Question 1

Question 1(x)A metal piece of mass 5 g has thermal capacity 2.5 J K⁻¹. If the mass of the metal is tripled, then its specific heat capacity will be :7.5 J K⁻¹2.5 J K⁻¹1.5 J g⁻¹ K⁻¹0.5 J g⁻¹ K⁻¹

Accepted Answer

0.5 J g⁻¹ K⁻¹Reason — As,Specific heat capacity=Thermal capacitymass=2.5 JK−15 g=0.5 J g−1 K−1	ext {Specific heat capacity} = \dfrac {	ext {Thermal capacity}}{	ext {mass}} = \dfrac{2.5\ 	ext J 	ext K⁻¹}{5\ 	ext g} = 0.5\ 	ext J\ 	ext g⁻¹\ 	ext K⁻¹Specific heat capacity=massThermal capacity​=5 g2.5 JK−1​=0.5 J g−1 K−1

Question 2

Question 1(xii)In the given circuits Y and Z, the resistors, R1 and R2, are connected in :series in both the circuitsparallel in both the circuitsparallel in Y and series in Zseries in Y and parallel in Z

Accepted Answer

parallel in both the circuitsReason — Circuit Y may be redrawn as :Both the resistors are connected across points A and B. Hence, those are connected in parallel.Circuit Z may be redrawn as :Both the resistors are connected across points A and B. Hence, those are connected in parallel.

Question 3

Question 1(xiii)A radioactive element P emits one α-particle and transforms to a new element Q. What will be the position of the element Q in the periodic table?One group to the left of POne group to the right of PTwo groups to the right of PTwo groups to the left of P

Accepted Answer

Two groups to the left of PReason — The reaction is :ZAP ⟶Z−2A−4Q +24He_{	ext Z}^{	ext A}	ext{P}\ \longrightarrow \ _{	ext Z - 2}^{	ext A - 4}	ext{Q}\ + \ _{2}^{4}	ext{He}ZA​P ⟶Z−2A−4​Q +24​HeIn periodic table elements are arranged in ascending order of atomic number and the atomic number of Q is 2 less than that of P. Hence, position of Q will 2 groups left of P.

Question 4

Question 3(i)A ray of light enters a rectangular glass slab submerged in water at an angle of incidence 55°. Does this ray undergo total internal reflection when it moves from water to glass? Justify your answer. (The critical angle for glass–water interface is 54°)

Accepted Answer

This ray will not undergo total internal reflection because for total internal reflection, the light ray should travel from denser to rarer medium but in the given situation light ray travels from rarer to denser medium since refractive index of glass is more than that of water. So, there will be no total internal reflection.

Question 5

Question 3(ii)According to the NEW colour convention which colour of wire is connected to :(a) the metal body of the appliance(b) the switch of the appliance?

Accepted Answer

(a) Earth wire is connected to the metal body of electrical appliances and according to new convention the colour of earth wire is green or yellow.(b) Live wire is connected to the switch of an appliance and according to new convention the colour of live wire is brown.

Question 6

Question 3(iii)(a) Which of the two, alternating current or direct current, produces a varying magnetic field when it flows through a conductor?(b) State the frequency of the alternating current supply in India

Accepted Answer

(a) Alternating current produces a varying magnetic field when it flows through a conductor because alternating current changes its magnitude as well as direction periodically and thus, it provides a varying current and hence varying magnetic field as magnetic field is directly proportional to the applied current.(b) 50 Hz.

Question 7

Question 3(iv)Calculate the amount of heat absorbed by 200 g of paraffin wax to melt completely at its melting point.[Specific latent heat of fusion of paraffin wax = 146 Jg⁻¹]

Accepted Answer

Given,Mass of paraffin (m) = 200 gSpecific latent heat of fusion of paraffin (L) = 146 J g⁻¹Let, required heat be Q.Then,Q=mL=200×146=29200 J=29.2 kJ	ext Q = 	ext {mL} = 200 	imes 146 = 29200\ 	ext J = 29.2\ 	ext {kJ}Q=mL=200×146=29200 J=29.2 kJ

Question 8

Question 3(v)Copper wire is wound around a steel bar FT. Current is allowed to pass through the coil for some time and then the bar is removed.(a) Draw only the magnetised bar FT and mark its poles.(b) Trace two magnetic lines of force around FT clearly indicating the direction.

