ICSE Class 10 Physics Board Exam Question Paper 2020 with Solutions

ICSE Class 10 Physics Board Exam Question Paper 2020 with Solutions

SECTION I (40 Marks)

Attempt all questions from this Section.

Question 1

(a) (i) Define moment of force [1 Mark]

Answer:
The moment of force is equal to the product of the magnitude of the force and the perpendicular distance of the line of action of the force from the axis of rotation.

Teacher's Note:
a) Mention the formula \( \tau = F \times d \) along with the definition for complete clarity.
b) Do not forget to state that the distance considered is perpendicular to the line of action of force.

(a) (ii) Write the relationship between the SI and CGS unit of moment of force. [1 Mark]

Answer:
The SI unit of moment of force is Newton-metre (\( \text{N m} \)) and the CGS unit is dyne-centimetre (\( \text{dyn cm} \)). The relationship is \( 1\text{ N m} = 10^7\text{ dyn cm} \).

Teacher's Note:
a) Convert 1 Newton to \( 10^5 \) dynes and 1 metre to \( 10^2 \) centimetres to derive the relation.
b) Ensure proper units are written alongside the numerical values.

(b) Define a kilowatt hour. How is it related to joule? [2 Marks]

Answer:
1 kilowatt hour (\( \text{kWh} \)) is the electrical energy consumed by an electrical appliance of power 1 kilowatt operating in 1 hour.
Relationship with Joule: \( 1\text{ kWh} = 3.6 \times 10^6\text{ J} \).

Teacher's Note:
a) Define kilowatt-hour as commercial unit of electrical energy.
b) Show the conversion \( 1\text{ kW} \times 1\text{ h} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ J} \) to secure full marks.

(c) A satellite revolves round a planet in a circular orbit. What is the work done by the satellite at any instant? Give a reason. [2 Marks]

Answer:
The work done by the satellite at any instant is zero.
Reason: The gravitational force acting on the satellite is directed towards the centre of the circular orbit, which is perpendicular to the instantaneous displacement (velocity) of the satellite along the tangent.

Teacher's Note:
a) State the work formula \( W = F s \cos\theta \).
b) Mention that \( \theta = 90^{\circ} \) between the centripetal force and displacement, hence \( \cos 90^{\circ} = 0 \).

(d) (i) Identify the class of the lever shown in the diagram below: [1 Mark]

[Figure: A lever diagram with Fulcrum (F) at one end, Load (L) at the other end, and Effort (E) in between them.]

Answer:
Class III lever.

Teacher's Note:
a) In a class III lever, the effort lies between the fulcrum and the load.
b) Common examples include a foot-treadle, fire tongs, and human forearm.

(d) (ii) How is it possible to increase the M.A of the above lever without increasing its length? [1 Mark]

Answer:
The mechanical advantage (M.A) can be increased by shifting the load closer to the fulcrum (or shifting the effort away from the fulcrum, keeping total length constant by altering arm lengths).

Teacher's Note:
a) Recall that for a class III lever, \( \text{M.A} = \frac{\text{Effort arm}}{\text{Load arm}} \).
b) Decreasing the load arm length increases the mechanical advantage.

(e) Give one example of each when: [2 Marks]
(i) Chemical energy changes into electrical energy.
(ii) Electrical energy changes into sound energy

Answer:
(i) Electric cell (or Battery) in use.
(ii) Loudspeaker.

Teacher's Note:
a) Ensure simple and precise devices are named.
b) Do not confuse with inverse transformations like charging a battery.

Question 2

(a) A crane "A" lifts a heavy load in 5 seconds. Whereas another crane "B" does the same work in 2 seconds. Compare the power of crane "A" to that of crane "B" [2 Marks]

Answer:
Power \( P = \frac{W}{t} \).
Since work done \( W \) is the same for both cranes:
\( \frac{P_A}{P_B} = \frac{t_B}{t_A} = \frac{2}{5} = 2:5 \).

Teacher's Note:
a) Power is inversely proportional to time taken when work done is constant.
b) Write the ratio clearly in the final form \( 2:5 \).

(b) A ray of light falls normally on a rectangular glass slab. Draw a ray diagram showing the path of the ray till it emerges out of the slab [2 Marks]

Answer:
When a ray of light falls normally on a rectangular glass slab, it passes straight without any deviation because the angle of incidence \( i = 0^{\circ} \), hence the angle of refraction \( r = 0^{\circ} \).
[Figure: A ray of light striking normally (perpendicularly) to the surface of a rectangular glass slab, passing straight through it without bending, and emerging from the opposite face along the same straight line path.]

