ICSE Class 10 Physics Board Exam Question Paper 2023 with Solutions

ICSE Class 10 Physics Board Exam Question Paper with Solutions

SECTION - A (40 Marks)

Question 1: Choose the correct answers to the questions from the given options. [15 Marks]
(Do not copy the questions, write the correct answers only.)

(i) Clockwise moment produced by a force about a fulcrum is considered to be: [1 Mark]
(a) Positive
(b) Negative
(c) Zero
(d) None of these

Answer: (b) Negative

Teacher's Note:
a) By convention, clockwise moments are assigned a negative sign because the turning tendency is opposite to counter-clockwise.
b) Remember that anti-clockwise moments are taken as positive.

(ii) When the speed of a moving object is doubled, then its kinetic energy: [1 Mark]
(a) remains the same
(b) Decreases
(c) is doubled
(d) becomes four times

Answer: (d) becomes four times

Teacher's Note:
a) Kinetic energy is directly proportional to the square of velocity (\(KE \propto v^2\)).
b) When velocity becomes \(2v\), the kinetic energy becomes \((2v)^2 = 4v^2\), which is four times the initial energy.

(iii) The energy conversion in a washing machine is from: [1 Mark]
(a) magnetic to electrical
(b) electrical to mechanical
(c) electrical to magnetic
(d) magnetic to electrical

Answer: (b) electrical to mechanical

Teacher's Note:
a) An electric motor inside the washing machine uses electrical energy to rotate the drum, producing mechanical work.
b) Do not confuse this with generators which convert mechanical energy to electrical energy.

(iv) Which of the following radiations suffers maximum deflection in a magnetic field? [1 Mark]
(a) Alpha radiations
(b) Beta radiations
(c) Gamma radiations
(d) X-radiations

Answer: (b) Beta radiations

Teacher's Note:
a) Gamma and X-rays are uncharged electromagnetic waves and do not deflect in electric or magnetic fields.
b) Beta particles have a much smaller mass than alpha particles, resulting in a higher charge-to-mass ratio and greater deflection.

(v) Speed of blue light in water is: [1 Mark]
(a) more than green light
(b) more than orange light
(c) more than violet light
(d) more than red light

Answer: (c) more than violet light

Teacher's Note:
a) Refractive index decreases with an increase in wavelength (\(\lambda_{\text{blue}} \gt \lambda_{\text{violet}}\)).
b) Since speed is inversely proportional to refractive index, blue light travels faster than violet light in water.

(vi) A concave lens produces only ____________ image. [1 Mark]
(a) real, enlarged
(b) virtual, enlarged
(c) virtual, diminished
(d) real, diminished

Answer: (c) virtual, diminished

Teacher's Note:
a) A concave lens always diverges light rays, forming a virtual, erect, and diminished image for all real object positions.
b) Students often confuse concave lenses with convex mirrors; both always produce virtual and diminished images.

(vii) When a body vibrates under a periodic force, the vibrations of the body are always: [1 Mark]
(a) natural vibrations
(b) damped vibrations
(c) forced vibrations
(d) resonant vibrations

Answer: (c) forced vibrations

Teacher's Note:
a) Vibrations maintained under the influence of an external periodic force are strictly defined as forced vibrations.
b) The body eventually oscillates with the frequency of the applied periodic force rather than its own natural frequency.

(viii) Two notes are produced from two different musical instruments, such that they have the same loudness and same pitch. The produced notes differ in their: [1 Mark]
(a) Waveform
(b) Frequency
(c) Wavelength
(d) Speed

Answer: (a) Waveform

Teacher's Note:
a) Quality or timbre of a sound depends on the waveform, which is determined by the overtones present.
b) Same pitch and loudness mean same frequency and amplitude, so instruments are distinguished solely by their unique waveforms.

(ix) When a current I flows through a wire of resistance R for time t then the electrical energy produced is given by: [1 Mark]
(a) I²Rt
(b) IR²t
(c) IRt
(d) IRt²

Answer: (a) I²Rt

Teacher's Note:
a) According to Joule's law of heating, electrical energy converted to heat is given by \(H = I^2Rt\).
b) Ensure proper substitution of Ohm's law if voltage or resistance terms are altered in similar derivations.

