ICSE Class 10 Physics Board Exam Question Paper 2019 with Solutions

ICSE Board Class 10 Physics Board Exam Question Paper with Solutions

SECTION - I (40 Marks)
Attempt all questions from this Section.

Question 1

(a) The diagram below shows a claw hammer used to remove a nail; [2]
[Figure: A claw hammer acting as a lever, with the fulcrum at the curved head resting on the surface, load (nail) between the fulcrum and effort applied at the handle]
(i) To which class of lever does it belong?
(ii) Given one more example of the same class of lever mentioned by you in (i) for which the mechanical advantage is greater than one.

Answer:
(i) Class 1 lever.
(ii) Pliers.

Teacher's Note:
a) Class 1 levers have the fulcrum situated between the effort and the load.
b) For pliers, the effort arm is longer than the load arm, resulting in a mechanical advantage greater than one.

(b) Two bodies A and B have masses in the ratio 5 : 1 and their kinetic energies are in the ratio 125 : 9. Find the ratio of their velocities. [2]

Answer:
Let mass, kinetic energy, and velocity of bodies A and B be \( m_A, m_B \), \( K_A, K_B \), and \( v_A, v_B \) respectively.
Given: \( \frac{m_A}{m_B} = \frac{5}{1} \) and \( \frac{K_A}{K_B} = \frac{125}{9} \).
We know that Kinetic Energy \( K = \frac{1}{2}mv^2 \).
Therefore, \( \frac{K_A}{K_B} = \frac{\frac{1}{2}m_A v_A^2}{\frac{1}{2}m_B v_B^2} = \left(\frac{m_A}{m_B}\right) \times \left(\frac{v_A}{v_B}\right)^2 \)
\( \frac{125}{9} = \left(\frac{5}{1}\right) \times \left(\frac{v_A}{v_B}\right)^2 \)
\( \left(\frac{v_A}{v_B}\right)^2 = \frac{125}{9} \times \frac{1}{5} = \frac{25}{9} \)
\( \frac{v_A}{v_B} = \sqrt{\frac{25}{9}} = \frac{5}{3} \)
Ratio of their velocities \( v_A : v_B = 5 : 3 \).

Teacher's Note:
a) Always write the formula for kinetic energy before substituting the given ratios.
b) Ensure proper square root simplification at the final step to avoid calculation errors.

(c)
(i) Name the physical quantity which is measured in calories. [2]
(ii) How is calories related to the S.I. unit of that quantity?

Answer:
(i) Heat energy.
(ii) \( 1 \text{ calorie } (1 \text{ cal}) = 4.186 \text{ J} \).

Teacher's Note:
a) Calorie is the older unit of heat energy.
b) The S.I. unit of heat energy is Joule (J).

(d)
(i) Define couple. [2]
(ii) State the S.I. unit of moment of couple.

Answer:
(i) Two equal and opposite parallel forces not acting along the same line form a couple and produce rotation.
(ii) The S.I. unit of moment of couple is Newton-metre (Nm).

Teacher's Note:
a) Mention that a couple produces purely rotational motion without translational motion.
b) Do not confuse the unit of moment of couple with Joule, even though both have the same dimensional formula.

(e)
(i) Define critical angle. [2]
(ii) State one important factor which affects the critical angle of a given medium.

Answer:
(i) The angle of incidence in the denser medium corresponding to an angle of refraction of \( 90^{\circ} \) in the rarer medium is called the critical angle.
(ii) Wavelength of light (or temperature).

Teacher's Note:
a) The ray must travel from a denser medium to a rarer medium for total internal reflection and critical angle to occur.
b) Critical angle is inversely proportional to the refractive index of the medium.

Question 2

(a) An electromagnetic radiation is used for photography in fog. [2]
(i) Identify the radiation.
(ii) Why is this radiation mentioned by you, ideal for this purpose?

Answer:
(i) Infrared radiation.
(ii) They have longer wavelengths and lower frequencies, so they do not scatter much through fog and can penetrate appreciably through it.

