hailstorm than during or before the hailstorm? [3]
(b) ‘If two bodies have the same specific heat capacities, then they will always absorb the same amount of heat if their temperature increases by the same amount.’ State whether the given statement is true or false.
Answer:
(ii) (a) Before and during the hailstorm there is plenty of heat present in the atmosphere. But after the hailstorm ice absorbs heat energy required for melting from the surrounding, so the temperature of surrounding falls and we feel cold.
(b) The statement is true.

(iii) A metal piece of mass 420 g present at 80°C is dropped in 80 g of water present at 20°C in a calorimeter of mass 84 g. If the final temperature of the mixture is 30°C then calculate the specific heat capacity of the metal piece. [4]
[Specific heat capacity of water = 4.2 Jg^-1 °C^-1,
Specific heat capacity of the calorimeter = 200 Jkg^-1 °C^-1]
Answer:
[By the principle of calorimetry]
Heat released by the metal piece Heat absorbed by the calorimeter + Heat absorbed by the water.
Final temperature of the mixture = 30°C
Mass of metal piece = 420 g
Temperature of metal piece = 80°C
Specific heat capacity of metal piece
= c (to be found)
Heat energy released by metal piece
= mcΔt = 420 × c × (80 – 50) = 21000c J
Mass of calorimeter = 84 g
Temperature of calorimeter = 20°C
Specific heat capacity of calorimeter
= 200 J kg^-1°C^-1 = 0.200 J g^-1°C^-1
Heat energy absorbed by the calorimeter
= mcΔt = 84 × 0.2 × (30 – 20) = 168J
Mass of water = 80 g
Temperature of water = 20°C
Specific heat capacity of water = 4.2Jg^-1°C^-1
Heat energy absorbed by water
= mcΔt
= 80 × 4.2 × (30 – 20) = 3360 J
By the principle of calorimetry,
21000c = 3360 + 168
21000c = 3528
c = \(\frac{3528}{21000}\)
= 0.168 J g^-1 K^-1

Question 4.
(i) Rohit playing a flute and Anita playing a piano emit sounds of same pitch and loudness. [3]
(a) Name one characteristic that is different for waves from the two different instruments.
Answer:
(a) Timbre

(b) If now the loudness of the sound from flute becomes four times that of the sound from piano, then write the value of the ratio Ar: Ar. (Ar – amplitude of sound wave from flute, Ar – amplitude of sound wave from piano)
Answer:
A_F : A_P = 2 : 1

(c) Define ‘Pitch’ of a sound.
Answer:
Pitch is defined as the characteristic of sound which is used for differentiating between the shrill and flat sound.

(ii) (a) Name two factors on which the force experienced by a conductor carrying current, placed in a magnetic field, depends. Also state how these factors affect the force. [3]
(b) With the help of which rule you can determine the direction of force acting on a current carrying conductor placed in a magnetic field ?
Answer:
(a) Magnetic field and current. Force is proportional to both the magnetic field and the current.
(b) Fleming’s left hand rule.

(iii) (a) What is nuclear energy ? [4]
(b) After emission of a nuclear radiation, the atomic number of the daughter nucleus increases by 1. Identify the nuclear radiation.
(c) Write a nuclear reaction indicating the nuclear change mentioned in {b).
(d) What is the special name given to the parent and daughter nucleus when this radiation is emitted ?
Answer:
(a) The energy released during nuclear fission or
fusion.
(b) Beta decay
(c)

(d) Isobars

Question 5.
(i) An appliance rated 440 W, 220 V is connected across 220 V supply. [3]
(a) Calculate the maximum current that the appliance can draw.
(b) Calculate the resistance of the appliance.
Answer:
(a) I = \(\frac{P}{V}\) = \(\frac{440}{220}\) = 2A
(b) R = \(\frac{\mathrm{V}^2}{\mathrm{P}}\) = \(\frac{(220)^2}{440}\) = 110Ω

(ii) The diagram below shows a vibrating tuning fork E mounted on a sound box X. When the vibrating tuning forks A, B, C and D are placed on the sound box Y one by one, it is observed that a louder sound is produced when the turning fork B is placed on Y. [3]

(a) What is the frequency of tuning fork E ?
(b) Why does B produce a louder sound ?
Answer:
(a) 320 Hz
(b) Resonance

(iii) (a) From the graph of heating curve given below state the melting point and boiling point of the substance. [4]

(b) Complete and rewrite the following nuclear reaction by filling the blanks.
²³⁵U → ₉₀Th + ⁴₂He
Answer:
(a) Melting point 40°C, boiling point 160°C
(b) \({ }_{92}^{235} U\) → \(9_{90}^{231} \mathrm{Th}\) + \({ }_2^4 \mathrm{He}\)

Question 6.

