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

ICSE Class 10 Physics Question Paper 2019 Solved

Time: 1½ hours
Maximum Marks: 80

General Instructions:

• Answers to this paper must be written on ihe 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) The diagram below shows a claw hammer used to remove a nail; [2]

(i) To which class of lever does it belong?
(ii) Given one more example of the same class of lever mentioned by you in (i) for which the mechanical advantage is greater than one.
Answer:
(i) Class One.
(ii) Shears used for cutting the thin metal sheets.

(b) Two bodies A and B have masses in the ratio 5:1 and their kinetic energies are in the ratio 125:9. Find the ratio of their velocities. [2]
Answer:
Given \(\frac{m_{\mathrm{A}}}{m_{\mathrm{B}}}\) = \(\frac{5}{1}\)
Kinetic energy is given by K = \(\frac{1}{2} m v^2\)

(c) (i) Name the physical quantity which is measured in calories. [2]
(ii) How is calories related to the S.I. unit of that quantity?
Answer:
(i) Heat energy.
(ii) 1 cal = 4.2 J.

(d) (i) Define couple. [2]
(ii) State the S.I. unit of moment of couple.
Answer:
(i) It is defined as a set of two equal and opposite forces acting along different lines of action.
(ii) SI unit is N m.

(e) (i) Define critical angle. [2]
(ii) State one important factor which affects the critical angle of a given medium.
Answer:
(i) It is the angle of incidence in the denser medium for which angle of refraction in the rarer medium is 90°.
(ii) Nature of the two media in contact.

Question 2.
(a) An electromagnetic radiation is used for photography in fog. [2]
(i) Identify the radiation.
(ii) Why is this radiation mentioned by you, ideal for this purpose?
Answer:
(i) infrared radiation.
(ii) Because it can penetrate fog.

(b) (i) What is the relation between the refractive index of water with respect to air (_aµ_w) and the refractive index of air with respect to water (_wµ_a). [2]

(ii) If the refractive index of water with respect to air (_aµ_w) is \(\frac{5}{3}\)
Calculate the refractive index of air with respect to water (_wµ_a).
Answer:
(i) The relation is _aμ_w = \(\frac{1}{{ }_w \mu_a}\)
(ii) Given _wμ_a = \(\frac{5}{3}\), therefore,
we have _wμ_a = \(\frac{1}{a \mu_w}\) = \(\frac{1}{5 / 3}\) = \(\frac{3}{5}\)

(c) The specific heat capacity of a substance A is 3,800 Jkg^-1K^-1 and that of a substance B is 400 Jkg^-1K^-1. Which of the two substances is a good conductor of heat? Give a reason for your answer. [2]
Answer:
Substance B. Because it requires relatively little heat to raise its temperature.

(d) A man playing a flute is able to produce notes of different frequencies. If he closes the holders near his mouth, will the pitch of the note produced, increase or decrease? Given a reason. [2]
Answer:
The pitch will decrease, as the pitch is inversely proportional to the length of the vibrating air column.

(e) The diagram below shows a light source P embedded in a rectangular glass block ABCD of critical angle 42°. Complete the path of the ray PQ till it emerges out of the block. [Write necessary angles.] [2]

Answer:
The path of ray is as shown:

Question 3.
(a) (i) If the lens is placed in water instead of air, how does its focal length change?
(ii) Which lens, thick or thin has greater focal length?
Answer:
(i) Its focal length increases.
(ii) Thin lens.

(b) Two waves of the same pitch have amplitudes in the ratio 1:3.
What will be the ratio of their:
(i) Intensities and
(ii) Frequencies?
Answer:
(i) Ratio of intensities will be 1: 9.
(ii) Ratio of frequencies will be 1 : 1.

(c) How does an increase in the temperature affect the specific resistance of a : [2]
(i) Metal and
(ii) Semiconductor?
Answer:
(i) Increases.
(ii) Decreases.

(d) (i) Define resonant vibrations. [2]
(ii) Which characteristic of sound, makes it possible to recognize a person by his voice without seeing him?
Answer:
(i) If a body is made to vibrate with its fundamental frequency by an external source, then the vibrations are called resonant vibrations.
(ii) Quality.

(e) It is possible for a hydrogen (H) nucleus to emit an alpha particle? [2]
Answer:
No, as it does not possess a helium nucleus.

Question 4.
(a) Calculate the effective resistance across AB : [2]

Answer:
Resistors of 5Ω and 4Ω are in senes, therefore, their net resistance is R_S = 5 + 4 = 9Ω.
Now, this 9Ω and 3Ω are in parallel, hence net resistance
R_P = \(\frac{9 \times 3}{9+3}\) = \(\frac{18}{12}\) = 1.5Ω
Now, this is in series with the 8Ω resistor, hence net resistance of the circuit is
R_T = 8 + 1.5 = 9.5Ω

(b) (i) State whether the specific heat capacity of a substance remains the same when its state changes from solid to liquid.
(ii) Given one example to support your answer. [2]
Answer:
(i) No, specific heat capacity does not remain the same when it changes from solid to liquid.
(ii) Let us take the example of water.
Specific heat capacity of water
= 4.2 J kg^-1K^-1
and specific heat capacity of ice
= 2.1 J kg^-1K^-1

(c) A magnet kept at the centre of two coils A and B is moved to and fro as shown in the diagram. The two galvanometers show deflection. [2]

Answer:
x

(d) State with a reason whether:
x > y or x < y [x and y are magnitudes of deflection.]
(i) Why a nuclear fusion reaction is called a thermos nuclear reaction?
(ii) Complete the reaction:
³He₂ + ²H₁ → ⁴He₂ + ………… + Energy
Answer:
(i) Because this occurs at a very high temperature.
(ii) ³He₂ + ²H₂ → ⁴He₁ + ¹H₁ + Energy

(e) State two ways to increase the speed of rotation of a D.C. motor.
Answer:
The two ways are
(i) Strong magnet and
(ii) High current.

