ICSE Class 10 Physics Board Exam Question Paper 2024 with Solutions

CISCE Class 10 Physics Board Exam Question Paper 2024 with Solutions

SECTION-A (40 Marks)

Attempt all questions from this Section.

Q1. Choose the correct answer to the questions from the given options. (Do not copy the questions. Write the correct answer only.) [15 Marks]

1.1. When a bell fixed on a cycle rings, then the energy conversion that takes place is ______ [1 Mark]
1. Gravitational potential energy to sound energy.
2. Kinetic energy to sound energy.
3. Sound energy to electrical energy.
4. Sound energy to mechanical energy.

Answer: (2) Kinetic energy to sound energy.

Mechanical force moves the clapper, converting it into kinetic energy which then strikes the bell to produce sound.

Teacher's Note:
a) Identify the initial form of energy associated with the moving parts of the mechanical bell and the final output energy.
b) Students often confuse mechanical energy with sound energy; remember that mechanical energy of the moving clapper is converted into sound.

1.2. A door lock is opened by turning the lever (handle) of length 0.2 m. If the moment of force produced is 1 Nm, then the minimum force required is ______ [1 Mark]
1. 5 N
2. 10 N
3. 20 N
4. 0.2 N

Answer: (1) 5 N

\( \text{Force} = \frac{\text{Moment of force}}{\text{Length of lever}} = \frac{1\text{ Nm}}{0.2\text{ m}} = 5\text{ N} \).

Teacher's Note:
a) Use the formula: Moment of Force = Force \(\times\) perpendicular distance.
b) Ensure proper SI units are used in calculation.

1.3. A force 'F' moves a load from A to C as shown in the figure below. For the calculation of the work done, which of these lengths would you use as the displacement? [1 Mark]
1. 3m
2. 4m
3. 5m
4. 7m

[Figure: Right-angled triangle ABC with base AB = 4 m, perpendicular BC = 3 m, and hypotenuse AC = 5 m. Force F acts along the hypotenuse AC at an angle of 37 degrees to the base.]

Answer: (3) 5m

Work done depends on the displacement in the direction of the force, which is the total distance AC along the hypotenuse.

Teacher's Note:
a) Displacement is the actual straight-line distance moved by the body in the direction of the applied force.
b) Do not confuse displacement with the vertical or horizontal components unless specified.

1.4. A radioactive nucleus containing 128 nucleons emits a \(\beta\)-particle. After \(\beta\)- emission the number of nucleons present in the nucleus will be ______ [1 Mark]
1. 128
2. 129
3. 124
4. 127

Answer: (1) 128

During beta emission, a neutron converts into a proton, changing the atomic number by +1, but the total number of nucleons (mass number) remains unchanged.

Teacher's Note:
a) Recall that beta emission does not alter the mass number of a nucleus.
b) Students frequently mistake beta decay as a change in mass number; always remember mass number remains constant.

1.5. Assertion (A): Ultraviolet radiations is scattered more as compared to the microwave radiations.
Reason (R): Wavelength of ultraviolet radiation is more than the wavelength of microwave radiation. [1 Mark]
1. Both A and R are true.
2. A is true but R is false.
3. A is false but R is true.
4. Both A and R are false.

Answer: (2) A is true but R is false.

Ultraviolet radiation has a shorter wavelength than microwave radiation, hence it is scattered much more according to Rayleigh scattering.

Teacher's Note:
a) Scattering is inversely proportional to the fourth power of wavelength (\( \lambda^{-4} \)).
b) UV radiation has a much shorter wavelength than microwaves, making reason (R) incorrect.

1.6. When the stem of vibrating tuning fork is pressed on a table, the tabletop starts vibrating. These vibrations are definitely an example of ______ [1 Mark]
1. Resonance
2. Natural vibrations
3. Forced vibrations
4. Damped vibrations

Answer: (3) Forced vibrations

The tabletop is forced to vibrate under the influence of the periodic frequency of the tuning fork.

Teacher's Note:
a) Forced vibrations occur when a body vibrates under the influence of an external periodic force.
b) Do not confuse this with resonance, which is a special case of forced vibrations where frequencies match.

1.7. Which of the following is a class III lever? [1 Mark]
1. Pair of scissors
2. Wheelbarrow
3. Crowbar
4. Human forearm

Answer: (4) Human forearm

In a human forearm, the effort lies between the fulcrum (elbow) and the load (hand).