(i) Study the above figure and Answer the following : [3]
(a) What type of vibration does the above figure represent ?
Answer:
Damped vibrations.

(b) State one reason for which the amplitude of the vibration decreases with time.
Answer:
Loss of energy because of friction.

(c) Write an example of natural vibrations.
Answer:
Vibration of the bob of a simple pendulum.

(ii) A certain beam of a particles, p particles and y radiation travel through a region of electric field produced between two oppositely charged parallel plates A (+) and B (-). [3]
(a) Which of the above three has the maximum speed ?
(b) Which one deviates the most from its original path ?
(c) Which one does not deviate at all when passing through a region of electric or magnetic field ?
Answer:
(a) Gamma radiation
(b) Beta particle
(c) Gamma

(iii) If a wire of resistance 2 Ω gets stretched to thrice its original length : [4]
(a) Calculate the new resistance of the wire.
(b) What happens to the specific resistance of the wire ?
Answer:
(a) New resistance
RN = n²R = 3² × 2= 18Ω
(b) No change

Solving ICSE Class 10 Physics Previous Year Question Papers ICSE Class 10 Physics Question Paper 2013 is the best way to boost your preparation for the board exams.

ICSE Class 10 Physics Question Paper 2013 Solved

(Time : 1 ½ hours)
Maximum Marks: 80

General Instructions:

• Answers to this Paper must be written on the paper provided separately.
• You will not be allowed to write during the first 15 minutes.
• This time is to be spent in reading the Question Paper,
• The time given at the head of this Paper is the time allowed for writing the answers.
• Section I is compulsory. Attempt any four questions from Section II.
• The intended marks for questions or parts of questions are given in brackets [ ].

Section I (40 Marks)
Attempt all questions from this Section.

Question 1.
(a) Give any two effects of a force on a non-rigid body. (2)
Answer:
Force when applied on a non-rigid body can :

1. change the dimensions (shape/size) of body.
2. cause motion in the body.

(b) One end of a spring is kept fixed while the other end is stretched by a force as shown in the diagram.

1. Copy the diagram and mark on it the direction of the restoring force.
2. Name one instrument which works on the above principle. (2)

Answer:
Restoring force always acts in a direction opposite to the direction of force.

1. 
2. Spring balance

(c) (i) Where is the centre of gravity of a uniform ring situated ?
(ii) ‘The position of the centre of gravity of a body remains unchanged even when the body is deformed.’ State whether the statement is true or false. (2)
Answer:
(i) At its geometric centre.
(ii) False (Because position of centre of gravity of a body depends on distribution of mass in it so if body is deformed mass distribution varies and so position of centre of gravity also varies).

(d) A force is applied on a body of mass 20 kg moving with a velocity of 40 ms^-1. The body attains a velocity of 50 ms^-1 in 2 seconds. Calculate the work done by the body. (2)
Answer:
m = 20 kg
u =40 ms^-1
v = 50 ms^-1
t = 2 s

W.D. = Change in kinetic energy (According to work energy theorem)
= \(\frac{1}{2}\) mv²– \(\frac{1}{2}\) mu²
W.D. = \(\frac{1}{2}\) × 20 x (50)²– \(\frac{1}{2}\) × 20 × (40)²
= \(\frac{1}{2}\) × 20 × 2500 – \(\frac{1}{2}\) × 20 × 1600                  .
= 25000 – 16000 = 9000 J or 9 × io³ J

(e) A type of single pulley is very often used as a machine even though it does not give any gain in mechanical advantage.
(i) Name the type of pulley used.
(ii) For what purpose is such a pulley used ? (2)
Answer.
(i) Single fixed pulley.
(ii) It is used only to change the direction of force applied i.e., in a more convenient direction.

