ICSE Class 10 Physics Board Exam Question Paper 2018 with Solutions

ICSE Class 10 Physics Board Exam Question Paper with Solutions

SECTION - I (40 Marks)

Attempt all questions from this Section

Question 1

(a) (i) State and define the S.I. unit of power. [1 Mark]
(ii) How is the unit horse power related to the S.I. unit of power? [1 Mark]

Answer:
(i) The rate of doing work is called power. S.I. unit of power is watt (W). One watt is defined as the power of an agent which does work at the rate of one joule per second.
(ii) 1 horsepower (hp) = \( 746\text{ W} \).

Teacher's Note:
a) Students must mention both the name of the unit (watt) and its definition in terms of work per unit time.
b) The conversion factor of \( 746\text{ W} \) for 1 horsepower is a frequently tested standard value in numerical problems.

(b) State the energy changes in the following cases while in use: [2 Marks]
(i) An electric iron
(ii) A ceiling fan

Answer:
(i) Electrical energy is converted into heat energy.
(ii) Electrical energy is converted into mechanical energy.

Teacher's Note:
a) Identify the input form of energy supplied to the appliance and the final useful output form of energy.
b) Ensure clarity by stating the energy transformation sequence precisely without omitting intermediate stages if implied, though direct conversion is expected here.

(c) The diagram below shows a lever in use :
(i) To which class of levers does it belong? [1 Mark]
(ii) Without changing the dimensions of the lever, if the load is shifted towards the fulcrum what happens to the mechanical advantage of the lever? [1 Mark]

[Figure: A lever diagram showing fulcrum at one end, load L pointing downwards between fulcrum and effort, and effort E pointing upwards at the other end.]

Answer:
(i) It belongs to a Class-II lever.
(ii) The mechanical advantage of the lever will increase.

Teacher's Note:
a) In a Class-II lever, the load lies between the fulcrum and the effort, giving a mechanical advantage greater than 1.
b) Decreasing the load arm while keeping the effort arm constant increases the mechanical advantage (\( M.A. = \text{Effort arm} / \text{Load arm} \)).

(d) (i) Why is the ratio of the velocities of light of wavelengths \( 4000\text{ \AA} \) and \( 8000\text{ \AA} \) in vacuum 1 : 1 ? [1 Mark]
(ii) Which of the above wavelengths has a higher frequency ? [1 Mark]

Answer:
(i) In vacuum, light of all wavelengths travels with the same constant speed of \( 3 \times 10^{8}\text{ m/s} \), which is independent of wavelength or frequency.
(ii) Wavelength of \( 4000\text{ \AA} \) has a higher frequency because frequency is inversely proportional to wavelength (\( v = n\lambda \)).

Teacher's Note:
a) Remember that all electromagnetic waves travel at the same speed in a vacuum regardless of their frequency or wavelength.
b) Use the relation \( c = n\lambda \) to justify that a shorter wavelength corresponds to a higher frequency.

(e) (i) Why is the motion of a body moving with a constant speed around a circular path said to be accelerated ? [1 Mark]
(ii) Name the unit of physical quantity obtained by the formula \( \frac{2K}{V^{2}} \), where \( K \) : kinetic energy, \( V \) : linear velocity. [1 Mark]

Answer:
(i) In circular motion, even though the speed is constant, the direction of motion changes continuously at every point, which changes the velocity and thus causes acceleration.
(ii) The physical quantity is mass, and its S.I. unit is kilogram (\( \text{kg} \)).

Teacher's Note:
a) Velocity is a vector quantity; hence, a change in direction constitutes a change in velocity, producing centripetal acceleration.
b) Substitute \( K = \frac{1}{2}mv^{2} \) into the given formula to derive \( \frac{2(\frac{1}{2}mv^{2})}{v^{2}} = m \), yielding mass.

Question 2

(a) The power of a lens is \( -5\text{ D} \).
(i) Find its focal length. [1 Mark]
(ii) Name the type of lens. [1 Mark]

Answer:
(i) Focal length \( f = \frac{1}{P} = \frac{1}{-5} = -0.2\text{ m} \) (or \( -20\text{ cm} \)).
(ii) Since the power (and focal length) is negative, the lens is a concave lens.

Teacher's Note:
a) The formula connecting power and focal length is \( P = \frac{1}{f(\text{in metres})} \).
b) Always include the proper sign conventions for power and focal length in lenses.

(b) State the position of the object in front of a converging lens if :
(i) It produces a real and same size image of the object. [1 Mark]
(ii) It is used as a magnifying lens. [1 Mark]

Answer:
(i) The object is placed at \( 2F \) (at a distance of twice the focal length) in front of the lens.
(ii) The object is placed between the optical centre and the principal focus (\( F_{1} \)) of the lens.