Teacher's Note:
a) Class III levers have the effort situated between the fulcrum and the load.
b) Memorize common examples of each class of lever to answer quickly.

1.8. The specific resistance of a conductor depends on its ______ [1 Mark]
1. Length
2. Material
3. Area of cross section
4. Radius

Answer: (2) Material

Specific resistance (resistivity) is a characteristic property of the material and does not depend on dimensions.

Teacher's Note:
a) Resistivity depends only on the nature of the material and temperature.
b) Students often confuse resistance with resistivity; resistance depends on dimensions, but resistivity does not.

1.9. Identify the option that displays the correct wiring with correct colour code: [1 Mark]
1. [Figure: Diagram showing appliance connected to mains with switch on live brown wire on left and neutral blue wire on right]
2. [Figure: Diagram showing appliance connected with brown wire and yellow wire]
3. [Figure: Diagram showing switch on the neutral wire]
4. [Figure: Diagram showing switch on the yellow wire]

Answer: (1)

The switch must always be connected to the live wire (brown), and the neutral wire is blue.

Teacher's Note:
a) The live wire is brown (previously red) and must always contain the switch and fuse.
b) Neutral wire is blue (previously black).

1.10. The potential difference between terminals of a cell in a closed electric circuit is ______. [1 Mark]
1. Terminal voltage
2. Electro motive force
3. Voltage drop
4. None of these

Answer: (1) Terminal voltage

When current is drawn in a closed circuit, the potential difference across the terminals is termed terminal voltage.

Teacher's Note:
a) Terminal voltage is less than EMF due to the internal resistance of the cell.
b) EMF is measured only when the circuit is open.

1.11. During melting of ice at 0°C the ______. [1 Mark]
1. Energy is released and temperature remains constant.
2. Energy is absorbed and temperature remains constant.
3. Energy is released and temperature decreases.
4. Energy is absorbed and temperature increases.

Answer: (2) Energy is absorbed and temperature remains constant.

Latent heat of fusion is absorbed during melting without any rise in temperature.

Teacher's Note:
a) Phase change occurs at a constant temperature by absorbing latent heat.
b) Make sure to specify that temperature remains unchanged during phase transitions.

1.12. Linear magnification (m) produced by a concave lens is ______ [1 Mark]
1. m < 1
2. m > 1
3. m = 1
4. m = 2

Answer: (1) m < 1

A concave lens always forms a virtual, diminished image, hence magnification is always less than 1.

Teacher's Note:
a) Concave lenses produce diminished images for all real positions of the object.
b) Magnification for concave lenses is positive and less than 1.

1.13. A radioactive element is placed in an evacuated chamber. Then the rate of radioactive decay will ______. [1 Mark]
1. Decrease
2. Increase
3. Remain unchanged
4. Depend on the surrounding temperature

Answer: (3) Remain unchanged

Radioactivity is a spontaneous nuclear phenomenon unaffected by external physical conditions such as pressure or vacuum.

Teacher's Note:
a) Radioactive decay is independent of external factors like temperature, pressure, or environment.
b) It is solely governed by the properties of the atomic nucleus.

1.14. The graph given below shows heat energy supplied against change in temperature when no energy is lost to the surroundings. The slope of this graph will give: [1 Mark]
1. Specific heat capacity
2. Latent heat of fusion
3. Latent heat of vaporization
4. Heat capacity

[Figure: Line graph with Heat Energy on y-axis and Change in Temperature on x-axis starting from origin (0,0).]

Answer: (4) Heat capacity

Slope = Heat energy supplied / Change in temperature = Heat capacity.

Teacher's Note:
a) Heat capacity is defined as the amount of heat energy required to raise the temperature of the entire body by 1°C.
b) Be careful to distinguish between heat capacity (slope of this graph) and specific heat capacity (which divides by mass).

1.15. A block of glass is pushed into the path of the light as shown below. Then the converging point x will: [1 Mark]
1. Move away from the slab.
2. Move towards the slab.
3. Not shift.
4. Move towards the left side of the lens.

[Figure: Convex lens converging light rays at point x, with a rectangular glass block placed in the path of rays between the lens and point x.]

Answer: (1) Move away from the slab.

A glass slab shifts the converging point of light rays away from itself due to refraction.

Teacher's Note:
a) Introducing a glass slab causes lateral shift, delaying the convergence of rays.
b) Consequently, the focal point or point of convergence shifts further away.

Q2.