Answer:
Since the ray AB strikes normally on surface PQ, it enters the prism without deviation. It then strikes the second face PR at an angle greater than the critical angle, undergoes total internal reflection, and emerges normally through the base QR.
[Figure: Ray AB enters perpendicularly into face PQ, travels straight to hit face PR inside the prism, reflects completely inside due to total internal reflection, and emerges perpendicularly out of base QR.]

Teacher's Note:
a) Check for correct normals at each refracting surface.
b) Ensure total internal reflection is correctly shown at the internal face for an equilateral prism.

(d) Where should an object be placed in front of a convex lens in order to get :
(i) an enlarged real image
(ii) enlarged virtual image? [2 Marks]

Answer:
(i) Between \( F \) and \( 2F \) (or between focus and twice the focal length) in front of the convex lens.
(ii) Between the optical centre (\( O \)) and the principal focus (\( F \)) of the convex lens.

Teacher's Note:
a) Clearly specify the positions with respect to focus \( F \) and \( 2F \).
b) Real image is formed on the other side, while virtual image is formed on the same side as the object.

(e) A pond appears to be 2.7 m deep. If the refractive index of water is \( \frac{4}{3} \), find the actual depth of the pond [2 Marks]

Answer:
Refractive index \( \mu = \frac{\text{Real depth}}{\text{Apparent depth}} \)
\( \frac{4}{3} = \frac{\text{Real Depth}}{2.7} \)
Real Depth = \( \frac{4}{3} \times 2.7 = 4 \times 0.9 = 3.6\text{ m} \).

Teacher's Note:
a) State the formula relating refractive index, real depth, and apparent depth clearly.
b) Include proper SI units in the final numerical answer.

Question 3

(a) The wave lengths for the light of red and blue colours are nearly \( 7.8 \times 10^{-7}\text{ m} \) and \( 4.8 \times 10^{-7}\text{ m} \) respectively.
(i) Which colour has the greater speed in a vacuum?
(ii) Which colour has a greater speed in glass? [2 Marks]

Answer:
(i) Both red and blue colours have the same speed in a vacuum (speed of light, \( c \)).
(ii) Red colour has a greater speed in glass.

Teacher's Note:
a) Light of all wavelengths travels at the same speed in a vacuum.
b) In a denser medium like glass, speed depends on refractive index; since red has a longer wavelength and lower refractive index than blue, red travels faster.

(b) Draw a graph between displacement from mean position and time for a body executing free vibration in a vacuum [2 Marks]

Answer:
[Figure: A sine wave graph representing undamped simple harmonic motion where displacement is plotted on the y-axis and time on the x-axis, showing a continuous sinusoidal curve with constant amplitude over time.]

Teacher's Note:
a) Ensure the amplitude remains constant throughout since it is free vibration in a vacuum (no damping).
b) Label axes correctly as Displacement and Time.

(c) A sound wave travelling in water has wavelength \( 0.4\text{ m} \). Is this wave audible in air? (The speed of sound in water = \( 1400\text{ ms}^{-1} \)) [2 Marks]

Answer:
Frequency \( f = \frac{v}{\lambda} = \frac{1400}{0.4} = 3500\text{ Hz} \).
Yes, this wave is audible in air because the frequency lies within the audible range of human hearing (\( 20\text{ Hz} \) to \( 20,000\text{ Hz} \)).

Teacher's Note:
a) Frequency remains unchanged when sound passes from one medium to another.
b) State the audible frequency range clearly to justify the conclusion.

(d) Why does stone lying in the sun get heated up much more than water lying for the same duration of time? [2 Marks]

Answer:
The specific heat capacity of stone is much less than the specific heat capacity of water. Therefore, for the same amount of heat absorbed and same mass, the rise in temperature of stone is much higher than that of water.

Teacher's Note:
a) Mention specific heat capacity explicitly as the core concept.
b) Water has one of the highest specific heat capacities, making it resist temperature changes.

(e) Why is it not advisable to use a piece of copper wire as fuse wire in an electric circuit? [1 Mark]

Answer:
Copper has a very low resistivity and a high melting point. As a result, it will not melt easily when a heavy current flows through the circuit, failing to protect electrical appliances from damage.