Teacher's Note:
a) For a converging lens, a real, inverted image of the same size is formed only when the object is at \( 2F \).
b) A virtual, erect, and magnified image (magnifying glass effect) is obtained when the object is within the focal length.

(c) (i) State the relation between the critical angle and the absolute refractive index of a medium. [1 Mark]
(ii) Which colour of light has a higher critical angle ? Red light or Green light. [1 Mark]

Answer:
(i) \( \mu = \frac{1}{\sin c} \) or \( \sin c = \frac{1}{\mu} \), where \( c \) is the critical angle and \( \mu \) is the absolute refractive index.
(ii) Red light has a higher critical angle.

Teacher's Note:
a) Refractive index is inversely proportional to the sine of the critical angle.
b) Since the refractive index for red light is less than that for green light, red light has a larger critical angle.

(d) (i) Define scattering. [1 Mark]
(ii) The smoke from a fire looks white. Which of the following statements is true ?
1. Molecules of the smoke are bigger than the wavelength of light.
2. Molecules of the smoke are smaller than the wavelength of light [1 Mark]

Answer:
(i) Scattering is the process of absorption and subsequent re-emission of light energy in all directions by dust particles or molecules in the atmosphere.
(ii) Statement 1 is true (Molecules of the smoke are bigger than the wavelength of light).

Teacher's Note:
a) Scattering involves light striking particles and being redirected without a change in wavelength for elastic scattering.
b) Particles larger than the wavelength scatter all colors equally (non-selective scattering), making smoke appear white.

(e) The following diagram shows a \( 60^{\circ}, 30^{\circ}, 90^{\circ} \) glass prism of critical angle \( 42^{\circ} \). Copy the diagram and complete the path of incident ray AB emerging out of the prism marking the angle of incidence on each surface. [2 Marks]

[Figure: A right-angled triangle prism with angles \( 60^{\circ} \) at X, \( 90^{\circ} \) at Y, and \( 30^{\circ} \) at Z. Incident ray AB enters normally to face YZ, strikes face XZ at angle of incidence \( i = 30^{\circ} \), and emerges out refracting away from the normal.]

Answer:
The incident ray AB enters normally to the surface YZ without deviation, strikes the hypotenuse surface XZ at an angle of incidence \( i = 30^{\circ} \) (which is less than the critical angle \( 42^{\circ} \)), and refracts into the air bending away from the normal.

Teacher's Note:
a) Normal incidence results in undeviated entry into the prism.
b) Since the angle of incidence (\( 30^{\circ} \)) is less than the critical angle (\( 42^{\circ} \)), the ray refracts out into the air rather than undergoing total internal reflection.

Question 3

(a) Displacement distance graph of two sound waves A and B, travelling in a medium, are as shown in the diagram below. Study the two sound waves and compare their :
(i) Amplitude [1 Mark]
(ii) Wavelengths [1 Mark]

[Figure: Displacement-distance graph showing wave A with amplitude \( 20\text{ cm} \) and a certain wavelength, and wave B with amplitude \( 10\text{ cm} \) and double the wavelength of A.]

Answer:
(i) The amplitude of sound wave B is half of the amplitude of sound wave A.
(ii) The wavelength of sound wave B is double the wavelength of sound wave A.

Teacher's Note:
a) Amplitude is the maximum displacement from the mean position read on the vertical axis.
b) Wavelength is the distance for one full wave cycle measured along the horizontal distance axis.

(b) You have three resistors of values \( 2\text{ }\Omega \), \( 3\text{ }\Omega \) and \( 5\text{ }\Omega \). How will you join them so that the total resistance is more than \( 7\text{ }\Omega \)?
(i) Draw a diagram for the arrangement. [1 Mark]
(ii) Calculate the equivalent resistance. [1 Mark]

Answer:
To get a total resistance more than \( 7\text{ }\Omega \), all three resistors must be connected in series.
(i) Diagram: Three resistors of \( 2\text{ }\Omega \), \( 3\text{ }\Omega \), and \( 5\text{ }\Omega \) connected in a single linear series path.
(ii) \( R_{\text{equivalent}} = R_{1} + R_{2} + R_{3} = 2 + 3 + 5 = 10\text{ }\Omega \).

Teacher's Note:
a) Connecting resistors in series adds their individual resistances directly to produce the maximum possible total resistance.
b) Always draw clear circuit diagrams showing series connections with proper resistor symbols and terminal labels.

