ICSE Class 10 Physics Board Exam Question Paper 2025 with Solutions

ICSE Class 10 Physics Board Exam Question Paper with Solutions

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

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

(i) A body is acted upon by two equal and opposite forces, that are NOT along the same straight line. The body will: [1 Mark]
(A) remain stationary
(B) have only rotational motion
(C) have only rectilinear motion
(D) have both rectilinear and rotational motion

Answer: (B) have only rotational motion

Equal and opposite forces not acting along the same straight line constitute a couple, which produces purely rotational motion without any translatory motion.

Teacher's Note:
a) Remember that a couple causes rotation about a fixed axis or center.
b) Students often confuse a couple with balanced forces that result in equilibrium.

(ii) Which among the following is a vector quantity? [1 Mark]
(A) work
(B) power
(C) energy
(D) moment of couple

Answer: (D) moment of couple

Moment of a couple (or torque) has both magnitude and a specific rotational direction, making it a vector quantity.

Teacher's Note:
a) Work, power, and energy are scalar quantities.
b) Always check if a physical quantity requires a direction for its complete specification.

(iii) What is the correct energy transformation during burning of a candle? [1 Mark]
(A) heat \(\rightarrow\) kinetic + potential
(B) heat \(\rightarrow\) chemical + light
(C) chemical \(\rightarrow\) heat + light
(D) mechanical \(\rightarrow\) chemical + heat

Answer: (C) chemical \(\rightarrow\) heat + light

Candle wax stores chemical energy which upon burning releases heat energy and light energy.

Teacher's Note:
a) Combustion reactions convert stored chemical energy into thermal and luminous energy.
b) Ensure students do not confuse the source of energy with the products.

(iv) When a ray of light passes from one optical medium to another, which of the following physical quantities does NOT change? [1 Mark]
(A) Amplitude of the wave
(B) Frequency of the wave
(C) Wavelength of the wave
(D) Speed of the wave

Answer: (B) Frequency of the wave

Frequency is a characteristic of the source and remains unchanged when light travels from one medium to another.

Teacher's Note:
a) Speed and wavelength change when light crosses an optical boundary.
b) Frequency depends entirely on the source emitting the light wave.

(v) Which one of the following combinations is the correct ascending order of electromagnetic waves in terms of wavelength? [1 Mark]
(A) Gamma-rays, visible light, microwaves
(B) Microwaves, visible light, gamma-rays
(C) Gamma-rays, microwaves, visible light
(D) Microwaves, gamma-rays, visible light

Answer: (A) Gamma-rays, visible light, microwaves

Gamma-rays have the shortest wavelength, followed by visible light, and microwaves have much longer wavelengths.

Teacher's Note:
a) Memorize the full electromagnetic spectrum in order of increasing wavelength or frequency.
b) Gamma-rays have wavelengths less than \(0.01 \text{ nm}\), while microwaves range from \(1 \text{ mm}\) to \(1 \text{ m}\).

(vi) For a lever, a graph is plotted with load on Y-axis and effort on X-axis. Which of the following represents the slope of the graph? [1 Mark]
(A) Mechanical advantage
(B) Velocity ratio
(C) \(\frac{1}{\text{Velocity ratio}}\)
(D) \(\frac{1}{\text{Mechanical advantage}}\)

Answer: (A) Mechanical advantage

Slope = \(\frac{\text{Y-axis}}{\text{X-axis}} = \frac{\text{Load}}{\text{Effort}} = \text{Mechanical Advantage}\).

Teacher's Note:
a) Slope is always change in Y divided by change in X.
b) Mechanical advantage is defined as Load divided by Effort.

(vii) For a real image formed by a convex lens, the ratio of \(I : O = 2 : 5\), then the object is: [1 Mark]
(\(I\) is the height of the image and \(O\) is the height of the object)
(A) between \(O\) and \(F\)
(B) beyond \(2F\)
(C) at \(F\)
(D) between \(F\) and \(2F\)

Answer: (B) beyond \(2F\)

Since \(I \lt O\) (\(2 \lt 5\)), the image is diminished. A convex lens forms a real and diminished image only when the object is placed beyond \(2F\).

Teacher's Note:
a) Magnification \(m = \frac{I}{O} = \frac{2}{5} = 0.4\) (less than 1).
b) Magnification is less than 1 when the object is beyond \(2F_1\).

(viii) A ray of light is incident normally on a face of an equilateral prism. The ray gets totally reflected at the second refracting surface. The total deviation produced in the path of the ray is: [1 Mark]
(A) \(30^{\circ}\)
(B) \(60^{\circ}\)
(C) \(90^{\circ}\)
(D) \(120^{\circ}\)

Answer: (D) \(120^{\circ}\)

The angle of incidence on the second face is \(60^{\circ}\). Total deviation \(\delta = \delta_1 + \delta_2 = 30^{\circ} + 90^{\circ} = 120^{\circ}\).

Teacher's Note:
a) Trace the ray carefully through each face of the equilateral prism.
b) Sum up the individual deviations at each refracting surface.