2.1. (a) In the following atoms, which one is a radioisotope? Give one use of this isotope.
\( O^{16}, C^{14}, N^{14}, He^{4} \) [2 Marks]

Answer:
\( C^{14} \) is the radioisotope.
Use: It is used in carbon dating to determine the age of archaeological artifacts.

Teacher's Note:
a) Identify carbon-14 as a well-known radioactive isotope of carbon.
b) Mention any valid application such as radio-carbon dating or biomedical tracing.

2.1. (b) Name the class of the lever shown in the picture below: [1 Mark]

[Figure: A bottle opener with the load at one end, effort at the other end, and fulcrum at the opposite edge.]

Answer:
Class II lever.

Teacher's Note:
a) In a bottle opener, the load is in the middle between the fulcrum and the effort.
b) Class II levers always act as force multipliers.

2.2. (a) When a stone tied to a string is rotated in a horizontal plane, the tension in the string provides ______ force necessary for circular motion. [1 Mark]

Answer:
Centripetal force.

Teacher's Note:
a) Centripetal force is directed towards the center of the circular path.
b) Tension acts as the necessary inward pull.

2.2. (b) Work done by this force at any instant is ______. [1 Mark]

Answer:
Zero.

Teacher's Note:
a) Work done is zero because the centripetal force is always perpendicular to the instantaneous displacement.
b) Recall that \( W = F \cdot s \cdot \cos(90^{\circ}) = 0 \).

2.3. A non uniform beam of weight 120 N pivoted at one end is shown in the diagram below. Calculate the value of F to keep the beam in equilibrium. [2 Marks]

[Figure: A beam of length 1.0 m pivoted at one end, with weight 120 N acting downwards at 0.20 m from the pivot, and upward force F applied at the free end at 0.80 m from the pivot.]

Answer:
By the principle of moments:
Anticlockwise moment = Clockwise moment
\( F \times 0.80 = 120 \times 0.20 \)
\( F = \frac{120 \times 0.20}{0.80} = 30\text{ N} \).

Teacher's Note:
a) Apply the principle of moments: sum of anticlockwise moments equals sum of clockwise moments about the pivot.
b) Check distances carefully from the pivot point.

2.4. (a) Meera chose to use a block and tackle system of '9' pulleys instead of a single movable pulley to lift a heavy load.
What is the advantage of using a block and tackle system over a single movable pulley? [1 Mark]

Answer:
The block and tackle system provides a much higher mechanical advantage (equal to 9), multiplying the effort effectively, whereas a single movable pulley has a theoretical mechanical advantage of 2.

Teacher's Note:
a) Mechanical advantage increases with the number of pulleys in a block and tackle system.
b) It makes lifting heavy loads significantly easier.

2.4. (b) Meera chose to use a block and tackle system of '9' pulleys instead of a single movable pulley to lift a heavy load.
Why should she connect more number of pulleys in the upper fixed block? [1 Mark]

Answer:
To support the extra pulleys and maintain the structural stability while maximizing the mechanical advantage of the system.

Teacher's Note:
a) The upper block is fixed to a rigid support and houses the majority of the pulleys in standard configurations.
b) This arrangement helps in efficient rope routing and load distribution.

2.5. Sumit does 600 J of work in 10 min and Amit does 300 J of work in 20 min. Calculate the ratio of the powers delivered by them. [2 Marks]

Answer:
For Sumit: \( P_1 = \frac{W_1}{t_1} = \frac{600\text{ J}}{600\text{ s}} = 1\text{ W} \)
For Amit: \( P_2 = \frac{W_2}{t_2} = \frac{300\text{ J}}{1200\text{ s}} = 0.25\text{ W} \)
Ratio of powers \( \frac{P_1}{P_2} = \frac{1}{0.25} = 4:1 \).

Teacher's Note:
a) Always convert time into SI units (seconds) before calculating power.
b) Power is defined as work done per unit time.

2.6. 5 bulbs are connected in series in a room. One bulb is fused. It is removed and remaining 4 bulbs are again connected in series to the same circuit. What will be the effect on the following physical quantities?
(Increases, Decreases, Remain Same).
a. Resistance
b. Intensity of light [2 Marks]

Answer:
a. Decreases
b. Increases

Teacher's Note:
a) Total resistance in a series combination is the sum of individual resistances; removing one bulb reduces the total resistance.
b) Decreased resistance increases current, thereby increasing the brightness (intensity) of the remaining bulbs.