(c) (i) What do you understand by the term nuclear fusion? [1 Mark]
(ii) Nuclear power plants use nuclear fission reaction to produce electricity. What is the advantage of producing electricity by fusion reaction? [1 Mark]

Answer:
(i) Nuclear fusion is the process in which two light nuclei combine together to form a single heavy nucleus with the release of a large amount of energy.
(ii) Nuclear fusion releases much higher energy per unit mass than nuclear fission, and it does not produce long-lived radioactive toxic nuclear waste.

Teacher's Note:
a) Clearly distinguish between fusion (combining light nuclei) and fission (splitting a heavy nucleus).
b) Highlight that fusion is environmentally cleaner and yields higher energy output per unit mass.

(d) (i) What do you understand by free vibrations of a body? [1 Mark]
(ii) Why does the amplitude of a vibrating body continuously decrease during damped vibrations? [1 Mark]

Answer:
(i) Free vibrations are the periodic vibrations of a body of constant amplitude in the absence of any external friction or resistive force, executed at its natural frequency.
(ii) The amplitude of a vibrating body decreases because the body continuously loses energy to the surrounding medium due to frictional and air resistance forces.

Teacher's Note:
a) Emphasize that free vibrations continue indefinitely only in an ideal vacuum without damping forces.
b) Damped vibrations experience energy dissipation, leading to a progressive reduction in oscillation amplitude.

(e) (i) How is the e.m.f. across primary and secondary coils of a transformer related with the number of turns of coil in them? [1 Mark]
(ii) On which type of current do transformers work? [1 Mark]

Answer:
(i) The relation is given by \( \frac{E_{s}}{E_{p}} = \frac{N_{s}}{N_{p}} \), where induced e.m.f. is directly proportional to the number of turns in the respective coil.
(ii) Transformers work only on alternating current (A.C.).

Teacher's Note:
a) The transformation ratio depends directly on the ratio of secondary turns to primary turns.
b) Direct current (D.C.) cannot produce a changing magnetic flux, which is why transformers do not operate on D.C.

Question 4

(a) (i) How can a temperature in degree Celsius be converted into S.I. unit of temperature? [1 Mark]
(ii) A liquid X has the maximum specific heat capacity and is used as a coolant in Car radiators. Name the liquid X. [1 Mark]

Answer:
(i) To convert a temperature in degree Celsius to Kelvin (S.I. unit), add \( 273 \) (or \( 273.15 \)) to the Celsius value.
(ii) Liquid X is water.

Teacher's Note:
a) The S.I. unit of temperature is Kelvin, not degree Celsius.
b) Water is used as a coolant because of its exceptionally high specific heat capacity, allowing it to absorb large amounts of heat with a small rise in temperature.

(b) A solid metal weighing \( 150\text{ g} \) melts at its melting point of \( 800^{\circ}\text{C} \) by providing heat at the rate of \( 100\text{ W} \). The time taken for it to completely melt at the same temperature is \( 4\text{ min} \). What is the specific latent heat of fusion of the metal? [3 Marks]

Answer:
Given data:
Mass (\( m \)) = \( 150\text{ g} = 0.15\text{ kg} \)
Power (\( P \)) = \( 100\text{ W} \)
Time (\( t \)) = \( 4\text{ min} = 4 \times 60 = 240\text{ s} \)
Heat supplied (\( Q \)) = \( P \times t = 100 \times 240 = 24,000\text{ J} \)
Using formula \( Q = m \times L \):
\( 24,000 = 0.15 \times L \)
\( L = \frac{24,000}{0.15} = 160,000\text{ J kg}^{-1} \) or \( 1.6 \times 10^{5}\text{ J kg}^{-1} \).

Teacher's Note:
a) Always convert mass into S.I. units (kilograms) and time into seconds before calculating heat energy in joules.
b) Specific latent heat of fusion is energy required per unit mass to change state at constant temperature.

(c) Identify the following wires used in a household circuit :
(i) The wire is also called as the phase wire. [1 Mark]
(ii) The wire is connected to the top terminal of a three pin socket. [1 Mark]

Answer:
(i) Live wire
(ii) Earth wire

Teacher's Note:
a) The live wire carries current at high potential to appliances in a household circuit.
b) The earth wire is thicker and connected to the top pin for safety by grounding metallic appliance bodies.

(d) (i) What are isobars ? [1 Mark]
(ii) Give one example of isobars. [1 Mark]

Answer:
(i) Isobars are atoms of different elements having the same mass number (\( A \)) but different atomic numbers (\( Z \)).
(ii) \( ^{23}_{\ 11}\text{Na} \) and \( ^{23}_{\ 12}\text{Mg} \) (or \( ^{40}_{\ 18}\text{Ar} \) and \( ^{40}_{\ 20}\text{Ca} \)).

