Q: Question 1(ii)The kinetic energy of a given body depends on the :positioncentre of gravity of the body.momentumdisplacement

momentumReason — As,Kinetic Energy=12mv2\text {Kinetic Energy} = \dfrac{1}{2}\text m \text v^2Kinetic Energy=21mv2And,Momentum (p) = mvThen,Kinetic Energy=12mm2v2=p22m\text {Kinetic Energy} = \dfrac{1}{2\text m}\text m^2 \text v^2 = \dfrac{p^2}{2\text m}Kinetic Energy=2m1m2v2=2mp2So, the kinetic energy of a given body depends on its momentum.

Q: Question 3(iii)A transformer is used to change a high alternating e.m.f. to a low alternating e.m.f. of the same frequency.(a) Identify the type of transformer used for the above purpose.(b) State whether the turns ratio of the above transformer is = 1 or > 1 or < 1.

(a) Step down transformer since it is used to change a high alternating voltage to a low alternating voltage.(b) Turns ratio is n < 1, i.e., number of turns in the secondary are less than the number of turns in the primary.

Q: Question 3(iv)A solid of mass 60 g at 100°C is placed in 150 g of water at 20°C. The final steady temperature is 25°C. Calculate the heat capacity of solid.[Specific heat capacity of water = 4.2 Jg⁻¹K⁻¹]

Given,Mass of solid (ms) = 60 gMass of water (mw) = 150 gFall in temperature of solid (△ts) = (100 - 25) = 75°CRise in temperature of water (△tw) = (25 - 20) = 5°CSpecific heat capacity of water (cw) = 4.2 Jg⁻¹K⁻¹Let specific heat capacity of solid be cs.Then,Heat energy given by solid = mscs△ts = 60 x c x 75 = 4500 x c …..(1)Heat energy taken by water = mwcw△tw = 150 × 4.2 × 5 = 3150 …..(2)Assuming that there is no loss of heat energy,Heat energy given by solid = Heat energy taken by water.Equating equations 1 & 2, we get,4500 × cs = 3150cs=31504500\text c_\text s = \dfrac{3150}{4500}cs=45003150cs = 0.7 J g⁻¹ K⁻¹Now,Heat capacity of solid = mass x specific heat capacity = 60 x 0.7 = 42 J K⁻¹Hence, Heat capacity of solid = 42 J K⁻¹.

Q: Question 3(v)(a) Name the principle of AC generator.(b) State its one use.

(a) Faraday's first law or principle of electromagnetic induction.(b) It is used as a back-up for the lifts in tall buildings in case of electricity failure.

Question 1

Question 1(ii)The kinetic energy of a given body depends on the :positioncentre of gravity of the body.momentumdisplacement

Accepted Answer

momentumReason — As,Kinetic Energy=12mv2	ext {Kinetic Energy} = \dfrac{1}{2}	ext m 	ext v^2Kinetic Energy=21​mv2And,Momentum (p) = mvThen,Kinetic Energy=12mm2v2=p22m	ext {Kinetic Energy} = \dfrac{1}{2	ext m}	ext m^2 	ext v^2 = \dfrac{p^2}{2	ext m}Kinetic Energy=2m1​m2v2=2mp2​So, the kinetic energy of a given body depends on its momentum.

Question 2

Question 1(iii)For burning of coal in a thermoelectric station, the energy conversion taking place is :chemical to heat to mechanicalchemical to heat to mechanical to electricalchemical to heat to lightheat to chemical to mechanical

Accepted Answer

chemical to heat to mechanical to electricalReason — For burning of coal in a thermoelectric station, the chemical energy of coal on burning changes into the heat energy of steam then in a steam engine, the heat energy of steam changes into the kinetic energy of the turbine in the form of rotational kinetic energy due to which it rotates the armature of the generator connected to it and thus kinetic energy gets transformed into electrical energy.

Question 3

Question 1(vii)A convex lens has focal length 12 cm with an object at a distance of 20 cm in front of the lens. He obtains a blurred image on the screen placed at a distance of 23 cm in front of the lens. In order to obtain the clear image, he has to move the screentowards the lens.away from the lens.to a position very far away from the lens.either towards or away from the lens.

