ICSE 2020 Physics Question Paper Solved for Class 10

Solving ICSE Class 10 Physics Previous Year Question Papers ICSE Class 10 Physics Question Paper 2020 is the best way to boost your preparation for the board exams.

ICSE Class 10 Physics Question Paper 2020 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) (i) Define moment of force. (2)
(ii) Write the relationship between the SI and CGS unit of moment of force.
Answer:
(i) It is defined as the product of force and the arm of the couple.
(ii) 1 n m = 10⁷ dyn cm

(b) Define a kilowatt hour. How is it related to joule ? (2)
Answer:
It is the defined as the energy consumed when an agent of 1 kilowatt is used for 1 hour.
1 kWh = 3.6 × 10⁶J

(c) A satellite revolves around a planet in a circular orbit. What is the work done by the satellite at any instant? Give a reason. (2)
Answer:
Zero. As the centripetal force acts on the satellite perpendicular to its motion.

(d) (i) Identify the class of the lever shown in the diagram below : (2)

(ii) How is it possible to increase the M.A. of the above lever without increasing its length ?
Answer:
(i) Third class.
(ii) By increasing its effort arm i.e., increasing FE

(e) Give one example of each when : (2)
(i) Chemical energy changes into electrical energy.
(ii) Electrical energy changes into sound energy.
Answer:
(i) Electrochemical cell
(ii) Loud speaker.

Question 2.
(a) A crane ‘A’ lifts a heavy load in 5 seconds, whereas another crane ‘B’ does the same work in 2 seconds. Compare the power of crane ‘A’ to that of crane ‘B’ (2)
Answer:
Power is inversely proportional to timc, thus from the expression P = \(\frac{\mathrm{W}}{t}\) we have
\(\frac{P_A}{P_B}\) = \(\frac{t_{\mathrm{B}}}{t_{\mathrm{A}}}\) = \(\frac{2}{5}\) = 0.4

(b) A ray of light falls normally on a rectangular glass slab. (2)
Draw a ray diagram showing the path of the ray till it emerges out of the slab.
Answer:
The ray diagram is as shown.

(c) Complete the path of the mono-chromatic light ray AB incident on the surface PQ of the equilateral glass prism PQR till it emerges out of the prism due to refraction. (2)

Answer:
The path of the rays is as shown.

(d) Where should an object be placed in front of a convex lens in order to get: (2)
(i) an enlarged real image
(ii) enlarged virtual image ?
Answer:
(i) Beyond 2F (i) Between optical centre and the focus.

(e) A pond appears to be 2.7 in deep. If the refractive index of water is \(\frac{4}{3}\), find the actual depth of the pond. (2)
Answer:
AD = 2.7, µ = 4/3, RD
We know that RD = µ × AD = 4/3 × 2.7
= 3.6m

Question 3.
(a) The wave lengths for the light of red and blue colours are nearly 7.8 × 10^-7 m and 4.8 × 10^-7 m respectively.
(i) Which colour has the greater speed in a vacuum ?
(ii) Which colour has a greater speed in glass ? (2)
Answer:
(i) Both have the same speed in vacuum.
(ii) Red tight has more speed in glass.

(b) Draw a graph between displacement from mean position and time for a body executing free vibration in a vacuum. (2)
Answer:
The graph is as shown

(c) A sound wave travelling in water has wavelength 0.4 m. (2)
Is this wave audible in air ? (The speed of sound in water = 1400 ms^-1)
Answer:
Given λ = 0.4 m,
Using the relation V = vλ
We have v = V/λ = 1400/0.4 = 3500 Hz
Since this lies in the audible range (20 to 20000 Hz) it will be audible.

(d) Why does stone lying in the sun get heated up much more than water lying for the same duration of time? (2)
Answer:
Because water has a large specific heat while stone has a small specific heat.

(e) Why is it not advisable to use a piece of copper wire as fuse wire in an electric circuit? (2)
Answer:
Because copper has a high melting point. It will not melt easily when a high electric current passes through it and may damage the electrical appliances.