Teacher's Note:
a) Ensure students specify that isobars belong to *different* elements with identical mass numbers.
b) Chemical properties differ because atomic numbers (number of protons) are distinct.

(e) State any two advantages of electromagnets over permanent magnets. [2 Marks]

Answer:
1. An electromagnet can produce a very strong magnetic field compared to a permanent magnet.
2. The strength of the magnetic field of an electromagnet can be easily changed by varying the current or number of turns in the solenoid, whereas a permanent magnet has a fixed field strength.

Teacher's Note:
a) Focus on controllability and strength as the primary practical benefits of electromagnets.
b) Mentioning that the polarity of an electromagnet can also be reversed adds clarity.

SECTION - II (40 Marks)

Attempt any four questions from this Section

Question 5

(a) (i) Derive a relationship between S.I. and C.G.S. unit of work. [2 Marks]
(ii) A force acts on a body and displaces it by a distance S in a direction at an angle \( \theta \) with the direction of force. What should be the value of \( \theta \) to get the maximum positive work? [1 Mark]

Answer:
(i) S.I. unit of work is Joule (\( \text{J} \)) and C.G.S. unit is erg.
\( 1\text{ Joule} = 1\text{ Newton} \times 1\text{ metre} \)
Since \( 1\text{ N} = 10^{5}\text{ dynes} \) and \( 1\text{ m} = 10^{2}\text{ cm} \):
\( 1\text{ J} = (10^{5}\text{ dynes}) \times (10^{2}\text{ cm}) = 10^{7}\text{ dynes cm} = 10^{7}\text{ ergs} \).
(ii) For maximum positive work, \( \theta = 0^{\circ} \) (since \( \cos 0^{\circ} = 1 \)).

Teacher's Note:
a) Show step-by-step substitution of fundamental unit conversions for Newton to dyne and metre to centimeter.
b) Emphasize that work is a scalar product of force and displacement vectors, depending on \( \cos\theta \).

(b) A half metre rod is pivoted at the centre with two weights of \( 20\text{ gf} \) and \( 12\text{ gf} \) suspended at a perpendicular distance of \( 6\text{ cm} \) and \( 10\text{ cm} \) from the pivot respectively as shown below.
(i) Which of the two forces acting on the rigid rod causes clockwise moment ? [1 Mark]
(ii) Is the rod in equilibrium ? [1 Mark]
(iii) The direction of \( 20\text{ gf} \) force is reversed. What is the magnitude of the resultant moment of the forces on the rod? [2 Marks]

[Figure: A half-metre rod pivoted at the centre with a \( 20\text{ gf} \) force at \( 6\text{ cm} \) to the left and a \( 12\text{ gf} \) force at \( 10\text{ cm} \) to the right.]

Answer:
(i) Force due to \( 12\text{ gf} \) (acting on the right side) causes a clockwise moment.
(ii) Yes, the rod is in equilibrium because anticlockwise moment (\( 20\text{ gf} \times 6\text{ cm} = 120\text{ gf cm} \)) equals clockwise moment (\( 12\text{ gf} \times 10\text{ cm} = 120\text{ gf cm} \)).
(iii) When the direction of \( 20\text{ gf} \) is reversed, both forces act in the same direction (clockwise).
Resultant moment = Moment of \( 20\text{ gf} \) + Moment of \( 12\text{ gf} \)
\( = (20\text{ gf} \times 6\text{ cm}) + (12\text{ gf} \times 10\text{ cm}) = 120 + 120 = 240\text{ gf cm} \) (or \( 0.2352\text{ Nm} \)).

Teacher's Note:
a) Apply the Principle of Moments: Clockwise moment = Anticlockwise moment for rotational equilibrium.
b) When one force is reversed, check whether moments add up or subtract based on their new directions.

(c) (i) Draw a diagram to show a block and tackle pulley system having a velocity ratio of 3 marking the direction of load(L), effort(E) and tension (T). [3 Marks]
(ii) The pulley system drawn lifts a load of \( 150\text{ N} \) when an effort of \( 60\text{ N} \) is applied. Find its mechanical advantage. [1 Mark]
(iii) Is the above pulley system an ideal machine or not ? [1 Mark]

Answer:
(i) Diagram: A block and tackle system with 3 pulleys (2 in the fixed block and 1 in the movable block, or vice versa for VR = 3), showing downward load L, downward effort E, and upward string tensions T.
(ii) Mechanical Advantage (\( M.A. \)) = \( \frac{\text{Load}}{\text{Effort}} = \frac{150\text{ N}}{60\text{ N}} = 2.5 \).
(iii) No, it is not an ideal machine because its mechanical advantage (\( 2.5 \)) is less than its velocity ratio (\( 3 \)) due to friction and weight of movable parts.

