Complete ICSE Class X Physics 2024 question paper with accurate, step-by-step solutions.
(Attempt all questions from this section)
When a bell fixed on a cycle rings, then the energy conversion that takes place is:
(b) kinetic energy to sound energy
Explanation:
When you ring a cycle bell, you apply a force (mechanical/kinetic energy) by pressing or rotating the lever. This mechanical movement causes a metal striker to hit the bell, producing sound energy.
A door lock is opened by turning the lever (handle) of length 0.2 m. If the moment of force produced is 1 Nm, then the minimum force required is:
(a) 5 N
Explanation:
Moment of force = Force × perpendicular distance
1 = Force × 0.2
Force = 1 ÷ 0.2 = 5 N
A force ‘F’ moves a load from A to C as shown in the figure below. For the calculation of the work done, which of these lengths would you use as the displacement?
(c) 5 m
Explanation:
Work done is calculated using displacement in the direction of force. The straight-line displacement from A to C is 5 m.
A radioactive nucleus containing 128 nucleons emits a β-particle. After β-emission the number of nucleons present in the nucleus will be:
(a) 128
Explanation:
During β-emission, a neutron changes into a proton or a proton changes into a neutron. Therefore, the total number of nucleons remains unchanged.
Assertion (A): Ultraviolet radiations are scattered more as compared to the microwave radiations.
Reason (R): Wavelength of ultraviolet radiation is more than the wavelength of microwave radiation.
(b) A is true but R is false.
Explanation:
Scattering varies inversely as the fourth power of wavelength. Ultraviolet radiation has a much shorter wavelength than microwave radiation, so it is scattered more.
When the stem of a vibrating tuning fork is pressed on a table, the tabletop starts vibrating. These vibrations are definitely an example of:
(c) forced vibrations
Explanation:
The vibrating tuning fork applies a periodic force to the tabletop and forces it to vibrate.
Which of the following is a class III lever?
(d) Human forearm
Explanation:
In the human forearm, the elbow is the fulcrum and the effort applied by the biceps lies between the fulcrum and the load.
The specific resistance of a wire depends on:
(b) the material of the wire
Explanation:
Specific resistance is a property of the material and, at a given temperature, does not depend on the dimensions of the wire.
Identify the option that displays the correct wiring with correct colour code:
(a) 
(b) 
(c) 
(d) 
(b)
Explanation:
The switch must be connected in the live wire so that the appliance is disconnected from the high-potential supply when switched off.
The potential difference between the terminals of a cell in a closed circuit is called:
(a) terminal voltage
Explanation:
When the cell supplies current, the potential difference measured across its terminals is called its terminal voltage.
During the melting of ice at 0°C:
(b) energy is absorbed and temperature remains constant
Explanation:
The absorbed heat is used as latent heat of fusion to change the state of ice into water, so the temperature remains at 0°C.
The linear magnification ‘m’ produced by a concave lens is:
(a) m < 1
Explanation:
A concave lens always forms a virtual, erect and diminished image. Therefore, its linear magnification is less than 1.
A radioactive element is kept in an evacuated chamber. The rate of radioactive emission will:
(c) remain unchanged
Explanation:
Radioactive decay is a nuclear phenomenon and is unaffected by external physical conditions such as pressure, temperature or vacuum.
The graph given below shows heat energy supplied against change in temperature when no energy is lost to the surrounding. The slope of this graph will give:
(d) heat capacity
Explanation:
Since Q = mcΔT, slope = \frac{Q}{\Delta T} = mc, which is the heat capacity of the body.
A block of glass is pushed into the path of light as shown below. The point X will:
(a) move away from the slab
Explanation:
The glass slab increases the optical path and shifts the point of convergence away from the slab.
(a) In the following atoms, which one is a radioisotope? Give one use of it.
O-16, C-14, N-14, He-4
(b) Name the class of the lever shown in the picture below:
(a) C-14
C14 is used to determine the age of ancient biological materials, such as fossils, bones, or archaeological samples (up to about 50,000 years old).
(b) Class II lever
Explanation:
In a class II lever, the load lies between the fulcrum and the effort.
Fill in the blanks:
(a) A stone tied to a string is rotated in a horizontal plane. The tension in the string provides the __________ force necessary for circular motion.
(b) The work done by this force at any instant is __________.
(a) centripetal
(b) zero
A non-uniform beam of weight 120 N pivoted at one end is shown in the diagram below. Calculate the force F required to keep the beam in equilibrium.