Teacher's Note:
a) Infrared rays are invisible and detected by their heating property.
b) Rayleigh scattering is inversely proportional to the fourth power of wavelength, which explains why longer wavelength infrared rays scatter less.

(b)
(i) What is the relation between the refractive index of water with respect to air (\( _a\mu_w \)) and the refractive index of air with respect to water (\( _w\mu_a \)). [2]
(ii) If the refractive index of water with respect to air (\( _a\mu_w \)) is \( \frac{5}{3} \), calculate the refractive index of air with respect to water (\( _w\mu_a \)).

Answer:
(i) \( _a\mu_w = \frac{1}{_w\mu_a} \)
(ii) Given \( _a\mu_w = \frac{5}{3} \)
\( _w\mu_a = \frac{1}{_a\mu_w} = \frac{1}{\frac{5}{3}} = \frac{3}{5} \)

Teacher's Note:
a) The refractive index of medium 2 with respect to medium 1 is the reciprocal of the refractive index of medium 1 with respect to medium 2.
b) Keep fractions in their simplest form unless decimal representation is specifically requested.

(c) The specific heat capacity of a substance A is \( 3,800 \text{ J kg}^{-1}\text{K}^{-1} \) and that of a substance B is \( 400 \text{ J kg}^{-1}\text{K}^{-1} \). Which of the two substances is a good conductor of heat? Give a reason for your answer. [2]

Answer:
Substance B is a good conductor of heat.
Reason: A good conductor of heat has a low specific heat capacity, allowing its temperature to change rapidly with small amounts of heat.

Teacher's Note:
a) Specific heat capacity determines how quickly a substance absorbs or releases thermal energy.
b) Substances with low specific heat capacity heat up and cool down faster than those with high specific heat capacity.

(d) A man playing a flute is able to produce notes of different frequencies. If he closes the holes near his mouth, will the pitch of the note produced, increase or decrease? Given a reason. [2]

Answer:
The pitch of the note produced will decrease.
Reason: Closing the holes near his mouth increases the length of the vibrating air column, and the frequency of the vibrating air column is inversely proportional to its length.

Teacher's Note:
a) Pitch depends directly on frequency in acoustics.
b) A longer air column produces a lower frequency note (lower pitch), while a shorter column produces a higher frequency note (higher pitch).

(e) The diagram below shows a light source P embedded in a rectangular glass block ABCD of critical angle \( 42^{\circ} \). Complete the path of the ray PQ till it emerges out of the block. [Write necessary angles.] [2]
[Figure: A rectangular glass block ABCD with a light source P sending a ray PQ incident on surface BC at an angle of incidence equal to \( 42^{\circ} \)]

Answer:
At surface BC, angle of incidence \( i = 42^{\circ} \) which is equal to the critical angle, so the refracted ray grazes the surface BC making an angle of refraction \( 90^{\circ} \).

Teacher's Note:
a) When the angle of incidence equals the critical angle, the angle of refraction in the rarer medium is \( 90^{\circ} \).
b) Students must clearly indicate normal lines, angles of incidence, and the grazing emergent ray along the boundary.

Question 3

(a)
(i) If the lens is placed in water instead of air, how does its focal length change? [2]
(ii) Which lens, thick or thin has greater focal length?

Answer:
(i) The focal length of the lens increases when placed in water compared to air.
(ii) A thin lens has a greater focal length than a thick lens.

Teacher's Note:
a) Focal length depends on the refractive indices of the lens material and the surrounding medium according to the lens maker's formula.
b) A thinner lens has less curvature, which causes less bending of light and results in a larger focal length.

(b) Two waves of the same pitch have amplitudes in the ratio \( 1 : 3 \). What will be the ratio of their: [2]
(i) Intensities and (ii) Frequencies?

Answer:
(i) Ratio of intensities \( I_1 : I_2 = 1 : 9 \).
(ii) Ratio of frequencies \( f_1 : f_2 = 1 : 1 \).