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

Question 5.
(a) A body of mass 10 Kg is kept at a height of 5 m. It is allowed to fall and reach the ground. [3]
(i) What is the total mechanical energy possessed by the body at the height of 2m assuming it is a friction less medium?
(ii) What is the kinetic energy possessed by the body just before hitting the ground?
Take g = 10m/s².
Answer:
(i) The total mechanical energy is the total potential energy it possesses before its fall.
E = mgh = 10 × 10 × 5 = 500 J
(ii) It is equal to the maximum potential energy E = mgh
= 10 × 10 × 5 = 500J.

(b) A uniform meter scale is in equilibrium as shown in the diagram: [3]

(i) Calculate the weight of the meter scale.
(ii) Which of the following options is correct to keep the ruler in equilibrium when 40 gf wt is shifted to 0 cm mark?
F is shifted towards 0 cm.
or
F is shifted towards 100 cm.
Answer:
(i) The weight of the metre scale acts at its centre of gravity i.e., at 50 cm mark.
By the concept of moments, we have
40 × (30 – 5) = W × (50 – 30)
or W = (40 × 25)/20 = 50 gf
(ii) F is shifted towards 0 cm.

(c) The diagram below shown a pulley arrangement: [4]
(i) Copy the diagram and mark the direction of tension on each stand of the string.
(ii) What is the velocity ratio of the arrangement?
(iii) If the tension acting on the string is T, then what is the relationship between T and effort E?
(iv) If the free end of the string moves through a distance x, find the distance by which the load is raised.

Answer:
(i) The direction of tension is as marked.

(ii) VR = Distance travelled by effort/Distance travelled by load = \(\frac{x}{2}\).
The distance travelled by load is half the distance moved by effort = \(\frac{x}{2}\)
(iii) T= E,
(iv) The distance travelled by load is half the distance moved by effort = \(\frac{x}{2}\).

Question 6.
(a) How does the angle of deviation formed by a prism change with the increase in the angle of incidence? [3]
Draw a graph showing the variation in the angle of deviation with the angle of incidence at a prism surface.
Answer:
As the angle of incidence increases the angle of deviation decreases becomes minimum and then increases. The graph is as shown:

(b) A virtual, diminished image is formed when an object is placed between the optical centre and the principal focus of a lens. [3]
(i) Name the type of lens which forms the above image,
(ii) Draw a ray diagram to show the formation of the image with the above stated characteristics.
Answer:
(i) Concave lens
(ii) The ray diagram is as shown:

(c) An object is placed at a distance of 24 cm from a convex lens of focal length 8 cm. [4]
(i) What is the nature of the image so formed?
(ii) Calculate the distance of the image from the lens,
(iii) Calculate the magnification of the image.
Answer:
Given u = -24 cm, f = +8 cm, v = ?
(i) Real, inverted and diminished.
(ii) Using the lens formula
\(\frac{1}{f}\) = \(\frac{1}{v}\) – \(\frac{1}{u}\) ⇒ \(\frac{1}{v}\) = \(\frac{1}{f}\) + \(\frac{1}{u}\)
\(\frac{1}{v}\) = \(\frac{1}{8}\) + \(\frac{1}{(-24)}\) = \(\frac{3-1}{24}\) = \(\frac{2}{24}\) = \(\frac{1}{12}\)
v = +12 cm
(iii) Using the expression m = \(\frac{v}{u}\) = \(\frac{12}{-24}\) = \(\frac{1}{2}\)

Question 7.
(a) It is observed that during march-past we hear a base drum distinctly from a distance compared to the side drums. [3]
(i) Name the characteristic of sound associated with the above observation,
(ii) Given a reason for the above observation.
Answer:
(i) Intensity of sound.
(ii) This is because the amplitude of vibration of a base drum is far greater than the amplitude of vibration of the side drum.

(b) A pendulum has a frequency of 4 vibrations per second. An observer starts the pendulum and fires a gun simultaneously. He hears the echo from the cliff after 6 vibrations of the pendulum. If the velocity of sound in air is 340 m/s, find the distance between the cliff and observer. [3]
Answer:
Given f = 4 vbs, T = \(\frac{1}{f}\) = \(\frac{1}{4}\) = 0.25 s, for a frequency of
6 vbs times t = \(\frac{0.25}{4}\) × 6 = 0.375 s, V = 340 ms^-1,
Using S = \(\frac{\mathrm{V} \times t}{2}\) = \(\frac{340 \times 0.375}{2}\) = 63.75 m

(c) Two pendulums C and D are suspended from a wire as shown in the figure given below. Pendulum C is made to oscillate by displacing it from its mean position. It is seen that D also starts oscillating.