Question 2.
(a) (i) In what way does an ‘Ideal machine’ differ from a ‘Practical machine’ ?
(ii) Can a simple machine act as a force multiplier and a speed multiplier at the same time? (2)
Answer:
(i) Ideal machine is free from friction or Output work = Input work or efficiency is 100% whereas for all practical machines due to friction there is always loss in energy and efficiency is less than 100% and V.R. is not equal to M. A. or output work is less than input work.

(ii) No, a simple machine at one time can act as a force multiplier where effort arm is bigger than load arm but at some other time can act as speed multiplier where effort arm is smaller than load arm so it is not possible at one time.

(b) A girl of mass 35 kg climbs up from the first floor of a building at a height 4 m above the ground to the third floor at a height 12 m above the ground. What will be the increase in her gravitational potential energy ? (g = 10 ms^-2). (2)
Answer:
m = 35 kg
h₁ = 4 m
h₂ = 12 m
g = 10 ms^-2
Increase in gravitational
P.E. = mg (h₂ – h_})
= 35 × 10(12 – 4)
= 35 × 10 × 8
= 2800 J

(c) Which class of lever found in the human body is being used by a boy (2)
(i) when he holds a load on the palm of his hand.
(ii) when he raises the weight of his body on his toes ?
Answer:
(i) Class III lever (in holding a load on the palm of his hand).
(ii) Class II lever (in raising the weight of body on toes).

(d) A ray of light is moving from a rarer medium to a denser medium and strikes a plane mirror placed at 90° to the direction of the ray as shown in the diagram.

(i) Copy the diagram and mark arrows to show the path of the ray of light after it is reflected from the mirror.
(ii) Name the principle you have used to mark the arrows to show the direction of the ray.(2)
Answer:
(i) It reverses the path.
(ii) Principle of reversibility of light.

(e) (i) The refractive index of glass with respect to air is 1.5. What is the value of the refractive index of air with respect to glass ?
(ii) A ray of light is incident as a normal ray on the surface of separation of two different mediums. What is the value of the angle of incidence in this case ? (2)
Answer:
(i) _airμ_glass
glassμ_air = \(\frac{1}{\text { air } \mu_{\text {glass }}}=\frac{1}{1.5}=\frac{2}{3}\) = 0.67
(ii) ∠i = ∠Zr = 0°

Question 3.
(a) A bucket kept under a running tap is getting filled with water. A person sitting at a distance is able to get an idea when the bucket is about to be filled.
(i) What change takes place in the sound to give this idea ?
(ii) What causes the change in the sound ? (2)
Answer:
(i) As the water level in the bucket increases, length of air column decreases, frequency increases so the sound becomes shriller and a person can get an idea of bucket to be filled.
(ii) Due to change in length of air column, frequency changes.

(b) A sound made on the surface of a lake takes 3 s to reach a boatman.
How much time will it take to reach a diver inside the water at the same depth ?
[Velocity of sound in air = 330 ms^-1 ; Velocity of sound in water = 1450 ms^-1] (2)
Answer:
Given depth is same or d is same.
So d = vt (in air)
= 330 × 3 = 990 m
So time required in water
t = \(\frac{d}{v}=\frac{990}{1450}\) = 0.68 s

(c) Calculate the equivalent resistance between the points A and B for the following combination of resistors: (2)

Answer:
As in the circuit at position C, current goes through three different paths which are parallel to each other :

1st path CDEF
2nd path CGHF
3rd path CF
Resistance of 1st path CDEF
= 4 Ω + 4 Ω + 4 Ω
= 12 Ω
Resistance of 2nd path CGHF
= 2Ω + 2Ω + 2Ω
= 6 Ω
Resistance of 3rd path CF = 4 Ω
So \(\frac{1}{\text { T.R. }}=\frac{1}{12}+\frac{1}{6}+\frac{1}{4}\)
= \(\frac{1+2+3}{12}=\frac{6}{12}=\frac{1}{2}\)
So T.R. in parallel = 2 Ω
Now, AC and FB are in series.
∴ Total resistance = 5 Ω+ 2 Ω + 6 Ω
= 13 Ω

(d) You have been provided with a solenoid AB.

