ICSE Class 10 Physics Board Exam Question Paper 2016 with Solutions

ICSE Class 10 Physics Board Exam Question Paper with Solutions

SECTION I

Question 1
(a) (i) Give an example of a non-contact force which is always of attractive nature.
(ii) How does the magnitude of this non-contact force on the two bodies depend on the distance of separation between them? [2 Marks]

Answer:
(i) Gravitational force.
Example: A ball placed on a table starts rolling down when the table is tilted due to gravity.
(ii) It varies inversely as the square of the distance of separation between them, i.e., as the distance between the two bodies increases, the magnitude of the force between them decreases.

Teacher's Note:
a) Remember that gravitational force is always attractive, unlike electrostatic or magnetic forces which can be repulsive.
b) State the inverse-square dependence clearly mentioning the distance of separation.

Question 1
(b) A boy weighing \( 40\text{ kgf} \) climbs up a stair of 30 steps each \( 20\text{ cm} \) high in 4 minutes and a girl weighing \( 30\text{ kgf} \) does the same in 3 minutes. Compare:
(i) The work done by them.
(ii) The power developed by them. [2 Marks]

Answer:
(i) Let \( m_1 = 40\text{ kgf} \), \( m_2 = 30\text{ kgf} \). Total height \( h_1 = h_2 = 30 \times 20\text{ cm} = 600\text{ cm} = 6\text{ m} \).
Work done by the boy \( W_1 = m_1 g h_1 = 40 \times g \times 6 \).
Work done by the girl \( W_2 = m_2 g h_2 = 30 \times g \times 6 \).
Ratio of work done \( \frac{W_1}{W_2} = \frac{40 \times g \times 6}{30 \times g \times 6} = \frac{40}{30} = 4:3 \).
(ii) Time taken by boy \( t_1 = 4\text{ minutes} = 240\text{ s} \); time taken by girl \( t_2 = 3\text{ minutes} = 180\text{ s} \).
Power of boy \( P_1 = \frac{W_1}{t_1} = \frac{40 \times 6}{240} \).
Power of girl \( P_2 = \frac{W_2}{t_2} = \frac{30 \times 6}{180} \).
Ratio of power \( \frac{P_1}{P_2} = \frac{40 \times 6 / 240}{30 \times 6 / 180} = \frac{1}{1} = 1:1 \).

Teacher's Note:
a) Work done depends only on weight and vertical height, not on time taken.
b) Power is work done per unit time; always convert minutes into seconds before calculating power.

Question 1
(c) With reference to the terms Mechanical Advantage, Velocity Ratio and efficiency of a machine, name and define the term that will not change for a machine of a given design. [2 Marks]

Answer:
Velocity Ratio (V.R.) will not change for a machine of a given design.
Velocity Ratio: It is defined as the ratio of the velocity of effort to the velocity of load, or the ratio of the displacement of effort to the displacement of load.

Teacher's Note:
a) Mechanical advantage and efficiency decrease due to friction and weight of moving parts, but velocity ratio remains constant for a fixed design.
b) Write the formula \( \text{V.R.} = \frac{d_E}{d_L} \) or \( \frac{v_E}{v_L} \) for completeness.

Question 1
(d) Calculate the mass of ice required to lower the temperature of \( 300\text{ g} \) of water at \( 40^{\circ}\text{C} \) to water at \( 0^{\circ}\text{C} \).
(Specific latent heat of ice = \( 336\text{ J/g} \), Specific heat capacity of water = \( 4.2\text{ J/g}^{\circ}\text{C} \)) [2 Marks]

Answer:
Heat lost by water = Heat gained by ice to melt at \( 0^{\circ}\text{C} \)
\( m c \Delta t = M L \)
\( 300 \times 4.2 \times (40 - 0) = M \times 336 \)
\( 300 \times 4.2 \times 40 = 336 M \)
\( 50400 = 336 M \)
\( M = \frac{50400}{336} = 150\text{ g} \).
Mass of ice required is \( 150\text{ g} \).