Teacher's Note:
a) A fuse wire must have low melting point and high resistivity.
b) Copper possesses properties opposite to what is required for an effective fuse wire.

Question 4

(a) Calculate the total resistance across AB : [2 Marks]

[Figure: A circuit diagram showing two parallel resistors of \( 3\text{ }\Omega \) and \( 6\text{ }\Omega \) connected in series with a resistor of \( 5\text{ }\Omega \) between terminals A and B.]

Answer:
Equivalent resistance of parallel combination of \( 3\text{ }\Omega \) and \( 6\text{ }\Omega \):
\( \frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2} \implies R_p = 2\text{ }\Omega \).
Total resistance across AB (in series with \( 5\text{ }\Omega \) resistor):
\( R_{total} = R_p + 5 = 2 + 5 = 7\text{ }\Omega \).

Teacher's Note:
a) First solve the parallel branch and then add the series resistance.
b) Include proper units (\( \Omega \)) in each step.

(b) Two metallic blocks P and Q have masses in ratio \( 2:1 \) and are supplied with the same amount of heat. If their temperatures rise by the same degree, compare their specific heat capacities. [2 Marks]

Answer:
Heat supplied \( Q = m c \Delta T \).
Since \( Q \) and \( \Delta T \) are the same for both blocks:
\( m_P c_P = m_Q c_Q \implies \frac{c_P}{c_Q} = \frac{m_Q}{m_P} = \frac{1}{2} \).
Ratio of specific heat capacities \( c_P : c_Q = 1:2 \).

Teacher's Note:
a) Use the standard heat equation \( Q = mc\Delta T \).
b) Note that specific heat capacity is inversely proportional to mass for equal heat and temperature rise.

(c) When a current carrying conductor is placed in a magnetic field, it experiences a mechanical force. What should be the angle between the magnetic field and the length of the conductor so that the force experienced is:
(i) Zero
(ii) Maximum? [2 Marks]

Answer:
(i) \( 0^{\circ} \) (or \( 180^{\circ} \))
(ii) \( 90^{\circ} \)

Teacher's Note:
a) Magnetic force depends on \( \sin\theta \) where \( \theta \) is the angle between the conductor and magnetic field.
b) \( \sin 0^{\circ} = 0 \) (minimum force) and \( \sin 90^{\circ} = 1 \) (maximum force).

(d) A nucleus \( _{84}\text{X}^{202} \) of an element emits an alpha particle followed by a beta particle. The final nucleus is \( _a\text{Y}^b \). Find a and b [2 Marks]

Answer:
Initial nucleus: \( _{84}\text{X}^{202} \).
Emission of an alpha particle (\( _2\text{He}^4 \)):
New nucleus has atomic number \( = 84 - 2 = 82 \) and mass number \( = 202 - 4 = 198 \).
Subsequent emission of a beta particle (\( _{-1}\text{e}^0 \)):
Final nucleus \( _a\text{Y}^b \) has atomic number \( a = 82 - (-1) = 83 \) and mass number \( b = 198 - 0 = 198 \).
Therefore, \( a = 83 \) and \( b = 198 \).

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

(e) The diagram below shows a loop of wire carrying current I: [2 Marks]

[Figure: A circular loop of wire connected to a DC source with a switch/key, carrying current I.]

(i) What is the magnetic polarity of the loop that faces us?
(ii) With respect to the diagram how can we increase the strength of the magnetic field produced by this loop?

Answer:
(i) Anticlockwise current corresponds to North (\( N \)) polarity.

(ii) By increasing the magnitude of electric current in the loop or by increasing the number of turns in the coil.

Teacher's Note:
a) Clockwise current indicates South polarity and anticlockwise indicates North polarity.
b) Field strength is directly proportional to current and number of turns.

SECTION II (40 Marks)

Attempt any four questions from this Section.

Question 5

(a) The figure below shows a simple pendulum of mass \( 200\text{ g} \). It is displaced from the mean position A to the extreme position B. The potential energy at the position A is Zero. At the position B the pendulum bob is raised by \( 5\text{ m} \). [3 Marks]

[Figure: A simple pendulum swinging from mean position A to extreme position B, passing through C, with height at B shown as 5 m above A.]