Accepted Answer

(a) Since, looking from the F side current is flowing in anti-clockwise direction and hence, north pole developed at that end and south pole at the T end.The labelled diagram is given below :(b)

Question 9

Question 3(vi)A current flows through a metallic conductor for a long period of time. State the change you would expect in its:(a) Resistance(b) Resistivity

Accepted Answer

(a) As current flows for a long period of time, heat is generated and temperature increases. So, the resistance of metallic wire increases as resistance is directly proportional to the temperature of a conductor.(b) As temperature increases, resistivity of metallic wire increases.

Question 10

Question 3(vii)Curium is a radioactive element with the symbol 96247Cm_{96}^{247}	ext{Cm}96247​Cm named in honour of Madam Curie. The graph of number of protons vs number of neutrons for some elements are shown below :(a) Which point on the graph indicates the element Cm?(b) Which point on the graph indicates daughter nucleus after Cm undergoes radioactive decay of 1α followed by 2β?(c) State the mass number of the daughter nucleus.

Accepted Answer

(a) As, Curium has 96 protons and 247 – 96 = 151 neutrons and point P represents 96 protons and 151 neutrons. Hence, P point indicates element Curium.(b) After the radioactive decay of 1 α and 2 β particles, the atomic number and mass number of daughter nucleus will be 243 and 96 respectively. So, it will have 243 – 96 = 147 neutrons.As, point R represents 96 protons and 147 neutrons. Hence R point indicates the daughter nucleus.(c) Mass number of daughter nucleus = Mass number of Curium – Mass number of α particle = 247 – 4 = 243

Question 11

Question 4(i)(a) Out of the three rays (I, J, H) shown in the diagram, which ray will suffer Total Internal Reflection while inside the prism? (Critical angle of the prism is 42°.)(b) Copy the diagram to complete the path of the ray which you have named in (a) till it comes out of the prism.

Accepted Answer

(a)From the above diagram, it is visible that inside the prism, the angles of incidence of rays I, J and H are respectively 30˚, 60˚ and 30˚. Hence only the ray J will suffer total internal reflection.(b)

Question 12

Question 4(ii)A rectangular glass block of refractive index 1.5 has an air bubble trapped inside it as shown in the diagram. When seen from the surface AB, it appears to be 4 cm deep.(a) Calculate the actual depth of the air bubble from the surface AB.(b) For which colour of light, blue or yellow, the apparent depth will be greater?(c) Turning the glass block upside down, DOES NOT change the apparent depth of the air bubble. State True or False.

Accepted Answer

(a) Given,Refractive index of glass = 1.5Apparent depth of air bubble = 4 cmAs,Refractive index of glass=Real depth of air bubbleApparent depth of air bubble⇒Real depth of air bubble=Refractive index of glass×Apparent depth of air bubble=1.5×4=6 cm\text {Refractive index of glass} = \dfrac{\text {Real depth of air bubble}}{\text {Apparent depth of air bubble}} \[1em] ⇒ \text {Real depth of air bubble} = \text {Refractive index of glass}\times \text {Apparent depth of air bubble} \[1em] = 1.5 \times 4 = 6 \text { cm}Refractive index of glass=Apparent depth of air bubbleReal depth of air bubble​⇒Real depth of air bubble=Refractive index of glass×Apparent depth of air bubble=1.5×4=6 cmSo, actual depth of the air bubble from the surface AB is 6 cm.(b) As,Refractive index of glass=Real depth of air bubbleApparent depth of air bubble⇒Real depth of air bubble=Refractive index of glass×Apparent depth of air bubble\text {Refractive index of glass} = \dfrac{\text {Real depth of air bubble}}{\text {Apparent depth of air bubble}} \[1em] ⇒ \text {Real depth of air bubble} = \text {Refractive index of glass}\times \text {Apparent depth of air bubble} \[1em]Refractive index of glass=Apparent depth of air bubbleReal depth of air bubble​⇒Real depth of air bubble=Refractive index of glass×Apparent depth of air bubbleNow, apparent depth will increase, if refractive index decreases and as wavelength increases, refractive index decreases. As wavelength of yellow light is greater than wavelength of blue light so, refractive index for yellow light is smaller than refractive index of blue light.Hence, for yellow light apparent depth will be greater.(c) When the block is turned upside down, then the real depth of the bubble is 15 – 6 = 9 cmAnd,Apparent depth of air bubble=Real depth of air bubbleRefractive index of glass=91.5=6 cm\text {Apparent depth of air bubble} = \dfrac{\text {Real depth of air bubble}}{\text {Refractive index of glass}}=\dfrac{9}{1.5} =6 \text { cm}\[1 em]Apparent depth of air bubble=Refractive index of glassReal depth of air bubble​=1.59​=6 cmSo, the statement is false.