Weight (W) = 120 N
From the figure,
Distance of C.G. from the pivot point O = r = 0.20 m
Distance of F from the pivot point O = R = 0.80 m
Now,
Moment of force about pivot point O due to C.G. = r x W
= 0.20 × 120
= 24 Nm (clocKwise direction)
and
Moment of force about pivot point O due to F = R x F
= 0.80 × F (anticlockwise direction)
For equilibrium, 0.80F = 24
F = 24 ÷ 0.80 = 30 N
Meera chose to use a block and tackle system of ‘9’ pulleys instead of a single movable pulley to lift a heavy load.
(a) What is the advantage of using a block and tackle system over a single movable pulley?
(b) Why should she connect more number of pulleys in the upper fixed block?
(a) The block and tackle provides a higher mechanical advantage, so a smaller effort can lift the load.
(b) More pulleys are placed in the upper fixed block to reduce the weight of the lower movable block.
Sumit does 600 J of work in 10 min and Amit does 300 J of work in 20 min. Calculate the ratio of the powers delivered by them.
Work done by Sumit (WS) = 600 J
Time taken by Sumit = 10 × 60 = 600 s
Power developed by Sumit = \frac{600}{600} = 1 W
Work done by Amit (WA) = 300 J
Time taken by Amit = 20 × 60 = 1200 s
Power developed by Amit = \frac{300}{1200} = 0.25 W
Required ratio = 1 : 0.25 = 4 : 1
5 bulbs are connected in series in a room. One bulb is fused. It is removed and remaining 4 bulbs are again connected in series to the same circuit. What will be the effect on the following physical quantities? (Increases, Decreases, Remain Same)
(a) Resistance
(b) Intensity of light
(a) In a series connection, total resistance is the sum of individual resistances.
Let resistance of each bulb be R.
Initially :
Rtotal = 5R
After removing one bulb:
Rtotal = 4R
So, resistance will decrease.
(b) When resistance decreases, total current in the circuit (assuming the same voltage supply) will increase :
I = V/R
So, more current → more power delivered to each bulb.
Since,
Power (P) = I2R,
Hence, intensity of light increases for each remaining bulb.
Rohan conducted experiments on echo in different media. He observed that a minimum distance of ‘x’ meters is required for the echo to be heard in oxygen and ‘y’ meters in benzene. Compare ‘x’ and ‘y’. Justify your answer.
Given,
Speed of sound in oxygen = 340 ms-1
Speed of sound in benzene = 200 ms-1
Minimum distance required to hear an echo = distance sound travels in 0.1 s (since the human ear can distinguish sound if the reflected sound reaches after 0.1 s)
So, Minimum distance for echo = \frac{\text{Speed of sound}\times\text{Minimum time to hear an echo}}{2}
and tmin = 0.1 sec
For medium containing oxygen:
x = \frac{340\times0.1}{2} = 17 m
For medium containing benzene:
y = \frac{200\times0.1}{2} = 10 m
So, from the above calculations x > y.
(a) In a reading glass what is the position of the object with respect to the convex lens used?
(b) Why can we not use concave lens for the same purpose?
(a) In a reading glass, the object is placed between the optical center and the principal focus of the convex lens to produce a virtual and enlarged image of the object so that a magnified view of the object can be seen.
(b) A concave lens always forms a virtual, erect, but diminished (smaller) image of the object so it cannot magnify the object.
A fuse is rated 5 A. Can it be used with a geyser rated 1540 W, 220 V. Write Yes or No. Give supporting calculations to justify your answer.
Maximum current passing through the fuse (Imax) = 5 A
Input voltage of the geyser (V) = 220 V
Power rating of the geyser (P) = 1540 W
Let, current passing through geyser be I.
Then, P = IV
I = \frac{P}{V} = \frac{1540}{220} = 7 A
As, Imax = 5 A < I = 7 A
So if the fuse is used with the geyser then it will draw more current greater than the maximum current and in turn fuse will melt.
No, the fuse can not be used with the geyser.
State two factors affecting the speed of rotation of the coil in a D.C. motor.
The speed of rotation of coil can be increased by the following two ways:
fHow much heat is required to convert 500 g of ice at 0°C to water at 0°C? The latent heat of fusion of ice is 330 J g−1.
Given, Mass (m) = 500 g
Latent heat of fusion of ice (L) = 330 J g−1
Let the required heat be Q.