Teacher's Note:
a) Equate heat lost by hot body to heat gained by cold body according to the principle of calorimetry.
b) Ensure all units are consistent (grams and Joules) before performing calculation.

Question 1
(e) What do you understand by the following statements:
(i) The heat capacity of the body is \( 60\text{ J K}^{-1} \).
(ii) The specific heat capacity of lead is \( 130\text{ J kg}^{-1}\text{ K}^{-1} \). [2 Marks]

Answer:
(i) The given body requires \( 60\text{ J} \) of heat energy to raise its temperature through \( 1\text{ K} \) (or \( 1^{\circ}\text{C} \)).
(ii) \( 1\text{ kg} \) of lead requires \( 130\text{ J} \) of heat energy to raise its temperature through \( 1\text{ K} \) (or \( 1^{\circ}\text{C} \)).

Teacher's Note:
a) Heat capacity is for the entire body, whereas specific heat capacity is defined per unit mass.
b) Mention the unit quantity (\( 1\text{ K} \) and \( 1\text{ kg} \)) clearly to score full marks.

Question 2
(a) State two factors upon which the heat absorbed by a body depends. [2 Marks]

Answer:
1. Mass of the body: Heat absorbed is directly proportional to the mass of the body (\( Q \propto M \)).
2. Rise in temperature: Heat absorbed is directly proportional to the change in temperature (\( Q \propto \Delta t \)).

Teacher's Note:
a) A third factor is the nature of the material (specific heat capacity), though stating mass and temperature rise is sufficient for two factors.
b) Express the relation mathematically for clarity.

Question 2
(b) A boy uses blue colour of light to find the refractive index of glass. He then repeats the experiment using red colour of light. Will the refractive index be the same or different in the two cases ? Give a reason to support your answer. [2 Marks]

Answer:
The refractive index will be different in the two cases. Specifically, refractive index for blue light is greater than that for red light (\( \mu_B \gt \mu_R \)).
Reason: In glass, the speed of red light is greater than the speed of blue light, and refractive index is inversely proportional to speed (\( \mu = c/v \)).

Teacher's Note:
a) Refractive index depends on the wavelength (or colour) of light.
b) Violet and blue light travel slower in glass than red light, hence denser medium effect is more pronounced for blue.

Question 2
(c) Copy the diagram given below and complete the path of light ray till it emerges out of the prism. The critical angle of glass is \( 42^{\circ} \). In your diagram mark the angles wherever necessary. [2 Marks]

[Figure: Equilateral triangular prism ABC with angle A = \( 60^{\circ} \), B = \( 60^{\circ} \), C = \( 60^{\circ} \). A ray PQ is incident on face AB at angle of incidence making \( 60^{\circ} \) with the normal, refracts inside hitting face AC, undergoes total internal reflection, and emerges out from face BC.]

Answer:
The ray enters face AB, refracts inside at an angle of refraction, strikes the second face AC at an angle greater than the critical angle (\( 42^{\circ} \)), suffers total internal reflection, and finally emerges out normally or refracting away from the normal through face BC with proper angle markings (\( 60^{\circ} \), \( 30^{\circ} \)) at each interface.

Teacher's Note:
a) Ensure the angle of incidence on the second face is calculated and shown to be greater than \( 42^{\circ} \).
b) Draw arrowheads on light rays to indicate the direction of propagation.

Question 2
(d) State the dependence of angle of deviation:
(i) On the refractive index of the material of the prism.
(ii) On the wavelength of light. [2 Marks]

Answer:
(i) Angle of deviation is directly proportional to the refractive index of the material of the prism (higher refractive index produces greater deviation).
(ii) Angle of deviation is inversely proportional to the wavelength of light (\( \delta \propto \frac{1}{\text{wavelength}} \)).