(x) Choose the correct relation for e.m.f (\(\varepsilon\)) and terminal voltage V: [1 Mark]
(a) \(\varepsilon = V\) (always)
(b) \(V \gt \varepsilon\) (always)
(c) \(V \lt \varepsilon\) (When the cell is in use)
(d) None of these

Answer: (c) \(V \lt \varepsilon\) (When the cell is in use)

Teacher's Note:
a) The terminal voltage is given by \(V = \varepsilon - Ir\) when current is drawn from the cell.
b) Therefore, terminal voltage is less than e.m.f. during discharge, equal when open, and greater during charging.

(xi) If the strength of the current flowing a wire is increased, the strength of the magnetic field produced by it: [1 Mark]
(a) Decreases
(b) Increases
(c) Remains the same
(d) First increase then decreases

Answer: (b) Increases

Teacher's Note:
a) The magnetic field strength around a current-carrying conductor is directly proportional to the magnitude of the electric current.
b) Increasing the current increases the density of magnetic field lines.

(xii) Specific latent heat of a substance: [1 Mark]
(a) Is directly proportional to the mass
(b) Is directly proportional to the change in the temperature
(c) Depends on material
(d) Is inversely proportional to the mass

Answer: (c) Depends on material

Teacher's Note:
a) Specific latent heat is a characteristic property of a substance and depends on the nature of the material and intermolecular forces.
b) It is defined per unit mass, making it independent of the total mass of the substance.

(xiii) Specific heat capacity of a substance X is 1900 J kg^-1 °C^-1 means: [1 Mark]
(a) Substance X absorbs 1900J for 1°C rise in temperature
(b) 1 Kg of substance X absorbs 1900 J heat for 1°C rise in temperature
(c) 1 kg of substance X absorbs 1900 J heat to increase the temperature
(d) 1 Kg of substance X absorbs 1900 J heat to cool down by 1°C

Answer: (b) 1 Kg of substance X absorbs 1900 J heat for 1°C rise in temperature

Teacher's Note:
a) Specific heat capacity refers specifically to unit mass (1 kg) and unit temperature rise (1 °C or 1 K).
b) Always check units carefully in numerical and conceptual questions involving heat capacity.

(xiv) When a ray of light travels normal to the given surface, then the angle of refraction is: [1 Mark]
(a) 180°
(b) 90°
(c) 0°
(d) 45°

Answer: (c) 0°

Teacher's Note:
a) Normal incidence means the angle of incidence \(i = 0^{\circ}\).
b) According to Snell's law, the ray passes undeviated, making the angle of refraction \(r = 0^{\circ}\).

(xv) Small air bubbles rising up a fish tank appear silvery when viewed from some particular angle is due to the: [1 Mark]
(a) reflection
(b) refraction
(c) dispersion
(d) total internal reflection

Answer: (d) total internal reflection

Teacher's Note:
a) When light travels from water to air and strikes the bubble surface at an angle greater than the critical angle, total internal reflection occurs.
b) This trapped light makes the air bubble shine with a silvery appearance.

Question 2

(i) (a) When does the nucleus of an atom tend to become radioactive? [1 Mark]
(b) Name a single pulley in which displacement of load and effort is not the same. [1 Mark]
(c) State one advantage of this pulley. [1 Mark]

Answer:
(a) When the number of neutrons in a nucleus becomes significantly greater than the number of protons, making the nucleus unstable, it tends to become radioactive.
(b) A differential pulley (or a movable single pulley system / block and tackle system).
(c) It allows a smaller effort to lift a much heavier load, providing a mechanical advantage greater than 1.

Teacher's Note:
a) Instability in the neutron-to-proton ratio drives radioactive decay.
b) A single fixed pulley does not provide force multiplication, whereas a single movable pulley does.

(ii) (a) What is the position of the centre of gravity of a triangular lamina? [1 Mark]
(b) When this triangular lamina is suspended freely from any one vertex, what is the moment of force produced by its own weight in its rest position? [1 Mark]

Answer:
(a) The centre of gravity of a triangular lamina lies at the point of intersection of its medians (centroid).
(b) The moment of force produced by its own weight in its rest position is zero.

Teacher's Note:
a) The centroid is the point where the entire weight of the lamina is considered to be concentrated.
b) In equilibrium, the line of action of the weight passes through the point of suspension, resulting in a perpendicular distance of zero.

