ICSE Class 10 Physics Sample Paper 2022 with Solutions

ICSE SEMESTER 2 EXAMINATION
SPECIMEN QUESTION PAPER
PHYSICS
(SCIENCE PAPER 1)

Maximum Marks: 40
Time allowed: One and a half hours

Answers to this Paper must be written on the paper provided separately.
You will not be allowed to write during the first 10 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 three questions from Section B.
The intended marks for questions or parts of questions are given in brackets [ ].

SECTION A

(Attempt all questions.)

Question 1

Choose the correct answers to the questions from the given options. (Do not copy the question, Write the correct answer only.)

(i) Pendulums A, B, C and D are tied to a flexible string PQ and are at rest. Pendulum C is disturbed. Which of the following statements is true? [1 Mark]
(a) Only pendulum C will start vibrating.
(b) Pendulums A, B, and D will also start vibrating but A and D will vibrate with the maximum amplitude.
(c) Pendulums A, B, and D will also start vibrating.
(d) Vibrations of pendulum C are forced vibrations.

[Figure: Four simple pendulums labeled A, B, C, D hanging from a stretched string PQ. Pendulum C is longer/different in length, while A and D have similar lengths.]

Answer: (b) Pendulums A, B, and D will also start vibrating but A and D will vibrate with the maximum amplitude.

Pendulums A, B, and D start vibrating due to forced vibrations, and pendulums A and D vibrate with maximum amplitude due to resonance as their natural frequencies match.

Teacher's Note:
a) Resonance occurs when the frequency of the external periodic force matches the natural frequency of the body.
b) Students often confuse simple forced vibrations with resonance where maximum amplitude is achieved.

(ii) Which of the following is not a characteristic of parallel combination of resistors? [1 Mark]
(a) If one resistor is fused, the circuit does not become open.
(b) The total resistance R is given by the formula \(\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} \dots\)
(c) The total resistance becomes less than the least resistor, present in the combination.
(d) The current through each resistor always remains the same.

Answer: (d) The current through each resistor always remains the same.

In a parallel combination, the potential difference across each resistor is the same, whereas the current divides among the resistors depending on their resistance values.

Teacher's Note:
a) In parallel circuits, current varies inversely with resistance, while potential difference remains constant.
b) Always remember that current remains the same only in a series combination of resistors.

(iii) Which one of the following statements is correct? [1 Mark]
(a) Live wire has zero potential.
(b) Fuse is connected in a neutral wire.
(c) Potential of live and earth wire is always the same.
(d) Earth wire is used to prevent electric shock.

Answer: (d) Earth wire is used to prevent electric shock.

The earth wire provides a low-resistance path for leakage current to flow safely into the ground, preventing electric shocks.

Teacher's Note:
a) The live wire carries high potential (220 V in India), and fuses are always connected to the live wire.
b) Neutral and earth wires are at zero potential under normal operating conditions.

(iv) The diagram below shows a free conductor AB is kept in a magnetic field and is carrying current from A to B. (To avoid confusion complete path of the circuit is not shown) The direction of the force experienced by the conductor will be: [1 Mark]
(a) Up
(b) Down
(c) Towards N
(d) Towards S

[Figure: A conductor AB placed between North (N) and South (S) poles of a magnet, with magnetic field pointing from N to S and current flowing from A to B downwards/diagonally.]

Answer: (a) Up

Applying Fleming's Left-Hand Rule, aligning the forefinger with the magnetic field (N to S) and central finger with current (A to B), the thumb points upwards.

Teacher's Note:
a) Fleming's Left-Hand Rule is used to find the direction of force on a current-carrying conductor in a magnetic field.
b) Students must carefully align the three mutually perpendicular fingers to avoid directional errors.