Question 4.
(a) Calculate the total resistance across AB: (2)

Answer:
Resistors of resistance 6Ω and 3Ω are in parallel therefore, we have R_p = \(\frac{6 \times 3}{6+3}\) = 2Ω. This is in series with the 5Ω resistors, hence total resistance is 2 + 5 = 7Ω

(b) Two metallic blocks P and Q having masses in ratio 2:1 are supplied with the same amount of heat. If their temperatures rise by same degree, compare their specific heat capacities. (2)
Answer:
Given \(\frac{m_{\mathrm{P}}}{m_{\mathrm{Q}}}\) = \(\frac{2}{1}\), We know that Q = m c θ, since Q and q are same therefore, we have \(\frac{c_{\mathrm{P}}}{c_{\mathrm{Q}}}\) = \(\frac{m_{\mathrm{Q}}}{m_{\mathrm{P}}}\) = \(\frac{1}{2}\)

(c) When a current carrying conductor is placed in a magnetic field, it experiences a mechanical force. What should be the angle between the magnetic field and the length of the conductor so that the force experienced is : (2)
(i) Zero
(ii) Maximum ?
Answer:
The expression for force is F = B I L sin θ
(i) Zero force will be for θ = 0°
(ii) Maximum force will be for θ = 90°

(d) A nucleus ₈₄X²⁰² of an element emits an alpha particle followed by a beta particle. The final nucleus is _aY^b. Find a and b. (2)
Answer:
The scheme of decay is
,
therefore a = 83, b = 198

(e) The diagram below shows a loop of wire carrying current I: (2)

(i) What is the magnetic polarity of the loop that faces us?
(ii) With respect to the diagram how can we increase the strength of the magnetic field produced by this loop?
Answer:
(i) South Pole.
(ii) By decreasing the resistance of the circuit. This will increase the current and hence the magnetic field of the loop.

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

Question 5.
(a) The figure below shows a simple pendulum of mass 200 g. It is displaced from the mean position A to the extreme position B. The potential energy at the position A is zero. At the position B the pendulum bob is raised by 5 m. (3)

(i) What is the potential energy of the pendulum at the position B ?
(ii) What is the total mechanical energy at point C ?
(iii) What is the speed of the bob at the position A when released from B ?
(Take g= 10 ms^-2 and there is no loss of energy.)
Answer:
(i) Potential energy at position B is = mgh = 0.2 × 10 × 5 = 10 J
(ii) Total mechanical energy = potential energy at point B = 10 J
(iii) For position A: 1/2 mv² = mgh, v = \(\sqrt{2 g h}\) = \(\sqrt{2 \times 10 \times 5}\)
= 10 ms^-1

(b) (i) With reference to the direction of action, how does a centripetal force differ from a centrifugal force during uniform circular motion ? (3)
(ii) Is centrifugal force the force of reaction of centripetal force ?
(iii) Compare the magnitudes of centripetal and centrifugal force. (3)
Answer:
(i) Centripetal force is directed inwards towards the centre of the circular path while centriftigal force is directed outwards/ away from the centre of the circular path.
(ii) No
(iii) They are equal in magnitude.

(c) A block and tackle system of pulleys has velocity ratio 4. (4)
(i) Draw a neat labelled diagram of the system indicating clearly the points of application and direction of load and effort.
(ii) What will be its V.R. if the weight of the movable block is doubled ?
Answer:
(i) The labelled diagram is as shown.

(ii) There is no change in the velocity ratio i.e., it will remain 4

Question 6.
(a) A diver in water looks obliquely at an object AB in air. [3]

(i) Does the object appear taller, shorter or of the same size to the diver ?
(ii) Show the path of two rays AC and AD starting from the tip of the object as it travels towards the diver in water and hence obtain the image of the object.
Answer:
(i) Taller
(ii) The diagram is as shown.

(b) Complete the path of the ray AB through the glass prism in PQR till it emerges out of the prism. Given the critical angle of the glass as 42°. [3]

Answer:
The path of the ray is as shown.