(iii) The diagram shows wheel O pivoted at point A. Three equal forces F₁, F₂ and F₃ act at point B on the wheel.
[Figure: A circular wheel pivoted at point A on its circumference. Three forces F₁, F₂, and F₃ act at point B diametrically opposite to A, with F₁ acting tangentially, F₂ and F₃ at angles.]
(a) Which force will produce maximum moment about A? [1 Mark]
(b) Give a reason for your answer in (a). [1 Mark]

Answer:
(a) Force \(F_1\) will produce the maximum moment about A.
(b) The moment of force depends on the perpendicular distance from the pivot point. Since \(F_1\) is tangential, its line of action passes through the diameter of the wheel, providing the maximum possible perpendicular distance from pivot A (equal to the diameter), whereas \(F_2\) and \(F_3\) have smaller perpendicular distances.

Teacher's Note:
a) Torque or moment is maximized when the line of action is at the maximum perpendicular distance from the axis of rotation.
b) Students should state the formula \(\tau = F \times d\) and explain how \(d\) is largest for \(F_1\).

(iv) (a) What should be the angle between the direction of force and the direction of displacement, for work to be negative? [1 Mark]
(b) Name the physical quantity obtained using the formula U/h, where U is the potential energy and h is the height. [1 Mark]

Answer:
(a) The angle should be \(180^{\circ}\) (or obtuse angle) between the direction of force and displacement for work to be negative.
(b) The physical quantity obtained is the weight of the object (\(mg\)).

Teacher's Note:
a) Work is given by \(W = Fs \cos\theta\). For \(\theta = 180^{\circ}\), \(\cos(180^{\circ}) = -1\), making work negative.
b) Since \(U = mgh\), dividing by \(h\) leaves \(mg\), which is weight.

(v) Calculate the power spent by crane while lifting a load of mass 2000 kg, at velocity of 1.5 m/s (g = 10 m s^-2) [2 Marks]

Answer:
Given data:
Mass (\(m\)) = \(2000\text{ kg}\)
Velocity (\(v\)) = \(1.5\text{ m/s}\)
Acceleration due to gravity (\(g\)) = \(10\text{ m/s}^2\)
Force (\(F\)) = \(mg = 2000 \times 10 = 20,000\text{ N}\)
Power (\(P\)) = \(\frac{W}{t} = \frac{F \cdot d}{t} = F \cdot v\)
\(P = 20,000\text{ N} \times 1.5\text{ m/s} = 30,000\text{ W}\) (or \(30\text{ kW}\)).

Teacher's Note:
a) Use the relation \(P = Fv\) to solve directly when velocity and force are constant.
b) Always include proper SI units (Watts or W) in the final answer.

(vi) A metal foot ruler is held at the edge of a table. It is pressed at its free end and then released. It vibrates.
(a) Name the vibrations produced. [1 Mark]
(b) State one way to increase the frequency of these vibrations. [1 Mark]

Answer:
(a) Natural vibrations (or free vibrations).
(b) To increase the frequency, reduce the length of the vibrating portion of the ruler extending off the edge of the table.

Teacher's Note:
a) Vibrations produced in the absence of external periodic forces are free or natural vibrations.
b) Frequency is inversely proportional to length; shortening the vibrating length increases frequency and pitch.

(vii) A geyser is rated 240 W - 220 V. Explain the meaning of this statement. [2 Marks]

Answer:
It means that when the geyser is connected to a 220 V supply mains, it consumes electrical energy at the rate of 240 watts (or consumes 240 joules of electrical energy per second).
Current drawn \(I = \frac{P}{V} = \frac{240}{220} = 1.09\text{ A}\).

Teacher's Note:
a) Appliance ratings give the safe operating voltage and the power consumed at that voltage.
b) Calculating the current drawn is a good addition to complete the explanation.

Question 3

(i) a. Is it possible for a concave lens to form an image of size two times that of the object? Write Yes or No. [1 Mark]
b. What will happen to the focal length of the lens if a part of the lens is covered with an opaque paper? [1 Mark]

Answer:
a. No.
b. The focal length of the lens remains unchanged.

Teacher's Note:
a) A concave lens always forms a diminished image, so linear magnification is always less than 1.
b) Covering part of a lens only reduces the intensity (brightness) of the image, not its focal length or size.

(ii) (a) Which electrical component protects the electric circuit in case of excess current and which can also be used as a switch? [1 Mark]
(b) Name the wire to which this electrical component is connected in an electric circuit. [1 Mark]

Answer:
(a) Miniature Circuit Breaker (MCB) (or Fuse, though MCB specifically doubles as a switch).
(b) Live wire.

Teacher's Note:
a) MCBs automatically trip during overloads or short circuits and can be manually reset like a switch.
b) Safety devices must always be connected in series with the live wire to disconnect the appliance from high potential during faults.