Accepted Answer

away from the lens.Reason —Given,Focal length (f) = +12 cmObject distance (u) = -20 cmBlurred image distance (v') = +23 cmLet, clear image distance be v.From lens formula,1f=1v–1u1v=1f+1u1v=112+1(−20)1v=112−1201v=5−360=260v=602=+30 cm\dfrac{1}{\text f} = \dfrac{1}{\text v} – \dfrac{1}{\text u} \[1 em] \dfrac{1}{\text v} = \dfrac{1}{\text f} + \dfrac{1}{\text u} \[1 em] \dfrac{1}{\text v}=\dfrac{1}{12} + \dfrac{1}{(-20)}\[1 em] \dfrac{1}{\text v}=\dfrac{1}{12} - \dfrac{1}{20}\[1 em] \dfrac{1}{\text v}=\dfrac{5-3}{60 }= \dfrac{2}{60}\[1 em] \text v = \dfrac{60}{2} = +30 \text { cm}f1​=v1​–u1​v1​=f1​+u1​v1​=121​+(−20)1​v1​=121​−201​v1​=605−3​=602​v=260​=+30 cmNow,Clear image distance (v) - Blurred image distance (v') = 30 - 23 = 7 cmSo, in order to obtain the clear image, he has to move the screen 7 cm away from the lens.

Question 4

Question 1(xii)Current is flowing through a coil as shown in the figure. Which one of the given figures will correctly depict the magnetic polarity and the direction of the lines of force along the axis of the coil.

Accepted Answer

Reason — As current is flowing in clockwise direction in the coil and according to clock rule if the current in wire around that face is in anticlockwise direction, the face has the north polarity, while if the current at that face is in clockwise direction, the face has the south polarity.

Question 5

Question 2(ii)Draw a graph of potential energy vs height for a body thrown vertically upwards. [Assume no friction is present.]

Accepted Answer

The graph of potential energy vs height for a body thrown vertically upwards is shown below where u(J) represents potential energy in joules and h(m) refers to height in metres.

Question 6

Question 3(i)(a) Refer to the diagram given below. A lens with two different refractive indices is shown. If the rays are coming from a distant object, then how many images will be seen?(b) A glass lens always forms a virtual, erect and diminished image of an object kept in front of it. Identify the lens.

Accepted Answer

(a) According to the lens maker formula :1f=(μ−1)(1R1−R2)\dfrac{1}{	ext f} = (	ext μ - 1)\left(\dfrac{1}{	ext R_1-	ext R_2}ight)f1​=(μ−1)(R1​−R2​1​)Focal length is a function of refractive index, hence for two different refractive indices there will be two different focal lengths.Two different focal length implies two different images for same object.Hence, 2 images will be formed.(b) Concave lens because irrespective of the position of the object, the image formed by a thin concave (or divergent) lens is always virtual, upright, diminished, and it is situated on the side of the object between the focus and the lens.

Question 7

Question 3(ii)If live wire makes an accidental contact with the metal case, which circuit (A or B) in the diagram, illustrating an electric iron, is considered safe for the user (Assuming the fuse is present in the live wire in both circuits)? Justify your answer.

Accepted Answer

The electric iron, is considered safe for the user in circuit A because in this circuit metal case is earthed so the person won't get an electric shock but in circuit B metal case is not earthed so the circuit will be completed through the body of the person giving him a shock.

Question 8

Question 3(iii)A transformer is used to change a high alternating e.m.f. to a low alternating e.m.f. of the same frequency.(a) Identify the type of transformer used for the above purpose.(b) State whether the turns ratio of the above transformer is = 1 or > 1 or < 1.

Accepted Answer

(a) Step down transformer since it is used to change a high alternating voltage to a low alternating voltage.(b) Turns ratio is n < 1, i.e., number of turns in the secondary are less than the number of turns in the primary.