(c) A lens of focal length 20 cm forms an inverted image at a distance 60 cm from the lens. [4]
(i) Identify the lens.
(ii) How far is the lens present in front of the object’ ?
(iii) Calculate the magnification of the image.
Answer:
Given f = 20 cm, y = 60 cm
(i) Convex lens
(ii) Using the expression \(\frac{1}{f}\) = \(\frac{1}{v}\) = \(\frac{1}{u}\) we have
\(\frac{1}{u}\) = \(\frac{1}{v}\) – \(\frac{1}{f}\) = \(\frac{1}{60}\) – \(\frac{1}{20}\)
= \(\frac{1-3}{60}\) = –\(\frac{2}{60}\) = –\(\frac{1}{30}\)
or u = -30 cm
(iii) Magnification is given by
m = \(\frac{v}{u}\) = \(\frac{60}{-30}\) = -2

Question 7.
(i) Give reasons for the following : [3]
During the day :
(i) Clouds appear white.
(ii) Sky appears blue,
Answer:
(i) Clouds are at much lower height. They are seen due to scattering of light due to lower parts of earth’s atmosphere, which contains large dust particles, water droplets, ice particles etc. As the size of scatterer (these molecules) is much greater than the wavelength of light, Rayleigh scattering is not valid. Rather, all wavelengths are scattered nearly equally. Therefore all equally scattered colours merge to give us the sensation of white. Hence, clouds generally appear white.

(ii) The molecules of air and other fine particles in the atmosphere have size comparable to the wavelength of visible light. These are therefore, more effective in scattering light of shorter wavelengths at the blue end than light of longer wavelengths at the red end. Thus, when sunlight passes through the atmosphere, the fine particles in air scatter the blue colour (shorter wavelengths) more strongly than red. The scattered blue light enters our eyes and makes the sky appear blue.

(b) (i) Name the system which enables us to locate underwater objects by transmitting ultrasonic waves and detecting the reflecting impulse. [3 ]
(ii) What are acoustically measurable quantities related to pitch and loudness ?
Answer:
(i) SONAR
(ii) Frequency and Intensity of sound.

(c) (i) When a tuning fork [vibrating] is held close to ear, one hears a faint hum. The same [vibrating tuning fork] is held such that its stem is in contact with the table surface, then one hears a loud sound. Explain. [4]
(ii) A man standing in front of a vertical cliff fires a gun. He hears the echo after 3.5 seconds. On moving closer to the cliff by 84 m, he hears the echo after 3 seconds. Calculate the distance of the cliff from the initial position of the man.
Answer:
(i) This is due to the phenomenon of resonance, the table also begins to vibrate with the same frequency as that of the tuning fork and produces large sound.
(ii) In the first case, we have
V = \(\frac{2 d}{t}\) = \(\frac{2 d}{3.5}\) … (i)
In the second case, we have
V = \(\frac{2 d}{t}\) = \(\frac{2(d-84)}{3}\) … (ii)
From (i) and (ii) we have
\(\frac{2 d}{3.5}\) = \(\frac{2(d-84)}{3}\)
Or 6d = 7d – 588 or d = 588 m

Question 8.
(a) The diagram below shows the core of a transformer and its input and output connections. [3]

(i) State the material used for the core.
(ii) Copy and complete the diagram of the transformer by drawing input and output coils.
Answer:
(i) Soft iron
(i) The diagram is as shown.

(b) (i) What are superconductors ? [3]
(ii) Calculate the current drawn by an appliance rated 110 W, 220 V when connected across 220 V supply.
(iii) Name a substance whose resis-tance decreases with the increase in temperature.
Answer:
(i) These are the substances which have zero resistance at low temperatures.
(ii) I = P/V= 110/ 220 = 0.5 A
(iii) Mercury

(iii)We know that semiconductor’s resistance decreases with the increase in temperature.
For examples : Silicon, Graphite, etc.