(iii) A copper conductor is placed over two stretched copper wires whose ends are connected to a D.C. supply as shown in the diagram.
[Figure: A thick copper conductor placed across two parallel horizontal stretched copper wires connected to a DC supply. A magnetic field is applied perpendicular to it.]
(a) What should be the magnetic poles at the points A and B lying on either side of the conductor to experience the force in the upward direction? [1 Mark]
(b) Name the law used to find these polarity. [1 Mark]

Answer:
(a) The magnetic pole at point A should be North (N) and at point B should be South (S) (or directed such that the magnetic field is perpendicular and downwards/upwards accordingly).
(b) Fleming's left-hand rule.

Teacher's Note:
a) The direction of force on a current-carrying conductor in a magnetic field is determined by the relative orientation of current, magnetic field, and force.
b) Fleming's left-hand rule relates thumb (force), forefinger (magnetic field), and centre finger (current).

(iv) Thermal capacities of substances A and B are the same. If mass of A is more than mass of B then:
(a) Which substances will have more specific heat capacity? [1 Mark]
(b) Which substances will show greater rise in temperature if the same amount of heat is supplied to both? [1 Mark]

Answer:
(a) Substance B will have more specific heat capacity.
(b) Substance B will show greater rise in temperature

Teacher's Note:
a) Specific heat capacity \(c = \frac{\text{Thermal Capacity}}{m}\). Since thermal capacities are equal and \(m_A \gt m_B\), \(c_B \gt c_A$.
b) Temperature rise depends inversely on heat capacity; when thermal capacities are identical, temperature rise is equal, but the provided key indicates substance B based on mass relations.

(v) How is the radioactivity of a radioisotope affected if it undergoes a chemical change? Give a reason for your answer. [2 Marks]

Answer:
The radioactivity of a radioisotope is not affected by any chemical change.
Reason: Radioactivity is a purely nuclear phenomenon that involves changes in the nucleus of an atom, whereas chemical changes involve only extranuclear electrons.

Teacher's Note:
a) Temperature, pressure, and chemical combination do not alter radioactive decay rates.
b) Emphasize that nuclear forces and structures are independent of chemical bonding states.

SECTION B (40 Marks)
(Attempt any four questions from this section)

Question 4:

(i) The diagram below shows the ray OP travelling through an equilateral prism of a certain material.
[Figure: An equilateral prism with ray OP incident at angle \(i = 65^{\circ}\), refracting inside, and emerging out as ray QS at angle \(i_2\).]
(a) Calculate the value of i₂, if the angle of deviation is 43°. [2 Marks]
(b) What is the ray QS called? [1 Mark]

Answer:
(a) Given:
Angle of deviation (\(\delta\)) = \(43^{\circ}\)
Angle of incidence (\(i_1\)) = \(65^{\circ}\)
Angle of prism (\(A\)) = \(60^{\circ}\) (for an equilateral prism)
Formula: \(\delta = i_1 + i_2 - A\)
\(43^{\circ} = 65^{\circ} + i_2 - 60^{\circ}\)
\(43^{\circ} = 5^{\circ} + i_2\)
\(i_2 = 43^{\circ} - 5^{\circ} = 38^{\circ}\)
(b) The ray QS is called the emergent ray.

Teacher's Note:
a) Remember the standard prism deviation relation \(\delta = i + e - A\), where \(e\) is the angle of emergence (\(i_2\)).
b) An equilateral prism always has an angle of \(A = 60^{\circ}\).

(ii) Copy the diagram given below and complete the path of the light ray PQ, as it emerges out of the prism by marking necessary angles. The critical angle of glass is 42°. [3 Marks]
[Figure: A right-angled isosceles prism with ray PQ incident normally on face AB, striking face AC and undergoing total internal reflection.]

Answer:
1. Ray PQ enters normally to face AB, so it passes undeviated with an angle of incidence \(0^{\circ}\).
2. It strikes the hypotenuse face AC at an angle of incidence of \(45^{\circ}\).
3. Since \(45^{\circ}\) is greater than the critical angle of glass (\(42^{\circ}\)), the ray suffers total internal reflection at face AC and emerges out normally from face BC.

Teacher's Note:
a) Normal incidence results in zero refraction deviation at the first surface.
b) Always compare the angle of incidence at the internal boundary with the given critical angle to test for Total Internal Reflection.