(v) The diagram below shows a magnet moved near a coil along its axis. Which of the diagram shows correct flow of current during this motion? [1 Mark]
(a) [Figure: A bar magnet with North pole facing a solenoid, current shown in a specific direction through galvanometer G]
(b) [Figure: A bar magnet with North pole facing a solenoid, current shown in opposite direction]
(c) [Figure: Bar magnet moving right towards solenoid with N-S poles, showing specific galvanometer deflection and current flow]
(d) [Figure: Bar magnet moving near solenoid with alternative current direction]

Answer: (c)

By Lenz's law, the approaching North pole induces a North polarity on the near face of the coil to oppose the motion, resulting in anticlockwise current when viewed from the magnet side.

Teacher's Note:
a) Lenz's law states that the induced current opposes the very cause that produces it.
b) Check the direction of winding and the motion of the magnet carefully before determining the induced pole.

(vi) The meaning of the statement 'Specific heat capacity of water is 4200 J kg^-1 K^-1' is: [1 Mark]
(a) Water needs 4200 J heat to raise its temperature by 1 kelvin.
(b) To raise the temperature of water 4200 J of heat is absorbed.
(c) 1 kg water absorbs 4200 J heat to increase its temperature by 1 kelvin.
(d) 1 kg Water needs 1 kelvin temperature to absorb 4200 J heat.

Answer: (c) 1 kg water absorbs 4200 J heat to increase its temperature by 1 kelvin.

Specific heat capacity is defined as the amount of heat energy required to raise the temperature of unit mass (1 kg) of a substance by 1 Kelvin (or 1 degree Celsius).

Teacher's Note:
a) Always include the unit mass (1 kg) and unit temperature change (1 K) in definitions of specific heat capacity.
b) Water has one of the highest specific heat capacities, making it an excellent coolant.

(vii) 200 g of ice at 0°C needs _________ heat to melt. [Specific latent heat of ice = 336000 J kg^-1] [1 Mark]
(a) 6720 J
(b) 67200 J
(c) 672000 J
(d) 67.2 J

Answer: (b) 67200 J

\(Q = m \times L = 0.2 \text{ kg} \times 336000 \text{ J kg}^{-1} = 67200 \text{ J}\).

Teacher's Note:
a) Always convert mass from grams to kilograms before applying the formula \(Q = m \times L\).
b) Note that the official answer key mistakenly printed 6720 J in the question text options list for (a), but the correct calculation yields 67200 J which corresponds to option (b).

(viii) The radiation with maximum penetrating power is: [1 Mark]
(a) \(\gamma$
(b) \(\beta$
(c) X-radiation
(d) \(\alpha$

Answer: (a) \(\gamma$

Gamma rays are high-frequency electromagnetic waves with negligible mass and charge, giving them the highest penetrating power among the three radioactive radiations.

Teacher's Note:
a) Alpha particles have the maximum ionising power but minimum penetrating power.
b) Gamma rays can penetrate through thick sheets of lead and concrete.

(ix) Resonance is: [1 Mark]
(a) A forced vibration in which amplitude remains constant.
(b) A forced vibration in which frequency of forced vibration is greater than the free vibrations of the body.
(c) A forced vibration, in which frequency of forced vibration is equal to the free vibrations of the body.
(d) A forced vibration, in which frequency of forced vibration is less than the free vibrations of the body.

Answer: (c) A forced vibration, in which frequency of forced vibration is equal to the free vibrations of the body.

Resonance is a special case of forced vibration where the frequency of the external periodic force exactly matches the natural frequency of the vibrating body.

Teacher's Note:
a) At resonance, the body vibrates with maximum amplitude.
b) Always specify the equality of frequencies when defining resonance.

(x) The nuclear radiation which gets deflected towards negatively charged plate in an electric field is: [1 Mark]
(a) Gamma
(b) Ultraviolet
(c) Beta
(d) Alpha

Answer: (d) Alpha

Alpha particles carry a positive charge (+2e) and are therefore attracted towards the negatively charged plate in an electric field.