(ix) In a closed circuit containing a bulb and a cell, the electromotive force (\(\varepsilon\)) and the terminal voltage (\(V\)) is related as. [1 Mark]
(Given \(I\) is current and \(r\) is internal resistance.)
(A) \(V = \varepsilon + Ir$
(B) \(V = \varepsilon - Ir$
(C) \(V = \varepsilon \div Ir$
(D) \(V = \varepsilon \times Ir$

Answer: (B) \(V = \varepsilon - Ir\)

When a cell supplies current, there is a potential drop across its internal resistance, hence \(V = \varepsilon - Ir\).

Teacher's Note:
a) Use \(V = \varepsilon + Ir\) only during charging of a cell.
b) For a discharging cell in a closed circuit, terminal voltage is less than emf by \(Ir\).

(x) A metal piece of mass \(5 \text{ g}\) has thermal capacity \(2.5 \text{ J K}^{-1}\). If the mass of the metal is tripled, then its specific heat capacity will be: [1 Mark]
(A) \(7.5 \text{ J K}^{-1}$
(B) \(2.5 \text{ J K}^{-1}$
(C) \(1.5 \text{ J g}^{-1} \text{K}^{-1}$
(D) \(0.5 \text{ J g}^{-1} \text{K}^{-1}\)

Answer: (D) \(0.5 \text{ J g}^{-1} \text{K}^{-1}\)

Specific heat capacity is a characteristic property of a substance and does not depend on its mass. Initial specific heat capacity \(= \frac{2.5}{5} = 0.5 \text{ J g}^{-1}\text{K}^{-1}\).

Teacher's Note:
a) Thermal capacity depends on mass, but specific heat capacity is independent of mass.
b) Always check the units provided in the options.

(xi) Assertion (A): As the level of water in a tall measuring cylinder kept under running tap rises, the pitch of sound gradually increases.
Reason (R): Frequency of sound is inversely proportional to the length of the water column. [1 Mark]
(A) Both (A) and (R) are true and (R) is correct explanation of (A).
(B) Both (A) and (R) are true and (R) is not the correct explanation of (A).
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.

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

As water level rises, air column length decreases, increasing frequency and pitch. Thus (R) is false because frequency is inversely proportional to length of air column, not water column.

Teacher's Note:
a) Pitch depends on the vibrating air column length in a measuring cylinder.
b) As air column length decreases, frequency increases.

(xii) In the given circuits Y and Z, the resistors, \(R_1\) and \(R_2\), are connected in: [1 Mark]

[Figure: Two circuits Y and Z. Circuit Y shows two parallel branches with a cell and resistors. Circuit Z shows two resistors in a parallel loop connected to a cell.]

(A) series in both the circuits
(B) parallel in both the circuits
(C) parallel in Y and series in Z
(D) series in Y and parallel in Z

Answer: (B) parallel in both the circuits

Tracing the connections in both circuits Y and Z reveals that both resistors have their corresponding ends connected together across the same two common points.

Teacher's Note:
a) Redraw complex-looking schematics to identify parallel or series combinations easily.
b) Both circuits represent standard parallel connections of two resistors.

(xiii) A radioactive element \(P\) emits one \(\alpha\)-particle and transforms to a new element \(Q\). What will be the position of the element \(Q\) in the periodic table? [1 Mark]
(A) One group to the left of \(P$
(B) One group to the right of \(P$
(C) Two groups to the right of \(P$
(D) Two groups to the left of \(P$

Answer: (D) Two groups to the left of \(P\)

Emission of an \(\alpha\)-particle decreases the atomic number by \(2\), shifting the element two groups to the left in the periodic table.

Teacher's Note:
a) An alpha particle is a helium nucleus (\(^4_2\text{He}\)).
b) Atomic number decreases by 2 during alpha decay.

(xiv) Each of the substances given below is supplied with same amount of heat. Which one will attain the highest temperature? [1 Mark]

(A) Aluminium
(B) Copper
(C) Iron
(D) Lead

Answer: (D) Lead

Rise in temperature \(\Delta t = \frac{Q}{m c}\). The substance with the lowest specific heat capacity \(c\) will attain the highest temperature.

Teacher's Note:
a) Lower specific heat capacity means smaller heat requirement for unit temperature rise.
b) Lead has the lowest specific heat capacity among the given choices.

(xv) The following figure shows a small bar magnet falling freely through a copper ring. For the observer at A, the direction of the induced current will be: [1 Mark]

[Figure: A bar magnet with north pole pointing down falling through a horizontal copper ring, viewed by an observer A from above.]

(A) clockwise when magnet is above and below the ring
(B) anticlockwise when magnet is above and below the ring
(C) anticlockwise when magnet is above the ring and clockwise when the magnet is below the ring
(D) clockwise when magnet is above the ring and anticlockwise when the magnet is below the ring

Answer: (C) anticlockwise when magnet is above the ring and clockwise when the magnet is below the ring

When the north pole approaches from above, the top face acquires a north polarity (anticlockwise current). When it recedes below, the top face acquires a south polarity (clockwise current).

Teacher's Note:
a) Apply Lenz's law: induced current opposes the motion of the magnet.
b) Opposition means repulsion on approach and attraction on recession.