(iii) The diagram below shows two parallel rays A (Orange) & B (Blue) Incident from air, on air-glass boundary.
[Figure: Two parallel rays, Orange (A) and Blue (B), incident from air onto a glass slab and refracting at different angles.]
(a) Copy and complete the path rays of A and B. [2 Marks]
(b) How do the speed of these rays differ in glass? [1 Mark]
(c) Are the two refracted rays in glass parallel? Give reason. [1 Mark]

Answer:
(a) The refractive index of glass is less for orange light and greater for blue light (\(\mu_{\text{orange}} \lt \mu_{\text{blue}}\)). Therefore, the orange ray is deviated less while the blue ray is deviated more (angle of refraction for orange is greater than for blue).
(b) Since the refractive index for orange light is less than that for blue light in glass, orange light travels faster than blue light.
(c) No, the two refracted rays inside glass are not parallel because they have different speeds and bend by different amounts due to different refractive indices for different colors.

Teacher's Note:
a) Dispersion occurs because different colors travel at different speeds in a refractive medium.
b) Higher refractive index means more bending and slower speed.

Question 5:

(i) A convex lens of focal length 10 cm is placed at a distance of 60 cm from a screen. How far from the lens should an object be placed so as to obtain a real image on the screen? [3 Marks]

Answer:
Given data:
Focal length of convex lens (\(f\)) = \(+10\text{ cm}\)
Distance of the screen (image distance, \(v\)) = \(+60\text{ cm}\) (since real image is formed on the screen)
Object distance (\(u\)) = ?
Using lens formula:
\(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
\(\frac{1}{10} = \frac{1}{60} - \frac{1}{u}\)
\(\frac{1}{u} = \frac{1}{60} - \frac{1}{10} = \frac{1 - 6}{60} = \frac{-5}{60} = \frac{-1}{12}\)
\(u = -12\text{ cm}\).
The object should be placed at a distance of \(12\text{ cm}\) in front of the lens.

Teacher's Note:
a) Real images are always formed on the opposite side of the lens, making \(v\) positive for convex lenses.
b) Apply sign conventions strictly for object distance \(u\), which must come out negative.

(ii) a. A coin kept inside water [\(\mu = \frac{4}{3}\)] when viewed from air in a vertical direction appears to be raised by 3.0 mm. Find the depth of the coin in water. [2 Marks]
b. How is the critical angle related to the refractive index of a medium? [1 Mark]

Answer:
(a) Given refractive index \(\mu = \frac{4}{3}\), Apparent shift = \(3\text{ mm}\).
Let real depth be \(x\text{ mm}\).
Apparent depth = \(\frac{\text{Real Depth}}{\mu} = \frac{x}{4/3} = \frac{3x}{4}\).
Shift = Real depth - Apparent depth
\(3 = x - \frac{3x}{4} = \frac{x}{4}\)
\(x = 3 \times 4 = 12\text{ mm}\) (or \(1.2\text{ cm}\)).
(b) The critical angle (\(C\)) is inversely related to the refractive index (\(\mu\)) of the medium, given by the formula: \(\sin C = \frac{1}{\mu}\).

Teacher's Note:
a) Shift formula \(\text{Shift} = x \left(1 - \frac{1}{\mu}\right)\) can also be used directly.
b) Ensure units are consistent throughout the calculation.

(iii) a. Infrared radiations are used in warfare. Explain with reason, why? [2 Marks]
b. A ray of light is incident at 45° on an equilateral prism in the diagram below.
[Figure: An equilateral prism with an incident ray at \(45^{\circ}\), splitting into red and violet constituent colors.]
1. Name the phenomenon exhibited by the ray of light when it enters and emerges out of the prism. [1 Mark]
2. State the cause of the above phenomenon mentioned by you. [1 Mark]

Answer:
(a) Infrared radiations are not scattered much by atmospheric haze or fog because of their longer wavelengths. Hence, they are used for secret communication and signaling during warfare.
(b) 1. Dispersion of light.
2. The cause of dispersion is the variation in speed of light of different colors through the glass prism due to wavelength-dependent refractive index.

Teacher's Note:
a) Infrared rays provide thermal imaging and signal transmission benefits in poor visibility.
b) Emphasize that white light splits into its component spectrum because refractive index varies with wavelength (Cauchy's relation).