2.7. Rohan conducted experiments on echo in different media. He observed that a minimum distance of 'x' meters is required for the echo to be heard in oxygen and 'y' meters in benzene. Compare 'x' and 'y'. Justify your answer.
Speed of sound in oxygen: \( 340\text{ ms}^{-1} \)
Speed of sound in benzene: \( 200\text{ ms}^{-1} \) [2 Marks]

Answer:
\( x > y \).
Justification: Minimum distance required for an echo is directly proportional to the speed of sound in that medium (\( d = \frac{v \times t}{2} \)). Since the speed of sound in oxygen is greater than in benzene, the required distance 'x' is greater than 'y'.

Teacher's Note:
a) The minimum distance for an echo depends directly on the speed of sound in the medium, keeping the persistence of hearing time (0.1 s) constant.
b) State the formula \( d = \frac{v \times t}{2} \) clearly to earn full marks.

Q3.

3.1. (a) In a reading glass what is the position of the object with respect to the convex lens used? [1 Mark]

Answer:
The object is placed between the optical center (pole) and the principal focus of the convex lens.

Teacher's Note:
a) A reading glass uses a convex lens to form an enlarged, erect, and virtual image.
b) This requires the object to be within the focal length.

3.1. (b) In a reading glass what is the position of the object with respect to the concave lens used?
Why can't we use the concave lens to position the object in the reading glass? [1 Mark]

Answer:
A concave lens always forms a virtual, diminished image for any position of the object, so it cannot produce a magnified image required for reading.

Teacher's Note:
a) Concave lenses diverge light and never produce magnified upright images.
b) Hence, they are unsuitable for use as magnifying reading glasses.

3.2. A fuse is rated 5 A. Can it be used with a geyser rated 1540 W, 220 V Write Yes or No. Give supporting calculations to justify your answer. [2 Marks]

Answer:
No.
Current drawn by the geyser \( I = \frac{P}{V} = \frac{1540\text{ W}}{220\text{ V}} = 7\text{ A} \).
Since the current drawn (7 A) exceeds the fuse rating (5 A), the fuse will melt and blow.

Teacher's Note:
a) Calculate current using \( I = P / V \).
b) Conclude clearly that a 5 A fuse is inadequate for an appliance drawing 7 A.

3.3. State two factors affecting the coil's rotation speed in a D.C. motor. [2 Marks]

Answer:
1. Strength of the magnetic field.
2. Magnitude of current flowing through the coil.

Teacher's Note:
a) Other factors include the number of turns in the coil and the area of the coil.
b) Mention any two clear and scientifically accurate factors.

3.4. How much heat is required to convert 500 g of ice at 0°C to water at 0°C? The latent heat of fusion of ice is \( 330\text{ Jg}^{-1} \). [2 Marks]

Answer:
\( Q = m \times L \)
\( Q = 500\text{ g} \times 330\text{ Jg}^{-1} = 1,65,000\text{ J} = 165\text{ kJ} \).

Teacher's Note:
a) Use the formula \( Q = mL \) for phase change at constant temperature.
b) Convert joules to kilojoules if desired, stating units clearly.

3.5. Copy and complete the nuclear reaction by filling in the blanks.
\( {}_{92}U^{235} + {}_{0}n^{1} \rightarrow {}_{56}Ba^{\underline{\quad}} + {}_{\underline{\quad}}Kr^{92} + 3 {}_{0}n^{1} \) [2 Marks]

Answer:
\( {}_{92}U^{235} + {}_{0}n^{1} \rightarrow {}_{56}Ba^{141} + {}_{36}Kr^{92} + 3 {}_{0}n^{1} \)

Teacher's Note:
a) Balance both mass numbers and atomic numbers on both sides of the nuclear equation.
b) Mass number: \( 235 + 1 = 141 + 92 + 3(1) = 236 \); Atomic number: \( 92 = 56 + 36 \).

SECTION-B (40 Marks)

Attempt any four questions from this Section.

Q4.

4.1. The image of a candle flame placed at a distance of 36 cm from a spherical lens, is formed on a screen placed at a distance of 72 cm from the lens. Calculate the focal length of the lens and its power. [3 Marks]

Answer:
Object distance \( u = -36\text{ cm} \)
Image distance \( v = +72\text{ cm} \) (since it is formed on a screen, lens is convex)
Using lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
\( \frac{1}{f} = \frac{1}{72} - \frac{1}{-36} = \frac{1}{72} + \frac{1}{36} = \frac{1 + 2}{72} = \frac{3}{72} = \frac{1}{24} \)
Focal length \( f = +24\text{ cm} = +0.24\text{ m} \)
Power \( P = \frac{1}{f(\text{in meters})} = \frac{1}{0.24} = +4.17\text{ D} \).