Teacher's Note:
a) For a block and tackle system with velocity ratio 3, there are 3 supporting strands of string holding the load block.
b) An ideal machine has \( M.A. = V.R. \); if \( M.A. \lt V.R. \), energy losses are present.

Question 6

(a) A ray light XY passes through a right angled isosceles prism as shown below.
(i) What is the angle through which the incident ray deviates and emerges out of the prism? [1 Mark]
(ii) Name the instrument where this action of prism is put into use. [1 Mark]
(iii) Which prism surface will behave as a mirror ? [1 Mark]

[Figure: A right-angled isosceles prism with a ray XY entering normally into face AC, striking AB at \( 45^{\circ} \), totally internally reflecting by \( 90^{\circ} \), and emerging out through BC.]

Answer:
(i) The ray deviates through an angle of \( 90^{\circ} \).
(ii) Periscope.
(iii) Surface AB (hypotenuse).

Teacher's Note:
a) Total internal reflection occurs at the hypotenuse because the angle of incidence (\( 45^{\circ} \)) exceeds the critical angle for glass (\( 42^{\circ} \)).
b) This total internal reflection acts as a perfect 100% efficient mirror without silvering.

(b) An object AB is placed between O and \( F_{1} \) on the principal axis of converging lens as shown in the diagram. Copy the diagram and by using three standard rays starting from point A, obtain an image of the object AB. [3 Marks]

[Figure: A converging lens with principal axis, optical centre O, foci \( F_{1} \), \( 2F_{1} \), \( F_{2} \), \( 2F_{2} \), and an object AB placed between O and \( F_{1} \).]

Answer:
Ray diagram showing object AB between optical centre O and focus \( F_{1} \):
1. A ray parallel to the principal axis passes through focus \( F_{2} \) after refraction.
2. A ray passing through the optical centre O goes straight without deviation.
3. When extended backwards, these refracted rays diverge and meet at point \( A^{\prime} \), forming a virtual, erect, and magnified image \( A^{\prime}B^{\prime} \) on the same side of the lens as the object.

Teacher's Note:
a) Ensure all arrowheads on incident and refracted rays are correctly drawn.
b) Virtual rays must be represented using dashed lines behind the lens.

(c) An object is placed at a distance of \( 12\text{ cm} \) from a convex lens of focal length \( 8\text{ cm} \). Find :
(i) the position of the image [3 Marks]
(ii) nature of the image [1 Mark]

Answer:
Given data:
Object distance (\( u \)) = \( -12\text{ cm} \)
Focal length (\( f \)) = \( +8\text{ cm} \)
Using lens formula:
\( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
\( \frac{1}{v} - \frac{1}{-12} = \frac{1}{8} \)
\( \frac{1}{v} + \frac{1}{12} = \frac{1}{8} \)
\( \frac{1}{v} = \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24} \)
\( v = +24\text{ cm} \)
(i) Position of image: At a distance of \( 24\text{ cm} \) on the other side of the lens.
(ii) Nature of the image: Real, inverted, and magnified.

Teacher's Note:
a) Apply standard sign conventions strictly: object distance \( u \) is always negative, and focal length of a convex lens is positive.
b) A positive sign for image distance \( v \) indicates that the image is formed on the side opposite to the object (real image).

Question 7

(a) Draw the diagram of a right angled isosceles prism which is used to make an inverted image erect. [3 Marks]

Answer:
Diagram showing an erecting prism (Porro prism or right-angled isosceles prism used for inversion):
An inverted object PQ enters normally through face AC, undergoes total internal reflection by \( 90^{\circ} \) twice (or \( 180^{\circ} \) total deviation) at the hypotenuse face AB, and emerges out through BC as an erect image \( P^{\prime}Q^{\prime} \).

Teacher's Note:
a) This prism arrangement deviates the light ray by \( 180^{\circ} \), thereby inverting an upside-down image to make it erect.
b) Clearly label the object, image, and the \( 45^{\circ}-90^{\circ}-45^{\circ} \) angles of the prism.

(b) The diagram above shows a wire stretched over a sonometer. Stems of two vibrating tuning forks A and B are touched to the wooden box of the sonometer. It is observed that the paper rider (a small piece of paper folded at the centre) present on the wire flies off when the stem of vibrating tuning fork B is touched to the wooden box but the paper just vibrates when the stem of vibrating tuning fork A is touched to the wooden box.
(i) Name the phenomenon when the paper rider just vibrates. [1 Mark]
(ii) Name the phenomenon when the paper rider flies off. [1 Mark]
(iii) Why does the paper rider fly off when the stem of tuning fork B is touched to the box? [1 Mark]

[Figure: Sonometer apparatus with wooden box, stretched wire, paper rider, and two tuning forks A and B.]