Question 2

(i) Complete the following by choosing the correct answers from the bracket: [6 Marks]

(a) In uniform circular motion the centrifugal force acts ________ (towards the centre/away from the centre/along the tangential direction). [1 Mark]

Answer: away from the centre.

Teacher's Note:
a) Centrifugal force is a pseudo force acting outwards in a rotating frame.
b) Centripetal force acts towards the centre.

(b) Refractive index of a medium is independent of ________ (temperature/angle of incidence/wavelength of light). [1 Mark]

Answer: angle of incidence.

Teacher's Note:
a) Refractive index depends on temperature and wavelength (color) of light.
b) It is constant for a given pair of media and wavelength regardless of incidence angle.

(c) Heat absorbed during change of phase depends on ________ (mass/change in temperature/specific heat capacity) of the substance. [1 Mark]

Answer: mass.

Teacher's Note:
a) Latent heat formula is \(Q = m L\).
b) Temperature remains constant during phase change.

(d) Emf of a cell is ________ (greater than / less than / equal to) the terminal voltage when the cell is in open circuit. [1 Mark]

Answer: greater than.

Teacher's Note:
a) In open circuit, current is zero, so terminal voltage equals emf (\(V = \varepsilon\)).
b) Under closed circuit load, terminal voltage is less than emf.

(e) In a step-up transformer the turns ratio is ________ (more than 1/ less than 1/ equal to 1). [1 Mark]

Answer: more than 1.

Teacher's Note:
a) Turns ratio \(N_s / N_p\) is greater than 1 for a step-up transformer.
b) Secondary turns exceed primary turns in a step-up transformer.

(f) The nuclear radiation with lowest ionising power is ________ (\(\alpha / \beta / \gamma\)). [1 Mark]

Answer: \(\gamma\)

Teacher's Note:
a) Gamma rays are high-frequency electromagnetic waves with lowest ionizing power.
b) Alpha particles possess the highest ionizing power.

(ii) A non-uniform kite is hanging freely from the branch of a tree as shown. Study the figure and answer the following: [2 Marks]

[Figure: A kite hanging from a branch of a tree with marked points P, Q, R, S along its vertical suspension line.]

(a) Fill in the blank.
________ (P, Q, R or S) is the most probable position of its centre of gravity. [1 Mark]

Answer: Q

Teacher's Note:
a) The center of gravity lies vertically below the point of suspension.
b) Point Q lies on the line of suspension within the body of the non-uniform lamina.

(b) Support your answer to (a) with a reason. [1 Mark]

Answer: When a body is freely suspended, its center of gravity always lies vertically below the point of suspension.

Teacher's Note:
a) Plumb line method locates the center of gravity.
b) Point Q represents this condition.

(iii) The displacement-time graph of a sound wave produced by a vibrating wire is shown below. [2 Marks]

[Figure: A sinusoidal displacement-time graph with peaks labeled P, Q, R, S.]

(a) How will you adjust the tension in the wire, to reduce the length of PR? [1 Mark]

Answer: Increase the tension in the wire.

Teacher's Note:
a) Reducing the length of PR means decreasing the time period, which increases frequency.
b) Frequency is directly proportional to the square root of tension.

(b) Which characteristic of sound is affected by the reduction in the length of PR? [1 Mark]

Answer: Pitch (or frequency).

Teacher's Note:
a) Pitch depends directly on frequency.
b) Shorter time period implies higher frequency and higher pitch.

Question 3

(i) A ray of light enters a rectangular glass slab submerged in water at an angle of incidence \(55^{\circ}\). Does this ray undergo total internal reflection when it moves from water to glass? Justify your answer. (The critical angle for glass-water interface is \(54^{\circ}\)) [2 Marks]

Answer: No, the ray will not undergo total internal reflection.
For total internal reflection, the light ray must travel from denser to rarer medium. Here, light travels from water (rarer) to glass (denser), so total internal reflection is not possible.

Teacher's Note:
a) Two essential conditions for total internal reflection are light travelling from denser to rarer medium and angle of incidence greater than critical angle.
b) Water is optically rarer compared to glass.

(ii) According to the NEW colour convention which colour of wire is connected to: [2 Marks]
(a) the metal body of the appliance [1 Mark]

Answer: Green or yellow (Earth wire).

Teacher's Note:
a) Earth wire provides a safe low-resistance path for leakage current.
b) New convention uses green or yellow.

(b) the switch of the appliance? [1 Mark]

Answer: Brown (Live wire).

Teacher's Note:
a) Switches are always connected to the live wire for safety.
b) Brown is the new colour code for live wire.

(iii) (a) Which of the two, alternating current or direct current, produces a varying magnetic field when it flows through a conductor? [1 Mark]

Answer: Alternating current (AC).

Teacher's Note:
a) Alternating current changes its magnitude and direction periodically.
b) This variation produces a changing magnetic field.

(b) State the frequency of the alternating current supply in India. [1 Mark]

Answer: \(50 \text{ Hz}\).

Teacher's Note:
a) Standard domestic frequency in India is \(50 \text{ Hz}\).
b) This means the current reverses direction \(100\) times per second.