Teacher's Note:
a) Flint glass has a higher refractive index than crown glass and thus produces more deviation.
b) Red light has a longer wavelength and bends least, while violet has a shorter wavelength and bends most.

Question 2
(e) The ratio of amplitude of two waves is \( 3:4 \). What is the ratio of their:
(i) loudness ? (ii) frequencies ? [2 Marks]

Answer:
(i) Loudness is directly proportional to the square of the amplitude (\( L \propto a^2 \)).
Ratio of loudness \( = \frac{a_1^2}{a_2^2} = \frac{3^2}{4^2} = \frac{9}{16} = 9:16 \).
(ii) Frequency does not depend on the amplitude of the wave.
Ratio of frequencies remains \( 1:1 \).

Teacher's Note:
a) Do not confuse amplitude ratio directly with loudness ratio; square the amplitudes.
b) Frequency is determined by the source and is unaffected by amplitude changes.

Question 3
(a) State two ways by which the frequency of transverse vibrations of a stretched string can be increased. [2 Marks]

Answer:
1. By decreasing the length of the vibrating string (\( f \propto \frac{1}{l} \)).
2. By decreasing the radius (or thickness) of the string (\( f \propto \frac{1}{r} \)).
(Alternatively: By increasing the tension in the string, \( f \propto \sqrt{T} \)).

Teacher's Note:
a) Mention any two correct factors from length, tension, or radius.
b) State the inverse or direct proportionality clearly for full credit.

Question 3
(b) What is meant by noise pollution ? Name one source of sound causing noise pollution. [2 Marks]

Answer:
Noise pollution is a sound produced by an irregular succession of disturbances that is loud, unwanted, unpleasant, and harmful to human ears.
Source: Sound produced by bursting of firecrackers / loudspeakers / heavy traffic.

Teacher's Note:
a) Define noise as unwanted or unpleasant sound resulting from irregular vibrations.
b) Provide any common practical source of noise pollution.

Question 3
(c) The V-I graph for a series combination and for a parallel combination of two resistors is shown in the figure below.
Which of the two A or B, represents the parallel combination ? Give a reason for your answer. [2 Marks]

[Figure: V-I graph with voltage V on y-axis and current I on x-axis, showing two straight lines labeled A and B. Line A has a smaller slope than line B.]

Answer:
Line A represents the parallel combination.
Reason: The slope of the V-I graph represents resistance (\( R = \frac{V}{I} \)). Line A is less steep than line B, meaning resistance of A is less than resistance of B. Since the equivalent resistance in a parallel combination is always less than in a series combination, A represents the parallel combination.

Teacher's Note:
a) Slope of V-I graph equals resistance (\( \Delta V / \Delta I \)).
b) Parallel combination offers lesser net resistance than series combination for the same resistors.

Question 3
(d) A music system draws a current of \( 400\text{ mA} \) when connected to a \( 12\text{V} \) battery.
(i) What is the resistance of the music system ?
(ii) The music system is left playing for several hours and finally the battery voltage drops and the music system stops playing when the current drops to \( 320\text{ mA} \). At what battery voltage does the music system stop playing ? [2 Marks]

Answer:
(i) Given \( I = 400\text{ mA} = 0.4\text{ A} \), \( V = 12\text{V} \).
Resistance \( R = \frac{V}{I} = \frac{12}{0.4} = 30\,\Omega \).
(ii) New current \( I' = 320\text{ mA} = 0.32\text{ A} \), resistance \( R = 30\,\Omega \).
New voltage \( V' = I' \times R = 0.32 \times 30 = 9.6\text{ V} \).

Teacher's Note:
a) Always convert milliamperes into amperes (\( \div 1000 \)) before applying Ohm's law.
b) Resistance of the music system remains constant even when the battery voltage drops.