Question 13

Question 4(iii)(a) An object is placed at 2F position of a convex lens. Draw a ray diagram showing the formation of the image.(b) How will the size of the image change if we, ONLY replace the lens in the above arrangement with another lens of a greater focal length?

Accepted Answer

(a) Ray diagram showing the formation of the image when the object is placed at 2F position of a convex lens is shown below:(b) As,Magnification=IO=vu	ext {Magnification} = \dfrac{	ext I}{	ext O} = \dfrac{	ext v}{	ext u}Magnification=OI​=uv​So, size of the image remains unchanged since, magnification is not related to the focal length of the lens.

Question 14

Question 5(i)An object is placed in front of a concave lens at a distance of 45 cm from it. If its image is formed at a distance of 30 cm from the lens, calculate the focal length of the lens.

Accepted Answer

(a) Given,Object distance (u) = -45 cmImage distance (v) = -30 cmFrom lens formula,1f=1v−1u1f=1−30−1(−45)1f=−130+1451f=−3+2901f=−190f=−901f=−90 cm\dfrac{1}{\text f} = \dfrac{1}{\text v} - \dfrac{1}{\text u} \[1em] \dfrac{1}{\text f} = \dfrac{1}{-30}-\dfrac{1}{(-45)}\[1em] \dfrac{1}{\text f}=-\dfrac{1}{30}+\dfrac{1}{45}\[1em] \dfrac{1}{\text f} = \dfrac{-3+2}{90}\[1em] \dfrac{1}{\text f} = -\dfrac{1}{90} \[1 em] \text f = -\dfrac{90}{1}\[1 em] \text f= - 90 \text { cm}f1​=v1​−u1​f1​=−301​−(−45)1​f1​=−301​+451​f1​=90−3+2​f1​=−901​f=−190​f=−90 cm

Question 15

Question 5(ii)Two rays PQ and RS are incident on a rectangular glass block as shown in the diagram. Observe the diagram and answer the questions that follow.Which of these two rays will :(a) have greater lateral displacement on emerging out of the block?(b) travel with greater speed in the block?(c) scatter more in the atmosphere?

Accepted Answer

(a) Angle of incidence for both the rays being same, the directions of both the rays remain same. As the refracted ray corresponding to RS bends towards the normal is more than that corresponding to ray PQ hence, the lateral displacement of the emergent ray corresponding to incident ray RS will also be more.Diagram shows, x1 and x2 are the lateral shifts of the emergent rays corresponding to incident rays PQ and RS.So, here x2 > x1(b) Since, the refracted ray corresponding to incident ray PQ bends less towards the normal, hence this travels with greater speed.(c) Speed corresponding to incident ray PQ is more than the speed corresponding to incident ray RS. So, wavelength corresponding to incident ray PQ is more than the wavelength corresponding to incident ray RS. So, RS will scatter more in atmosphere.

Question 16

Question 5(iii)(a) Name the radiations :for which a quartz prism is used to study the spectrum.which are used in remote sensing devices.which are used in traffic signals in India.(b) Name one property common to all electromagnetic radiations.

Accepted Answer

(a)For ultraviolet radiation, a quartz prism is used since, ordinary glass absorbs UV radiation.Infra red radiation.Visible light radiation (Red, Yellow, Green)(b) All electromagnetic radiations travels with the same speed ie., 3 × 108 ms-1.

Question 17

Question 6(i)Akash takes a uniform metre scale and suspends a weight of 2 N at one end ‘X’, and a weight of 5 N on the other end ‘Y”. He then balances the ruler horizontally on a knife edge placed at 70 cm from X. Draw a diagram of the arrangement and calculate the weight of the ruler.