(iv) Calculate the amount of heat absorbed by \(200 \text{ g}\) of paraffin wax to melt completely at its melting point. [Specific latent heat of fusion of paraffin wax \(= 146 \text{ J g}^{-1}\)] [2 Marks]

Answer:
\(Q = m \times L = 200 \text{ g} \times 146 \text{ J g}^{-1} = 29,200 \text{ J}\).

Teacher's Note:
a) Use formula \(Q = m L\) for phase change without temperature variation.
b) Ensure correct units are substituted.

(v) Copper wire is wound around a steel bar FT. Current is allowed to pass through the coil for some time and then the bar is removed. [2 Marks]
(a) Draw only the magnetised bar FT and mark its poles. [1 Mark]

Answer:
Bar FT with North Pole marked at F and South Pole marked at T (depending on current direction shown in the diagram).

Teacher's Note:
a) Steel retains magnetism to form a permanent magnet.
b) Use clock face rule or right-hand thumb rule to determine poles.

(b) Trace two magnetic lines of force around FT clearly indicating the direction. [1 Mark]

[Figure: Steel bar FT with closed magnetic field lines emerging from North and entering South outside the bar.]

Answer: Magnetic field lines emerge from the North pole and enter the South pole outside the magnet, directed from North to South.

Teacher's Note:
a) Field lines form continuous closed loops.
b) Arrowheads must indicate direction from N to S externally.

(vi) A current flows through a metallic conductor for a long period of time. State the change you would expect in its: [2 Marks]
(a) Resistance [1 Mark]

Answer: Resistance increases.

Teacher's Note:
a) Current flowing for a long time causes heating (\(I^2 Rt\)).
b) Rise in temperature increases the resistance of metallic conductors.

(b) Resistivity [1 Mark]

Answer: Resistivity increases.

Teacher's Note:
a) Resistivity of a metallic conductor is directly proportional to temperature.
b) As temperature rises due to heating, resistivity increases.

(vii) Curium is a radioactive element with the symbol \(^{247}_{96}\text{Cm}\) named in honour of Madam Curie. The graph of number of protons vs number of neutrons for some elements are shown below: [3 Marks]

[Figure: A graph plotting number of protons (Y-axis from 92 to 98) against number of neutrons (X-axis from 146 to 154), showing plotted points P, Q, R, S.]

(a) Which point on the graph indicates the element Cm? [1 Mark]

Answer: Point P.

Teacher's Note:
a) Atomic number of Curium is \(96\). Number of neutrons \(= 247 - 96 = 151\).
b) Point P corresponds to 96 protons and 151 neutrons.

(b) Which point on the graph indicates daughter nucleus after Cm undergoes radioactive decay of \(1\alpha\) followed by \(2\beta\)? [1 Mark]

Answer: Point R.

Teacher's Note:
a) Alpha decay decreases proton number by 2 and neutron number by 2. Beta decay increases proton number by 1 and decreases neutron number by 1.\br />b) Net change for \(1\alpha\) and \(2\beta\) leaves proton number unchanged at 96 and neutrons reduced by 4 (\(151 - 4 = 147\)), which corresponds to point R.

(c) State the mass number of the daughter nucleus. [1 Mark]

Answer: \(243\).

Teacher's Note:
a) Mass number = Protons + Neutrons \(= 96 + 147 = 243\).
b) Alternatively, initial mass number \(247\) minus \(4\) (due to alpha particle) equals \(243\).

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

Question 4

(i) Out of the three rays (I, J, H) shown in the diagram, which ray will suffer Total Internal Reflection while inside the prism? (Critical angle of the prism is \(42^{\circ}\).) [3 Marks]

[Figure: A triangular prism with incident rays I, J, H striking different faces at various angles of incidence.]

(a) Out of the three rays (I, J, H) shown in the diagram, which ray will suffer Total Internal Reflection while inside the prism? (Critical angle of the prism is \(42^{\circ}\).) [1 Mark]

Answer: Ray J.

Teacher's Note:
a) Calculate the angle of incidence at the inner face for each ray.
b) Ray J strikes at an angle greater than the critical angle (\(42^{\circ}\)).

(b) Copy the diagram to complete the path of the ray which you have named in (a) till it comes out of the prism. [2 Marks]

Answer:
The ray J reflects internally according to the laws of reflection and emerges out of the second face normally or at an angle less than critical angle.

Teacher's Note:
a) Ensure angle of incidence equals angle of reflection at the internal boundary.
b) Draw normal correctly at the point of emergence.

(ii) A rectangular glass block of refractive index \(1.5\) has an air bubble trapped inside it as shown in the diagram. When seen from the surface AB, it appears to be \(4 \text{ cm}\) deep. [3 Marks]

[Figure: A rectangular glass block of thickness \(15 \text{ cm}\) with an air bubble inside, viewed from top surface AB.]

(a) Calculate the actual depth of the air bubble from the surface AB. [2 Marks]

Answer:
Refractive index \(n = \frac{\text{Real Depth}}{\text{Apparent Depth}}\)
\(1.5 = \frac{\text{Real Depth}}{4 \text{ cm}}$
Real Depth \(= 1.5 \times 4 \text{ cm} = 6 \text{ cm}\).