Then,
Q = mL = 500 × 330 = 165000 J = 165 kJ
Copy and complete the nuclear reaction by filling in the blanks.
{}_{92}^{235}U+{}_{0}^{1}n\longrightarrow{}_{56}^{\ldots}Ba+{}_{\ldots}^{92}Kr+3{}_{0}^{1}n
As only neutrons are involved in the reaction, the number of protons and total nucleon number remain constant.
Initially, number of protons = 92 and number of nucleons = 235 + 1 = 236.
Let the atomic number of Kr be Z and the mass number of Ba be A.
92 = 56 + Z, so Z = 36.
236 = A + 92 + 3, so A = 141.
Therefore,
{}_{92}^{235}U+{}_{0}^{1}n\longrightarrow{}_{56}^{141}Ba+{}_{36}^{92}Kr+3{}_{0}^{1}n
(Attempt any four questions)
The image of a candle flame placed at a distance of 36 cm from a spherical lens, is formed on a screen placed at a distance of 72 cm from the lens. Calculate the focal length of the lens and its power.
Given, Object distance (u) = −36 cm
Image distance (v) = +72 cm
Let, focal length be f.
From lens formula, \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}
⇒ \frac{1}{\mathrm{f}}=\frac{1}{72}-\frac{1}{(-36)}=\frac{1}{72}+\frac{1}{36}=\frac{1+2}{72}
⇒ \frac{1}{\mathrm{f}}=\frac{3}{72}
⇒ f = \frac{72}{3} = +24 cm
Now, Power = \frac{1}{\mathrm{f}\text{ (in metre)}}=\frac{1}{24\times10^{-2}}=\frac{100}{24}\approx4.17 D
Below is an incomplete table showing the arrangement of electromagnetic spectrum in the increasing order of their wavelength. Complete the table.
| Gamma ray | X-ray | U V rays | Visible rays | Infrared | A | Radio waves |
(a) Identify the radiation A.
(b) Name the radiation used to detect fracture in bones.
(c) Name one property common to both A and Radio waves.
| Gamma ray | X-ray | U V rays | Visible rays | Infrared | Micro Wave | Radio waves |
(a) Radiation A: Micro Wave
(b) X-rays are used to detect fracture in bones.
(c) Micro waves and radio waves are part of electromagnetic waves.
(a) Why do we use red colour as a danger signal on the top of a skyscraper?
(b) The diagram below shows the path of a blue ray through the prism:
(i) Calculate the critical angle of the material of the prism for blue colour.
(ii) What is the measure of the angle of this prism (A)?
(iii) Which colour should replace the blue ray, for the ray to undergo Total Internal Reflection?
(a) Red colour is used because it has the longest wavelength among visible light, so it is scattered the least and remains visible from a long distance.
(b) (i) At Q, angle of incidence = 133° − 90° = 43°. Hence, the critical angle is 43°.
(ii) In △APQ, 90° + 47° + ∠PAQ = 180°. Hence, ∠A = 43°.
(iii) Indigo should replace the blue ray because its wavelength is lower.
(a) Refractive index of glass with respect to water is \frac{9}{8}. Find the refractive index of water with respect to glass.
(b) Name the principle used to find the value in part (a).
(c) If we change the temperature of water, then will the ratio \frac{9}{8} remain the same? Write yes or no.
(a) Refractive index of water with respect to glass = \frac{1}{9/8}=\frac{8}{9}
(b) Principle of reversibility of light.
(c) No. Refractive index changes with temperature, so the ratio will not remain the same.
Light travels a distance of ‘10x’ units in time ‘t1’ in vacuum and it travels a distance of ‘x’ units in time ‘t2’ in a denser medium. Using this information answer the question that follows:
(a) ‘Light covers a distance of ‘20x’ units in time ‘t1’ in diamond.’ State true or false.
(b) Calculate the refractive index of the medium in terms of ‘t1’ and ‘t2’.
(a) False because speed of light in vacuum is highest and diamond is a denser medium.
(b) Speed in vacuum = \frac{10x}{t_1}
Speed in medium = \frac{x}{t_2}
Refractive index = \frac{10x/t_1}{x/t_2}=\frac{10t_2}{t_1}
A monochromatic ray of light is incident on an equilateral prism placed at minimum deviation position with an angle of incidence 45° as shown in the diagram.
(a) Copy the diagram and complete the path of the ray PQ.