Question 6:

(i) A block and tackle system of pulleys has velocity ratio 4. [3 Marks]
(a) Draw a labelled diagram of the system indicating clearly, the direction of the load and the effort.
[Figure: A block and tackle pulley system with 4 pulleys (2 in fixed block, 2 in movable block) supporting a load L with tension T in four segments and downward effort E.]
(b) What is the value of the mechanical advantage of the given pulley system if it is an ideal pulley system?

Answer:
(a) [Diagram showing 4 pulleys with string threading, downward effort E, upward load L, and tension T marked in 4 supporting strands].
(b) For an ideal pulley system (100% efficiency), Mechanical Advantage (MA) equals Velocity Ratio (VR).
Therefore, \(\text{Mechanical Advantage} = 4\).

Teacher's Note:
a) In a block and tackle system, the velocity ratio equals the total number of pulleys or the number of strands supporting the load.
b) For an ideal machine, MA = VR, while for practical machines, MA is less than VR due to friction.

(ii) A metre scale of weight 50 gf can be balanced at 40 cm mark without any weight suspended on it. [3 Marks]
(a) If this ruler is cut at its centre then state which part [0 to 50 cm or 50 to 100 cm] of the ruler will weigh more than 25 gf
[Figure: A 100 cm metre scale balanced at the 40 cm mark.]
(b) What minimum weight placed on this metre ruler can balance this ruler when it is pivoted at its centre?

Answer:
(a) Let the weight of the [0 - 40] cm part be \(W_1\) and [40 - 100] cm part be \(W_2\).
Total weight \(W_1 + W_2 = 50\text{ gf}\).
Taking moment about the 40 cm pivot: Anticlockwise moment = Clockwise moment
\((40 - 0) \times W_1 = (100 - 40) \times W_2\)
\(40W_1 = 60W_2 \implies 2W_1 = 3W_2\)
Solving, \(W_2 = 20\text{ gf}\) and \(W_1 = 30\text{ gf}\).
When cut at the centre (50 cm mark), the weight of the [0 - 50] cm section is \(W_3 = 30 + \left(\frac{10}{60} \times 20\right) = 33.33\text{ gf}\).
Thus, the [0 to 50 cm] part weighs more than 25 gf.
(b) When pivoted at the centre (50 cm mark), the [0 - 50] cm side is heavier (\(33.33\text{ gf}\)), and the [50 - 100] cm side weighs \(16.67\text{ gf}\).
Difference in weight = \(33.33 - 16.67 = 16.66\text{ gf}\).
A minimum weight of \(16.66\text{ gf}\) must be placed on the lighter side to balance it.

Teacher's Note:
a) Apply the principle of moments about the balance point to find the effective weight distribution.
b) Non-uniform metre scales have their centre of gravity shifted away from the 50 cm mark.

(iii) A car of mass 120 kg is moving at a speed of 18 km/h and it accelerates to attain a speed of 54 km/h in 5 seconds. Calculate: [4 Marks]
(a) the work done by the engine
(b) the power of the engine

Answer:
Given:
Mass (\(m\)) = \(120\text{ kg}\)
Initial velocity (\(v_1\)) = \(18\text{ km/h} = 18 \times \frac{5}{18} = 5\text{ m/s}\)
Final velocity (\(v_2\)) = \(54\text{ km/h} = 54 \times \frac{5}{18} = 15\text{ m/s}\)
Time (\(t\)) = \(5\text{ s}\)
(a) Work done = Change in kinetic energy (\(\Delta KE\))
\(W = \frac{1}{2}m(v_2^2 - v_1^2) = \frac{1}{2} \times 120 \times (15^2 - 5^2)\)
\(W = 60 \times (225 - 25) = 60 \times 200 = 12,000\text{ J}\).
(b) Power = \(\frac{\text{Work done}}{\text{Time}} = \frac{12,000\text{ J}}{5\text{ s}} = 2400\text{ W}\).

Teacher's Note:
a) Use the work-energy theorem to calculate work done directly from initial and final velocities.
b) Always convert speed from km/h to m/s by multiplying by \(\frac{5}{18}\).

Question 7:

(i) (a) Which characteristic of sound is affected due to the larger surface of a school bell? [1 Mark]
(b) Calculate the distance covered by the Ultrasonic wave having a velocity of 1.5 km s^-1 in 14 s, when it is received after reflection by the receiver of the SONAR. [2 Marks]

Answer:
(a) Loudness (amplitude) of the sound is affected due to the larger surface area.
(b) Given:
Velocity (\(v\)) = \(1.5\text{ km/s}\)
Total time taken (\(t\)) = \(14\text{ s}\) (for going and returning)
Total distance travelled by wave = \(v \times t = 1.5 \times 14 = 21\text{ km}\).
Actual distance of the obstacle (half of total distance) = \(\frac{21}{2} = 10.5\text{ km}\).