(i) What is the potential energy of the pendulum at the position B?
(ii) What is the total mechanical energy at point C?
(iii) What is the speed of the bob at the position A when released from B? (Take \( g = 10\text{ ms}^{-2} \) and that is no loss of energy.)

Answer:
Mass \( m = 200\text{ g} = 0.2\text{ kg} \), height \( h = 5\text{ m} \), \( g = 10\text{ ms}^{-2} \).
(i) Potential Energy at B \( = mgh = 0.2 \times 10 \times 5 = 10\text{ J} \).
(ii) Total Mechanical Energy at C \( = \) Total Energy at B \( = 10\text{ J} \) (by conservation of mechanical energy).
(iii) At position A, all potential energy converts to kinetic energy:
\( \frac{1}{2}mv^2 = 10 \implies \frac{1}{2} \times 0.2 \times v^2 = 10 \implies 0.1 v^2 = 10 \implies v^2 = 100 \implies v = 10\text{ ms}^{-1} \).

Teacher's Note:
a) Convert mass from grams to kilograms before calculation.
b) Total mechanical energy remains conserved throughout the motion in the absence of friction.

(b) (i) With reference to the direction of action how does a centripetal force differ from a centrifugal force during uniform circular motion.
(ii) Is centrifugal force the form of reaction of centripetal force?
(iii) Compare the magnitude of centripetal and centrifugal force [3 Marks]

Answer:
(i) Centripetal force acts towards the centre of the circular path, whereas centrifugal force acts away from the centre along the radius.
(ii) No, centrifugal force is not the reaction of centripetal force because both forces act on the same body (centripetal acts on the revolving body towards the centre, while centrifugal is a pseudo force experienced in a rotating frame).
(iii) Both forces are equal in magnitude (\( F = \frac{mv^2}{r} \)).

Teacher's Note:
a) Clearly distinguish the direction of both forces.
b) Emphasize that action and reaction forces act on two different bodies, which clarifies why centrifugal force is not a reaction force.

(c) A block and tackle system of pulleys has velocity ratio 4.
(i) Draw a neat labeled diagram of the system indicating clearly the points of application, and direction of load and effort.
(ii) What will be its V.R if the weight of the movable block is doubled? [4 Marks]

Answer:
(i) [Figure: A block and tackle pulley system consisting of 4 pulleys (2 in the upper fixed block and 2 in the lower movable block), with a single continuous string passing around them, effort acting downwards, load acting downwards from the movable block, and upper block attached to a rigid support.]
(ii) The velocity ratio (V.R) of a pulley system depends only on the total number of pulleys (or strands of tackle supporting the movable block) and is independent of the weight of the movable block. Therefore, the V.R remains 4.

Teacher's Note:
a) For V.R = 4, there are 4 pulleys in total with 2 in each block (or equivalent configuration).
b) Weight of the movable block affects mechanical advantage and efficiency, but never velocity ratio.

Question 6

(a) A diver in water looks obliquely at an object AB in air [3 Marks]

[Figure: An object AB in air above water surface, with rays AC and AD traveling towards the eye of a diver submerged in water.]

(i) Does the object appear taller, shorter or of the same size to the diver?
(ii) Show the path of two rays AC & AD starting from the tip of the object as it travels towards the diver in water and hence obtain the image of the object

Answer:
(i) The object appears taller (or magnified) to the diver.
(ii) [Figure: Rays AC and AD starting from tip A of the object in the rarer medium (air) refract away from the normal as they enter the denser medium (water), and when produced backwards, they intersect at a point higher than A, forming an enlarged virtual image.]

Teacher's Note:
a) When light travels from a rarer to a denser medium, it bends towards the normal, but here light travels from air to water towards the diver's eye, causing rays to bend towards the normal upon entry into water.
b) Back-tracing refracted rays shows the virtual image formed at a higher position.

(b) Complete the path of the ray AB through the glass prism PQR till it emerges out of the prism. Given the critical angle of the glass as \( 42^{\circ} \) [3 Marks]

[Figure: A right-angled isosceles glass prism PQR with angle \( 30^{\circ} \) at the top apex, incident ray AB striking normally to the base QR.]

Answer:
Ray AB enters normally into the prism through base QR without deviation, strikes the face PQ at an angle of incidence of \( 60^{\circ} \) (which is greater than the critical angle of \( 42^{\circ} \)), undergoes total internal reflection, and emerges normally through face PR.
[Figure: Ray AB enters base QR without bending, hits face PQ at \( 60^{\circ} \), reflects completely into the prism, and exits perpendicularly through face PR.]