(i) Name the type of oscillation, C will execute.
(ii) Name the type of oscillation, D will execute.
(iii) If the length of D is made equal to C then what difference will you notice in the oscillations of D?
(iv) What is the name of the phenomenon when the length of D is made equal to C?
Answer:
(i) Free or natural oscillation.
(ii) Forced oscillation.
(iii) The amplitude of the oscillation of D will be equal to the amplitude of the oscillation of C.
(iv) Resonance.

Question 8.
(a) (i) Write one advantage of connecting electrical appliance in parallel combination. [3]
(ii) What characteristic should a fuse wire have?
(iii) Which wire in a power circuit is connected to the metallic body of the appliance?
Answer:
(i) All appliances will continue to work even if one of the appliance does not work.
(ii) High resistivity and low melting point.
(iii) The ground wire.

(b) The diagram below shows a dual control switch circuit connected to a bulb.

(i) Copy the diagram and complete it so that bulb is switch ON.
(ii) Out of A & B which one is the live wire and which one is the neutral wire?
Answer:
(i) The diagram is as shown:

(ii) A is connected to the live wire and B to neutral wire.

(c) The diagram below shows a circuit with the key k open. Calculate :

(i) the resistance of the circuit when the key k is open.
(ii) the current drawn from the cell when the key k is open.
(iii) the resistance of the circuit when the key k is closed.
(iv) the current drawn from the cell when the key k is closed.
Answer:
(i) R_S = 5 + 0.5 = 5.5Ω
(ii) I = \(\frac{V}{R}\) = \(\frac{3.3}{5.5}\) = 0.6A
(iii) R_P = \(\frac{5 \times 5}{5+5}\) = 2.5Ω, thus total resistance is
R_T = 2.5 + 0.5 = 30Ω
(iv) I = \(\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{T}}}\) = \(\frac{3.3}{3.0}\) = 1.1 A

Question 9.
(a) (i) Define Calorimetry. [3]
(ii) Name the material used for making a Calorimeter.
(iii) Why is a Calorimeter made up of thin sheets of the above material answered in (ii)?
Answer:
(i) Calorimetry is the measurement of heat.
(ii) Copper.
(iii) Copper has small specific heat capacity. The thin sheets ensure that the box has small heat capacity.

(b) The melting point of naphthalene is 80°C and the room temperature is 30°C. A sample of liquid naphthalene at 100°C is cooled down to the room temperature.
Draw a temperature time graph to represent this cooling. In the graph, mark the region which corresponds to the freezing process. [3]
Answer:
The graph is as shown:

(c) 104 g of water at 30°C is taken in a calorimeter made of copper of mass 42g. When a certain mass of ice at 0°C is added to it, the final temperature of the mixture after the ice has melted, was found to be 10° C. Find the mass of ice added.
[Specific heat capacity of water = 4.2 Jg^-loC^-1; Specific latent heat of fusion of ice = 336 Jg^-1; Specific heat capacity of copper = 0.4 Jg^-loC¹] [4]
Answer:
Mass of water m₁ = 104g
Temperature of water T₁ = 30°C
Final temperature T₂ = 10°C
Mass of calorimeter m₂ = 42 g
Temperature of ice T₃ = 0°C
Let the mass of ice added be = x g
Total heat energy lost
= (104 × 4.2 × 20) + (42 × 0.4 × 20)
= 8736 + 336 = 9072 J
Total heat energy gained
= (336 × x) + (x × 4.2 × 10)
= 336x + 42x = 378x
By principle of calorimetry,
Heat lost = Heat gained
378x = 9072
x = \(\frac{9072}{378}\) = 24 g

Question 10.
(a) Draw a neat labeled diagram of an A.C. generator. [3]
Answer:
The diagram is shown :

(b) (i) Define nuclear fission. [3]
(ii) Rewrite and complete the following nuclear reaction by filling in the atomic number of
Ba and mass number of Kr :

Answer:
(i) It is the splitting of a heavy nucleus into two or more smaller nuclei with the release of tremendous energy.
(ii)

(c) The diagram below shows a magnetic needle kept just below the conductor AB which is kept in North South direction. [4]
(i) In which direction will the needle deflect when the key is closed?
(ii) Why is the deflection produced?
(iii) What will be the change in the deflection if the magnetic needle is taken just above the conductor AB?
(iv) Name one device which work on this principle.

Answer:
(i) East.
(ii) A current carrying conductor produces a magnetic field around itself.
(iii) The direction of deflection will reverse. (Towards west)
(iv) An electromagnet.

Solving ICSE Class 10 Maths Previous Year Question Papers ICSE Class 10 Maths Question Paper 2012 is the best way to boost your preparation for the board exams.

ICSE Class 10 Maths Question Paper 2012 Solved

Time Allowed: 2 1/2 hours
Max. Marks: 80

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.

Attempt all questions from Section A and any four questions from Section B. All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answer. Omission of essential working will result in loss of marks. The intended marks for questions or parts of questions are given in brackets [ ]

Mathematical tables are provided.