Question 3
(e) Calculate the quantity of heat produced in a \( 20\,\Omega \) resistor carrying \( 2.5\text{ A} \) current in 5 minutes. [2 Marks]

Answer:
Given \( R = 20\,\Omega \), \( I = 2.5\text{ A} \), time \( t = 5\text{ minutes} = 5 \times 60 = 300\text{ s} \).
Heat produced \( H = I^2 R t \)
\( H = (2.5)^2 \times 20 \times 300 \)
\( H = 6.25 \times 20 \times 300 = 125 \times 300 = 37,500\text{ J} = 37.5\text{ kJ} \).

Teacher's Note:
a) Use Joule's law of heating: \( H = I^2 R t \).
b) Convert time from minutes to seconds and express final answer in Joules or kiloJoules clearly.

Question 4
(a) State the characteristics required of a good thermion emitter. [2 Marks]

Answer:
1. Low work function: So that electrons can be easily emitted even at relatively lower temperatures.
2. High melting point: So that the substance does not melt when heated to high temperatures required for thermionic emission.

Teacher's Note:
a) Tungsten coated with thoria is a common example satisfying these conditions.
b) Both low work function and high melting point are essential properties.

Question 4
(b) An element \( _Z^S A \) decays to \( _{85}R^{222} \) after emitting \( 2\alpha \) particles and \( 1\beta \) particle. Find the atomic number and atomic mass of the element S. [2 Marks]

Answer:
Emission of one \( \alpha \) particle decreases mass number by 4 and atomic number by 2. Emission of two \( \alpha \) particles decreases mass number by 8 and atomic number by 4.
Emission of one \( \beta \) particle increases atomic number by 1 while mass number remains unchanged.
Let initial element be \( _Z^S A \). After decay:
Mass number: \( S - 8 = 222 \implies S = 230 \).
Atomic number: \( Z - 4 + 1 = 85 \implies Z - 3 = 85 \implies Z = 88 \).
Atomic mass = 230, Atomic number = 88.

Teacher's Note:
a) Keep track of changes in mass number (\( A \)) and atomic number (\( Z \)) during alpha and beta decays.
b) Set up linear equations based on the final daughter nucleus values.

Question 4
(c) A radioactive substance is oxidized. Will there be any change in a nature of its radioactivity ? Give a reason for your answer. [2 Marks]

Answer:
No, there will be no change in the nature of its radioactivity.
Reason: Radioactivity is a nuclear phenomenon caused by the instability of the nucleus, whereas oxidation is a chemical process involving valence electrons. Physical or chemical changes do not affect the nucleus.

Teacher's Note:
a) Emphasize that radioactivity is strictly independent of external physical and chemical conditions.
b) State that it originates from the nucleus, not the electron shells.

Question 4
(d) State the characteristics required in a material to be used as an effective fuse wire. [2 Marks]

Answer:
1. High electrical resistivity.
2. Low melting point.

Teacher's Note:
a) High resistivity ensures it gets heated quickly when excess current flows.
b) Low melting point ensures it melts readily to break the circuit and protect appliances.

Question 4
(e) Which coil of a step up transformer is made thicker and why ? [2 Marks]

Answer:
The primary coil of a step-up transformer is made thicker.
Reason: A step-up transformer increases voltage and decreases current. Therefore, the primary coil carries a higher current than the secondary coil, requiring a thicker wire to prevent overheating.

Teacher's Note:
a) Thicker wires have lower resistance and are used to safely carry larger currents.
b) In a step-up transformer, primary current is greater than secondary current (\( I_p \gt I_s \)).

SECTION II

Question 5
(a) A stone of mass 'm' is rotated in a circular path with a uniform speed by tying a strong string with the help of your hand. Answer the following questions :
(i) Is the stone moving with a uniform or variable speed ?
(ii) Is the stone moving with a uniform acceleration ? In which direction does the acceleration act ?
(iii) What kind of force acts on the hand and state its direction ? [3 Marks]

Answer:
(i) Uniform speed (magnitude of velocity is constant, but direction changes continuously).
(ii) It is moving with variable velocity, hence it has acceleration (centripetal acceleration). The acceleration acts radially towards the center of the circular path.
(iii) Centripetal force acts on the hand, directed outwards (reaction to the centripetal force acting on the stone, directed along the string towards the center).