Question 9

Question 3(iv)A solid of mass 60 g at 100°C is placed in 150 g of water at 20°C. The final steady temperature is 25°C. Calculate the heat capacity of solid.[Specific heat capacity of water = 4.2 Jg⁻¹K⁻¹]

Accepted Answer

Given,Mass of solid (ms) = 60 gMass of water (mw) = 150 gFall in temperature of solid (△ts) = (100 - 25) = 75°CRise in temperature of water (△tw) = (25 - 20) = 5°CSpecific heat capacity of water (cw) = 4.2 Jg⁻¹K⁻¹Let specific heat capacity of solid be cs.Then,Heat energy given by solid = mscs△ts = 60 x c x 75 = 4500 x c …..(1)Heat energy taken by water = mwcw△tw = 150 × 4.2 × 5 = 3150 …..(2)Assuming that there is no loss of heat energy,Heat energy given by solid = Heat energy taken by water.Equating equations 1 & 2, we get,4500 × cs = 3150cs=31504500	ext c_	ext s = \dfrac{3150}{4500}cs​=45003150​cs = 0.7 J g⁻¹ K⁻¹Now,Heat capacity of solid = mass x specific heat capacity = 60 x 0.7 = 42 J K⁻¹Hence, Heat capacity of solid = 42 J K⁻¹.

Question 10

Question 3(v)(a) Name the principle of AC generator.(b) State its one use.

Accepted Answer

(a) Faraday's first law or principle of electromagnetic induction.(b) It is used as a back-up for the lifts in tall buildings in case of electricity failure.

Question 11

Question 3(vi)(a) Name the radiations that are emitted during the decay of a nucleus, which has highest penetrating power?(b) Does the emission of the above-mentioned radiation result in a change in the mass number?

Accepted Answer

(a) Gamma (γ) radiation.(b) No, the emission of gamma radiation does not result in a change in the mass number as they are a form of electromagnetic radiation, essentially high-energy photons, and they have no mass or charge. Since they are massless, their emission does not affect the number of protons or neutrons in the nucleus, and therefore the mass number remains unchanged.

Question 12

Question 3(vii)The graph (fig A) illustrates the correlation between the number of protons (x-axis) and the number of neutrons (y-axis) for elements A, В, C, D, and E in the periodic table. These elements are denoted by the letters rather than their conventional symbols. When the element C, depicted in the graph, undergoes radioactive decay, it releases radioactive rays. When these rays are directed into the plane of the paper in the presence of a magnetic field, as indicated in the fig B, they experience deflection, causing them to move upwards.(a) Name the radioactive radiations emitted by the element C.(b) Identify the daughter element from the graph.(c) Name the law used to identify the radioactive radiations emitted by the element.

Accepted Answer

(a) Beta radiations.(b) The daughter element is E because emission of beta particle reduces one neutron from C and increases it's one proton so the element with one neutron less than C and one excess proton will be the daughter element which is clearly E as seen from the graph.(c) Fleming's left hand rule.

Question 13

Question 4(i)The diagram below shows a fish in the tank and its image seen in the surface of water.(a) Name the phenomenon responsible for the formation of this image.(b) A double convex lens with refractive index μ1 inside two liquids of refractive indices μ2 and μ3 are shown in the diagrams below. The refractive indices are such that μ2 > μ1 and μ1 > μ3How would a parallel incident beam of light refract when it comes out of the lens in each of the cases shown above?(1) in figure a.(2) in figure b.

Accepted Answer

(a) Total internal reflection.(b) According to the lens maker formula :1f∝(refractive index of 2nd mediumrefractive index of 1st medium−1)\dfrac{1}{\text f} ∝ \left(\dfrac {\text {refractive index of 2nd medium}}{\text {refractive index of 1st medium} } - 1\right)f1∝(refractive index of 1st mediumrefractive index of 2nd medium−1) ......(i)(1) In figure a :Refractive index of 1st medium = μ3Refractive index of 2nd medium = μ1Then ,from relation (i)1f∝(μ1μ3−1)\dfrac{1}{\text f} ∝ \left(\dfrac {\text μ_1 }{\text μ_3} - 1\right)f1∝(μ3μ1−1)As,μ1 > μ3⇒μ1μ3>1⇒1f>0⇒f>0\Rightarrow \dfrac{\text μ_1}{\text μ_3} \gt 1 \[1 em] \Rightarrow \dfrac{1}{\text f} \gt 0 \[1 em] \Rightarrow \text f \gt 0⇒μ3μ1>1⇒f1>0⇒f>0Since, the resulting focal length is positive so this lens behaves as a converging lens.Hence, a parallel incident beam of light converges when it comes out of the lens im figure a.(2) In figure b :Refractive index of 1st medium = μ2Refractive index of 2nd medium = μ1Then, from relation (i)1f∝(μ1μ2−1)\dfrac{1}{\text f} ∝ \left(\dfrac {\text μ_1 }{\text μ_2} - 1\right)f1∝(μ2μ1−1)As,μ1 < μ2⇒μ1μ2<1⇒1f<0⇒f<0\Rightarrow \dfrac{\text μ_1}{\text μ_2} \lt 1 \[1 em] \Rightarrow \dfrac{1}{\text f} \lt 0 \[1 em] \Rightarrow \text f \lt 0⇒μ2μ1<1⇒f1<0⇒f<0Since, the resulting focal length is negative so this lens behaves as a diverging lens.Hence, a parallel incident beam of light diverges when it comes out of the lens im figure b.