(i) What is the polarity at end A ?
(ii) Give one advantage of an electromagnet over a permanent magnet. (2)
Answer:
(i) North.
(ii) Strength of an electromagnet can be changed – according to its use and it will be a magnet till the time current passes through it whereas strength of permanent magnet cannot be increased and cannot be magnetised and demagnetised in an instance.

(e) (i) Name the device used to protect the electric circuits from overloading and short circuits.
(ii) On what effect of electricity does the above device work ? (2)
Answer:
(i) Fuse.
(ii) Fleating effect of electric current.

Question 4
(a) Define the term ‘Heat capacity’ and state its S.I. unit. (2)
Answer:
Heat capacity is the amount of heat energy required to change the temperature of a given mass of a body by 1°C or by 1 kelvin. S.I. unit is JC^_1 or JK^-1.

(b) What is meant by Global warming ? (2)
Answer:
Trapping of infrared radiations in atmosphere by certain gases like methane and carbon dioxide causes global warming.

(c) How much heat energy is released when 5 g of water at 20°C changes to ice at 0°C?
[Specific heat capacity of water = 4.2 Jg^-1 °C^-1; Specific latent heat of fusion of ice = 336 g^-1] (2)
Answer:
Heat energy released :

1. 5 g of water at 20°C changes to water at 0°C (mcθ)
   = 5 × 4.2 × 20 = 420 J.
2. water at 0°C changes to ice at 0°C (mL)
   = 5 × 336= 1680 J
   ∴ Total heat energy required = 1680 + 420 = 2100 J

(d) Which of the radioactive radiations
(i) can cause severe genetical disorders.
(ii) are deflected by an electric field ?
Answer:
(i) γ
(ii) α as well as β radiations.

(e) A radioactive nucleus undergoes a series of decays according to the sequence

If the mass number and atomic number of X₃ are 172 and 69 respectively, what is the mass number and atomic number of X ?
Answer:

If X₃ has At. No. 69 and Mass No. 172.

(As emission of α decreases At. No. by 2 and Mass No. by 4)
∴ X has At No. 72, Mass No. 180
(As emission of β increases At. No. by 1 and no change  in Mass No.)

Section II (40 Marks)
Attempt any four questions from this Section.

Question 5.
(a) (i) With reference to their direction of action, how does a centripetal force differ from a centrifugal force ?
(ii) State the Principle of conservation of energy.
(iii) Name the form of energy which a body may possess even when it is not in motion. (3)
Answer:
(i) Centripetal force is always directed towards the centre of circle whereas centrifugal force acts on a body away from the centre.
(ii) Energy can neither be created nor can it be destroyed. It only changes from one form to another.
(iii) Potential energy.

(b) A coolie is pushing a box weighing 1500 N up an inclined plane 7.5 m long on to a platform, 2.5 m above the ground.
(i) Calculate the mechanical advantage of the inclined plane.
(ii) Calculate the effort applied by the coolie.
(iii) In actual practice, the coolie needs to apply more effort than what is calculated. Give one reason why you think the coolie needs to apply more effort. (3)
Answer:

(i) M.A. = \(\frac{\text { Length }}{\text { Height }}=\frac{7.5}{2.5}\) = 3

(ii) Effort = ?
M.A. = \(\frac{\mathrm{L}}{\mathrm{E}}\)
3 = \(\frac{1500}{E}\)
E = 500 N

(iii) In actual practice due to friction etc. efficiency is not 100% and V.R. is more than M.A. or output is less than work input. So he needs to apply more effort than calculated.

(c) A block and tackle system of pulley’s a velocity ratio 4.
(i) Draw a labelled diagram of the system indicating clearly the points of application and directions of a load and effort.
(ii) What is the value of the mechanical advantage of the given pulley system if it is an ideal
pulley system ? (4)
Answer:
(i) Load = 4T
E = T
∴ V.R. = 4
Block and tackle system of pulley having V.R. = 4

(ii) M.A. = V.R. if it is ideal pulley
= 4

Question 6
(a) Name the radiations :
(i) that are used for photography at night,
(ii) used for detection of fracture in bones.
(ii) whose wavelength range is from 100 Å to 4000 Å (or 10 nm to 400 nm). (3)
Answer:
(i) Infrared radiations.
(ii) X- rays.
(iii) Ultraviolet