Teacher's Note:
a) Uniform speed does not mean uniform velocity because direction changes at every point in circular motion.
b) Distinguish clearly between the force acting on the stone (towards center) and the reaction force acting on the hand (away from center).

Question 5
(b) From the diagram given below, answer the questions that follow :
(i) What kind of pulleys are A and B ?
(ii) State the purpose of pulley B.
(iii) What effort has to be applied at C to just raise the load L = 20 kgf ? (Neglect the weight of pulley A and friction). [3 Marks]

[Figure: Pulley system showing a single movable pulley A connected to load L and a single fixed pulley B at the top over which string passes to effort C.]

Answer:
(i) A is a Single Movable Pulley and B is a Single Fixed Pulley.
(ii) The purpose of pulley B is to change the direction of force applied so that effort can be applied in a more convenient downward direction.
(iii) For equilibrium, \( L = 2T \) and effort \( E = T \). Therefore, \( E = \frac{L}{2} = \frac{20\text{ kgf}}{2} = 10\text{ kgf} \).

Teacher's Note:
a) A single fixed pulley has mechanical advantage equal to 1, while a single movable pulley has theoretical mechanical advantage equal to 2.
b) State the relation between load, tension, and effort clearly.

Question 5
(c) (i) An effort is applied on the bigger wheel of a gear having 32 teeth. It is used to turn a wheel of 8 teeth. Where is it used ?
(ii) A pulley system has three pulleys. A load of 120N is overcome by applying an effort of 50N. Calculate the Mechanical Advantage and Efficiency of this system. [4 Marks]

Answer:
(i) Since effort is applied on the driving gear with more teeth (32) and turns a wheel with fewer teeth (8), this is a gear system used to gain in speed (speed multiplier). It is used in bicycles / clocks.
(ii) Given number of pulleys \( n = 3 \), Load \( L = 120\text{ N} \), Effort \( E = 50\text{ N} \).
Velocity Ratio (\( \text{V.R.} \)) for 3 pulleys \( = 3 \).
Mechanical Advantage (\( \text{M.A.} \)) \( = \frac{L}{E} = \frac{120}{50} = 2.4 \).
Efficiency (\( \eta \)) \( = \frac{\text{M.A.}}{\text{V.R.}} \times 100 = \frac{2.4}{3} \times 100 = 80\% \).

Teacher's Note:
a) When driving gear has more teeth than driven gear, there is a gain in speed and loss in torque.
b) Always calculate efficiency as percentage by multiplying by 100.

Question 6
(a) (i) What is the principle of method of mixtures ?
(ii) What is the other name given to it ?
(iii) Name the law on which the principle is based. [3 Marks]

Answer:
(i) When a hot body is mixed or kept in thermal contact with a cold body, heat energy lost by the hot body is equal to the heat energy gained by the cold body, provided no heat is lost to the surroundings.
(ii) Principle of Calorimetry.
(iii) Law of Conservation of Energy.

Teacher's Note:
a) Mention the condition of full insulation (no heat exchange with surroundings) for validity.
b) The principle is a direct consequence of the law of conservation of energy.

Question 6
(b) Some ice is heated at a constant rate, and its temperature is recorded after every few seconds, till steam is formed at \( 100^{\circ}\text{C} \). Draw a temperature time graph to represent the change. Label the two phase changes on the graph. [3 Marks]

[Figure: Temperature-time graph starting below \( 0^{\circ}\text{C} \) for ice, rising to \( 0^{\circ}\text{C} \) (melting phase change), rising to \( 100^{\circ}\text{C} \), and staying at \( 100^{\circ}\text{C} \) (boiling phase change into steam).]