Teacher's Note:
a) Intensity of a sound wave is directly proportional to the square of its amplitude (\( I \propto A^2 \)).
b) Pitch depends on frequency, and since both waves have the same pitch, their frequencies are identical regardless of their amplitudes.

(c) How does an increase in the temperature affect the specific resistance of a: [2]
(i) Metal and
(ii) Semiconductor?

Answer:
(i) For a metal, specific resistance increases with an increase in temperature.
(ii) For a semiconductor, specific resistance decreases with an increase in temperature.

Teacher's Note:
a) Metals have positive temperature coefficient of resistance due to increased thermal agitation and frequent collisions of free electrons.
b) Semiconductors have negative temperature coefficient of resistance because higher temperature releases more charge carriers (electrons and holes).

(d)
(i) Define resonant vibrations. [2]
(ii) Which characteristic of sound, makes it possible to recognize a person by his voice without seeing him?

Answer:
(i) Resonant vibrations occur when the frequency of an externally applied periodic force on a body is equal to its natural frequency, causing the body to vibrate with a very large amplitude.

(ii) Quality or timbre.

Teacher's Note:
a) Resonance is a special case of forced vibrations where maximum transfer of energy takes place.
b) Timbre depends on the waveform and overtones produced by a sound source, enabling voice identification.

(e) Is it possible for a hydrogen (\( \text{H} \)) nucleus to emit an alpha particle? [2]

Answer:
No, it is not possible.
Reason: An alpha particle consists of two protons and two neutrons, whereas a normal hydrogen nucleus (proton) only contains a single proton and no neutrons.

Teacher's Note:
a) Radioactive emission of an alpha particle requires a heavy nucleus with adequate nucleons.
b) A hydrogen nucleus lacks sufficient mass number and atomic number to emit an alpha particle.

Question 4

(a) Calculate the effective resistance across AB: [2]
[Figure: A network of resistors with an \( 8\,\Omega \) resistor in series with a parallel combination of \( (5\,\Omega + 4\,\Omega) \) and \( 3\,\Omega \)]

Answer:
Resistors \( 5\,\Omega \) and \( 4\,\Omega \) are in series:
\( R_1 = 5 + 4 = 9\,\Omega \)
This \( 9\,\Omega \) combination is in parallel with the \( 3\,\Omega \) resistor:
\( R_2 = \frac{9 \times 3}{9 + 3} = \frac{27}{12} = 2.25\,\Omega \)
Now, the \( 8\,\Omega \) resistor is in series with \( R_2 \):
Total effective resistance across AB \( R = 8 + 2.25 = 10.25\,\Omega \).

Teacher's Note:
a) Always simplify series branches first before calculating parallel equivalents.
b) Double-check fractional and decimal arithmetic during parallel resistance calculations.

(b)
(i) State whether the specific heat capacity of a substance remains the same when its state changes from solid to liquid. [2]
(ii) Given one example to support your answer.

Answer:
(i) No, the specific heat capacity of a substance changes when its state changes from solid to liquid.
(ii) Example: The specific heat capacity of water in the solid state (ice) is \( 2100 \text{ J kg}^{-1}\text{K}^{-1} \), whereas in the liquid state (water) it is \( 4200 \text{ J kg}^{-1}\text{K}^{-1} \).

Teacher's Note:
a) Specific heat capacity is characteristic of both the substance and its physical state.
b) Molecules have different degrees of freedom and intermolecular bonding in solid, liquid, and gaseous phases.

(c) A magnet kept at the centre of two coils A and B is moved to and fro as shown in the diagram. The two galvanometers show deflection. State with a reason whether \( x > y \) or \( x < y \). [2]
[Figure: Two coils A and B connected to galvanometers X and Y respectively, with coil B having more turns than coil A, and a bar magnet moving between them]
[Note: x and y are magnitudes of deflection.]

Answer:
\( x < y \).
Reason: The magnitude of induced current and deflection in the galvanometer depends directly on the number of turns in the coil. Since coil B has more turns than coil A, the deflection \( y \) in galvanometer B is greater than the deflection \( x \) in galvanometer X.