Answer:
(i) Forced vibrations.
(ii) Resonance.
(iii) The paper rider flies off because the natural frequency of the vibrating wire matches the frequency of tuning fork B, causing resonance and large amplitude oscillations.

Teacher's Note:
a) Forced vibrations occur when a body vibrates under the influence of an external periodic force of different frequency.
b) Resonance is a special case of forced vibrations where frequencies match, resulting in maximum energy transfer and large amplitudes.

(c) A person is standing at the sea shore. An observer on the ship which is anchored in between a vertical cliff and the person on the shore fires a gun. The person on the shore hears two sounds, 2 seconds and 3 seconds after seeing the smoke of the fired gun. If the speed of sound in the air is \( 320\text{ ms}^{-1} \) then calculate :
(i) the distance between the observer on the ship and the person on the shore. [2 Marks]
(ii) the distance between the cliff and the observer on the ship. [2 Marks]

Answer:
Let speed of sound \( v = 320\text{ m/s} \).
(i) Time taken for sound to reach the person on the shore from the ship = \( 2\text{ s} \).
Distance between observer on ship and person on shore (\( d_{1} \)) = \( v \times t_{1} = 320 \times 2 = 640\text{ m} \).
(ii) The person on shore hears the second sound (echo from cliff) after \( 3\text{ s} \).
Total time taken for echo sound to travel from ship to cliff and back to ship plus direct sound time, or total path travelled by echo sound = \( 3\text{ s} \times 320\text{ m/s} = 960\text{ m} \).
Alternatively, time taken for echo = \( 3\text{ s} \). Distance from ship to cliff and back = \( 320 \times 3 = 960\text{ m} \).
Distance between cliff and observer on ship (\( d_{2} \)) = \( \frac{960}{2} = 480\text{ m} \).
[Note: Official marking scheme calculates \( d_{2} = \frac{320}{2} = 160\text{ m} \), but based on standard echo timing differences where sound travels to cliff and back, \( d_{2} = 480\text{ m} \). The key gives \( 160\text{ m} \)]

Teacher's Note:
a) Direct sound travels from ship to shore in 2 seconds, giving the distance directly as speed multiplied by time.
b) The echo from the cliff takes longer; subtract or account for the total reflection path correctly when determining cliff distance.

Question 8

(a) (i) A fuse is rated \( 8\text{ A} \). Can it be used with an electrical appliance rated \( 5\text{ kW}, 200\text{ V} \)? Give a reason. [2 Marks]
(ii) Name two safety devices which are connected to the live wire of a household electric circuit. [1 Mark]

Answer:
(i) Given: Power \( P = 5\text{ kW} = 5000\text{ W} \), Voltage \( V = 200\text{ V} \).
Current \( I = \frac{P}{V} = \frac{5000}{200} = 25\text{ A} \).
Since the operating current of the appliance (\( 25\text{ A} \)) is greater than the fuse rating (\( 8\text{ A} \)), the fuse will melt and blow out immediately. Therefore, it *cannot* be used.
(ii) Fuse and MCB (Miniature Circuit Breaker) [Switch can also be accepted as per context].

Teacher's Note:
a) Always calculate the operating current using \( I = \frac{P}{V} \) and compare it with the fuse rating before concluding.
b) Safety devices must always be connected to the live wire to disconnect the appliance from high potential during faults.

(b) (i) Find the equivalent resistance between A and B. [3 Marks]
(ii) State whether the resistivity of a wire changes with the change in the thickness of the wire. [1 Mark]

[Figure: A circuit diagram showing \( 6\text{ }\Omega \) and \( 3\text{ }\Omega \) resistors in parallel, connected in series with a parallel combination of \( 4\text{ }\Omega \) and \( 2\text{ }\Omega \) resistors between terminals A and B.]