Accepted Answer

The diagram of the arrangement is shown below :Given,Weight at the end X (W1) = 2 NWeight at the end Y (W2) = 5 NDistance of X from the knife (R1) = 70 cmDistance of Y from the knife (R2) = 100 cm - 70 cm = 30 cmLet,W = weight of the scaleNow,Anticlockwise moment (due to 2N) = R1 x W1 = 2 × 70 = 140 NmClockwise moment (due to 5N) = R2 x W2 = 5 × 30 = 150 NmAnticlockwise moment (due to weight of the scale) = W × 20 = 20 WFor equilibrium,Anticlockwise moment of force about knife = Clockwise moment of force about knife150=140+20 W20W=150−140=10W=1020=0.5 N150 = 140 + 20\text { W} \[1em] 20 \text W = 150 - 140 = 10 \[1em] \text W = \dfrac{10}{20} = 0.5\ \text N150=140+20 W20W=150−140=10W=2010​=0.5 NSo, weight of the scale is 0.5 N.

Question 18

Question 6(ii)Three levers X, Y, Z of equal lengths are shown in the diagram.(a) Which class of lever do these belong to?(b) Among these (X, Y or Z) which one will give the maximum mechanical advantage? Justify your answer.

Accepted Answer

(a) All are class I levers since, the fulcrum is in between the load and effort.(b) As,Mechanical Advantage=Effort ArmLoad Arm	ext {Mechanical Advantage} = \dfrac{	ext {Effort Arm}}{	ext {Load Arm}}Mechanical Advantage=Load ArmEffort Arm​For X, effort arm is greater than the load arm. Hence, it has the maximum mechanical advantage.

Question 19

Question 6(iii)Richa weighing 40 kgf leaves point P on her skateboard and reaches point Q on the ground with velocity 10 ms-1. Calculate.(a) the kinetic energy of Richa at point Q(b) the vertical height of point P above the ground. (use g as 10 m/s2 and neglect friction)(c) the kinetic energy of Richa at point R. (While moving from Q to R, she loses 500 J of energy against friction.)

Accepted Answer

(a) Given,Speed at Q (v) = 10 ms-1Then,Kinetic energy at Q = 12\dfrac{1}{2}21​ mv2 = 12\dfrac{1}{2}21​ × 40 × 102 = 2000 J(b) If the height of P is H, then :Potential energy at P = mgH = 40 x 10 x H = 400HAs, friction is negligible so total mechanical energy is conserved.From law of conservation of total mechanical energy,Potential energy at P = Kinetic energy at Q⇒ 400H = 2000 J⇒ H = 2000400\dfrac{2000}{400}4002000​ = 5 m(c) Given,Height of point R (h) = 3 mEnergy lost from Q to R through friction = 500 JNowPotential energy at R = mgh = 40 × 10 × 3 = 1200 JSo, from law of conservation of energyKinetic energy at R + Potential energy at R + Energy lost from Q to R through friction = Potential energy at P⇒ Kinetic energy at R = Potential energy at P - Potential energy at R - Energy lost from Q to R through friction⇒ Kinetic energy at R = 2000 - 1200 - 500 = 300 J

Question 20

Question 7(ii)A submarine in the sea, sends ultrasonic ping and a stopwatch is started simultaneously. The stopwatch stops on receiving the reflected wave from an obstacle and reads 1 minute 40 seconds. Calculate the distance of the obstacle from the submarine (Speed of sound in water 1500 ms-1)

Accepted Answer

Given,Time taken = 1 minute 40 seconds = 100 sSpeed of sound = 1500 ms-1)So,Distance of the obstacle=Speed of sound×Time taken2=1500×1002=1500002=75000 m=75 km\text {Distance of the obstacle} = \dfrac {\text {Speed of sound} \times \text {Time taken}}{2} \[1em] =\dfrac{1500 \times 100}{2} = \dfrac{150000}{2} \[1em] = 75000\ \text m = 75\ \text {km}Distance of the obstacle=2Speed of sound×Time taken​=21500×100​=2150000​=75000 m=75 km

Solved 2025 Question Paper ICSE Class 10 Physics

Section A

Section B