Answer:
The temperature-time graph shows two horizontal flat regions: one at \( 0^{\circ}\text{C} \) representing melting (fusion of ice into water) and another at \( 100^{\circ}\text{C} \) representing boiling (vaporisation of water into steam), with properly labeled axes and phase change zones.

Teacher's Note:
a) During phase change, heat supplied is latent heat used to change state, so temperature remains constant.
b) Ensure both \( 0^{\circ}\text{C} \) and \( 100^{\circ}\text{C} \) plateaus are clearly labeled.

Question 6
(c) A copper vessel of mass \( 100\text{ g} \) contains \( 150\text{ g} \) of water at \( 50^{\circ}\text{C} \). How much ice is needed to cool it to \( 5^{\circ}\text{C} \) ?
Given: Specific heat capacity of copper = \( 0.4\text{ J g}^{-1}\text{ }^{\circ}\text{C}^{-1} \), Specific heat capacity of water = \( 4.2\text{ J g}^{-1}\text{ }^{\circ}\text{C}^{-1} \), Specific latent heat of fusion of ice = \( 336\text{ J g}^{-1} \). [4 Marks]

Answer:
Heat lost by water + Heat lost by copper vessel = Heat gained by ice to melt at \( 0^{\circ}\text{C} \) + Heat gained by melted ice to reach \( 5^{\circ}\text{C} \).
\( m_w c_w \Delta t + m_c c_c \Delta t = M L + M c_w \Delta t' \)
\( [150 \times 4.2 \times (50 - 5)] + [100 \times 0.4 \times (50 - 5)] = (M \times 336) + (M \times 4.2 \times 5) \)
\( [150 \times 4.2 \times 45] + [100 \times 0.4 \times 45] = 336M + 21M \)
\( 28350 + 1800 = 357 M \)
\( 30150 = 357 M \)
\( M = \frac{30150}{357} = 84.45\text{ g} \) of ice.

Teacher's Note:
a) Account for both the water and the copper container when calculating heat lost.
b) Remember that the ice first melts at \( 0^{\circ}\text{C} \) and then the resulting water warms up from \( 0^{\circ}\text{C} \) to \( 5^{\circ}\text{C} \).

Question 7
(a) (i) Write a relationship between angle of incidence and angle of refraction for a given pair of media.
(ii) When a ray of light enters from one medium to another having different optical densities it bends. Why does this phenomenon occur ?
(iii) Write one condition where it does not bend when entering a medium of different optical density. [3 Marks]

Answer:
(i) Snell's law: The ratio of the sine of angle of incidence (\( i \)) to the sine of angle of refraction (\( r \)) is constant for a given pair of media (\( \frac{\sin i}{\sin r} = {}^1\mu_2 \)).
(ii) It occurs due to the change in speed of light when it passes from one medium to another of different optical density.
(iii) When a ray of light is incident normally (\( i = 0^{\circ} \)) on the boundary of the second medium, it passes undeviated.

Teacher's Note:
a) State Snell's law clearly using sine ratios.
b) Normal incidence means angle of incidence is zero, hence angle of refraction is also zero.

Question 7
(b) A lens produces a virtual image between the object and the lens.
(i) Name the lens.
(ii) Draw a ray diagram to show the formation of this image. [3 Marks]

[Figure: Ray diagram for a concave lens showing an object placed in front of it, forming a diminished, virtual, and erect image on the same side between the object and the optical centre.]

Answer:
(i) Concave lens.
(ii) The ray diagram shows a concave lens with an object placed anywhere in front of it. A ray parallel to the principal axis diverges after refraction and appears to come from the focus, while a ray passing through the optical center goes undeviated. Their intersection forms a virtual, erect, and diminished image between the object and the lens.