Question 14

Question 5(i)An object is placed at a distance 24 cm in front of a convex lens of focal length 8 cm.(a) What is the nature of the image so formed?(b) Calculate the distance of the image from the lens.

Accepted Answer

(a) As, the object distance (= 24 cm) is greater than the focal length (= 8 cm) of the convex lens which means the object placed beyond it's focus so the lens will form a real and inverted image.(b)Given,Object distance (u) = - 24 cmFocal length (f) = +8 cmFrom lens formula,1f=1v−1u1v=1f+1u1v=18+1(−24)[1em]1v=18−1241v=3−1241v=224v=242v=12 cm\dfrac{1}{\text f} = \dfrac{1}{\text v} - \dfrac{1}{\text u} \[1 em] \dfrac{1}{\text v} = \dfrac{1}{\text f} + \dfrac{1}{\text u} \[1 em] \dfrac{1}{\text v} = \dfrac{1}{8}+\dfrac{1}{(-24)} \ [1 em] \dfrac{1}{\text v} =\dfrac{1}{8}-\dfrac{1}{24} \[1 em] \dfrac{1}{\text v} = \dfrac{3-1}{24}\[1 em] \dfrac{1}{\text v} = \dfrac{2}{24} \[1 em] \text v = \dfrac{24}{2} \[1 em] \text v =12 \text { cm}f1​=v1​−u1​v1​=f1​+u1​v1​=81​+(−24)1​[1em]v1​=81​−241​v1​=243−1​v1​=242​v=224​v=12 cm∴ The image is formed at a distance of 12 cm behind the lens.

Question 15

Question 5(ii)When sunlight passes through water droplets in the atmosphere it gets dispersed into its constituent colours forming a rainbow. A similar phenomenon is observed when white light passes through a prism.(a) Which colour will show the maximum angle of deviation and which colour will show the minimum angle of deviation?(b) If Instead of sunlight, a green-coloured ray is passed through a glass prism. What will be the colour of the emergent ray?

Accepted Answer

(a) As the angle of deviation depends directly on the refractive index of a given transparent medium which is itself inversely proportional to the wavelength of light so the angle of deviation decreases with the increase in the wavelength of light. Thus, for visible light the angle of deviation of a prism is maximum for violet colour and minimum for red colour.(b) A green-coloured ray is a single colour light and hence a prism can not split a single light so the emergent ray will be of green colour.

Question 16

Question 5(iii)O is a luminescent particle trapped inside a glass block. A student traces the path of rays coming out of it and reflecting over a plane mirror as shown in the diagram below.Complete the table, using the labels from the figure. The first label is done for you.S. No.DescriptionLabela.an angle of reflection on the mirrorab.a partially reflected ray in the glass slabc.a critical angled.a refracted raye.an angle of refraction of the ray R

Accepted Answer

S. No.DescriptionLabela.an angle of reflection on the mirrorab.a partially reflected ray in the glass slabUc.a critical anglecd.a refracted rayQe.an angle of refraction of the ray R90° - e

Question 17

Question 6(ii)For each of the following scenarios, state whether the work done by gravity is positive, negative, or zero.(a) a person walks on a levelled road.(b) a person climbs a ladder.(c) a car in neutral gear is coming down the slope.