Answer:
(i) For \( 6\text{ }\Omega \) and \( 3\text{ }\Omega \) in parallel:
\( \frac{1}{R_{p1}} = \frac{1}{6} + \frac{1}{3} = \frac{1 + 2}{6} = \frac{3}{6} = \frac{1}{2} \implies R_{p1} = 2\text{ }\Omega \).
For \( 4\text{ }\Omega \) and \( 2\text{ }\Omega \) in parallel:
\( \frac{1}{R_{p2}} = \frac{1}{4} + \frac{1}{2} = \frac{1 + 2}{4} = \frac{3}{4} \implies R_{p2} = \frac{4}{3}\text{ }\Omega \).
Total equivalent resistance \( R_{T} = R_{p1} + R_{p2} = 2 + \frac{4}{3} = \frac{6 + 4}{3} = \frac{10}{3}\text{ }\Omega \) (or \( 3.33\text{ }\Omega \)).
[Note: Official marking scheme treats \( 4\text{ }\Omega \) and \( 12\text{ }\Omega \) or sums differently leading to \( 5\text{ }\Omega \); standard calculation for the shown parallel branch is \( 3.33\text{ }\Omega \)]
(ii) No, resistivity is a characteristic property of the material and does not change with the thickness or dimensions of the wire.

Teacher's Note:
a) Solve complex resistor networks by identifying individual parallel and series groups step-by-step.
b) Remember that resistivity (\( \rho \)) depends only on the material and temperature, not on length or cross-sectional area.

(c) An electric iron is rated \( 220\text{ V}, 2\text{ kW} \).
(i) If the iron is used for \( 2\text{ h} \) daily find the cost of running it for one week if it costs Rs. 4.25 per kWh. [3 Marks]
(ii) Why is the fuse absolutely necessary in a power circuit? [1 Mark]

Answer:
(i) Power \( P = 2\text{ kW} \), Time daily \( t = 2\text{ h} \).
Total energy consumed in 1 week (7 days) = \( \text{Power} \times \text{Total Time} \)
\( E = 2\text{ kW} \times (2\text{ h/day} \times 7\text{ days}) = 2 \times 14 = 28\text{ kWh} \).
Cost = \( 28\text{ kWh} \times \text{Rs. } 4.25/\text{kWh} = \text{Rs. } 119 \).\br (ii) A fuse is necessary to protect electrical appliances and wiring from damage caused by excessive current due to short-circuiting or overloading.

Teacher's Note:
a) Energy in kWh is calculated as Power in kilowatts multiplied by time in hours.
b) Always multiply daily consumption by the total number of days (7 for a week) before computing total cost.

Question 9

(a) (i) Heat supplied to a solid changes it into liquid. What is this change in phase called? [1 Mark]
(ii) During the phase change does the average kinetic energy of the molecules of the substance increase? [1 Mark]
(iii) What is the energy absorbed during the phase change called? [1 Mark]

Answer:
(i) Melting (or fusion).
(ii) No, average kinetic energy remains constant during a phase change.
(iii) Latent heat of fusion.

Teacher's Note:
a) During phase change, heat supplied is utilized entirely to change the intermolecular potential energy, keeping temperature and kinetic energy constant.
b) Latent heat does not cause a rise in temperature.

(b) (i) State two differences between "Heat Capacity" and "Specific Heat Capacity". [2 Marks]
(ii) Give a mathematical relation between Heat Capacity and Specific Heat Capacity. [1 Mark]

Answer:
(i) Differences:
1. Heat capacity is the amount of heat required to raise the temperature of an entire body by \( 1^{\circ}\text{C} \), whereas specific heat capacity is for unit mass of a substance.
2. Heat capacity is an extensive property (depends on mass), whereas specific heat capacity is an intensive property (independent of mass).
(ii) Heat Capacity (\( C^{\prime} \)) = Mass (\( m \)) \( \times \) Specific Heat Capacity (\( c \)).

Teacher's Note:
a) Clearly distinguish between whole-body property (heat capacity) and unit-mass property (specific heat capacity).
b) Note their S.I. units as \( \text{J K}^{-1} \) and \( \text{J kg}^{-1}\text{K}^{-1} \) respectively.

(c) The temperature of \( 170\text{ g} \) of water at \( 50^{\circ}\text{C} \) is lowered to \( 5^{\circ}\text{C} \) by adding certain amount of ice to it. Find the mass of ice added. [3 Marks]
Given: Specific heat capacity of water = \( 4200\text{ J kg}^{-1}\text{K}^{-1} \) (or \( 1\text{ cal g}^{-1}\text{}^{\circ}\text{C}^{-1} \)), Latent heat of ice = \( 336,000\text{ J kg}^{-1} \) (or \( 80\text{ cal g}^{-1} \)).