Teacher's Note:
a) A concave lens always produces a virtual, erect, and diminished image for any real object position.
b) Label principal axis, optical centre, focus, object, and image clearly.

Question 7
(c) What do you understand by the term 'Scattering of light' ? Which colour of white light is scattered the least and why ? [4 Marks]

Answer:
Scattering of light is the process of absorption and subsequent re-emission of light energy in various directions by particles or molecules of the medium whose size is comparable to the wavelength of light.
Red colour is scattered the least because scattering intensity is inversely proportional to the fourth power of wavelength (\( I \propto \frac{1}{\lambda^4} \)), and red light has the longest wavelength among visible colours.

Teacher's Note:
a) Mention Rayleigh's scattering law linking scattering intensity with wavelength (\( \lambda^4 \)).
b) Violet is scattered the most and red the least.

Question 8
(a) (i) Name the waves used for echo depth sounding.
(ii) Give one reason for their use for the above purpose.
(iii) Why are the waves mentioned by you not audible to us ? [3 Marks]

Answer:
(i) Ultrasonic waves.
(ii) They can travel undeviated through long distances and can be confined to a narrow beam without significant diffraction.
(iii) Because their frequency is greater than \( 20,000\text{ Hz} \), which lies above the upper limit of human hearing (\( 20\text{ Hz} \) to \( 20,000\text{ Hz} \)).

Teacher's Note:
a) Ultrasonic waves have high frequency and short wavelength, making them ideal for obstacle detection.
b) State the human audible frequency range clearly.

Question 8
(b) (i) What is an echo ?
(ii) State two conditions for an echo to take place. [3 Marks]

Answer:
(i) An echo is the sound heard after reflection from a distant obstacle after the original sound has ceased.
(ii) Conditions for an echo:
1. The minimum distance between the source of sound and the reflecting surface must be \( 17\text{ m} \).
2. The time interval between the original sound and the reflected sound must be at least \( 0.1\text{ s} \).

Teacher's Note:
a) Persistence of human hearing is \( 0.1\text{ s} \).
b) Take speed of sound in air as \( 340\text{ m/s} \) to verify the \( 17\text{ m} \) distance (\( d = \frac{340 \times 0.1}{2} = 17\text{ m} \)).

Question 8
(c) (i) Name the phenomenon involved in tuning a radio set to a particular station.
(ii) Define the phenomenon named by you in part (i) above.
(iii) What do you understand by loudness of sound ?
(iv) In which units is the loudness of sound measured ? [4 Marks]

Answer:
(i) Resonance.
(ii) Resonance is a condition of forced vibrations when the frequency of an externally applied periodic force matches the natural frequency of a body, causing the body to vibrate with a very large amplitude.
(iii) Loudness is the characteristic of sound by which a loud sound can be distinguished from a faint sound, both having the same pitch and frequency. It depends on the amplitude of the wave.
(iv) Phon (or decibel - dB).

Teacher's Note:
a) Tuning a radio relies on electrical resonance matching the circuit frequency to the incoming broadcast frequency.
b) Loudness is a subjective property related to intensity and amplitude.

Question 9
(a) (i) Which particles are responsible for current in conductors ?
(ii) To which wire of a cable in a power circuit should the metal case of a geyser be connected ?
(iii) To which wire should the fuse be connected ? [3 Marks]

Answer:
(i) Free electrons.
(ii) Earth wire (Green/Yellow).
(iii) Live wire (Red/Brown).

Teacher's Note:
a) Free electrons constitute electric current in metallic conductors.
b) Fuses must always be placed in the live wire to disconnect the appliance from high potential during faults.

Question 9
(b) (i) Name the transformer used in the power transmitting station of a power plant.
(ii) What type of current is transmitted from the power station ?
(iii) At what voltage is this current available to our household ? [3 Marks]

Answer:
(i) Step-up transformer.
(ii) Alternating Current (AC).
(iii) \( 220\text{ V} \) (at \( 50\text{ Hz} \)).