(c)

The diagram above shows three resistors connected across a cell of e.m.f. 1.8 V and internal resistance r. Calculate :
(i) Current through 3 Ω resistor.
(ii) The internal resistance r.
Answer:
(i) I₃ = \(\frac{\mathrm{R}_{1.5} \mathrm{I}}{\mathrm{R}_{1.5}+\mathrm{R}_3}\) = \(\frac{1.5 \times 0.3}{1.5+3}\) = 0.1A
(ii) Resistors 3Ω and 1.5Ω are connected in parallel.
Therefore their net resistance is
R_P = \(\frac{1.5 \times 3}{1.5+3}\) = \(\frac{4.5}{4.5}\) = 1Ω
This is in series with the 4Ω resistor and internal resistance of the cell r, hence total resistance of the circuit is
R_S = 1 + 4 + r = 5 + r
Now V = IR, therefore we have
1.8 = 0.3(5 + r)
Or r = 1Ω

Question 9.
(a) (i) Define heat capacity of a substance. [3]
(ii) Write the SI unit of heat capacity.
(iii) What is the relationship between heat capacity and specific beat capacity of a substance ?
Answer:
(i) Specific heat (c) of a material is defined as the amount of heat required in raising the temperature of a unit mass of a substance through 1 °C or 1 K.
(ii) In SI unit of heat capacity is J K^-1

(b) The diagram below shows the change’ of phases of a substance on a temperature vs time graph on heating the substance at a constant rate. [3]

(i) Why is the slope of CD less than slope of AB ?
(ii) What is the boiling and melting point of the substance ?
Answer:
(i) From section CD the specific heat is high as compared to section AB
(ii) t₂ and t₁ respectively

(c) A piece of ice of mass 60 g is dropped into 140 g of water at 50°C. Calculate the final temperature of water when all. the ice has melted. [4]
(Assume no heat is Inst to the surrounding)
Specific heat capacity of water = 4.2 Jg^-1 k^-1
Specific latent heat of fusion of ice = 336 Jg^-1
Answer:
Given m_ice = 60 g = 60 × 10^-3 kg, m_w = 140g
= 140 × 10^-3 kg, T₁ = 50°C, Specific heat capacity of water
= 4.2Jg^-1k^-1
Specific latent heat of fusion of ice = 336 J g^-1
Let the final temperature be T.
Now
Heat lost by water = Heat gained by ice
m_wc_w Δt = m_iceL_{i ice} + m_ice c_w Δt
140 × 10^-3 × 4.2 × 10³ × (50 – T)
= 60 × 10^-3 × 336 × 10³ + 60 × 10^-3 × 4.2 × 10³ × (T – 0)
588 × (50 – T) = 20160 + 252T
29400 – 588 T = 20160 + 252 T
840 T = 9240
T = 11°C

Question 10.
(a) (i) Draw a neat labelled diagram of a d.c. motor. (3)
(ii) Write any one use of a d.c. motor.
Answer:
(i) A neat and labelled diagram is as shown.

(ii) It is used in tape recorder CD players.

(b) (i) Differentiate between nuclear fusion and nuclear fission. (3)
(ii) State one safety precaution in the disposal of nuclear waste.
Answer:
(i) In nuclear fission a heavy nucleus splits into two or more smaller nuclei while in nuclear fusion two small nuclei fuse to form a heavy nucleus.
(ii) Nuclear waste should be disposed off in lead containers.

(c) An atomic nucleus A is composed of 84 protons and 128 neutrons. The nucleus A emits an alpha particle and is transformed into a nucleus B. (4)
(i) What is the composition of B ?
(ii)The nucleus B emits a beta particle and is transformed into a nucleus C. What is the composition of C ?
(iii) What is mass number of the nucleus A ?
(iv) Does the composition of C change if it emits gamma radiations ?
Answer:
(i) The reaction is as follows

The composition of B is 82 protons and 126 neutrons
(ii) The reaction is as follows.

The composition of B is 83 protons and 126 neutrons
(iii) Mass number of A = 212
(iv) No