(b) State two factors on which the angle of deviation depends.
(a) The path of ray PQ is completed as shown in the diagram.
(b) The angle of deviation depends on:
(a) Define Centre of Gravity.
(b) A hollow ice cream cone has height 6 cm.
(i) Where is the position of its centre of gravity from the broad base?
(ii) Will its position change when it is filled completely with honey? Write Yes or No.
(a) The centre of gravity of a body is the point about which the algebraic sum of moments of weights of all the particles constituting the body is zero and the entire weight of the body can be considered to act at this point, howsoever the body is placed.
(b) (i) For a hollow cone, centre of gravity = \frac{h}{3}=\frac63 = 2 cm from the base.
(ii) Yes. When filled completely, it becomes a solid cone and the new centre of gravity is \frac{h}{4}=\frac64 = 1.5 cm from the base.
Two identical marbles A and B are rolled down along Path 1 and Path 2 respectively. Path 1 is frictionless and Path 2 is rough.
(a) Which marble will surely reach the next peak?
(b) Along which path/s the mechanical energy will be conserved?
(c) Along which path/s is the law of conservation of energy obeyed?
(a) Marble A will surely reach the next peak because there will be no energy loss on the frictionless path.
(b) Mechanical energy will be conserved along Path 1.
(c) Since energy cannot be created or destroyed, the law of conservation of energy applies to both scenarios.
Given are two pulleys.
(a) Copy and complete the labelled diagram connecting the two pulleys with a tackle to obtain Velocity Ratio = 2.
(b) If Load = 48 kgf and efficiency is 80% then calculate:
(i) Mechanical Advantage.
(ii) Effort needed to lift the load.
(a) The completed labelled diagram is shown.
(b) Given, Velocity Ratio = 2, Load = 48 kgf and efficiency = 80% = 0.8.
Mechanical Advantage = Efficiency × Velocity Ratio = 0.8 × 2 = 1.6
Effort = \frac{\text{Load}}{\text{Mechanical Advantage}}=\frac{48}{1.6} = 30 kgf
(a) Name the waves used in SONAR.
(b) In the below diagram Lata stands between two cliffs and claps her hands. Determine the time taken by her to hear the first echo. Speed of sound in air = 320 ms−1.
(a) Ultrasonic Waves.
(b) Cliff A is only 10 m away, while a minimum of 17 m is required to hear a distinct echo. Hence, the echo will be heard from Cliff B.
Given, speed of sound = 320 ms−1
and distance from Cliff B = 160 m.
Total distance travelled = 2 × 160 = 320 m.
Time = \frac{320}{320} = 1 sec.
Hence, the first echo will be heard after 1 second.
(a) Complete the following radioactive reaction:
{}^{\square}_{\square}\mathrm{X}\longrightarrow{}^{\square}_{\square}\mathrm{Y}+{}^{4}_{2}\mathrm{He}\longrightarrow{}^{234}_{91}\mathrm{Z}+{}^{0}_{-1}\mathrm{e}
(b) Uranium is available in two forms U-235 and U-238. Which of the two isotopes of Uranium is more fissionable?
(a) {}^{238}_{92}X\longrightarrow{}^{234}_{90}Y+{}^{4}_{2}He\longrightarrow{}^{234}_{91}Z+{}^{0}_{-1}e
(b) U-235 is more fissionable because only fast neutrons can cause fission of U-238, whereas slow neutrons can trigger fission of U-235.
In the given diagram, a vibrating tuning fork is kept near the mouth of a burette filled with water. The length of the air column is adjusted by opening the tap of the burette. At a length of 5 cm of the air column, a loud sound is heard.
(a) Name the phenomenon illustrated by the above experiment.
(b) Why is a loud sound heard at this particular length?
(c) If the present tuning fork is replaced with a tuning fork of higher frequency, should the length of the air column increase or decrease to produce a loud sound? Give a reason.
(a) Resonance.
(b) The tuning fork frequency and the air column frequency are equal at this length.
(c) Since, frequency of the air column is inversely proportional to the length of the column therefore, in order to create a loud sound, the length of the air column should decrease as the frequency increases.
The voltage-current readings of a certain material are shown in the table given below:
| Voltage (V) | Current (I) |
|---|---|
| 10 V | 2 A |
| 20 V | 3 A |
| 30 V | 4 A |
Study the table.
(a) State whether the conductor used is ohmic or non-ohmic.
(b) Justify your answer.