Teacher's Note:
a) Use the standard relation between real depth, apparent depth, and refractive index.
b) Units must be written clearly in the final step.

(b) For which colour of light, blue or yellow, the apparent depth will be greater? [1 Mark]

Answer: Yellow light.

Teacher's Note:
a) Refractive index decreases with an increase in wavelength (\(\mu_{\text{yellow}} \lt \mu_{\text{blue}}\)).
b) Since apparent depth is inversely proportional to refractive index, apparent depth is greater for yellow light.

(c) Turning the glass block upside down, DOES NOT change the apparent depth of the air bubble. State True or False. [1 Mark]

Answer: False.

Teacher's Note:
a) Turning the block upside down changes the real depth from surface AB (from \(6 \text{ cm}\) to \(15 - 6 = 9 \text{ cm}\)).
b) Consequently, the apparent depth will change.

(iii) An object is placed at \(2F\) position of a convex lens. Draw a ray diagram showing the formation of the image. [4 Marks]

[Figure: Ray diagram showing an object at \(2F_1\) of a convex lens producing a real, inverted, and equal-sized image at \(2F_2\).]

(a) An object is placed at \(2F\) position of a convex lens. Draw a ray diagram showing the formation of the image. [2 Marks]

Answer:
1. Draw principal axis, optical center \(O\), foci \(F_1\), \(2F_1\) and \(F_2\), \(2F_2\).
2. Place object at \(2F_1\).
3. Draw a ray parallel to principal axis passing through \(F_2\) after refraction.
4. Draw another ray passing through optical center \(O\) without deviation to intersect at \(2F_2\).

Teacher's Note:
a) Image formed is real, inverted, and of the same size as the object.
b) Arrowheads on rays are mandatory.

(b) How will the size of the image change if we, ONLY replace the lens in the above arrangement with another lens of a greater focal length? [2 Marks]

Answer:
Size of the image remains unchanged.
Magnification depends on the ratio of image distance to object distance (\(m = v / u\)), and for an object at \(2F\), \(v = u\), making magnification \(-1\) regardless of focal length.

Teacher's Note:
a) Focal length does not alter the size ratio when the object is at \(2F\).
b) The image remains equal in size to the object.

Question 5

(i) An object is placed in front of a concave lens at a distance of \(45 \text{ cm}\) from it. If its image is formed at a distance of \(30 \text{ cm}\) from the lens, calculate the focal length of the lens. [3 Marks]

Answer:
Object distance \(u = -45 \text{ cm}\)
Image distance \(v = -30 \text{ cm}\)
Using lens formula:
\(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{-30} - \frac{1}{-45} = \frac{1}{f}\)
\(-\frac{1}{30} + \frac{1}{45} = \frac{1}{f}\)
\(\frac{-3 + 2}{90} = \frac{1}{f}\)
\(-\frac{1}{90} = \frac{1}{f}\)
\(f = -90 \text{ cm}\).

Teacher's Note:
a) Sign convention: both \(u\) and \(v\) are negative for a concave lens forming a virtual image on the same side.
b) Focal length of a concave lens is always negative.

(ii) Two rays PQ and RS are incident on a rectangular glass block as shown in the diagram. Observe the diagram and answer the questions that follow. [3 Marks]

[Figure: Two parallel light rays PQ and RS incident at different points on a glass block with corresponding lateral shifts \(x_1\) and \(x_2\).]

(a) Which of these two rays will have greater lateral displacement on emerging out of the block? [1 Mark]

Answer: Ray PQ (\(x_2 \gt x_1\)).

Teacher's Note:
a) Lateral displacement depends on angle of incidence and thickness.
b) Ray PQ bends less towards normal due to its specific geometry, resulting in greater lateral shift.

(b) travel with greater speed in the block? [1 Mark]

Answer: Both travel with the same speed.

Teacher's Note:
a) Speed of light in a given medium depends only on the refractive index of the medium.
b) Since both rays are in the same glass block, their speeds are equal.

(c) scatter more in the atmosphere? [1 Mark]

Answer: Ray corresponding to PQ (shorter wavelength / higher frequency).

Teacher's Note:
a) Rayleigh scattering is inversely proportional to the fourth power of wavelength.
b) Shorter wavelength scatters more.

(iii) (a) Name the radiations: [3 Marks]
1. for which a quartz prism is used to study the spectrum.
2. which are used in remote sensing devices.
3. which are used in traffic signals in India.

Answer:
1. Ultraviolet radiations.
2. Infrared radiations.
3. Visible light (Red light).

Teacher's Note:
a) Quartz absorbs ordinary glass UV rays, hence quartz prisms are used.
b) Infrared radiations penetrate haze well for remote sensing.

(b) Name one property common to all electromagnetic radiations. [1 Mark]

Answer: All electromagnetic radiations travel with the same speed in vacuum (\(3 \times 10^8 \text{ m s}^{-1}\)).

Teacher's Note:
a) They do not require a material medium for propagation.
b) They are transverse waves.