Teacher's Note:
a) Apply sign conventions correctly (object distance is always negative, real image distance is positive for convex lens).
b) Power must be calculated with focal length expressed in meters.

4.2. Below is an incomplete table showing the arrangement of electromagnetic spectrum in the increasing order of their wavelength. Complete the table:
Gamma ray | X-ray | UV rays | Visible rays | Infrared | A | Radio waves
a. Identify the radiation A.
b. Name the radiation used to detect fracture in bones.
c. Name one property common to both A and Radio waves. [3 Marks]

Answer:
a. Microwaves.
b. X-rays.
c. Both travel at the speed of light in vacuum and do not require a material medium for propagation.

Teacher's Note:
a) Know the sequence of electromagnetic waves in order of increasing wavelength or decreasing frequency.
b) State common properties of EM waves clearly.

4.3. (a) Why do we use red colour as a danger signal on the top of a skyscraper? [2 Marks]

Answer:
Red light has the longest wavelength in the visible spectrum, so it suffers minimum scattering by air molecules and dust particles, allowing it to be seen clearly from long distances.

Teacher's Note:
a) Link the use of red color to its maximum wavelength and minimum scattering.
b) This ensures high visibility through fog, smoke, and haze.

4.3. (b) The diagram below shows the path of a blue ray through the prism:
1. Calculate the critical angle of the material of the prism for blue colour.
2. What is the measure of the angle of this prism (A)?
3. Which colour should replace the blue ray, for the ray to undergo Total Internal Reflection? [3 Marks]

[Figure: Right-angled prism with an incident ray striking normally on surface AB, refracting and striking face AC at an angle of incidence such that the angle with normal is calculated, emerging along AC.]

Answer:
1. Angle of incidence at face AC is \( 43^{\circ} \) (since ray grazes the surface at \( 90^{\circ} \)), so critical angle \( c = 43^{\circ} \).
2. By triangle geometry, the angle of the prism \( A = 43^{\circ} \).
3. A colour with a shorter wavelength and higher refractive index than blue, such as violet or indigo light, should be used to decrease the critical angle and achieve total internal reflection.

Teacher's Note:
a) Use Snell's law and geometry of the prism to find angles.
b) Shorter wavelengths have higher refractive indices and smaller critical angles.

Q5.

5.1.
a. Refractive index of glass with respect to water is \( \frac{9}{8} \). Find the refractive index of water with respect to glass.
b. Name the principle used to find the value in part (a).
c. If we change the temperature of water, then will the ratio \( \frac{9}{8} \) remain the same? Write Yes or No. [3 Marks]

Answer:
a. \( {}^{g}\mu_{w} = \frac{1}{{}^{w}\mu_{g}} = \frac{1}{\frac{9}{8}} = \frac{8}{9} \).
b. Principle of reversibility of light.
c. No.

Teacher's Note:
a) Use the reciprocal relation for refractive indices of two media.
b) Refractive index depends on temperature because density and optical properties change with temperature.

5.2. (a) Light travels a distance of '10x' units in time 't1' in vacuum and it travels a distance of 'x' units in time 't2' in a denser medium. Using this information answer the question that follows:
'Light covers a distance of '20x' units in time 't1' in diamond'. State true or false. [1 Mark]
1. True
2. False

Answer: (2) False

Light travels slower in a denser medium like diamond than in a vacuum, so it cannot cover a greater distance in the same time interval.

Teacher's Note:
a) Speed of light in any medium is always less than in vacuum.
b) Therefore, the distance covered in vacuum in time t1 is the maximum possible distance.

5.2. (b) Light travels a distance of '10x' units in time 't1' in vacuum and it travels a distance of 'x' units in time 't2' in a denser medium. Using this information answer the question that follows:
Calculate the refractive index of the medium in terms of 't1' and 't2'. [3 Marks]

Answer:
Speed of light in vacuum \( c = \frac{10x}{t_1} \)
Speed of light in medium \( v = \frac{x}{t_2} \)
Refractive index \( \mu = \frac{c}{v} = \frac{\frac{10x}{t_1}}{\frac{x}{t_2}} = \frac{10t_2}{t_1} \).

Teacher's Note:
a) Refractive index is the ratio of the speed of light in vacuum to the speed of light in the medium.
b) Substitute speed expressions carefully and simplify.