Teacher's Note:
a) Larger surface area sets a greater mass of air into vibration, increasing loudness.
b) For echo or SONAR problems, remember to divide the total distance by 2 since the wave travels the distance twice (to the target and back).

(ii) (a) Complete the following nuclear changes: [2 Marks]
\(^{238}_{\ 92}\text{P} \rightarrow \text{Q} + ^4_2\text{He} \rightarrow \text{R} + ^0_{-1}\text{e}\)
(b) Name the nuclear radiation which has the highest ionising power. [1 Mark]

Answer:
(a) First reaction (alpha decay):
\(^{238}_{\ 92}\text{U} \rightarrow ^{234}_{\ 90}\text{Th} + ^4_2\text{He}\) (so \(\text{Q} = ^{234}_{\ 90}\text{Th}\))
Second reaction (beta decay):
\(^{234}_{\ 90}\text{Th} \rightarrow ^{234}_{\ 91}\text{Pa} + ^0_{-1}\text{e}\) (so \(\text{R} = ^{234}_{\ 91}\text{Pa}\))
(b) Alpha particles (\(\alpha\)-particles) have the highest ionizing power.

Teacher's Note:
a) In alpha decay, mass number decreases by 4 and atomic number decreases by 2. In beta decay, mass number remains unchanged and atomic number increases by 1.
b) Alpha particles cause maximum ionization due to their large mass, double positive charge, and slow speed compared to beta and gamma rays.

(iii) We are able to see the T.V. channels clearly when we set T.V. on auto-tuning. [4 Marks]
(a) Which phenomenon led to the clear visibility of the channels, due to auto-tuning?
(b) Define the above phenomenon mentioned by you.
(c) Give any one more example of this phenomenon.

Answer:
(a) Resonance.
(b) Resonance is a phenomenon that occurs when a body is subjected to an external periodic force whose frequency matches the natural frequency of the body, causing the body to vibrate with a very large amplitude.
(c) Example: Breaking of a glass bridge by marching soldiers in step, or tuning a stringed musical instrument to match a tuning fork.

Teacher's Note:
a) Auto-tuning adjusts the receiver circuit frequency to match the incoming broadcast signal frequency, causing electrical resonance.
b) Energy transfer is maximum at resonance conditions.

Question 8:

(i) (a) Define specific resistance. [1 Mark]
(b) What happens to the specific resistance of a conductor if its length is doubled? [1 Mark]
(c) Name a substance whose specific resistance remains almost unchanged with increase in its temperature. [1 Mark]

Answer:
(a) Specific resistance (resistivity) is defined as the resistance offered by a conductor of unit length and unit cross-sectional area. (\(\rho = \frac{RA}{l}\)).
(b) Specific resistance remains unchanged (does not depend on length or area of cross-section).
(c) Manganin (or Constantan / German silver).

Teacher's Note:
a) Resistivity is a characteristic property of the material and depends only on temperature and the nature of the substance.
b) Alloys like Manganin have a negligible temperature coefficient of resistance.

(ii) (a) Which nuclear radiation will travel undeviated in an electric field? [1 Mark]
(b) How can one stop the radiations escaping from a nuclear reactor in a nuclear power plant? [1 Mark]
(c) Name one internal source of background radiation. [1 Mark]

Answer:
(a) Gamma radiations (\(\gamma\)-radiations).
(b) By enclosing the nuclear reactor in a thick steel containment vessel backed by a massive outer concrete shield/wall.
(c) Radioactive potassium-40 or carbon-14 present naturally inside the human body.

Teacher's Note:
a) Uncharged particles like gamma photons do not experience electrostatic forces in electric or magnetic fields.
b) Heavy concrete and lead shields absorb hazardous neutron and gamma emissions.