Teacher's Note:
a) Calculate the angle of incidence on the internal face using geometry of the prism.
b) Since angle of incidence \( (60^{\circ}) \gt \text{critical angle } (42^{\circ}) \), total internal reflection takes place.

(c) A lens of focal length \( 20\text{ cm} \) forms an inverted image at a distance \( 60\text{ cm} \) from the lens.
(i) Identify the lens
(ii) How far is the lens present in front of the object?
(iii) Calculate the magnification of the image [4 Marks]

Answer:
(i) Since the image formed is inverted, the lens is a convex lens. Focal length \( f = +20\text{ cm} \), image distance \( v = +60\text{ cm} \).
(ii) Using lens formula \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \):
\( \frac{1}{20} = \frac{1}{60} - \frac{1}{u} \implies \frac{1}{u} = \frac{1}{60} - \frac{1}{20} = \frac{1 - 3}{60} = \frac{-2}{60} = \frac{-1}{30} \implies u = -30\text{ cm} \).
The lens is placed \( 30\text{ cm} \) in front of the object.
(iii) Magnification \( m = \frac{v}{u} = \frac{60}{-30} = -2 \).

Teacher's Note:
a) Apply proper sign conventions: convex lens focal length is positive, real image distance is positive, and object distance is negative.
b) Negative sign of magnification confirms the image is real and inverted.

Question 7

(a) Give reasons for the following during the day:
(i) Clouds appear white
(ii) Sky appears blue. [2 Marks]

Answer:
(i) Clouds contain water droplets and dust particles whose size is much larger than the wavelength of visible light. They scatter all wavelengths of white light equally (non-selective scattering), making clouds appear white.
(ii) Molecules in the atmosphere have a size smaller than the wavelength of light. According to Rayleigh scattering, shorter wavelengths (blue) are scattered much more strongly than longer wavelengths, making the sky appear blue.

Teacher's Note:
a) Distinguish clearly between particle size in clouds (large) and air molecules (small).
b) Mention equal scattering for clouds and selective (Rayleigh) scattering for the blue sky.

(b) (i) Name the system which enables us to locate underwater objects by transmitting ultrasonic waves and detecting the reflecting impulse.
(ii) What are acoustically measurable quantities related to pitch and loudness? [2 Marks]

Answer:
(i) SONAR (Sound Navigation and Ranging).
(ii) Frequency is related to pitch, and amplitude is related to loudness.

Teacher's Note:
a) SONAR uses ultrasonic waves due to their high frequency and ability to travel long distances in water.
b) Pitch depends on frequency, while loudness depends on the amplitude of the sound wave.

(c) (i) When a tuning fork (vibrating) is held close to ear, one hears a faint hum. The same (vibrating tuning fork) is held such that its stem is in contact with the table surface, then one hears a loud sound. Explain.
(ii) A man standing in front of a vertical cliff fires a gun. He hears the echo after \( 3.5\text{ seconds} \). On moving closer to the cliff by \( 84\text{ m} \), he hears the echo after \( 3\text{ seconds} \). Calculate the distance of incubator/cliff from the initial position of the man. [6 Marks]

Answer:
(i) When held in air, the vibrating tuning fork has a small surface area, setting a small mass of air into vibration, producing a faint sound. When placed on the table surface, forced vibrations are set up in the large surface area of the table, making a larger volume of air vibrate and producing a loud sound.
(ii) Let initial distance of man from cliff be \( d \). Speed of sound \( v \).
First case: \( 2d = v \times 3.5 = 3.5v \implies d = 1.75v \).
Second case: New distance \( = (d - 84) \). Time \( = 3\text{ s} \).
\( 2(d - 84) = v \times 3 = 3v \implies 2d - 168 = 3v \).
Substitute \( 2d = 3.5v \):
\( 3.5v - 168 = 3v \implies 0.5v = 168 \implies v = 336\text{ ms}^{-1} \).
Distance \( d = 1.75 \times 336 = 588\text{ m} \).

Teacher's Note:
a) Explain forced vibrations clearly for part (i).
b) Use the echo formula \( 2d = v \times t \) and solve simultaneous equations for part (ii).