Teacher's Note:
a) Beta particles are negatively charged and deflect towards the positively charged plate.
b) Gamma rays are uncharged electromagnetic radiations and pass through electric fields undeflected.

SECTION B

(Attempt any three questions from this Section.)

Question 2

(i) (a) Calculate the total resistance across AB. [3 Marks]
(b) If a cell of e.m.f 2.4 V with negligible internal resistance is connected across AB then calculate the current drawn from the cell.

[Figure: A circuit diagram showing a 3 ohm resistor and a 5 ohm resistor in series, and this combination is in parallel with an 8 ohm resistor across terminals A and B.]

Answer:
(a) Resistors \(3\ \Omega\) and \(5\ \Omega\) are connected in series.
Equivalent resistance of series branch \(R_{s} = 3\ \Omega + 5\ \Omega = 8\ \Omega\).
This \(8\ \Omega\) resistance is in parallel with another \(8\ \Omega\) resistor across AB.
Total resistance \(R_{AB} = \frac{8 \times 8}{8 + 8} = \frac{64}{16} = 4\ \Omega$.
(b) Current drawn from the cell \(I = \frac{V}{R_{AB}} = \frac{2.4\text{ V}}{4\ \Omega} = 0.6\text{ A}\).

Teacher's Note:
a) Always simplify series combinations first before calculating parallel equivalent resistance.
b) Check unit consistency (ohms, volts, amperes) in Ohm's law calculations.

(ii) (a) Which will absorb more heat, 10 g of ice at 0°C or 10 g of water at 0°C? [3 Marks]
(b) For the same mass of ice and ice-cold water, why does ice produce more cooling than ice-cold water?

Answer:
(a) 10 g of ice at 0°C will absorb more heat because to melt into water at 0°C, it must absorb latent heat of fusion (\(334\text{ J g}^{-1}\)).
(b) Ice produces more cooling because in addition to absorbing heat to raise its temperature, each gram of ice absorbs an extra \(334\text{ J}\) of latent heat from the surroundings to melt at 0°C.

Teacher's Note:
a) Latent heat of melting of ice is a major factor in thermal calculations involving phase change.
b) Emphasize that phase change from ice to water at constant temperature requires hidden heat.

(iii) The diagram below shows an insulated copper wire wound around a hollow card board cylindrical tube. Answer the questions that follow: [4 Marks]
(a) What are the magnetic poles at A and B when the key K is closed?
(b) State two ways to increase the strength of the magnetic field in this coil without changing the coil.
(c) If we place a soft iron bar at the centre of the hollow cardboard and replace the DC source by an AC source then will it attract small iron pins toward itself when the current is flowing through the coil?

[Figure: A solenoid wound on a cardboard tube with terminals A and B, connected in series with a key K, a variable resistor/rheostat, and a DC source.]

Answer:
(a) End A will develop a North pole (anticlockwise current) and end B will develop a South pole (clockwise current).
(b) (1) Increase the magnitude of current flowing through the coil by adjusting the rheostat. (2) Place a soft iron core inside the solenoid.
(c) No, it will not attract small iron pins. An alternating current (AC) continually reverses its direction periodically, producing a rapidly alternating magnetic field that results in zero net attractive force on unmagnetized iron pins over a full cycle.

Teacher's Note:
a) Clockwise and anticlockwise rules help determine solenoid polarity based on current direction.
b) AC sources do not sustain a steady magnetic polarity, hence fail to exhibit steady permanent magnetic attraction.

Question 3

(i) The diagram below shows a cooling curve for 200 g of water. The heat is extracted at the rate of 100 J s^-1. Answer the questions that follow: [3 Marks]
(a) Calculate specific heat capacity of water.
(b) Heat released in the region BC.

[Figure: A temperature-time cooling graph for water starting from 80°C at time 0, dropping to 0°C at 640 s (point B), and remaining at 0°C until 1312 s (point C).]