Section – A (40 Marks)
(Attempt all questions)

Question 1.
(a) If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\), find A² – 5A + 7I
(b) The monthly pocket money of Ravi and Sanjeev are in the ratio 5 :7. The expenditures are in the ratio 3 : 5. If each save Rs. 80 every month, find their monthly pocket money. [3]
(c) Using a Remainder Theorem factorise completely the following polynomial.
3x² + 2x² – 19x + 6 [4]
Solution :
(a)

(b) Given monthly pocket money of Ravi and Sanjeev are in the ratio of 5 : 7
Let Ravi’s pocket money be 5x and Sanjeev’s pocket money be 7 x
Given Ravi’s and Sanjeev’s expenditures are in the ratio of 3 : 5
Let Ravi’s expenditure be 3y and Sanjeev’s expenditure be 5y
According to Question
Monthly savings of Ravi = Monthly savings of Sanjeev = ₹ 80
Saving = Monthly Pocket money – Monthly expenditure Ravi’s monthly savings = 5x – 3y
Sanjeev’s monthly savings = 7x – 5y
By the question
5x – 3y = 80 …(1)
7x – 5y = 80 …(2)
Multiply (1) by 7 and (2) by 5, we get
35x – 21y = 560 …(3)
35x – 25y = 400 …(4)
Subtract (4) from (3)
4y = 160
or y = 40
Putting the value of y in (1), we get
5x – 3 × 40 = 80
or 5x – 120 = 80
or 5x = 120 + 80
or 5x = 200
∴ x = 40
Ravi’s monthly pocket money = 5 × 40 = ₹ 200 and Sanjeev’s monthly pocket money = 7 × 40 = ₹ 280

(c) Given f(x) = 3x³ + 2x² – 19x + 6
Put x = 2
∴ f(2) = 3(2)³ + 2(2)² – 19(2) + 6
= 24 + 8 – 38 + 6 = 0
∴ By Remainder Theorem (x – 2) is a factor off (x) Now, (x – 2) is a factor of 3x³ + 2x² – 19x + 6

∴ 3x³ + 2x² – 19x + 6 = (x – 2) (3x² + 8x – 3) … (1)
= (x – 2) (3x² + 9x – x – 3)
= (x – 2) {3x(x + 3) – 1(x + 3)}
= (x – 2) (x + 3) (3x – 1)

Question 2.
(a) On what sum of money will the difference between the compound interest and simple interest for 2 years be equal to Rs. 25, if the rate of interest charged for both is 5% p.a. ? [3]
(b) ABC is an isosceles right angled triangle with ∠ABC = 90°. A semi-circle is drawn with AC as the diameter. If AB = BC = 7 cm, find the area of the shaded region. (Take π = \(\frac{22}{7}\)) [3]

(c) Given a line segment AB joining the points A (-4, 6) and B (8, -3). Find : [4]
(i) the ratio in which AB is divided by the y- axis.
(ii) the coordinates of the point of intersection.
(iii) the length of AB.
Solution :
(a) Given
CI. – Si. = ₹ 25 for 2
Rate of interest = 5% p.a.
Let Principal be ₹ x, Time = 2 years,
Rate of interest 5% p.a.

(b) Given ABC is an isosceles triangle with AB = 7cm and BC = 7cm
∠ABC = 90°
Area of shaded portion = Area of semi circle – Area of ∆ABC
In ∆ABC, ∠ABC = 90°
∴ By Pythagoras Theorem
AC² = AB² + BC² = 7² + 7²
or AB = \(\sqrt{49+49}\)
= \(\sqrt{98}\) = 7\(\sqrt{2}\) cm
∴ Radius of semicircle = \(\frac{1}{2}\)AC
= \(\frac{1}{2}\) × 7\(\sqrt{2}\)
Now, Area of semicircle = \(\frac{1}{2}\)(Area of Circle)

Area of ∆ABC = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 7 × 7 = \(\frac{49}{2}\) cm²

(c) Here, line segment AB is joining the points IA(-4, 6) and B(8, -3).
Let C(0, y) be the point of intersection of y-axis and the line segment AB.

Let the required ratio be k : 1
∴ c(\(\frac{8 k-4}{k+1}, \frac{-3+6}{k+1}\)) = C(0, y)
⇒ \(\frac{8 k-4}{k+1}\) = 0
⇒ 8k – 4 = 0
⇒ k = \(\frac{1}{2}\)
Thus, the required ratio is 1: 2
Also, y = \(\frac{-3 \times \frac{1}{2}+6}{\frac{1}{2}+1}\) = \(\frac{-3+12}{1+2}\) = \(\frac{9}{3}\) = 3
Thus, the coordinates of the point of intersection is C(0, 3)
Now, AB = \(\sqrt{(8+4)^2+(-3-6)^2}\)

Question 3.
(a) In the given figure O is the centre of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle. [3]

(b) Evaluate without using trigonometric tables
cos² 26° + cos 64°sin 26° + \(\frac{\tan 36^{\circ}}{\cot 54^{\circ}}\) [3]
(c) Marks obtained by 40 students in a short assessment are given below, where a and b are two missing data.