5.3. A monochromatic ray of light is incident on an equilateral prism placed at minimum deviation position with an angle of incidence 45° as shown in the diagram?
a. Copy the diagram and complete the path of the ray PQ.
b. State two factors on which the angle of deviation depends. [3 Marks]

[Figure: Equilateral prism with ray PQ incident at angle 45° with normal N, passing through the prism and emerging out.]

Answer:
a. [Diagram showing the ray refracting inside the prism parallel to the base and emerging symmetrically at the second face with angle of emergence equal to angle of incidence].
b. Factors on which angle of deviation depends:
1. Angle of incidence.
2. Material of the prism (refractive index).

(The official key also accepts angle of the prism and wavelength of light).

Teacher's Note:
a) At minimum deviation, the angle of incidence equals the angle of emergence.
b) Mention any two standard factors affecting deviation.

Q6.

6.1. (a) Define the centre of gravity of a body. [1 Mark]

Answer:
The centre of gravity of a body is the point through which the entire weight of the body is supposed to act, regardless of its orientation.

Teacher's Note:
a) Emphasize that it is a point where the resultant of all gravitational forces acting on particles of the body is concentrated.
b) Definition must be precise to get full marks.

6.1. (b) A hollow ice cream cone has height 6 cm.
1. Where is the position of its center of gravity from the broad base?
2. Will its position change when it is filled completely with honey? Write Yes or No. [2 Marks]

Answer:
1. Position of center of gravity from the broad base is \( \frac{h}{3} = \frac{6}{3} = 2\text{ cm} \).
2. Yes.

Teacher's Note:
a) Center of gravity of a hollow cone lies at a height of \( h/3 \) from its base.
b) Filling it with honey changes it to a solid cone, shifting the center of gravity to \( h/4 \) from the base.

6.2. Two identical marbles A and B are rolled down along Path 1 and Path 2 respectively.
Path 1 is frictionless and Path 2 is rough.
a. Which marble will surely reach the next peak?
b. Along which path/s the mechanical energy will be conserved?
c. Along which path/s is the law of conservation of energy obeyed? [3 Marks]

[Figure: Wave-like tracks Path 1 and Path 2 with marbles A and B rolling over peaks X, Y, Z.]

Answer:
a. Marble A.
b. Along Path 1.
c. Along both Path 1 and Path 2.

Teacher's Note:
a) Friction causes energy loss as heat, so marble B on the rough path may fail to reach the peak.
b) Total mechanical energy is conserved only in the absence of non-conservative forces like friction, but the universal law of conservation of energy holds true everywhere.

6.3. (a) Copy and complete the labelled diagram connecting the two pulleys with a tackle to obtain Velocity Ratio= 2. [2 Marks]

[Figure: Pulley arrangement with one fixed and one movable pulley.]

Answer:
[Diagram showing a block and tackle system with two pulleys - one in the fixed block and one in the movable block, connected by a single string with effort acting downwards].

Teacher's Note:
a) A velocity ratio of 2 requires a total of 2 pulleys in the system.
b) Draw arrows and label Load (L), Effort (E), and Tension (T) clearly.

6.3. (b) If Load = 48 kgf and efficiency is 80% then calculate:
1. Mechanical Advantage.
2. Effort needed to lift the load. [2 Marks]

Answer:
1. Efficiency \( \eta = \frac{\text{M.A.}}{\text{V.R.}} \Rightarrow \text{M.A.} = \eta \times \text{V.R.} = 0.8 \times 2 = 1.6 \).
2. Effort \( E = \frac{\text{Load}}{\text{M.A.}} = \frac{48\text{ kgf}}{1.6} = 30\text{ kgf} \).

Teacher's Note:
a) Use the relation between efficiency, mechanical advantage, and velocity ratio.
b) Substitute values carefully to compute the effort.

Q7.

7.1. (a) Name the waves used in SONAR. [1 Mark]

Answer:
Ultrasonic waves.

Teacher's Note:
a) Ultrasonic waves have frequencies above 20,000 Hz.
b) They are used because they can travel long distances without significant diffraction.

7.1. (b) In the above diagram Lata stands between two cliffs and claps her hands. Determine the time taken by her to hear the first echo. Speed of sound in air \( 320\text{ ms}^{-1} \). [3 Marks]

[Figure: Lata standing between Cliff A (10 m away) and Cliff B (160 m away).]