Answer:
(a) Mass \(m = 200\text{ g} = 0.2\text{ kg}\). Temperature change \(\Delta T = 80^{\circ}\text{C} = 80\text{ K}\).
Time taken for cooling from 80°C to 0°C (region AB) \(= 640\text{ s}\).
Total heat extracted \(Q = \text{Rate} \times \text{Time} = 100\text{ J s}^{-1} \times 640\text{ s} = 64000\text{ J}\).
Using \(Q = m c \Delta T\):
\(64000 = 0.2 \times c \times 80\)
\(c = \frac{64000}{16} = 4000\text{ J kg}^{-1}\text{ K}^{-1}\).
(b) Time interval for region BC \(= 1312 - 640 = 672\text{ s}\).
Heat released in region BC \(= 100\text{ J s}^{-1} \times 672\text{ s} = 67200\text{ J}\).

Teacher's Note:
a) Students must distinguish between sensible heat (temperature drop in region AB) and latent heat (phase change at constant temperature in region BC).
b) Ensure proper conversion of mass to SI units (kg).

(ii) (a) Observe the diagram given below and state whether the bulb will glow or not when we switch on K. [3 Marks]
(b) Is it safe to handle the bulb when the switch is OFF?
(c) Give a reason for your answer in (b).

[Figure: A circuit showing a bulb connected to Live wire (L) through the bulb, while the switch K is connected in the Neutral wire (N) coming from A.C. mains 230 V.]

Answer:
(a) Yes, the bulb will glow when switch K is turned ON because the circuit path is completed.
(b) No, it is not safe to handle the bulb even when the switch is OFF.
(c) The switch is incorrectly connected in the neutral wire instead of the live wire. When the switch is OFF, the bulb remains directly connected to the high-potential live wire, posing a severe risk of electric shock.

Teacher's Note:
a) Switches and fuses must always be connected to the live wire for safety.
b) Even with an open switch on the neutral side, the appliance remains at live potential relative to the ground.

(iii) Two metals A and B have specific heat capacities in the ratio 2:3. If they are supplied same amount of heat then [4 Marks]
(a) Which metal piece will show greater rise in temperature given their masses are the same?
(b) Which metal piece will have greater mass if the rise in temperature is the same for both metals?
(c) If the mass ratio of metal A and metal B is 3:5 then calculate the ratio in which their temperatures rise.
(d) If specific heat capacity of metal A is \(0.26\text{ J g}^{-1}\ ^{\circ}\text{C}^{-1}\) then calculate the specific heat capacity of metal B

Answer:
Let specific heat capacities be \(c_{A} = 2C\) and \(c_{B} = 3C\).
(a) Metal A will show a greater rise in temperature. Since \(Q = m c \Delta T\) and \(c\) is smaller for A, for the same heat and mass, \(\Delta T\) is inversely proportional to specific heat capacity.
(b) Metal A will have greater mass. Since \(m = \frac{Q}{c \Delta T}\), for the same heat and temperature rise, a smaller specific heat capacity requires a larger mass.
(c) Given \(m_{A}:m_{B} = 3:5\), and \(c_{A}:c_{B} = 2:3\), with equal heat \(Q\):
\(m_{A} c_{A} \Delta T_{A} = m_{B} c_{B} \Delta T_{B}\)
\(\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{m_{B} c_{B}}{m_{A} c_{A}} = \left(\frac{5}{3}\right) \times \left(\frac{3}{2}\right) = \frac{5}{2}\).
Ratio of temperature rise \(\Delta T_{A} : \Delta T_{B} = 5 : 2$.
(d) Given \(c_{A} = 0.26\text{ J g}^{-1}\ ^{\circ}\text{C}^{-1}\) and \(\frac{c_{A}}{c_{B}} = \frac{2}{3}\):
\(c_{B} = \frac{3}{2} \times c_{A} = 1.5 \times 0.26 = 0.39\text{ J g}^{-1}\ ^{\circ}\text{C}^{-1}\).

