ICSE Class 9 Physics Sample Paper with Solutions Set 02

SECTION A (40 Marks)

Attempt all Questions from this Section

Question 1

(i) If car A is travelling at the speed of \( 80\text{ km/h} \) and car B at \( 60\text{ km/h} \) in the same direction, what is the velocity of the car A relative to the car B? [1 Mark]
a) \( 10\text{ km/h} \)
b) \( 20\text{ km/h} \)
c) \( 30\text{ km/h} \)
d) \( 40\text{ km/h} \)

Answer: (b) 20 km/h

The velocity of car A relative to car B is \( 80\text{ km/h} - 60\text{ km/h} = 20\text{ km/h} \).

Teacher's Note:
a) When two bodies move in the same direction, their relative velocity is the difference of their individual velocities.
b) Students often add the speeds instead of subtracting when bodies move in the same direction.

(ii) If you whirl a stone on the end of the string and the string suddenly breaks what will happen to the stone? [1 Mark]
a) Fly directly towards you.
b) Spiral in towards your hand.
c) Spiral away from your hand.
d) Fly off along the tangent to its circular path.

Answer: (d) Fly off along the tangent to its circular path

At any point on a circular path, the velocity vector is directed along the tangent. When the centripetal force ceases (string breaks), the stone flies off tangentially due to inertia of direction.

Teacher's Note:
a) Inertia of direction ensures that a body in circular motion continues along the tangent when the binding force disappears.
b) Do not confuse this with centrifugal force pulling the object outward.

(iii) The equations of motion can be derived by using [1 Mark]
a) Distance-time graph
b) Displacement-time graph
c) Velocity-time graph for uniform acceleration
d) Velocity-time graph for non-uniform acceleration

Answer: (c) Velocity-time graph for uniform acceleration

The standard graphical derivation of equations of motion uses a velocity-time graph under the condition of uniform acceleration.

Teacher's Note:
a) The slope of a velocity-time graph gives acceleration, and the area under it gives displacement.
b) Ensure students specify 'uniform acceleration' as non-uniform acceleration graphs cannot use these standard algebraic equations.

(iv) Newton's law of gravitation is applicable to [1 Mark]
a) Objects in solar system
b) Planets and stars only
c) Objects on Earth only
d) All bodies in the universe

Answer: (d) All bodies in the universe

Newton's law of gravitation is a universal law, applicable between any two masses anywhere in the universe.

Teacher's Note:
a) It is termed universal because it applies to terrestrial as well as celestial bodies irrespective of their size.
b) Remember that gravitational force is always attractive in nature.

(v) Assertion: When compared to a hollow tennis ball, a solid cricket ball is more difficult to stop.
Reason: Inertia of an object is inversely proportional to its mass. [1 Mark]
a) Both A and R are true and R is the correct explanation of A
b) Both A and R are true and R is not the correct explanation of A
c) Assertion is false but reason is true.
d) Assertion is true reason is false.

Answer: (d) Assertion is true reason is false.

A solid cricket ball has more mass and hence more inertia, making it harder to stop. Inertia is directly proportional to mass, not inversely proportional.

Teacher's Note:
a) Mass is a measure of the inertia of a body; greater mass implies greater inertia.
b) Read assertion-reason statements carefully to check proportionality relationships.

(vi) The same body is immersed in two liquids A and B in succession. The extent to which the body sinks in liquid B is less than in liquid A. What are the conclusions that could be derived from such an observation? [1 Mark]
a) Density of liquid B is more than liquid A
b) Density of liquid A is more than liquid B
c) No such conclusion can be made
d) Density of the solid is less than the liquid in both

Answer: (a) Density of liquid B is more than liquid A

Since the body sinks less in liquid B, a lesser volume is submerged, meaning liquid B exerts greater upthrust, which implies liquid B is denser than liquid A.

Teacher's Note:
a) For floating bodies, greater density of liquid results in lesser immersion depth.
b) Upthrust equals the weight of the displaced liquid.