Question 6

(i) Akash takes a uniform metre scale and suspends a weight of \(2 \text{ N}\) at one end 'X', and a weight of \(5 \text{ N}\) on the other end 'Y'. He then balances the ruler horizontally on a knife edge placed at \(70 \text{ cm}\) from X. Draw a diagram of the arrangement and calculate the weight of the ruler. [3 Marks]

[Figure: A metre scale pivoted at 70 cm mark from X, with 2 N at X (0 cm), weight W of ruler acting at 50 cm, and 5 N at Y (100 cm).]

Answer:
1. Draw a metre scale with fulcrum at \(70 \text{ cm}\).
2. Clockwise moment about fulcrum \(= 5 \text{ N} \times (100 - 70) \text{ cm} = 5 \times 30 = 150 \text{ N cm}\).
3. Anti-clockwise moments about fulcrum \(= (2 \text{ N} \times (70 - 0)) + (W \times (70 - 50)) = (2 \times 70) + (W \times 20) = 140 + 20W\).
4. By Principle of Moments: Clockwise moment = Anti-clockwise moment
\(150 = 140 + 20W\)
\(20W = 10\)
\(W = 0.5 \text{ N}\).

Teacher's Note:
a) Weight of a uniform metre scale acts at its centre of gravity (\(50 \text{ cm}\) mark).
b) Equate total clockwise torque to total anticlockwise torque about the pivot.

(ii) Three levers X, Y, Z of equal lengths are shown in the diagram. [3 Marks]

[Figure: Three levers X, Y, Z showing different positions of fulcrum, load, and effort.]

(a) Which class of lever do these belong to? [1 Mark]

Answer: Class I lever.

Teacher's Note:
a) Fulcrum lies between load and effort in all three levers.
b) This is the defining characteristic of a Class I lever.

(b) Among these (X, Y or Z) which one will give the maximum mechanical advantage? Justify your answer. [2 Marks]

Answer: Lever X.
Mechanical advantage \(= \frac{\text{Effort arm}}{\text{Load arm}}\). Lever X has the largest effort arm compared to its load arm, giving the maximum mechanical advantage.

Teacher's Note:
a) MA can be greater than 1, equal to 1, or less than 1 for Class I levers.
b) Longer effort arm increases mechanical advantage.

(iii) Richa weighing \(40 \text{ kgf}\) leaves point P on her skateboard and reaches point Q on the ground with velocity \(10 \text{ m s}^{-1}\). Calculate. [4 Marks]
(a) the kinetic energy of Richa at point Q [1 Mark]

Answer:
Mass \(m = 40 \text{ kg}\), Velocity \(v = 10 \text{ m s}^{-1}\)
Kinetic Energy \(= \frac{1}{2} m v^2 = \frac{1}{2} \times 40 \times (10)^2 = 20 \times 100 = 2000 \text{ J}\).

Teacher's Note:
a) Use mass in kg, not weight in kgf, for kinetic energy calculation (\(m = 40 \text{ kg}\)).
b) Formula is \(KE = \frac{1}{2}mv^2\).

(b) the vertical height of point P above the ground. (use \(g\) as \(10 \text{ m/s}^2\) and neglect friction) [2 Marks]

Answer:
Total mechanical energy is conserved.
Potential Energy at P = Kinetic Energy at Q
\(m g h = 2000 \text{ J}\)
\(40 \times 10 \times h = 2000$
\(400 h = 2000$
\(h = 5 \text{ m}\).

Teacher's Note:
a) Apply principle of conservation of mechanical energy.
b) Gravitational potential energy at top equals kinetic energy at bottom.

(c) the kinetic energy of Richa at point R. (While moving from Q to R, she loses \(500 \text{ J}\) of energy against friction.) [1 Mark]

Answer:
Energy at R = Kinetic Energy at Q - Energy lost against friction
\(= 2000 \text{ J} - 500 \text{ J} = 1500 \text{ J}\).

Teacher's Note:
a) Account for energy dissipated against non-conservative forces like friction.
b) Subtract energy loss from initial mechanical energy.

Question 7

(i) Draw a block and tackle system of pulleys with velocity ratio equal to 3. [3 Marks]

[Figure: Block and tackle pulley system with 2 pulleys in the upper fixed block and 1 pulley in the lower movable block, total 3 pulleys, with string threaded accordingly.]

Answer:
1. Draw a block with 2 pulleys fixed at the top and 1 pulley in the movable block at the bottom (total 3 pulleys).
2. Pass the string starting from the hook of the upper block or lower block around all pulleys such that number of strands supporting the movable block is 3.
3. Show effort acting downwards and load acting downwards from the lower block.

Teacher's Note:
a) Velocity ratio equals the total number of pulleys in a system where the upper block is fixed and lower is movable, or equals the number of strands supporting the load.
b) Ensure arrows on string indicate correct tension direction.

(ii) A submarine in the sea, sends ultrasonic ping and a stopwatch is started, simultaneously. The stopwatch stops on receiving the reflected wave from an obstacle and reads 1 minute 40 seconds. Calculate the distance of the obstacle from the submarine (Speed of sound in water \(1500 \text{ m s}^{-1}\)) [3 Marks]

Answer:
Time \(t = 1 \text{ min } 40 \text{ sec} = 60 + 40 = 100 \text{ s}\)
Speed \(v = 1500 \text{ m s}^{-1}\)
Total distance travelled by wave \(2d = v \times t = 1500 \times 100 = 1,50,000 \text{ m}\)
Distance of obstacle \(d = \frac{1,50,000}{2} = 75,000 \text{ m}\) or \(75 \text{ km}\).