Question 8

(a) The diagram below shows the core of a transformer and its input and output connections [2 Marks]

[Figure: A laminated soft iron core box with input AC mains on the left and output AC terminals on the right.]

(i) State the material used for the core.
(ii) Copy and complete the diagram of the transformer by drawing input and output coils

Answer:
(i) Soft iron (laminated soft iron core).
(ii) [Figure: Complete transformer diagram showing primary coil wound on the left limb connected to AC input, and secondary coil wound on the right limb connected to output terminals.]

Teacher's Note:
a) Soft iron is used because of its high permeability and low retentivity, reducing eddy current losses through lamination.
b) Ensure primary and secondary coils are wound on opposite or same limbs of the core clearly.

(b) (i) What are superconductors?
(ii) Calculate the current drawn by an appliance rated \( 110\text{ W}, 220\text{ V} \) when connected across \( 220\text{ V} \) supply.
(iii) Name a substance whose resistance decreases with the increase in temperature. [3 Marks]

Answer:
(i) Superconductors are substances whose electrical resistance becomes zero at very low temperatures.
(ii) Power \( P = V \times I \implies I = \frac{P}{V} = \frac{110}{220} = 0.5\text{ A} \).
(iii) Semiconductors (such as carbon, silicon, germanium) or electrolytes (or thermistors).

Teacher's Note:
a) Mention zero electrical resistance for superconductors.
b) Semiconductors possess a negative temperature coefficient of resistance.

(c) The diagram above shows three resistors connected across a cell of e.m.f. \( 1.8\text{ V} \) and internal resistance \( r \). Calculate: [5 Marks]

[Figure: A circuit diagram showing parallel combination of \( 3\text{ }\Omega \) and \( 1.5\text{ }\Omega \) resistors, in series with a \( 4\text{ }\Omega \) resistor, connected across a cell of e.m.f. 1.8 V and internal resistance r, with current of \( 0.3\text{ A} \) indicated.]

(i) Current through \( 3\text{ }\Omega \) resistor.
(ii) The internal resistance r.

Answer:
Parallel combination of \( 3\text{ }\Omega \) and \( 1.5\text{ }\Omega \):
\( R_p = \frac{3 \times 1.5}{3 + 1.5} = \frac{4.5}{4.5} = 1\text{ }\Omega \).
Total external resistance \( R = R_p + 4 = 1 + 4 = 5\text{ }\Omega \).
Total current in circuit \( I = 0.3\text{ A} \).
Potential difference across parallel combination \( V_p = I \times R_p = 0.3 \times 1 = 0.3\text{ V} \).
(i) Current through \( 3\text{ }\Omega \) resistor: \( I_3 = \frac{V_p}{3} = \frac{0.3}{3} = 0.1\text{ A} \).
(ii) Using relation \( E = I(R + r) \):
\( 1.8 = 0.3(5 + r) \)
\( 5 + r = \frac{1.8}{0.3} = 6 \)
\( r = 6 - 5 = 1\text{ }\Omega \).

Teacher's Note:
a) Calculate equivalent parallel resistance first before finding total resistance.
b) Use terminal voltage formula or emf equation \( E = V + Ir \) to determine internal resistance \( r \).

Question 9

(a) (i) Define heat capacity of a substance.
(ii) Write the SI unit of heat capacity.
(iii) What is the relationship between heat capacity and specific heat capacity of a substance? [3 Marks]

Answer:
(i) Heat capacity of a substance is the amount of heat energy required to raise the temperature of the entire body by \( 1^{\circ}\text{C} \) (or \( 1\text{ K} \)).
(ii) SI unit of heat capacity is Joule per Kelvin (\( \text{J K}^{-1} \)).
(iii) Heat Capacity \( C = m \times c \), where \( m \) is mass and \( c \) is specific heat capacity.

Teacher's Note:
a) Note the difference between heat capacity (for the whole body) and specific heat capacity (per unit mass).
b) Ensure correct units are written.

(b) The diagram below shows the change of phases of a substance on a temperature vs time graph on heating the substance at a constant rate [2 Marks]

[Figure: Temperature vs time graph showing regions AB (solid heating), BC (melting/fusion), CD (liquid heating), DE (vaporization/boiling), and EF (gas heating).]

(i) Why is the slope of CD less than slope of AB?
(ii) What is the boiling and melting point of the substance?