Teacher's Note:
a) Step-up transformers are used to transmit power at high voltage to minimize line energy losses.
b) Standard domestic supply in India is \( 220\text{ V} \) AC.

Question 9
(c) A battery of emf \( 12\text{ V} \) and internal resistance \( 2\,\Omega \) is connected with two resistors A and B of resistance \( 4\,\Omega \) and \( 6\,\Omega \) respectively joined in series.
Find
(i) Current in the circuit.
(ii) The terminal voltage of the cell.
(iii) The potential difference across \( 6\,\Omega \) Resistor.
(iv) Electrical energy spent per minute in \( 4\,\Omega \) Resistor. [4 Marks]

Answer:
Given \( E = 12\text{ V} \), internal resistance \( r = 2\,\Omega \), \( R_1 = 4\,\Omega \), \( R_2 = 6\,\Omega \).
Total resistance \( R_{\text{total}} = R_1 + R_2 + r = 4 + 6 + 2 = 12\,\Omega \).
(i) Current in the circuit \( I = \frac{E}{R_{\text{total}}} = \frac{12}{12} = 1\text{ A} \).
(ii) Terminal voltage \( V = E - Ir = 12 - (1 \times 2) = 10\text{ V} \).
(iii) Potential difference across \( 6\,\Omega \) resistor \( V_2 = I \times R_2 = 1 \times 6 = 6\text{ V} \).
(iv) Electrical energy spent per minute in \( 4\,\Omega \) resistor \( H = I^2 R_1 t = (1)^2 \times 4 \times 60 = 240\text{ J} \).

Teacher's Note:
a) Include internal resistance when calculating total circuit resistance.
b) Time \( t \) for one minute must be substituted as \( 60\text{ seconds} \).

Question 10
(a) Arrange \( \alpha \), \( \beta \) and \( \gamma \) rays in ascending order with respect to their:
(i) Penetrating power.
(ii) Ionising power.
(iii) Biological effect. [3 Marks]

Answer:
(i) Penetrating power: \( \alpha \lt \beta \lt \gamma \).
(ii) Ionising power: \( \gamma \lt \beta \lt \alpha \).
(iii) Biological effect: \( \alpha \lt \beta \lt \gamma \).

Teacher's Note:
a) Penetrating power and biological damage are highest for gamma rays and lowest for alpha particles.
b) Ionising power is inversely related to penetrating power, making alpha particles the most strongly ionising.

Question 10
(b) (i) In a cathode ray tube what is the function of anode ?
(ii) State the energy conversion taking place in a cathode ray tube.
(iii) Write one use of cathode ray tube. [3 Marks]

Answer:
(i) The function of anode is to accelerate the emitted electrons and focus them into a fine, high-speed beam.
(ii) Electrical energy is converted into kinetic energy of electrons and finally into light energy (fluorescence on screen).
(iii) Used as a display screen in oscilloscopes (CRO) to view waveforms.

Teacher's Note:
a) High positive potential on anode attracts and accelerates electrons.
b) Mention CRO or oscilloscope as a primary application.

Question 10
(c) (i) Represent the change in the nucleus of a radioactive element when a \( \beta \) particle is emitted.
(ii) What is the name given to elements with same mass number and different atomic number ?
(iii) Under which conditions does the nucleus of an atom tend to be radioactive ? [4 Marks]

Answer:
(i) \( _Z^A X \rightarrow _{Z+1}^A Y + _{-1}^0\beta \).
(ii) Isobars.
(iii) Nuclei with atomic number greater than 82 (\( Z \gt 82 \)), or nuclei where the neutron-to-proton ratio is unstable (excess of neutrons or protons), tend to be radioactive.

Teacher's Note:
a) During beta emission, mass number remains unchanged while atomic number increases by 1.
b) Elements beyond lead (\( Z = 82 \)) are naturally radioactive due to nuclear instability.