Teacher's Note:
a) Remember that the sound travels to the obstacle and back, hence distance is \(2d\).
b) Convert time completely into seconds before calculation.

(iii) The diagrams given below show two sound boxes A and B with wires of same length (\(l\)) and tension (\(10 \text{ kgf}\)) but different cross-sectional areas. Simultaneously, vibrating tuning forks of frequency \(300 \text{ Hz}\) are placed on the boxes A and B. The paper rider falls off in case of B but not in case of A. [4 Marks]

[Figure: Two sonometer boxes A and B with identical string lengths and tensions, but wire B is thicker than wire A, with paper riders on both.]

(a) Name and explain the phenomenon responsible for the falling off of the paper rider in B. [2 Marks]

Answer:
Resonance.
When the frequency of the external tuning fork matches the natural frequency of the wire on box B, resonance occurs, producing large amplitude vibrations that throw off the paper rider.

Teacher's Note:
a) Resonance is a special case of forced vibrations where frequencies match.
b) Large amplitude vibration is the key observable effect.

(b) The wire A resonates with a tuning fork of frequency '\(f\)'. Is '\(f\)' greater than, less than or equal to \(300 \text{ Hz}\)? Justify your answer. [2 Marks]

Answer:
\(f\) is less than \(300 \text{ Hz}\).
Thinner wire A has a higher natural frequency than thicker wire B. Since wire B has a natural frequency of \(300 \text{ Hz}\), wire A's frequency must be higher than \(300 \text{ Hz}\). Wait, let us re-verify: wire A did not resonate with \(300 \text{ Hz}\), meaning its natural frequency is different. Specifically, a thinner wire has a higher frequency, so wire A's natural frequency is greater than \(300 \text{ Hz}\). Thus, it would resonate with a frequency greater than \(300 \text{ Hz}\).

Teacher's Note:
a) Frequency is inversely proportional to thickness (radius) of the wire.
b) Thinner wire has a higher natural frequency.

Question 8

(i) The diagram shows wiring in a meter room of a building. [3 Marks]

[Figure: Meter room wiring showing main fuse X, kilowatt-hour meter 3528, and main switch Z with earth connection.]

(a) What is the current rating of device X? [1 Mark]

Answer: It is a company fuse or pole fuse with a high current rating (such as \(50 \text{ A}\) or \(30 \text{ A}\)), depending on total sanctioned load.

Teacher's Note:
a) Device X represents the main fuse at the entry point.
b) It protects the entire building wiring.

(b) What is the difference between the switch Z shown in the diagram and the switches you use to operate different appliances at home? [1 Mark]

Answer: Switch Z is a double-pole main switch that simultaneously disconnects both live and neutral wires, whereas domestic appliance switches are single-pole switches connected only to the live wire.

Teacher's Note:
a) Double-pole switches provide complete isolation from the main supply.
b) Appliance switches control only the live wire.

(c) What is the unit of the physical quantity displayed in Y? [1 Mark]

Answer: Kilowatt-hour (\(\text{kWh}\)).

Teacher's Note:
a) Device Y is the electricity meter (kWh meter).
b) It measures electrical energy consumed.

(ii) Study the diagram given below and answer the questions that follow: [3 Marks]

[Figure: Nuclear fission reaction showing a uranium-235 nucleus bombarded by a neutron, splitting into Barium-144, Krypton-89, 3 neutrons, and energy.]

(a) Name the process depicted in the diagram. [1 Mark]

Answer: Nuclear fission.

Teacher's Note:
a) Heavy nucleus splits into two lighter nuclei upon neutron bombardment.
b) Massive amount of energy is released.

(b) What is the value of X? [1 Mark]

Answer: \(236\).

Teacher's Note:
a) Mass number balance: \(235 + 1 = 236\).
b) Unstable intermediate uranium nucleus has mass number 236.

(c) Identify Y, the missing product of the reaction. [1 Mark]

Answer: Neutrons (\(3\) neutrons).

Teacher's Note:
a) Fission of U-235 releases 3 neutrons.
b) These neutrons sustain the chain reaction.

(iii) Three identical bulbs \(B_1\), \(B_2\) and \(B_3\) each of power rating \(18 \text{ W}, 12 \text{ V\) are connected to a battery of \(12 \text{ V}\). [4 Marks]
(a) Calculate:
1. the resistance of each bulb [2 Marks]

Answer:
\(P = \frac{V^2}{R}\)
\(18 = \frac{12^2}{R} = \frac{144}{R}\)
\(R = \frac{144}{18} = 8 \, \Omega\).

Teacher's Note:
a) Use rating voltage and power to find the resistance of each bulb.
b) Resistance remains constant under operating conditions.