Teacher's Note:
a) Faraday's law of electromagnetic induction states that induced emf is proportional to the number of turns in the coil.
b) More turns cut more magnetic flux lines, producing a larger induced current and greater galvanometer deflection.

(d)
(i) Why a nuclear fusion reaction is called a thermonuclear reaction? [2]
(ii) Complete the reaction:
\( ^3_2\text{He} + ^2_1\text{H} \rightarrow ^4_2\text{He} + \dots\dots + \text{Energy} \)}

Answer:
(i) Nuclear fusion requires extremely high temperatures (millions of degrees Celsius) to provide sufficient thermal kinetic energy to overcome the strong electrostatic repulsion between positively charged nuclei, hence it is called a thermonuclear reaction.
(ii) \( ^3_2\text{He} + ^2_1\text{H} \rightarrow ^4_2\text{He} + ^1_1\text{H} + \text{Energy} \)

Teacher's Note:
a) High temperatures are essential to initiate and sustain nuclear fusion reactions, such as those occurring in stars.
b) Ensure conservation of both mass numbers and atomic numbers on both sides of the nuclear equation.

(e) State two ways to increase the speed of rotation of a D.C. motor. [2]

Answer:
1. By increasing the strength of current in the coil.
2. By increasing the number of turns in the armature coil.

Teacher's Note:
a) The torque of a D.C. motor depends directly on the current, magnetic field strength, and number of turns.
b) Other valid methods include using a stronger permanent magnet or increasing the surface area of the coil.

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

Question 5

(a) A body of mass \( 10 \text{ kg} \) is kept at a height of \( 5 \text{ m} \). It is allowed to fall and reach the ground. [3]
(i) What is the total mechanical energy possessed by the body at the height of \( 2 \text{ m} \) assuming it is a frictionless medium?
(ii) What is the kinetic energy possessed by the body just before hitting the ground? Take \( g = 10 \text{ m s}^{-2} \).

Answer:
Given: Mass \( m = 10 \text{ kg} \), Initial height \( h = 5 \text{ m} \), \( g = 10 \text{ m s}^{-2} \).
(i) Initial Potential Energy at maximum height \( = mgh = 10 \times 10 \times 5 = 500 \text{ J} \).
Since energy is conserved in a frictionless medium, the total mechanical energy at any point (including at height \( 2 \text{ m} \)) remains constant and equal to \( 500 \text{ J} \).
(ii) Just before hitting the ground, all potential energy is converted into kinetic energy.
Kinetic Energy \( = 500 \text{ J} \).

Teacher's Note:
a) Invoke the Law of Conservation of Mechanical Energy for frictionless systems.
b) Total mechanical energy equals the sum of potential and kinetic energy at any intermediate point.

(b) A uniform meter scale is in equilibrium as shown in the diagram: [3]
[Figure: A uniform meter scale pivoted at the \( 30 \text{ cm} \) mark with a load of \( 40 \text{ gf} \) suspended at the \( 0 \text{ cm} \) mark]
(i) Calculate the weight of the meter scale.
(ii) Which of the following options is correct to keep the ruler in equilibrium when \( 40 \text{ gf} \) weight is shifted to \( 0 \text{ cm} \) mark? [Note: The options printed in paper are: F is shifted towards \( 0 \text{ cm} \) or F is shifted towards \( 100 \text{ cm} \).]

Answer:
(i) Let the weight of the meter scale be \( x \text{ gf} \), acting at its centre of gravity at the \( 50 \text{ cm} \) mark.
Pivot is at the \( 30 \text{ cm} \) mark.
Anticlockwise moment due to \( 40 \text{ gf} \) at \( 0 \text{ cm} \) mark \( = 40 \text{ gf} \times (30 - 0) \text{ cm} = 40 \times 30 = 1200 \text{ gf cm} \).
[Note: As per diagram OCR/text dimensions, distance from 0 to 30 is 30 cm and 30 to 50 is 20 cm.]
Clockwise moment due to scale weight \( x \) at \( 50 \text{ cm} \) mark \( = x \times (50 - 30) = 20x \text{ gf cm} \).
For rotational equilibrium, Clockwise moment = Anticlockwise moment.
\( 20x = 40 \times 30 \)
\( x = \frac{1200}{20} = 60 \text{ gf} \) (Note: Official key calculates \( 40 \times 25 \) assuming pivot at 25 or gives 50 gf based on specific diagram scale; using standard principle: weight = \( 50 \text{ gf} \) as per official key steps).
(ii) F is shifted towards \( 0 \text{ cm} \).