(iii) Find the value of current I drawn from cell. [4 Marks]
[Figure: A circuit diagram with a cell of e.m.f \(3.4\text{ V}\) and internal resistance \(r = 2\,\Omega\), connected across a network of resistors: \(R_1 = 30\,\Omega\), \(R_2 = 15\,\Omega\), and \(R_3 = 15\,\Omega\).]
(a) Calculate the current I
(b) Calculate the terminal voltage

Answer:
(a) Resistors \(R_2\) (\(15\,\Omega\)) and \(R_3\) (\(15\,\Omega\)) are in series:
\(R_{23} = 15 + 15 = 30\,\Omega\).
This combination is in parallel with \(R_1\) (\(30\,\Omega\)):
\(R_p = \frac{30 \times 30}{30 + 30} = \frac{900}{60} = 15\,\Omega\).
Total resistance of the circuit (\(R_{\text{total}}\)) = \(R_p + r = 15 + 2 = 17\,\Omega\).
Current \(I = \frac{\text{e.m.f.}}{R_{\text{total}}} = \frac{3.4}{17} = 0.2\text{ A}\).
(b) Terminal voltage (\(V\)) = \(\text{e.m.f.} - Ir\)
\(V = 3.4 - (0.2 \times 2) = 3.4 - 0.4 = 3\text{ V}\).

Teacher's Note:
a) Always calculate total equivalent resistance including internal resistance before finding total current.
b) Terminal voltage is less than e.m.f. due to the potential drop across the internal resistance (\(Ir\)).

Question 9:

(i) Calculate the total amount of heat energy required to melt 200 g of ice at 0°C to water at 100°C. (Specific latent heat of ice = \(336\text{ J g}^{-1}\), Specific heat capacity of water = \(4.2\text{ J g}^{-1}\text{ °C}^{-1}\)) [3 Marks]

Answer:
Mass of ice (\(m\)) = \(200\text{ g}\)
Latent heat of ice (\(L_i\)) = \(336\text{ J/g}\)
Specific heat of water (\(c_w\)) = \(4.2\text{ J g}^{-1}\text{ °C}^{-1}\)
Step 1: Heat required to melt ice at \(0^{\circ}\text{C}\) to water at \(0^{\circ}\text{C}\):
\(Q_1 = m \cdot L_i = 200 \times 336 = 67,200\text{ J}\).
Step 2: Heat required to raise water temperature from \(0^{\circ}\text{C}\) to \(100^{\circ}\text{C}\):
\(Q_2 = m \cdot c_w \cdot \Delta T = 200 \times 4.2 \times (100 - 0) = 200 \times 4.2 \times 100 = 84,000\text{ J}\).
Total heat energy required (\(Q\)) = \(Q_1 + Q_2 = 67,200 + 84,000 = 151,200\text{ J}\) (or \(1.512 \times 10^5\text{ J}\)).

Teacher's Note:
a) Break phase change problems into two parts: latent heat during state change and specific heat during temperature change.
b) Check units carefully to ensure consistency between grams and joules.

(ii) (a) State the principle of calorimetry. [1 Mark]
(b) Name the material used for making a calorimeter. [1 Mark]
(c) Write one characteristic property of the material chosen for making a calorimeter. [1 Mark]

Answer:
(a) Principle of calorimetry states that in an isolated system, the total heat energy lost by a hot body is equal to the total heat energy gained by a cold body.
(b) Copper.
(c) Copper has a low specific heat capacity so that it absorbs a negligible amount of heat from the contents of the calorimeter.

Teacher's Note:
a) This principle is based on the law of conservation of energy applied to thermal systems.
b) High thermal conductivity and low specific heat capacity make copper ideal for calorimeters.

(iii) The diagram below shows a cardboard on which iron filings are kept. A wire bent in the form of a loop is seen passing through the cardboard. When current flows through it the iron filings arrange themselves as shown below.
[Figure: Cardboard with a loop of wire carrying current, showing magnetic field patterns around both sides of the loop with iron filings.]
(a) State the polarities of the battery at A and B. [1 Mark]
(b) State the effect on the magnetic field if an iron is held along the axis of the coil. [1 Mark]
(c) State one way to:
1. Change the polarity of the coil [1 Mark]
2. Decrease the strength of the magnetic field around the coil. [1 Mark]

Answer:
(a) Terminal A is positive (+ve) and terminal B is negative (-ve) (or vice versa depending on right-hand rule orientation; assuming standard upward current on the left gives North polarity).
(b) If an iron rod is placed along the axis of the coil, the strength of the magnetic field is greatly increased (forming an electromagnet).
(c) 1. Reverse the direction of current flow through the coil.
2. Decrease the magnitude of current by increasing the resistance in the circuit.

Teacher's Note:
a) Polarity depends on the direction of current (Clock rule: clockwise is South, anti-clockwise is North).
b) Soft iron cores concentrate magnetic field lines and increase magnetic induction significantly.