(b) (i) Can the absolute refractive index of a medium be less than one ?
(ii) A coin placed at the bottom of a beaker appears to be raised by 4.0 cm.
If the refractive index of water is 4/3, find the depth of the water in the beaker. (3)
Answer:
(i) No, absolute refractive index of a medium is always greater than 1, as speed of light in any medium is always less than that in vacuum.
(ii) \(\frac{4}{3}=\frac{\text { Real depth }}{\text { Apparent depth }}\)
Let real depth = x
So apparent depth = x – 4
\(\frac{4}{3}=\frac{x}{x-4}\)
4x – 16 = 3x
4x – 3x = 16
x = 16
∴ Real depth = 16 cm

(c) An object AB is placed between 2F₁ and F₁ on the principal axis of a convex lens as shown in the diagram.

Copy the diagram and using three rays starting from point A, obtain the image of the object formed by the lens. (4)
Answer:

1st ray is ACA’
2nd ray is AOA’
3rd ray is BOB’
Image is formed beyond 2F₂ ; real, inverted and enlarged.

Question 7.
(a) (i) What is the principle on which SONAR is based.
(ii) An observer stands at a certain distance away from a cliff and produces a loud sound. He hears the echo of the sound after 1.8 s. Calculate the distance between the cliff and the observer if the velocity of sound in air is 340 ms’. (3)
Answer:
(i) Echo depth sounding.
Ultrasonic waves have the same speed as of audible sound but are not absorbed in the medium. So transmitter sends these waves receiver receives the waves back after striking the rigid obstacle so time taken is recorded. And we can calculate distance d = \(\frac{v t}{2}\)

(ii) t = 1.8 s
d = ?
v = 340 ms^-1
As d = \(\frac{v t}{2}\) = \(\frac{340 \times 2}{1.8}\) = 306 m

(b) A vibrating tuning fork is placed over the mouth of a burette filled with water. The tap of the burette is opened and the water level gradually starts falling. It is found that the sound from the tuning fork becomes very loud for a particular length of the water column.
(i) Name the phenomenon taking place when this happens.
(ii) Why does the sound become very loud for this length of the water column ? (3)
Answer:
(i) Resonance
(ii) When the frequency of air column becomes equal to the frequency of tuning fork i.e., vibrations of air column are in resonance with those of fork, loud sound is heard.

(c) (i) What is meant by the terms (1) amplitude (2) frequency of a wave ?
(ii) Explain why stringed musical instruments, like the guitar, are provided with a hollow box. (4)
Answer:
(i) (1) Maximum displacement of wave from its mean position on either side is known as amplitude.

(2) Number of vibrations produced in 1 second is frequency.

(ii) A vibrating string by itself produces a very weak sound which cannot be heard at a distance. Therefore, all stringed instruments are provided with sound boxes having air inside. So when string vibrates the air inside sets into forced vibrations which have same frequency. Due to large air column and resonance loud sound is heard.

Question 8.
(a) (i) It is observed that the temperature of the surrounding starts falling when the ice in a frozen lake starts melting. Give a reason for the observation.
(ii) How is the heat capacity of the body related to its specific heat capacity ? (3)
Answer:
(i) Ice for melting extracts 336 J/g heat energy from surroundings and as surroundings loose a large amount of heat energy, temperature falls.

(ii) Heat capacity is the amount of heat energy required to raise the temperature of given mass of a body by 1 °C or 1 K whereas specific heat capacity is the amount of heat energy required to raise a unit mass of a substance by 1 °C or 1 K.
Heat capacity = mc
Specific heat capacity= c
Units of S.H.C. = J/kg/K whereas of
H.C. = J/K

(b) (i) Why does a bottle of soft drink cool faster when surrounded by ice cubes than by ice cold water, both at 0° C ?
(ii) A certain amount of heat Q will warm 1 g of material X by 3°C and 1 g of material Y by 4°C. Which material has a higher specific heat capacity.
Answer:
(i) Ice cubes extract 336 J/g heat energy extra than ice cold water at 0°C. So it cools soft drink faster.
(ii) As mass of both is same, Q is same.
c = \(\frac{\mathrm{Q}}{m \theta}\), where Q is amount of heat energy given.
So a body having less increase in temperature will have more specific heat capacity.
∴ X has more S.H.C. than Y.