Answer:
(i) Slope of temperature-time graph is inversely proportional to specific heat capacity (\( \text{Slope} \propto \frac{1}{mc} \)). Since the specific heat capacity of the liquid (CD) is greater than that of the solid (AB), the slope of CD is less than the slope of AB.
(ii) Melting point is \( t_1 \) and boiling point is \( t_2 \).

Teacher's Note:
a) Steeper slope indicates lower specific heat capacity.
b) Flat portions (BC and DE) represent phase changes at constant temperatures (melting point and boiling point).

(c) A piece of ice of mass \( 60\text{ g} \) is dropped into \( 140\text{ g} \) of water at \( 50^{\circ}\text{C} \). Calculate the final temperature of water when all the ice has melted. (Assume no heat is lost to the surrounding).
Specific heat capacity of water \( = 4.2\text{ J g}^{-1}\text{K}^{-1} \)
Specific latent heat of fusion of ice \( = 336\text{ J g}^{-1} \) [5 Marks]

Answer:
Let the final temperature of water be \( T^{\circ}\text{C} \).
Heat gained by ice to melt at \( 0^{\circ}\text{C} \) and then raise its temperature to \( T^{\circ}\text{C} \):
\( Q_1 = m_i L_f + m_i c_w (T - 0) = 60 \times 336 + 60 \times 4.2 \times T = 20160 + 252T \).
Heat lost by warm water to cool down from \( 50^{\circ}\text{C} \) to \( T^{\circ}\text{C} \):
\( Q_2 = m_w c_w (50 - T) = 140 \times 4.2 \times (50 - T) = 588 \times (50 - T) = 29400 - 588T \).
By principle of conservation of energy (Heat lost = Heat gained):

\( 29400 - 588T = 20160 + 252T \)
\( 29400 - 20160 = 252T + 588T \)
\( 9240 = 840T \)
\( T = \frac{9240}{840} = 11^{\circ}\text{C} \).

Teacher's Note:
a) Equate total heat lost by water to total heat gained by ice (melting + warming the melted ice water).
b) Check arithmetic carefully across substitution steps.

Question 10

(a) (i) Draw a neat labeled diagram of a d.c. motor.
(ii) Write any one use of a d.c. motor [3 Marks]

Answer:
(i) [Figure: Labeled diagram of a D.C. motor showing permanent magnets (N and S poles), armature coil (ABCD), split-ring commutator (halves P and Q), carbon brushes (X and Y), and a DC source/battery.]
(ii) Used in electric fans, electric cars, water pumps, and toy motors.

Teacher's Note:
a) Ensure all major components like commutator and brushes are clearly labeled.
b) State any practical domestic or industrial application.

(b) (i) Differentiate between nuclear fusion and nuclear fission.
(ii) State one safety precaution in the disposal of nuclear waste. [3 Marks]

Answer:
(i) Differentiation table:

Teacher's Note:
a) Contrast fusion and fission based on combination versus splitting.
b) Emphasize safe shielding and deep burial for nuclear waste disposal.

(c) An atomic nucleus A is composed of 84 protons and 128 neutrons. The nucleus A emits an alpha particle and is transformed into a nucleus B.
(i) What is the composition of B?
(ii) The nucleus B emits a beta particle and is transformed into a nucleus C. What is the composition of C?
(iii) What is mass number of the nucleus A?
(iv) Does the composition of C change if it emits gamma radiations? [4 Marks]

Answer:
Initial nucleus A: Protons \( p = 84 \), Neutrons \( n = 128 \).
(i) Emission of an alpha particle (\( _2\text{He}^4 \)) means 2 protons and 2 neutrons are emitted.
Composition of B: Protons \( = 84 - 2 = 82 \); Neutrons \( = 128 - 2 = 126 \).
(ii) Emission of a beta particle (\( _{-1}\text{e}^0 \)) converts a neutron into a proton (atomic number increases by 1, mass number remains same).
Composition of C: Protons \( = 82 + 1 = 83 \); Neutrons \( = 126 - 1 = 125 \).
(iii) Mass number of A \( = \text{Protons} + \text{Neutrons} = 84 + 128 = 212 \).
(iv) No, emission of gamma radiation does not change the number of protons or neutrons; it only releases energy from an excited nucleus.

Teacher's Note:
a) Track proton and neutron numbers separately for each radioactive decay step.
b) Gamma radiation is electromagnetic emission and does not alter nucleon composition.