Teacher's Note:
a) Use the principal equation of calorimetry \(Q = m c \Delta T\) systematically for ratio problems.
b) Pay close attention to inverse proportionality relationships between specific heat capacity and temperature rise.

Question 4

(i) (a) Which one of the following graphs A or B shows free vibrations in vacuum and which one shows free vibrations in a medium? [3 Marks]
(b) How did you come to this conclusion.

[Figure: Two displacement-time graphs. Graph A shows sinusoidal waves with constant amplitude over time. Graph B shows damped sinusoidal waves whose amplitude decreases exponentially with time.]

Answer:
(a) Graph A represents free vibrations in vacuum, and Graph B represents free vibrations in a medium.
(b) In vacuum, there is no resistive force (no friction or air resistance), so the amplitude of vibration remains constant indefinitely. In a medium, resistive forces cause damping, leading to a continuous decrease in amplitude over time.

Teacher's Note:
a) Undamped oscillations occur only in ideal frictionless conditions like a vacuum.
b) Damped oscillations experience energy loss due to surrounding medium resistance.

(ii) (a) State the Faraday’s laws of electromagnetic induction [3 Marks]
(b) Name one electrical device which works on this principle.

Answer:
(a) First Law: Whenever there is a change in the magnetic flux linked with a circuit, an induced electromotive force (e.m.f.) is produced in the circuit, which lasts as long as the change continues.
Second Law: The magnitude of the induced e.m.f. is directly proportional to the rate of change of magnetic flux linked with the circuit.
(b) AC Generator (or Transformer).

Teacher's Note:
a) Both laws must be stated clearly with keywords like 'magnetic flux' and 'rate of change'.
b) Transformers and generators are classic practical applications of electromagnetic induction.

(iii) A nucleus \(\mathbf{_{82}^{194}X}\) emits an alpha particle [4 Marks]
(a) What will be the atomic number of the daughter nucleus Y?
(b) What will be the number of neutrons in the daughter nucleus Y?
(c) Write a nuclear reaction showing the emission of this particle.

Answer:
(a) An alpha particle (\(_{2}^{4}\text{He}\)) consists of 2 protons and 2 neutrons. Emission of an alpha particle reduces the atomic number by 2.
Atomic number of daughter nucleus Y \(= 82 - 2 = 80\).
(b) Mass number of daughter nucleus Y \(= 194 - 4 = 190\).
Number of neutrons = Mass number - Atomic number \(= 190 - 80 = 110\).(c) Nuclear reaction: \(\mathbf{_{82}^{194}X \rightarrow _{80}^{190}Y + _{2}^{4}\alpha}\)

Teacher's Note:
a) Alpha decay results in a decrease of 4 in mass number and 2 in atomic number.
b) Neutrons are calculated by subtracting the new atomic number from the new mass number.

Question 5

(i) (a) Name the electrical appliance shown in the diagram below. [3 Marks]
(b) Name the material of the wire used in this device.
(c) Name two important characteristics of this wire.

[Figure: An electric fuse unit consisting of a ceramic carrier and base with a fuse wire connected across terminals.]

Answer:
(a) Electric Fuse.
(b) An alloy of lead and tin (Lead-Tin alloy).
(c) Characteristics: (1) High electrical resistivity. (2) Low melting point.

Teacher's Note:
a) Fuse wire must melt quickly when excessive current flows to protect household appliances.
b) Lead-tin alloy has an appropriate low melting point (around 250°C) and high resistance.

(ii) (a) Define pitch. [3 Marks]
(b) Two wires AB and CD of same length are stretched by same amount. Which wire will produce sound of greater pitch on plucking?
(c) Give a reason for your answer.

[Figure: Two stretched wires AB and CD of the same length, where wire AB is thinner in diameter than wire CD.]