(vii) A typical solar cell can produce about _____ electricity. [1 Mark]
a) \( 7\text{ W} \)
b) \( 10\text{ W} \)
c) \( 0.7\text{ W} \)
d) \( 0.5\text{ W} \)

Answer: (c) 0.7 W

A standard single commercial solar cell generates a power output of about \( 0.7\text{ W} \).

Teacher's Note:
a) Solar cells convert solar energy directly into electrical energy based on the photovoltaic effect.
b) Students should memorize standard textbook factual data regarding energy sources.

(viii) What is the major problem in harnessing nuclear energy? [1 Mark]
a) Splitting of the nuclei
b) Duration for which the reaction has to be sustained.
c) Safe disposal of the spent fuel
d) Conversion of nuclear energy into electrical energy.

Answer: (c) Safe disposal of the spent fuel

The major hazard and problem of nuclear power generation is the safe storage and disposal of radioactive spent or used fuels.

Teacher's Note:
a) Spent nuclear fuel remains radioactive and hazardous for thousands of years.
b) Improper disposal causes severe environmental contamination and health risks.

(ix) An observer moves towards a stationary plane mirror at a speed of \( 4\text{ m/s} \). The speed of image with respect to mirror is, [1 Mark]
a) \( 4\text{ m/s} \) towards the mirror
b) \( 8\text{ m/s} \) towards the mirror
c) \( 4\text{ m/s} \) away from the mirror
d) \( 8\text{ m/s} \) away from the mirror

Answer: (a) 4 m/s towards the mirror

The speed of the image with respect to the mirror is equal to the speed of the object with respect to the mirror, which is \( 4\text{ m/s} \).

Teacher's Note:
a) Distinguish carefully between speed relative to the mirror and speed relative to the observer.
b) The relative speed of the image with respect to the observer would be \( 8\text{ m/s} \), but here it asks with respect to the mirror.

(x) The size of the image formed by the convex mirror is [1 Mark]
a) Always smaller than the size of the object
b) Always greater than the size of the object
c) Same as the size of the object
d) Depends on the position of the object

Answer: (a) Always smaller than the size of the object

A convex mirror always forms a virtual, erect, and diminished image for all real positions of the object.

Teacher's Note:
a) Because of this property, convex mirrors offer a wider field of view and are used as rear-view mirrors in vehicles.
b) Do not confuse convex mirrors with convex lenses.

(xi) Rahul is standing \( 35\text{ cm} \) in front of a plane mirror. The mirror is moved \( 2\text{ cm} \) towards Rahul. What will be the distance between Rahul and his image? [1 Mark]
a) \( 74\text{ cm} \)
b) \( 64\text{ cm} \)
c) \( 66\text{ cm} \)
d) \( 76\text{ cm} \)

Answer: (c) 66 cm

When the mirror moves \( 2\text{ cm} \) towards Rahul, the new distance between Rahul and the mirror is \( 35 - 2 = 33\text{ cm} \). The image is formed at \( 33\text{ cm} \) behind the mirror. Total distance between Rahul and his image = \( 33 + 33 = 66\text{ cm} \).

Teacher's Note:
a) Object distance from mirror equals image distance from mirror.
b) Recalculate new distances step-by-step whenever the mirror or object is shifted.

(xii) Which of the following is correctly arranged in the increasing order of speed of sound in steel, distilled water, sea water and oxygen? [1 Mark]
a) Steel > Distilled water > Sea water > Oxygen
b) Oxygen < Distilled water = Sea water < Steel
c) Oxygen < Distilled water < Sea water < Steel
d) Steel < Distilled water = Sea water < Oxygen

Answer: (c) Oxygen < Distilled water < Sea water < Steel

Sound travels slowest in gases (oxygen), faster in liquids (distilled water, with sea water being slightly faster due to dissolved salts), and fastest in solids (steel).

Teacher's Note:
a) Speed of sound depends on elasticity and density, generally increasing from gas to liquid to solid.
b) Sea water has dissolved impurities which increase its density and elasticity factors slightly compared to pure water.