(c) A calorimeter of mass 50 g and specific heat capacity 0.42 J g^-1 °C^-1 contains some mass of water at 20°C. A metal piece of mass 20 g at 100 °C is dropped into the calorimeter. After stirring, the final temperature of the mixture is found to be 22°C. Find the mass of water used in the calorimeter.
[specific heat capacity of the metal piece = 0.3 Jg 1 °C^-1 ; specific heat capacity of water = 4.2 Jg^-1 °C^-1] (4)
Answer

According to the principle of calorimetry
Heat lost by metal = Heat gained by calorimeter + Heat gained by water
20 × 0.3 × (100 – 22) = 50 × 0.42 × (22 – 20) + m × 4.2 × (22 – 20)
6 × 78 = 42+ 8.4m
468 – 42 = 8.4m
m = \(\frac{426}{8.4}=\frac{4260}{84}\)
= \(\frac{355}{7}\) = 50.71 g

Question 9.
(a) (i) State Ohm’s law.
(ii) A metal wire of resistance 6Ω is stretched so that its length is increased to twice its original length. Calculate its new resistance. (3)
Answer:
(i) Ohm’s law states that physical conditions remaining constant, potential difference is directly proportional . to the current i.e., V ∝ I, if R is kept constant.
So V = IR

(ii) R = 6 Ω,
when length = l, area = a
So R = ρ\(\frac{l}{a}\) ………………….. (i)
When the wire is stretched
New length = 21 d
New area = \(\frac{d}{2}\) , as ρ is same.
So, R’ = ρ \(\frac{2l}{a}\) × 2
= ρ \(\frac{l}{a}\) × 4  ………………. (ii)
Putting the value of ρ \(\frac{l}{a}\) = 6 in equation (ii), we have a
R’ = 6 × 4 = 24 Ω

(b) (i) An electrical gadget can give an electric shock to its user under certain circumstances. Mention any two of these circumstances.
(ii) What preventive measure provided in a gadget can protect a person from an electric shock ?
Answer:
(i) An electric shock may be caused either due to (1) poor insulation of wires (2) when electric appliances are touched with wet hands.
(ii) Insulation of wires must be of good quality and should be checked from time to time so that no wire is left naked and appliances should not be touched with wet hands.

(c) The figure shows a circuit

When the circuit is switched on, the ammeter reads 0.5 A.
(i) Calculate the value of the unknown resistor R.
(ii) Calculate the charge passing through the 3Ω resistor in 1205.
(iii) Calculate the power dissipated in the 3 Ω resistor. (4)
Answer:
(i) As V = IR (Total R)
So Total R = \(\frac{\mathrm{V}}{\mathrm{I}}=\frac{6}{0.5}\) = 12 Ω
∴  R in circuit = 12 – 3 = 9 Ω (as both are in series)
R = 9 Ω

(ii) Charge = Current × Time
= 0.5 × 120 = 60 C

(iii) Power dissipated in 3 × resistor
= I²R = 0.5 × 0.5 × 3 = 0.75 W

Question 10.
(a) Name the three main parts of a Cathode Ray Tube. (3)
Answer:

1. Electron gun.
2. Deflecting plates.
3. Fluorescent screen.

(b) (i) What is meant by Radioactivity ?
(ii) What is meant by nuclear waste ? (3)
(iii) Suggest one effective way for the safe disposal of nuclear waste.
Answer:
(i) Radioactivity is the self spontaneous disintegration of a heavy nucleus into a, (3 and y radiations in order to attain stability.

(ii) The radioactive material after its use is known as nuclear waste.

(iii) The nuclear waste should be kept in thick casks (thick lead containers) and then they must be buried in the specially constructed deep underground stores or in useless mines and these mines should be sealed after storing casks.

(c) (i) Draw a simple labelled diagram of d.c. electric motor.
(ii) What is the function of the split rings in a d.c. motor ?
(iii) State one advantage of a.c. over d.c. (4)
Answer:
(i)

(ii) Function of split ring is to change the direction of current flowing through the coil after each half rotation such that direction of motor is only one.

(iii) Alternating current can be stepped up or stepped down by using step up and step down transformers whereas direct current cannot be. So by doing this, heat losses are reduced e., it prevents huge loss of electrical energy into heat energy.