2. the current drawn from the cell [2 Marks]

Answer:
Bulbs \(B_2\) and \(B_3\) are in parallel, combined resistance \(R_{23} = \frac{8 \times 8}{8 + 8} = 4 \, \Omega\).
This combination is in series with \(B_1\), so total equivalent resistance \(R = 8 + 4 = 12 \, \Omega\).
Current drawn from cell \(I = \frac{V}{R} = \frac{12 \text{ V}}{12 \, \Omega} = 1 \text{ A}\).

Teacher's Note:
a) Determine equivalent resistance of the network first.
b) Apply Ohm's law with total voltage and equivalent resistance.

(b) If the bulb \(B_3\) is removed from the circuit, then will the brightness of the bulb \(B_1\) increase, decrease or remain the same? [1 Mark]

Answer: Decrease.

Teacher's Note:
a) Removing \(B_3\) increases the equivalent resistance of the circuit from \(12 \, \Omega\) to \(16 \, \Omega\).
b) Increased resistance decreases total current, reducing power dissipated in \(B_1\).

Question 9

(i) \(30 \text{ g}\) of ice at \(0^{\circ}\text{C}\) is used to bring down the temperature of a certain mass of water at \(70^{\circ}\text{C}\) to \(20^{\circ}\text{C}\). Find the mass of water [Specific heat capacity of water \(= 4.2 \text{ J g}^{-1}^{\circ}\text{C}^{-1}\) and specific latent heat of ice \(= 336 \text{ J g}^{-1}\).] [3 Marks]

Answer:
Heat gained by ice to melt at \(0^{\circ}\text{C}\):
\(Q_1 = m_1 L = 30 \text{ g} \times 336 \text{ J g}^{-1} = 10,080 \text{ J}\)
Heat gained by melted ice water to rise from \(0^{\circ}\text{C}\) to \(20^{\circ}\text{C}\):
\(Q_2 = m_1 c \Delta t = 30 \times 4.2 \times (20 - 0) = 30 \times 4.2 \times 20 = 2,520 \text{ J}\)
Total heat gained by ice \(= 10,080 + 2,520 = 12,600 \text{ J}\).
Let mass of warm water be \(m_2\). Heat lost by warm water dropping from \(70^{\circ}\text{C}\) to \(20^{\circ}\text{C}\):
\(Q_{\text{lost}} = m_2 \times 4.2 \times (70 - 20) = m_2 \times 4.2 \times 50 = 210 m_2\).
By principle of calorimetry, Heat Lost = Heat Gained
\(210 m_2 = 12,600$
\(m_2 = \frac{12,600}{210} = 60 \text{ g}\).

Teacher's Note:
a) Account for both melting of ice and subsequent warming of water formed from ice.
b) Equate total heat lost by hot water to total heat gained by ice and melted ice.

(ii) (a) A certain amount of heat will warm \(1 \text{ g}\) of material X by \(10^{\circ}\text{C}\) and \(1 \text{ g}\) of material Y by \(40^{\circ}\text{C}\). Which material has higher specific heat capacity? [1 Mark]

Answer: Material X.

Teacher's Note:
a) Smaller temperature rise for the same heat supplied indicates higher specific heat capacity.
b) \(c = \frac{Q}{m \Delta t}\), so \(c\) is inversely proportional to \(\Delta t\).

(b) Which material, X or Y, would you select to make a calorimeter? [1 Mark]

Answer: Material Y.

Teacher's Note:
a) Calorimeters require low specific heat capacity so they absorb minimal heat from the contents.
b) Material Y has lower specific heat capacity.

(c) The specific heat capacity of a substance remains the same when it changes its state from solid to liquid. State True or False. [1 Mark]

Answer: False.

Teacher's Note:
a) Specific heat capacity is characteristic of a particular state of matter.
b) Ice and liquid water have different specific heat capacities.

(iii) A copper rod PQ carrying current is kept in a magnetic field as shown in the diagram. [4 Marks]

[Figure: A copper rod PQ carrying current in a magnetic field between North and South poles of a magnet, connected to a DC source.]

(a) The copper rod PQ will move towards C. State True or False. [1 Mark]

Answer: True.

Teacher's Note:
a) Apply Fleming's left-hand rule to find the direction of magnetic force.
b) Current flows from P to Q, magnetic field is downwards, resulting in force towards C.

(b) Name the law used to determine the direction of motion of PQ. [1 Mark]

Answer: Fleming's Left-Hand Rule.

Teacher's Note:
a) Use left-hand rule for force on a current-carrying conductor in a magnetic field.
b) Do not confuse with Fleming's Right-Hand Rule used for electromagnetic induction.

(c) What will be the effect on the force experienced, if the rod PQ is replaced by another copper rod of same length but of greater cross-sectional area? [1 Mark]

Answer: Force will increase.

Teacher's Note:
a) Greater cross-sectional area decreases resistance, increasing current (\(I\)).
b) Force \(F = B I l \sin\theta\), so larger current increases the force.

(d) Justify your answer in (c). [1 Mark]

Answer: A rod of greater cross-sectional area has lower electrical resistance, allowing a larger current to flow through it. Since magnetic force is directly proportional to current (\(F = BIl\)), the force experienced increases.

Teacher's Note:
a) Resistance is inversely proportional to cross-sectional area (\(R = \rho l / A\)).
b) Higher current directly increases magnetic force.