Answer:
The distance to Cliff A is 10 m, which is less than the minimum required distance of 17 m for an echo, so no audible echo is heard from Cliff A.
The distance to Cliff B is 160 m. Distance traveled for echo \( d = 2 \times 160 = 320\text{ m} \).
Time \( t = \frac{d}{v} = \frac{320\text{ m}}{320\text{ ms}^{-1}} = 1\text{ second} \).

Teacher's Note:
a) Check the minimum distance condition (17 m) for hearing an echo before calculating time.
b) Consider the reflection from the farther cliff since the closer one does not satisfy the condition.

7.2. (a) Complete the following radioactive reaction:
\( {}_{92}^{238}\text{X} \rightarrow {}_{\underline{\quad}}^{234}\text{Y} + {}_{2}^{4}\text{He} \rightarrow {}_{91}^{234}\text{Z} + {}_{\underline{\quad}}^{\underline{\quad}}\text{e} \) [2 Marks]

Answer:
\( {}_{92}^{238}\text{X} \rightarrow {}_{90}^{234}\text{Y} + {}_{2}^{4}\text{He} \rightarrow {}_{91}^{234}\text{Z} + {}_{-1}^{0}\text{e} \)

Teacher's Note:
a) Alpha decay reduces mass number by 4 and atomic number by 2.
b) Beta decay increases atomic number by 1 while mass number remains unchanged.

7.2. (b) Uranium is available in two forms U-235 and U-238. Which of the two isotopes of Uranium is more fissionable? [1 Mark]

Answer:
U-235 is more fissionable.

Teacher's Note:
a) U-235 undergoes fission easily with slow (thermal) neutrons.
b) U-238 requires fast neutrons for fission and is much less reactive.

7.3. In the given diagram, a vibrating tuning fork is kept near the mouth of a burette filled with water. The length of the air column is adjusted by opening the tap of the burette. At a length of 5 cm of the air column, a loud sound is heard
a. Name the phenomenon illustrated by the above experiment.
b. Why is a loud sound heard at this particular length?
c. If the present tuning fork is replaced with a tuning fork of higher frequency, should the length of the air column increase or decrease to produce a loud sound? Give a reason. [3 Marks]

[Figure: Burette filled with water and tuning fork placed at the open top end, showing air column length of 5 cm.]

Answer:
a. Resonance.
b. At this length, the natural frequency of the air column matches the frequency of the tuning fork, causing resonance.
c. The length should decrease. Reason: Frequency is inversely proportional to the length of the air column (\( f \propto 1 / l \)), so a higher frequency requires a shorter air column.

Teacher's Note:
a) Resonance occurs when the forcing frequency equals the natural frequency of the vibrating system.
b) State the inverse relation between frequency and length of the air column clearly.

Q8.

8.1. The voltage - current readings of a certain material are shown in the table given below:
Voltage (V): 10 V | 20 V | 30 V
Current (I): 2 A | 3 A | 4 A
Study the table.
a. State whether the conductor used is ohmic or non-ohmic.
b. Justify your answer.
c. State Ohm's law. [3 Marks]

Answer:
a. Non-ohmic.
b. The ratio \( V / I \) is not constant (\( 10/2 = 5\ \Omega \), \( 20/3 = 6.67\ \Omega \), \( 30/4 = 7.5\ \Omega \)).
c. Ohm's Law: At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends.

Teacher's Note:
a) Check if the resistance (\( V/I \)) remains constant across all readings to determine ohmic nature.
b) State Ohm's law with the mandatory condition of constant temperature.

8.2. Below is the diagram of a transformer:
a. Identify the type of transformer.
b. In this type of transformer which of the wire is thicker, the primary or the secondary? Give a reason. [2 Marks]

[Figure: Step-down transformer with more turns in primary coil and fewer turns in secondary coil wound on an iron core.]

Answer:
a. Step-down transformer.
b. The secondary coil is thicker because it carries a higher current compared to the primary coil to handle lower voltage.

Teacher's Note:
a) Identify step-down transformer by observing fewer turns in the secondary coil.
b) Thicker wire is needed for carrying larger currents safely without overheating.

8.3. Study the diagram:
a. Calculate the total resistance of the circuit.
b. Calculate the current drawn from the cell.
c. State whether the current through \( 10\ \Omega \) resistor is greater than, less than or equal to the current through the \( 12\ \Omega \) resistor. [3 Marks]

[Figure: Circuit diagram with two parallel arms; upper arm has \( 10\ \Omega \) and \( 6\ \Omega \) in series; lower arm has \( 12\ \Omega \) and \( 4\ \Omega \) in series connected across a 4 V cell.]