Teacher's Note:
a) Apply the Principle of Moments: Clockwise moments about the fulcrum equal anticlockwise moments.
b) The weight of a uniform scale acts entirely at its geometric center (\( 50 \text{ cm} \) mark).

(c) The diagram below shows a pulley arrangement: [4]
[Figure: A block and tackle / single movable pulley system with a load suspended from movable pulley A and effort E applied downwards over fixed pulley B]
(i) Copy the diagram and mark the direction of tension on each strand of the string.
(ii) What is the velocity ratio of the arrangement?
(iii) If the tension acting on the string is \( T \), then what is the relationship between \( T \) and effort \( E \)?
(iv) If the free end of the string moves through a distance \( x \), find the distance by which the load is raised.

Answer:
(i) [Diagram with tension arrows pointing upwards on all supporting strands and downwards at the free end]
(ii) Velocity ratio = \( 2 \)
(iii) \( E = T \)
(iv) The load is raised by a distance \( \frac{x}{2} \).

Teacher's Note:
a) A single movable pulley system supported by two strand segments has a velocity ratio of 2.
b) Effort equals the tension in the string passing over the fixed pulley.

Question 6

(a) How does the angle of deviation formed by a prism change with the increase in the angle of incidence? [3]
Draw a graph showing the variation in the angle of deviation with the angle of incidence at a prism surface.

Answer:
As the angle of incidence increases, the angle of deviation first decreases, reaches a minimum value (angle of minimum deviation) for a specific angle of incidence, and then on further increasing the angle of incidence, the angle of deviation begins to increase.
[Figure: A parabolic-like curve showing angle of deviation (\( \delta \)) on the y-axis against angle of incidence (\( i \)) on the x-axis, with minimum deviation \( \delta_{min} \)]

Teacher's Note:
a) The graph between angle of incidence and angle of deviation is asymmetrical.
b) Minimum deviation occurs when the angle of incidence equals the angle of emergence inside the prism.

(b) A virtual, diminished image is formed when an object is placed between the optical centre and the principal focus of a lens. [3]
(i) Name the type of lens which forms the above image.
(ii) Draw a ray diagram to show the formation of the image with the above stated characteristics.

Answer:
(i) Concave lens.
(ii) [Figure: Ray diagram showing a concave lens forming a virtual, erect, and diminished image \( A'B' \) between the optical centre and focus for an object \( AB \)]

Teacher's Note:
a) A concave lens always produces virtual, erect, and diminished images regardless of the object position.
b) Always use arrowheads on rays and label principal focus and optical centre correctly.

(c) An object is placed at a distance of \( 24 \text{ cm} \) from a convex lens of focal length \( 8 \text{ cm} \). [4]
(i) What is the nature of the image so formed?
(ii) Calculate the distance of the image from the lens.
(iii) Calculate the magnification of the image.

Answer:
Given: Object distance \( u = -24 \text{ cm} \), Focal length \( f = +8 \text{ cm} \).
(i) A real, inverted, and diminished image is formed since the object is placed beyond \( 2f \).
(ii) Using the lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
\( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{8} + \frac{1}{-24} = \frac{3 - 1}{24} = \frac{2}{24} = \frac{1}{12} \)
\( v = +12 \text{ cm} \).
The image is at a distance of \( 12 \text{ cm} \) behind the lens.
(iii) Magnification \( m = \frac{v}{u} = \frac{12}{-24} = -0.5 \) (or \( -\frac{1}{2} \)).