Marks	5	6	7	8	9
No. of students	6	a	16	13	B

If the mean of the distribution is 7.2, find a and b. [4]
Solution :
(a) In the given figure AB is a tangent at B
AB = 15 cm and AC = 7.5cm

O is the centre of the circle
We know that
AB² = AC × AD
∴ 15 × 15 = 7.5 × AD
∴ AD = \(\frac{15 \times 15}{7.5}\) 30 cm
Now, CD = AD – AC = 30 cm – 7.5 cm
= 22.5 cm
∴ Radius of circle = \(\frac{1}{2}\)(diameter of the circle)
= \(\frac{1}{2}\) × 22.5 cm = 11.25 cm

(b) cos² 26° + cos 64° sin 26° + \(\frac{\tan 36^{\circ}}{\cot 54^{\circ}}\)
= cos² 26° + cos(90° – 26°)sin 26° + \(\frac{\tan \left(90^{\circ}-54^{\circ}\right)}{\cot 54^{\circ}}\)
= cos² 26° + sin 26° sin 26° + \(\frac{\cot 54^{\circ}}{\cot 54^{\circ}}\)
[∵ cos(90° – 26°) = sin 26°
tan (90° – 54°) = cot 54°]
= cos² 26° + sin² 26° + 1 [∵ sin²A + cos²A = 1]
= 1 + 1 = 2

(c)

Marks	5	6	7	8	9
Number of students	6	A	16	13	b

Given mean of distribution = 7.2

Σf = 35 + a + b = 40(given)
⇒ a + b = 5
Mean

= 7.2 (given)
⇒ 246 + 6a + 9b = 40 × 7.2 = 288
∴ 6a + 9b = 42
Multiply (1) by 6 and subtract from (2), we have

or b = 4
From(1) a + 4 = 5 ⇒ a = 1
∴ a = 1, b = 4

Question 4.
(a) Kiran deposited Rs. 200 per month for 36 months in a bank’s recurring deposit account. If the bank pays interest at the rate of 11% per annum, find the amount she gets on maturity. [3]
(b) Two coins are tossed once. Find the probability of getting:

(i) 2 heads
(ii) at least 1 tail.

(c) Using graph paper and taking 1 cm = 1 unit along both x-axis and y-axis :

(i) Plot the points A (-4, 4) and B (2, 2)
(ii) Reflect A and B in the origin to get the images A’ and B’ respectively.
(iii) Write down the co-ordinates of A’ and B’.
(iv) Give the geometrical name for the figure ABA’B’.
(v) Draw and name its lines of symmetry.

Solution :
(a) Monthly deposit = ₹ 200
Rate of interest = 11% p.a
Period of deposit (n) = 36 months
∴ Total deposited Amount = ₹ (200 × 36)
= ₹ 7200
Equivalent principal for 1 month
= \(\frac{n(n+1)}{2^{\prime}}\) × monthly deposit
= ₹\(\left(\frac{36 \times 37}{2} \times 200\right)\) = ₹(3600 × 37)
= ₹ 133200
Interest paid by the bank = ₹\(\frac{133200 \times 11}{100 \times 12}\)
= ₹111 × 11
= ₹ 1221
∴ Amount on maturity = Total Deposited Amount + Interest Amount
= ₹ (7200 – 1221)
= ₹ 8421

(b) The sample space of tossing 2 coins is (HH, HT, TH, T’T)
No. of outcomes in sample space = 4
(i) Probability of getting 2 heads = \(\frac{1}{4}\)
(ii) Probability of getting at least 1 tail. = \(\frac{3}{4}\)

(c) (i) and (ii)
(iii) Coordinates of A’ and B’ are A’( 4, -4) and B’(-2, -2)
(iv) AB A’B’ is a parallelogram
(v) AA’ and BB’ are the lines of symmetry.

Section-B (40 Marks)
(Answer any four questions from this Section)

Question 5.
(a) In the given figure, AB is the diameter of a circle with centre O.
∠BCD = 130°. Find:
(i) ∠DAB
(ii) ∠DRA

(b) Given \(\left[\begin{array}{cc}
2 & 1 \\
-3 & 4
\end{array}\right]\) X = \(\left[\begin{array}{l}
7 \\
6
\end{array}\right]\). Write:
(i) the order of the matrix X
(ii) the matrix X.

(c) A page from the Savings Bank Account of Mr. Prateek is given below :

If he receives Rs. 198 as interest on 1st January, 2007, find the rate of interest paid by the bank.
Solution :
(a) (i) Given AB = diameter of the circle, ∠BCD = 130°

Join BD
As ABCD is a cyclic quadrilateral
∴ ∠DCB + ∠DAB = 180°
or ∠DAB + 130° = 180°
or ∠DAB = 180° – 130° = 50°
Now, ∠ADB = 90° [angle in a semicircle]
(ii) ∴ In Δ ADB
∠DAB + ∠ADB + ∠DBA = 180°
∴ 50° + 90° + ∠DBA = 180°
or ∠DBA = 180° – 140° = 40°

(b) Given \(\left[\begin{array}{rr}
2 & 1 \\
-3 & 4
\end{array}\right]_{2 \times 2}\) X = \(\left[\begin{array}{l}
7 \\
6
\end{array}\right]_{2 \times 1}\) … (1)
(i) As \(\left[\begin{array}{rr}
2 & 1 \\
-3 & 4
\end{array}\right]\) is a matrix of order 2 × 2 and \(\left[\begin{array}{l}
7 \\
6
\end{array}\right]\) is a matrix of order 2 × 1
∴ X must have 2 rows and 1 column, the order of the Matrix X is 2 × 1

(ii)

∴ 2a + b = 7 … (3)
-3a + 4b = 6 …. (4)
Multiply (3) by 4 and subtract from (4), we have

⇒ a = 2
From (3), 2 × 2 + b = 7
⇒ b = 3
Thus, X = \(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)

(c)

Interest for one year = ₹ 198
Let rate of interest be r% p.a.
198 = \(\frac{52800 \times r}{100 \times 12}\) [I = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)]
⇒ r = \(\frac{198 \times 100 \times 12}{52800}\) = 4.5% p.a.