(c) State Ohm’s law.
(a) The conductor is non-ohmic.
(b) As, from Ohm’s Law
For V = 10 V and I = 2 A
R = \frac{\mathrm{V}}{\mathrm{I}} = \frac{10}{2} = 5 Ω ……. Equation 1
For V = 20 V and I = 3 A
R = \frac{\mathrm{V}}{\mathrm{I}} = \frac{20}{3} = 6.67 Ω ……. Equation 2
For V = 30 V and I = 4 A
R = \frac{\mathrm{V}}{\mathrm{I}} = \frac{30}{4} = 7.5 Ω ……. Equation 3
From equations 1, 2 and 3
Here, the ratio V/I is not constant and plot of V vs I graph will not be linear but if a conductor is an ohmic conductor then V/I should be constant to follow Ohm’s law and V vs I graph should be linear which is not the case here.
Hence, the conductor does not follow Ohm’s law and is a non-ohmic conductor.
(c) Ohm’s law states that the current flowing through the conductor is directly proportional to the potential difference across its ends provided that the physical conditions and the temperature of the conductor remain constant.
Below is the diagram of a transformer:
(a) Identify the type of transformer.
(b) In this type of transformer which of the wire is thicker, the primary or the secondary? Give a reason.
(a) It is a step-down transformer because the primary coil has more turns than the secondary coil.
(b) The secondary wire is thicker because the step-down transformer reduces voltage and increases current. A thicker wire reduces resistance and heat loss.
Study the diagram:
(a) Calculate the total resistance of the circuit.
(b) Calculate the current drawn from the cell.
(c) State whether the current through 10 Ω resistor is greater than, less than or equal to the current through the 12 Ω resistor.
Given, V = 4 V, R₁ = 10 Ω, R₂ = 6 Ω, R₃ = 12 Ω and R₄ = 4 Ω.
(a) R₁ + R₂ = 16 Ω and R₃ + R₄ = 16 Ω. These are in parallel.
\frac1{R_{eq}}=\frac1{16}+\frac1{16}=\frac2{16}
Req = 8 Ω.
(b) Current = \frac{4}{8} = 0.5 A.
(c) Each branch carries 0.25 A. Hence, current through the 10 Ω resistor equals current through the 12 Ω resistor.
85 g of water at 30°C is cooled to 5°C by adding certain mass of ice. Find the mass of ice required.
[Specific heat capacity of water = 4.2 J g−1 °C−1, Specific latent heat of fusion = 336 J g−1]
Heat lost by water = 85 × 4.2 × (30 − 5) = 8925 J.
Let mass of ice be ‘m’ g.
Heat gained by ice = mL + mc△t
= m × 336 + m × 4.2 × 5
= 357m.
By the principle of calorimetry, 8925 = 357m.
m = \frac{8925}{357} = 25 g.
Therefore, the mass of ice required is 25 g.
(a) Why does it become pleasantly warm when the lakes start freezing?
(b) Water freezes to form ice. What change would you expect in the average kinetic energy of the molecules?
(c) Which will contain more heat energy 1 g of ice at 0°C or 1 g water at 0°C?
(a) A large quantity of latent heat is released when lake water freezes, making the surroundings pleasantly warm.
(b) The average kinetic energy of the molecules decreases.
(c) 1 g of water at 0°C contains more heat energy because it releases 80 cal while forming 1 g of ice at 0°C.
(a) State one factor that affects the magnitude of induced current in an AC generator.
(b) Given below is a circuit to study the magnetic effect of electric current. ABCD is a cardboard kept perpendicular to the conductor XY. A magnetic compass is placed at the point P of the cardboard. P1 and P2 are the positions of the magnetic compass, before and after passing a current through XY respectively.
(i) Name the rule that is used to predict the direction of deflection of the magnetic compass.
(ii) State the direction of current in the conductor (X to Y or Y to X) when the circuit is complete.
(iii) If resistance R is increased, then what will be the effect on the magnetic lines of force around the conductor?
(a) Magnitude of magnetic field intensity affects the magnitude of induced current in an AC generator.
(b)(i) Right hand thumb rule.
(ii) Since the compass is pointing west, the current will flow from X to Y in an anticlockwise direction.
(iii) The amount of current flowing through the circuit will decrease if the resistance increases and the magnetic field then decreases as the magnetic lines of force surrounding the conductor is directly proportional to the current in the circuit.
No questions match these filters.