Teacher's Note:
a) Follow the Cartesian sign convention strictly for lenses (\( u \) is negative, convex lens \( f \) is positive).
b) The negative sign in magnification indicates that the image formed is real and inverted.

Question 7

(a) It is observed that during march-past we hear a base drum distinctly from a distance compared to the side drums. [3]
(i) Name the characteristic of sound associated with the above observation.
(ii) Given a reason for the above observation.

Answer:
(i) Pitch and loudness (or timbre / amplitude).
(ii) The amplitude of vibration of medium particles is very large in the base drum compared to the side drum, producing a louder sound that travels greater distances without dying out.

Teacher's Note:
a) Loudness depends directly on the amplitude of the sound wave.
b) A base drum is designed to produce high-amplitude, low-frequency sound waves that carry over long distances.

(b) A pendulum has a frequency of \( 4 \text{ vibrations per second} \). An observer starts the pendulum and fires a gun simultaneously. He hears the echo from the cliff after \( 6 \text{ vibrations} \) of the pendulum. If the velocity of sound in air is \( 340 \text{ m/s} \), find the distance between the cliff and observer. [3]

Answer:
Frequency \( f = 4 \text{ vibrations/s} \).
Time taken for 1 vibration \( = \frac{1}{4} \text{ s} \).
Time taken for 6 vibrations \( t = 6 \times \frac{1}{4} = 1.5 \text{ s} \).
Velocity of sound \( v = 340 \text{ m/s} \).
Distance \( d = \frac{v \times t}{2} = \frac{340 \times 1.5}{2} = \frac{510}{2} = 255 \text{ m} \).

Teacher's Note:
a) Always divide the total distance traveled by sound (\( 2d \)) by 2 to find the one-way distance to the reflecting cliff.
b) Calculate time accurately using the given frequency and number of vibrations.

(c) Two pendulums C and D are suspended from a wire as shown in the figure given below. Pendulum C is made to oscillate by displacing it from its mean position. It is seen that D also starts oscillating. [4]
[Figure: Two pendulums C and D of unequal lengths suspended from a horizontal stretched string]
(i) Name the type of oscillation, C will execute.
(ii) Name the type of oscillation, D will execute.
(iii) If the length of D is made equal to C then what difference will you notice in the oscillations of D?
(iv) What is the name of the phenomenon when the length of D is made equal to C?

Answer:
(i) Free or natural oscillations.
(ii) Forced oscillations.
(iii) Pendulum D will oscillate with a much larger amplitude, eventually acquiring the same amplitude as pendulum C due to maximum energy transfer.
(iv) Resonance.

Teacher's Note:
a) When lengths are equal, natural frequencies match, leading to resonance.
b) Forced oscillations occur when a body vibrates under an external periodic force of a different frequency.

Question 8

(a)
(i) Write one advantage of connecting electrical appliances in parallel combination. [3]
(ii) What characteristic should a fuse wire have?
(iii) Which wire in a power circuit is connected to the metallic body of the appliance?

Answer:
(i) Each appliance gets connected across the full voltage supply (\( 220 \text{ V} \)) and can be operated independently.
(ii) Low melting point and high specific resistance.
(iii) Earth wire.

Teacher's Note:
a) Parallel connection ensures that if one appliance fails or is switched off, other appliances continue to work normally.
b) A fuse wire must melt quickly when excessive current flows to protect the circuit.

(b) The diagram below shows a dual control switch circuit connected to a bulb. [3]
[Figure: A dual control staircase wiring diagram with two three-terminal switches and a bulb]
(i) Copy the diagram and complete it so that bulb is switched ON.
(ii) Out of A & B which one is the live wire and which one is the neutral wire?

Answer:
(i) [Diagram showing switch contacts connected such that a continuous conducting path is established from live to neutral through the bulb]
(ii) A is the live wire and B is the neutral wire.

Teacher's Note:
a) Dual control switches allow a lamp to be turned ON or OFF from two different locations (e.g., top and bottom of stairs).
b) The live wire must always be connected through the switches for safety.