Question 6.
(a) The printed price of an article is Rs. 60,000. The wholesaler allows a discount of 20% to the shopkeeper. The shopkeeper sells the article to the customer at the printed price. Sales tax (under VAT) is charged at the rate of 6% at every stage. Find:
(i) the cost to the shopkeeper inclusive of tax.
(ii) VAT paid by the shopkeeper to the Government.
(iii) the cost to the customer inclusive of tax. [3]

(b) Solve the following inequation and represent the solution set on the number line:
4x – 19 < \(\frac{3 x}{5}\) – 2 ≤ \(\frac{-2}{5}\) + x ∈ R [3]

(c) Without solving the following quadratic equation, find the value of m for which the given equation has real and equal roots.
x² + 2(m – 1)x + (m + 5) = 0 [4]
Solution :
(a) (i) Printed Price = ₹ 60,000
Discount given by wholesaler to shopkeeper = ₹ 20%
= ₹ \(\frac{60000 \times 20}{1000}\)
= ₹ 12000
∴ Net cost to shopkeeper
= ₹ (60000 – 12000)
= ₹ 4800
Sales Tax (VAT) changed by wholesaler
= ₹ \(\frac{48000 \times 6}{100}\)
= ₹ 2880
∴ Cost to shopkeeper inclusive of Tax
= ₹ (48000 + 2880)
= ₹ 50880

(ii) ∴ Sales Tax (VAT) charged by shopkeeper
= ₹ 60000 × \(\frac{6}{100}\)
= ₹ 3600
VAT Paid by shopkeeper to Government
= ₹ (3600 – 2880)
= ₹ 720

(iii) ∴ Cost to customer inclusive of Tax
= ₹ (60000 + 3600)
= ₹ 63600

(b)
4x – 19 < \(\frac{3 x}{5}\) – 2 ≤ \(\frac{-2}{5}\) + x
4x – 19 < \(\frac{3 x}{5}\) – 2
20x – 95 < 3x – 10
20x – 3x < 95 – 10
17x < 85
-4 ≤ x
\(\frac{3 x}{5}\) – 2 ≤ \(\frac{-2}{5}\) + x
3x – 10 ≤ -2 + 5x
-10 + 2 ≤ 5x – 3x
-8 ≤ 2x
x < 5
∴ Solution is -4 ≤ x < 5, x ∈ R
Representation

(c) Give equation is
x² + 2(m – 1)x + (m + 5) = 0
For the given equation to have real and equal roots
Discriminant =0
⇒ b² – 4ac = 0
Here, a = 1, b = 2(m – 1) = 2m – 2 and c = m + 5
∴ b² – 4ac = (2m – 2)² – 4 × 1 × (m + 5)
= 4m² + 4 – 8m – 4m – 20
= 4m² – 12m – 16
Now, b² – 4ac = 4m² – 12m – 16 = 0
⇒ m² – 3m – 4 = 0
⇒ (m – 4)(m + 1) = 0
⇒ m = 4 and m = -1

Question 7.
(a) A hollow sphere of internal and external radii 6 cm and 8 cm respectively is melted and recast into small cones of base radius 2 cm and height 8 cm. Find the number of cones. [3]
(b) Solve the following equation and give your answer correct to 3 significant figures :
5x² – 3x – 4 = 0 [3]
(c) As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships on the same side of the lighthouse in horizontal line with its base are 30° and 40° respectively. Find the distance between the two ships. Give your answer correct to the nearest metre. [4]
Solution :
(a) External radius of hollow sphere = 8 cm
Internal radius of hollow sphere = 6 cm
∴ Volume of hollow sphere = \(\frac{4}{3} \pi\)[R³ – r³]
= \(\frac{4}{3} \pi\)[8³ – 6³]
= \(\frac{4}{3} \pi\)[512 – 216]
= \(\frac{4}{3} \pi\)(296) cm³
A.T.Q. Volume of sphere = Volume of all small cones
Fora small cone r = 2 cm and h = 8 cm
∴ Volume of one small cone = \(\frac{1}{3} \pi r^2 h\)
= \(\frac{1}{3}\) × π × 2 × 2 × 8
= \(\frac{1}{3}\) × 32 × π cm³

= \(\frac{\frac{4}{3} \times \pi \times 296}{\frac{1}{3} \times 32 \times \pi}\) = \(\frac{296}{8}\) = 37

(b) Given equation is 5x² – 3x -4 = 0
Compare with ax² + bx + c = 0
Here, a = 5, b = 3 and c = -4