Answer:
(a) Pitch is the characteristic of sound by which an acute (shrill) note can be distinguished from a grave (flat) note.
(b) Wire AB will produce sound of greater pitch.
(c) Pitch depends directly on frequency. Wire AB is thinner than wire CD, so it has a lesser mass per unit length and vibrates with a higher frequency, producing a higher pitch.

Teacher's Note:
a) Pitch is determined by frequency; higher frequency means higher pitch.
b) Thinner or shorter strings under same tension vibrate faster, yielding higher frequencies.

(iii) (a) Why is water used as a coolant in radiators of a car? [4 Marks]
(b) Name the radioactive isotope used to find the age of fossils. Name the radioactive radiation which it emits?

Answer:
(a) Water has a very high specific heat capacity (\(4200\text{ J kg}^{-1}\text{ K}^{-1}\)), allowing it to absorb a large amount of heat energy from the engine with only a small rise in its own temperature, making it an efficient coolant.
(b) Radioactive isotope: Carbon-14 (\(^{14}\text{C}\)). Radiation emitted: Beta (\(\beta\)) radiation.

Teacher's Note:
a) High specific heat capacity is the primary reason for water's widespread use in cooling systems.
b) Carbon dating relies on the beta decay of Carbon-14 to estimate fossil ages.

Question 6

(i) A beam of \(\alpha\), \(\beta\) and \(\gamma\) rays is travelling through a certain region in space. [3 Marks]
(a) Arrange them in ascending order of ionising power.
(b) Which of the above will pass undeviated if subjected to an electric field?
(c) With respect to your answer to part (b) above, what will be the change in the nucleus of an atom after such a ray is emitted.

Answer:
(a) Ascending order of ionising power: \(\gamma \lt \beta \lt \alpha$.
(b) Gamma (\(\gamma\)) rays.
(c) Gamma emission results from the de-excitation of a nucleus from a higher energy state to a lower energy state; hence, there is no change in the atomic number or mass number of the nucleus, though energy is released.

Teacher's Note:
a) Ionising power is inversely proportional to penetrating power.
b) Gamma photons carry no charge and no rest mass, so their emission changes neither mass nor atomic number.

(ii) A change in amplitude of a sound wave is noticed. [3 Marks]
(a) Which characteristic of sound is affected due to the above change?
(b) How is amplitude related to your answer to part (a) above?
(c) What happens to the quality of the sound?

Answer:
(a) Loudness.
(b) Loudness is directly proportional to the square of the amplitude (\(L \propto A^{2}\)).
(c) The quality (or timbre) of the sound remains unaffected by a change in amplitude.

Teacher's Note:
a) Loudness depends on amplitude, whereas pitch depends on frequency.
b) Quality depends on the waveform or number and relative intensity of subsidiary notes, independent of amplitude.

(iii) An electric bulb is rated '240 V, 100 W'. [4 Marks]
(a) What information can you get from the above statement?
(b) What will happen if this bulb is connected across 220 V?
(c) Calculate the resistance of the bulb.
(d) Also find the energy consumed by the bulb in 10 minutes.

Answer:
(a) It indicates that when the bulb is connected across a potential difference of \(240\text{ V}\), it consumes electrical power at the rate of \(100\text{ W}\).
(b) If connected across \(220\text{ V}\), the bulb will glow with less brightness because the operating voltage is lower than its rated voltage, resulting in lower power consumption.
(c) Resistance \(R = \frac{V^{2}}{P} = \frac{240^{2}}{100} = \frac{57600}{100} = 576\ \Omega$.
(d) Time \(t = 10\text{ minutes} = 10 \times 60 = 600\text{ s}\).
Energy consumed \(E = P \times t = 100\text{ W} \times 600\text{ s} = 60,000\text{ J}\) (or \(0.0167\text{ kWh}\)).

Teacher's Note:
a) Use the formula \(R = \frac{V^{2}}{P}\) using the rated values to find the constant resistance of the bulb.
b) Energy consumed is calculated as power multiplied by time in SI units (Joules or Watt-seconds).