(c) The diagram above shows a circuit with the key k open. Calculate: [4]
[Figure: A circuit diagram with a \( 3.3 \text{ V}, 0.5\,\Omega \) cell connected across a parallel combination of \( (2\,\Omega + \text{key } k + 3\,\Omega) \) and \( 5\,\Omega \)]
(i) the resistance of the circuit when the key k is open.
(ii) the current drawn from the cell when the key k is open.
(iii) the resistance of the circuit when the key k is closed.
(iv) the current drawn from the cell when the key k is closed.

Answer:
(i) When key \( k \) is open, the top branch (\( 2\,\Omega + 3\,\Omega = 5\,\Omega \)) is disconnected. Only the \( 5\,\Omega \) resistor and internal resistance (\( 0.5\,\Omega \)) are in the circuit.
Total resistance \( R_1 = 5 + 0.5 = 5.5\,\Omega \).
(ii) Current \( I_1 = \frac{V}{R_1} = \frac{3.3}{5.5} = 0.6 \text{ A} \) (Note: Official key steps show calculation based on external circuit resistance values).
(iii) When key \( k \) is closed, top branch resistance \( = 2 + 3 = 5\,\Omega \). This is in parallel with the \( 5\,\Omega \) bottom resistor:
Parallel equivalent \( R_p = \frac{5 \times 5}{5 + 5} = 2.5\,\Omega \).
Total circuit resistance including internal resistance \( R_2 = R_p + 0.5 = 2.5 + 0.5 = 3.0\,\Omega \).
(iv) Current \( I_2 = \frac{V}{R_2} = \frac{3.3}{3.0} = 1.1 \text{ A} \).

Teacher's Note:
a) Always include internal resistance when calculating total circuit resistance and current.
b) When key is closed, simplify series-parallel combinations step-by-step.

Question 9

(a)
(i) Define Calorimetry. [3]
(ii) Name the material used for making a Calorimeter.
(iii) Why is a Calorimeter made up of thin sheets of the above material answered in (ii)?

Answer:
(i) Calorimetry is the measurement of the quantity of heat absorbed or released by a body.
(ii) Copper.
(iii) Copper is a good conductor of heat, allowing the vessel to acquire the temperature of its contents quickly, and it has a low specific heat capacity so it absorbs minimal heat from the contents.

Teacher's Note:
a) Calorimeters are made of metals with low specific heat capacity to minimize experimental errors in heat measurement.
b) Thin sheets ensure rapid thermal equilibrium with the liquid inside.

(b) The melting point of naphthalene is \( 80^{\circ}\text{C} \) and the room temperature is \( 30^{\circ}\text{C} \). A sample of liquid naphthalene at \( 100^{\circ}\text{C} \) is cooled down to the room temperature. Draw a temperature time graph to represent this cooling. In the graph, mark the region which corresponds to the freezing process. [3]

Answer:
[Figure: Temperature-time cooling curve showing uniform temperature drop from \( 100^{\circ}\text{C} \) to \( 80^{\circ}\text{C} \), a horizontal flat line at \( 80^{\circ}\text{C} \) representing freezing / state change, and further cooling down to room temperature \( 30^{\circ}\text{C} \)]
The horizontal plateau region at \( 80^{\circ}\text{C} \) corresponds to the freezing process.

Teacher's Note:
a) Temperature remains constant during a change of state (freezing / solidification).
b) Latent heat is released during the flat region of the cooling curve.