(c) Let A and B be the position of two ships
∴ Height of lighthouse OP = 80 m
Let distance between two ships, AB be x
Let OB be y
In Δ OAP ; tan 30° = \(\frac{\mathrm{OP}}{\mathrm{OA}}\) = \(\frac{80}{x+y}\)
∴ \(\frac{1}{\sqrt{3}}\) = \(\frac{80}{x+y}\)

or x + y = 80\(\sqrt{3}\)
= 80 × 1.732
= 138.56m
In Δ OBP
we have, tan 40° = \(\frac{\mathrm{OP}}{\mathrm{OB}}\) = \(\frac{80}{y}\)
∴ y = \(\frac{80}{\tan 40^{\circ}}\) = \(\frac{80}{0.8391}\)
= 95.34 m
Now, distance between two ships
= 138.56 m – 95.34 m = 43.22 m

Question 8.
(a) A man invests Rs. 9600 on Rs. 100 shares at Rs. 80. If the company pays him 18% dividend, find:
(i) the number of shares he buys.
(ii) his total dividend.
(iii) his percentage return on the shares. [3]
(b) In the given figure ΔABC and ΔAMP are right angled at B and M respectively.
Given AC = 10 cm, AP = 15 cm and PM = 12 cm.

(i) Prove ΔABC – ΔAMP
(ii) Find AB and BC. [3]
(c) If x = \(\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}\), using properties of proportion, show that x² – 2ax + I = 0.
Solution :
(a) (i) Amount invested = ₹ 9600
M.V. of each share = ₹ 80
∴ No. of shares held = \(\frac{₹ 9600}{₹ 80}\) = 120
(ii) Dividend on 1 share = 18% of N.V.
= 18% of ₹ 100 = ₹ 18
Total Dividend = No. of shares × Dividend on 1 share = ₹ 120 × 18 = ₹ 2160
(iii) Percentage return on shares
= \(\frac{₹ 21600}{₹ 9600}\) × 100 = 22.5%

(b)

(i) In the given figure,
∠AMP = 90° and ∠ABC = 90°
In ΔABC and ΔAMP
∴ ∠ABC = ∠AMP = 90°
∠A = ∠A [common]
∴ ΔABC \(\sim\) ΔAMP, [by AA similarity axiom]

(ii) Since ΔABC \(\sim\) ΔAMP, therefore their sides are proportional
\(\frac{\mathrm{AC}}{\mathrm{AP}}\) = \(\frac{\mathrm{BC}}{\mathrm{MP}}\)
∴ BC = \(\frac{10 \times 12}{15}\) = 8 cm
Now, in rt. ∠d ΔABC
AC² = AB² + BC² (By Pythagoras Theorem)
or AB² = AC² – BC²
or AB² = 10² – 8² = 100 – 64 = 36
∴ AB = 6 cm

(c)

Question 9.
(a) The line through A (-2, 3) and B (4, b) is perpendicular to the line 2x – 4y = 5. Find the value of b. [3]
(b) Prove that \(\frac{\tan ^2 \theta}{(\sec \theta-1)^2}\) = \(\frac{1+\cos \theta}{1-\cos \theta}\)
(c) A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car. [4]
Solution :
(a) Points A(-2, 3) and B(4, b)
Since the line through A and B is perpendicular to the line 2x – 4y = 5
∴ Product of their slopes = -1
\(\left(\frac{b-3}{4+1}\right)\left(\frac{2}{4}\right)\) = -1
\(\frac{b-3}{12}\) = -1
b – 3 = -12
b = -9

(b)

(c)
Case I
Distance = 400 km
Let speed of the car be x km/h

Case II
When speed of the car is increased by 12 km/h, then
New speed = (x + 12)km/h
Distañce = 400 km
Now, time taken (t₂) = \(\frac{400}{x+12}\) hours …. (2)
According to question
t₁ – t₂ = 1 hour 40 minutes = 1\(\frac{40}{60}\) hours
= 1\(\frac{2}{3}\) hours = \(\frac{5}{3}\) hours.
∴ From (1) and (2)
\(\frac{400}{x}\) – \(\frac{400}{x+12}\) = \(\frac{5}{3}\)
or 400\(\left[\frac{1}{x}-\frac{1}{x+12}\right]\) = \(\frac{5}{3}\)
or 400\(\left[\frac{x+12-x}{x(x+12)}\right]\) = \(\frac{5}{3}\)
or \(\frac{12 \times 400}{x^2+12 x s}\) = \(\frac{5}{3}\)
or x² + 12x = 12 × 400 × \(\frac{3}{5}\) = 2880
⇒ x² + 60x – 48x – 2880 = 0
(x + 60) (x – 48) = 0
⇒ x – 48 = 0
or x + 60 = 0
⇒ x = 48
or x = -60
But x cannot be -ve
Hence,
x = 48 km/h

Question 10.
(a) Construct a triangle ABC in which base BC = 6 cm, AB = 5.5 cm and ∠ABC = 120°.
(i) Construct a circle circumscribing the triangle ABC.
(ii) Draw a cyclic quadrilateral ABCD so that D is equidistant from B and C.
(b) The following distribution represents the height of 160 students of a school.