Accepted Answer

(a) Zero because when a person walks on a levelled, no work is done against the force of gravity as the displacement of the person is normal to the direction of force of gravity which is vertically downwards.(b) Negative because when a person climbs a ladder (upward), the displacement covered by the person is opposite to the direction of force of gravity (downwards), so the work done on the person by the force of gravity is negative.(c) Positive because when a car in neutral gear is coming down the slope, the direction of displacement covered by it is parallel to the direction of force of gravity (downwards), so the work done on the car by the force of gravity is positive.

Question 18

Question 6(iii)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.(a) What is the potential energy of the pendulum at the position В?(b) What is the total mechanical energy at point C?(c) What is the speed of the bob at the position A when released from B?(Take g = 10 ms-2 and given that there is no loss of energy.)

Accepted Answer

Given,h = 5 m,m = 200 g = 0.2 kg,g = 10 ms-2(a) Potential energy UB at B is given byUB = m x g x hSubstituting the values we get,UB = 0.2 x 10 x 5⇒ UB = 10 JHence, the potential energy of the pendulum at the position B = 10 J.(b) Total mechanical energy at point C = 10 JThe total mechanical energy is same at all points of the path due to conservation of mechanical energy.(c) At A, bob has only kinetic energy which is equal to potential energy at B,Therefore,12mvA2=UB⇒0.5×0.2×vA2=10⇒vA2=100.1⇒vA=100=10 ms−1\dfrac {1}{2}\text m \text v_\text A^2 = \text U_\text B \[1 em] ⇒ 0.5 \times 0.2 \times \text v_\text A^2 = 10 \[1 em] ⇒ \text v_\text A^2 = \dfrac{10}{0.1} \[1 em] ⇒ \text v_\text A = \sqrt {100} = 10\ \text m \text s^{-1}21​mvA2​=UB​⇒0.5×0.2×vA2​=10⇒vA2​=0.110​⇒vA​=100​=10 ms−1

Question 19

Question 7(i)A block and tackle system of pulleys has velocity ratio 4.(a) Draw a labelled diagram of the system indicating clearly, the direction of the load and effort.(b) Calculate the potential energy of the load 100 kgf lifted by this pulley to a height 5 m. (g = 10 ms-2)

Accepted Answer

(a)The labelled diagram of the system indicating clearly, the direction of the load and effort is shown below :(b) Given,m = 100 kgh = 5 mg = 10 ms-2Then,Potential Energy (U) = mgh = 100 × 10 × 5 = 5000 J

Question 20

Question 7(ii)A person standing in front of a cliff fires a gun and hears its echo after 3s. If the speed of sound in air is 336 ms-1.(a) Calculate the distance of the person from the cliff.(b) After moving a certain distance from the cliff, he fires the gun again and this time the echo is heard 1.5 s later than the first. Calculate the distance that the person moved.

Accepted Answer

(a) Given,Time taken to hear the echo = 3sSpeed of sound = 336 ms-1So,Distance of the person from the cliff=Speed of sound×Time taken2=336×32=10082=504 m\text {Distance of the person from the cliff} = \dfrac {\text {Speed of sound} \times \text {Time taken}}{2} \[1em] =\dfrac{336 \times 3}{2} = \dfrac{1008}{2} \[1em] = 504\ \text mDistance of the person from the cliff=2Speed of sound×Time taken​=2336×3​=21008​=504 mSo, distance of the person from the cliff is 504 m.(b) Given,Time taken to hear the new echo = 3 s + 1.5 s = 4.5 sSo,New distance of the person from the cliff=Speed of sound×Time taken2=336×4.52=15122=756 m\text {New distance of the person from the cliff} = \dfrac {\text {Speed of sound} \times \text {Time taken}}{2} \[1em] =\dfrac{336 \times 4.5}{2} = \dfrac{1512}{2} \[1em] = 756\ \text mNew distance of the person from the cliff=2Speed of sound×Time taken​=2336×4.5​=21512​=756 mDistance moved by the person = 756 - 504 = 252 mSo, distance moved by the person is 252 m.

Solved 2025 Specimen Paper ICSE Class 10 Physics

Section A

Section B