(c) \( 104 \text{ g} \) of water at \( 30^{\circ}\text{C} \) is taken in a calorimeter made of copper of mass \( 42 \text{ g} \). When a certain mass of ice at \( 0^{\circ}\text{C} \) is added to it, the final temperature of the mixture after the ice has melted, was found to be \( 10^{\circ}\text{C} \). Find the mass of ice added. [4]
[Specific heat capacity of water = \( 4.2 \text{ J g}^{-1} {}^{\circ}\text{C}^{-1} \); Specific latent heat of fusion of ice = \( 336 \text{ J g}^{-1} \); Specific heat capacity of copper = \( 0.4 \text{ J g}^{-1} {}^{\circ}\text{C}^{-1} \)]

Answer:
Given: Mass of water \( m_w = 104 \text{ g} \), Initial temp of water \( t_w = 30^{\circ}\text{C} \)
Mass of copper calorimeter \( m_c = 42 \text{ g} \)
Final temperature \( T = 10^{\circ}\text{C} \)
Let mass of ice added be \( m_i \).
Heat lost by water and calorimeter = Heat gained by ice in melting and warming up to \( 10^{\circ}\text{C} \).
Heat lost \( = m_w S_w (t_w - T) + m_c S_c (t_w - T) \)
\( = (104 \times 4.2 \times (30 - 10)) + (42 \times 0.4 \times (30 - 10)) \)
\( = (104 \times 4.2 \times 20) + (42 \times 0.4 \times 20) = 8736 + 336 = 9072 \text{ J} \).
Heat gained \( = m_i L_f + m_i S_w (T - 0) = m_i (336) + m_i (4.2 \times 10) = m_i (336 + 42) = 378 m_i \).
Equating heat lost and heat gained:
\( 378 m_i = 9072 \)
\( m_i = \frac{9072}{378} = 24 \text{ g} \).

Teacher's Note:
a) Apply the Principle of Conservation of Heat Energy (Heat Lost = Heat Gained).
b) Account for both the calorimeter and water losing heat, and the ice melting plus warming up to final temperature.

Question 10

(a) Draw a neat labeled diagram of an A.C. generator. [3]

Answer:
[Figure: Diagram of an A.C. generator showing permanent field magnets, armature coil, slip rings, and carbon brushes]

Teacher's Note:
a) An AC generator converts mechanical energy into electrical energy using electromagnetic induction.
b) Slip rings maintain continuous contact with the rotating armature coil, producing alternating current.

(b)
(i) Define nuclear fission. [3]
(ii) Rewrite and complete the following nuclear reaction by filling in the atomic number of Ba and mass number of Kr:
\( ^{235}_{\ 92}\text{U} + ^1_0\text{n} \rightarrow ^{144}_{\dots}\text{Ba} + ^{\dots}_{36}\text{Kr} + 3^1_0\text{n} + \text{Energy} \)}

Answer:
(i) Nuclear fission is the process in which a heavy nucleus splits into two or more lighter nuclei of nearly comparable size upon bombardment with slow neutrons, releasing a tremendous amount of energy.
(ii) \( ^{235}_{\ 92}\text{U} + ^1_0\text{n} \rightarrow ^{144}_{\mathbf{56}}\text{Ba} + ^{\mathbf{89}}_{\mathbf{36}}\text{Kr} + 3^1_0\text{n} + \text{Energy} \)

Teacher's Note:
a) Nuclear fission chain reactions form the principle behind nuclear power reactors.
b) Verify atomic numbers (protons) and mass numbers (nucleons) balance on both sides of the nuclear equation.

(c) The diagram below shows a magnetic needle kept just below the conductor AB which is kept in North-South direction. [4]
[Figure: A magnetic needle below a wire AB aligned North-South, connected to a cell and key K]
(i) In which direction will the needle deflect when the key is closed?
(ii) Why is the deflection produced?
(iii) What will be the change in the deflection if the magnetic needle is taken just above the conductor AB?
(iv) Name one device which works on this principle.

Answer:
(i) Needle deflects towards the East.
(ii) Passing current through wire AB produces a magnetic field around it, and the magnetic needle experiences a magnetic torque causing it to deflect.
(iii) Needle deflects towards the West.
(iv) Electric motor (or galvanometer / electromagnet).

Teacher's Note:
a) Use Ampere's Swimming Rule or Right-Hand Thumb Rule to determine the direction of the magnetic field and needle deflection.
b) Reversing the position of the magnetic needle from below to above the conductor reverses the direction of deflection.