Answer:
Using the Principle of Calorimetry:
Heat lost by water = Heat gained by ice
Let mass of ice be \( m \).
Heat lost by water to cool from \( 50^{\circ}\text{C} \) to \( 5^{\circ}\text{C} \):
\( Q_{1} = m_{w} \times c_{w} \times \Delta T = 170\text{ g} \times 1\text{ cal/g}^{\circ}\text{C} \times (50^{\circ}\text{C} - 5^{\circ}\text{C}) = 170 \times 45 = 7650\text{ cal} \).
Heat gained by ice to melt at \( 0^{\circ}\text{C} \) and warm up to \( 5^{\circ}\text{C} \):
\( Q_{2} = (m \times L_{f}) + (m \times c_{w} \times \Delta T_{\text{ice}}) = (m \times 80) + (m \times 1 \times 5) = 85m \).\br Equating heat lost and heat gained:
\( 85m = 7650 \implies m = \frac{7650}{85} = 90\text{ g} \).

Teacher's Note:
a) Account for both stages of ice absorption: melting at \( 0^{\circ}\text{C} \) and subsequent warming of the melted ice water up to \( 5^{\circ}\text{C} \).
b) Maintain consistent units (either all in calories/grams or all in Joules/kilograms) throughout the calculation.

Question 10

(a) The diagram shows a coil wound around a U shape soft iron bar AB.
(i) What is the polarity induced at the ends A and B when the switch is pressed ? [1 Mark]
(ii) Suggest one way to strengthen the magnetic field in the electromagnet. [1 Mark]
(iii) What will be the polarities at A & B if the direction of current is reversed in the circuit ? [1 Mark]

[Figure: A U-shaped soft iron core wound with a coil connected to a DC source and switch, with ends labelled A and B.]

Answer:
(i) End A is South pole, and End B is North pole (depending on conventional current direction from battery terminals).
(ii) Increase the magnitude of the electric current or increase the number of turns in the coil.
(iii) If the current direction is reversed, the polarities also get reversed: End A becomes North pole and End B becomes South pole.

Teacher's Note:
a) Use the Clock Face Rule or Right-Hand Thumb Rule to determine the magnetic polarity of a solenoid/core end.
b) Reversing current direction flips the magnetic field orientation completely.

(b) The ore of Uranium found in nature contains \( ^{238}_{\ 92}\text{U} \) and \( ^{235}_{\ 92}\text{U} \). Although both the isotopes are fissionable, it is found out experimentally that one of the two isotopes is more easily fissionable.
(i) Name the isotope of Uranium which is easily fissionable. [1 Mark]
(ii) Give a reason for your answer. [1 Mark]
(iii) Write a nuclear reaction when Uranium-238 emits an alpha particle to form a Thorium (Th) nucleus. [1 Mark]

Answer:
(i) \( ^{235}_{\ 92}\text{U} \)
(ii) \( ^{235}_{\ 92}\text{U} \) is less stable than \( ^{238}_{\ 92}\text{U} \), making it more susceptible to fission upon neutron bombardment.
(iii) Nuclear reaction:
\( ^{238}_{\ 92}\text{U} \longrightarrow \ ^{234}_{\ 90}\text{Th} + \ ^{4}_{\ 2}\text{He} \).

Teacher's Note:
a) Uranium-235 is fissile with thermal neutrons, whereas Uranium-238 requires fast neutrons.
b) Ensure mass numbers and atomic numbers balance correctly on both sides of any nuclear equation.

(c) Radiations given out from a source when subjected to an electric field in a direction perpendicular to their path are shown below in the diagram. The arrows show the path of the radiation A, B and C. Answer the following questions in terms of A, B and C.
(i) Name the radiation B which is unaffected by the electrostatic field. [1 Mark]
(ii) Why does the radiation C deflect more than A ? [1 Mark]
(iii) Which among the three causes the least biological damage externally? [1 Mark]
(iv) Name the radiation which is used in carbon dating. [1 Mark]

[Figure: Radioactive source in a lead box emitting three types of radiation A (deflected towards negative plate), B (undeflected straight), and C (deflected towards positive plate) in an electric field.]

Answer:
(i) Gamma radiations (\( \gamma \)-rays)
(ii) Radiation C (\( \beta \)-particles) deflects more than A (\( \alpha \)-particles) because beta particles have much smaller mass than alpha particles.
(iii) Gamma radiations (\( \gamma \)-rays / Radiation B) cause the least biological damage externally due to their high penetrating power and low ionizing power outside tissues.
(iv) Beta radiation (\( \beta \)-radiation): \( ^{14}_{\ 6}\text{C} \longrightarrow \ ^{14}_{\ 7}\text{N} + \ ^{\ \ 0}_{-1}\beta \).

Teacher's Note:
a) Alpha particles deflect towards the negative plate, beta particles towards the positive plate, and gamma rays pass undeflected.
b) Lighter particles experience greater acceleration and deflection in electric and magnetic fields for the same charge magnitude.