Answer:
a. Upper branch resistance \( R_1 = 10 + 6 = 16\ \Omega \)
Lower branch resistance \( R_2 = 12 + 4 = 16\ \Omega \)
Total equivalent resistance \( R_{eq} = \frac{16 \times 16}{16 + 16} = 8\ \Omega \)
b. Current drawn \( I = \frac{V}{R_{eq}} = \frac{4\text{ V}}{8\ \Omega} = 0.5\text{ A} \)
c. Equal to, because both parallel branches have the exact same total resistance (\( 16\ \Omega \)), dividing the total current equally.

Teacher's Note:
a) Calculate series combinations first for each branch, then combine them in parallel.
b) Since branch resistances are equal, current divides equally between them.

Q9.

9.1. 85 g of water at 30°C is cooled to 5°C by adding certain mass of ice. Find the mass of ice required.
[Specific heat capacity of water = \( 4.2\text{ Jg}^{-1}\text{°C}^{-1} \), Specific latent heat of fusion = \( 336\text{ Jg}^{-1} \)] [3 Marks]

Answer:
Heat lost by water \( Q_{\text{lost}} = m_w \times c_w \times \Delta t = 85 \times 4.2 \times (30 - 5) = 85 \times 4.2 \times 25 = 8,925\text{ J} \)
Heat gained by ice \( Q_{\text{gained}} = m_i \times L + m_i \times c_w \times \Delta t_i = m_i \times 336 + m_i \times 4.2 \times (5 - 0) = 336m_i + 21m_i = 357m_i \)
By principle of calorimetry: \( 357m_i = 8925 \Rightarrow m_i = \frac{8925}{357} = 25\text{ g} \).

Teacher's Note:
a) Equate heat lost by hot water to heat gained by ice (melting + raising temperature of melted ice from 0°C to 5°C).
b) Solve carefully for the mass of ice \( m_i \).

9.2. (a) Why does it become pleasantly warm when the lakes start freezing? [2 Marks]

Answer:
When water freezes into ice, it releases a large amount of latent heat (latent heat of fusion) into the surroundings. This exothermic process warms up the atmosphere around the lake.

Teacher's Note:
a) Freezing is an exothermic process where latent heat is liberated.
b) This explains why the temperature does not drop drastically immediately when freezing starts.

9.2. (b) Water freezes to form ice. What change would you expect in the average kinetic energy of the molecules? [1 Mark]

Answer:
The average kinetic energy of the molecules decreases as temperature drops during freezing.

Teacher's Note:
a) Kinetic energy is directly proportional to absolute temperature.
b) As water changes to ice at 0°C, molecular motion slows down.

9.3. (c) Which has more heat: 1 g ice at 0°C or 1 g water at 0°C? Give reason. [1 Mark]

Answer:
1 g of water at 0°C has more heat energy because it absorbs an additional amount of latent heat (336 J) during melting compared to ice at the same temperature.

Teacher's Note:
a) Water at 0°C possesses extra latent heat of fusion.
b) This makes water at 0°C more effective at cooling or holding thermal energy than ice at 0°C.

9.3. (a) State one factor that affects the magnitude of induced current in an AC generator. [1 Mark]

Answer:
Number of turns in the armature coil.

Teacher's Note:
a) Other factors include magnetic field strength and speed of rotation.
b) State any one valid factor clearly.

9.3. (b) Given below is a circuit to study the magnetic effect of electric current. ABCD is a cardboard kept perpendicular to the conductor XY. A magnetic compass is placed at the point P of the cardboard. \( P_1 \) and \( P_2 \) are the positions of the magnetic compass, before and after passing a current through XY respectively.
1. Name the rule that is used to predict the direction of deflection of the magnetic compass.
2. State the direction of current in the conductor (X to Y or Y to X) when the circuit is complete.
3. If resistance R is increased, then what will be the effect on the magnetic lines of force around the conductor? [3 Marks]

[Figure: Circuit with DC supply, rheostat R, vertical wire XY passing through horizontal cardboard ABCD, and magnetic compass.]

Answer:
1. Right-hand thumb rule.
2. From X to Y.
3. The magnetic field lines will become less dense because increasing resistance decreases the current flowing through the conductor.

Teacher's Note:
a) Use the right-hand thumb rule to correlate current direction with magnetic field orientation.
b) Magnetic field strength is directly proportional to current.