Height (in cm)	No. of students
140-145 145-150 150-155 155-160 160-165 165-170 170-175 175-180	12 20 30 38 24 16 12 8

Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine:
(i) The median height.
(ii) The inter quartile range.
(iii) The number of students whose height is above 172 cm.
Solution :

(a) Steps of Construction:
1. Take BC = 6 cm
2. Construct ∠ABC = 120° at point B, such that AB = 5.5 cm. Join AC.
3. Then, ABC is the required triangle.
4. Draw the perpendicular bisectors of sides BC and AB.
Let the perpendicular bisectors of AB and BC meet at point O.
5. Taking O as centre and radius equal to OA (or OB or OC), draw a circle
The circle so obtained is the required circle.
6. Let perpendicular bisector of BC intersect the circumference in D.
7. Join CD and DA, then ABCD is the required cyclic quadrilateral.

(b) The cumulative frequency table for the given continuous distribution is

Height (in cm)	No. of students	Cumulative frequency
140 – 145 145 – 150 150 – 155 155 – 160 160 – 165 165 – 170 170 – 175 175 – 180	12 20 30 38 24 16 12 8	12 32 62 100 124 140 152 160

Take 2 cm = 5 cm of height on x-axis & 2 cm = 20 students on y-axis
Plot the points (140, 0), (145, 12), (150, 32), (155, 62), (160, 100),(165, 124),(170, 140),(175, 152), (180, 160)
Join these points by a free hand drawing. The required ogive is shown in the graph.
Here n (no. of students) = 160, which is even

(i) To find median
Let A be a point on y-axis representing frequency =
\(\frac{1}{2}\left(\frac{n}{2}+\left(\frac{n}{2}+1\right)\right)\)
= \(\frac{1}{2}\)(80 + 81) = 80.5
Through A, draw a horizontal line to meet the ogive at A.
Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 157.4
∴ Required median = 157.4

(ii) To find lower quartile
Let B represent the point on y-axis representing frequency = \(\frac{\mathrm{N}}{4}\) = \(\frac{\mathrm{160}}{4}\) = 40
Through B, draw a horizontal line to meet the ogive at R. Through R, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 151.3
∴ Q₁ = lowerquartile = 151.3

To Find upper quartile

Let C be a point on y-axis representing frequency = \(\frac{3 \mathrm{~N}}{4}\) = 3 × \(\frac{160}{4}\) = 120
Through C, draw a horizontal line to meet the ogive at S. Through S, draw a vertical line to meet the x-axis at T. The abscissa of the point T represents 164.2
∴ Q₃ = Upper quartile = 164.2
∴ Interquartile Range = Q₃ – Q₁
= 164.2 – 151.3 = 12.9

(iii) Let D be a point on x-axis represent 172 cm. Through D, draw a vertical line to meet the ogive at the point E. Through E, draw a horizontal line to meet the y-axis at F. The ordinate of the point F represent 142
∴ No. of students whose height is above 172 cm = 160 – 142 = 18

Question 11.
(a) In triangle PQR, PQ = 24 cm, QR = 7 cm and ∠PQR = 90°. Find the radius of the inscribed circle.

(b) Find the mode and median of the following frequency distribution:

x	10	11	12	13	14	15
f	1	4	7	5	9	3

(c) The line through P (5, 3) intersects y axis at Q.

(i) Write the slope of the line.
(ii) Write the equation of the line.
(iii) Find the co-ordinates of Q.

Solution :
(a)

let the radius of the inscribed circle be x cm.
In rt. ∠d ΔPQR
PR² = PQ² + QR² = 24² + 7² = 576 + 49 = 625
∴ PR = 25 cm
Join OA, OC and OB
Now, from the figure AOCQ, AQ = x = QC = OC = AO
[∵ QC and QA are tangents to the circle from Q]
Now, CR = QR – QC = 7 – x
[Tangents to the circle]
Also, CR = RB = 7 – x
[Tangents to the circle]
Now, PA = PQ – AQ = 24 – x
Also, PA = PB = 24 – x … (2)
[Tangents to the circle]
From (1) and (2)
PR = PB + RB
25 = 24 – x + 7 – x
2x = 24 + 7 – 25
2x = 6
∴ x = 3 cm
Hence, radius of inscribed circle = 3 cm

(b)

x	10	11	12	13	14
f	1	4	7	5	9

Now

x	f	c.f.
10 11 12 13 14 15	1 4 7 8 9 3	1 5 12 17 26 29

∴ Σf = 29
∴ N = 29 (odd)
∴ Median = \(\left(\frac{n+1}{2}\right)\)th term = \(\left(\frac{29+1}{2}\right)\)2th term = 15^th term = 13
In the given distribution, the variate 14 has the maximum frequency i.e., 9.
∴ Mode = 14

(c)

(i) We know that slope of a line tan θ
[where θ is the angle made by the line with the positive direction of x-axis]
∴ slope of PQ = tan 45° = 1
⇒ m = 1

(ii) Equation of a line passing through P(5, 3) and slope
m = 1 is given by
y – 3 = 1(x – 5)
[y – y₁ = m(x – x₁)]
y – 3 = x – 5
or x – y = + 5 – 3 = 2
or x – y = 2

(iii) Since the line PQ meets y-axis at Q
∴ Coordinates of Q can be found by putting x = 0 in x – y = 2
y = -2
∴ Coordinates of Q are (0, -2)

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