(xiii) The instrument used to control current in an electric circuit is [1 Mark]
a) Ammeter
b) cell
c) plug key
d) Rheostat

Answer: (d) Rheostat

A rheostat is a variable resistor used to continuously regulate or control the magnitude of current in a circuit.

Teacher's Note:
a) Ammeter measures current, cell provides source EMF, and plug key makes or breaks the circuit.
b) Rheostat works by changing the length of resistance wire in the circuit path.

(xiv) What is the function of a key (or switch) in an electric circuit? [1 Mark]
a) To put on current in a circuit.
b) To put off current in a circuit.
c) To put on or off current in a circuit.
d) To control the rate of current flowing through the circuit.

Answer: (c) To put on or off current in a circuit.

A switch or key acts as a control device to open or close an electric circuit.

Teacher's Note:
a) An open key breaks the path, stopping the current; a closed key completes the path.
b) Option (c) is comprehensive as it covers both turning on and turning off.

(xv) If the field lines in a magnetic field are parallel and equidistant, the magnetic field is [1 Mark]
a) non uniform
b) uniform
c) zero
d) nothing can be decided

Answer: (b) uniform

Parallel and equidistant magnetic field lines represent a region of uniform magnetic field in both magnitude and direction.

Teacher's Note:
a) Non-uniform magnetic fields have converging, diverging, or non-equidistant field lines.
b) The Earth's magnetic field in a limited region can be considered nearly uniform.

Question 2

(i) Complete the following by choosing the correct answers from the bracket: [6 Marks]
a) Sound wave is a _____ [longitudinal/ electromagnetic/transverse] wave.
b) When ice melts, its temperature _____ [increases/ decreases/remains constant].
c) When a spring is held fixed at one end and the other end is stretched, the spring tends to regain its original shape due to _____ [Restoring Force/ Tensile Force/ Compressive Force].
d) Nuclear fusion reactions are also known as _____ [thermo-nuclear/chain-reaction] reactions.
e) A glass rod is rubbed with silk. The charge on the glass rod is _____ [Positive/Negative], and the charge on the silk is _____ [Positive/Negative].

Answer:
a) longitudinal
b) remains constant
c) Restoring Force
d) thermo-nuclear
e) Glass rod: Positive, Silk: Negative.

Teacher's Note:
a) Sound waves require a material medium and propagate via compressions and rarefactions (longitudinal).
b) During a phase change (melting), latent heat is absorbed without a change in temperature.

(ii) Amplitude of a transverse wave is \( 1.5\text{ m} \) and wavelength is \( 5\text{ m} \). If the velocity of wave is \( 300\text{ m/s} \), find the frequency of the wave. [2 Marks]

Answer:
Given: Velocity \( v = 300\text{ m/s} \), Wavelength \( \lambda = 5\text{ m} \).
Formula: \( v = f \times \lambda \)
Frequency \( f = \frac{v}{\lambda} = \frac{300}{5} = 60\text{ Hz} \).

Teacher's Note:
a) State the wave equation clearly before substituting values.
b) Frequency is measured in Hertz (Hz) or \( \text{s}^{-1} \).

(iii) In the given circuit diagram, you are asked to measure the current in the resistance wire XY and potential difference across it. Name the instruments that you would use. [2 Marks]

[Figure: Circuit diagram showing a resistor XY connected across a source marked with + and - terminals, with dots for meter connections]

Answer:
To measure current, an ammeter is connected in series with the resistor XY.
To measure potential difference, a voltmeter is connected in parallel across the resistor XY.

Teacher's Note:
a) Ammeters have low resistance and must always be connected in series.
b) Voltmeters have very high resistance and must always be connected in parallel.

Question 3

(i) Will the weight of \( 1\text{ kg} \) of iron and \( 1\text{ kg} \) of cotton be the same in air? Explain. [2 Marks]

Answer:
1. No, their true weights are the same (\( 9.8\text{ N} \)), but their apparent weights in air will differ.
2. Cotton has a much larger volume than iron, so it experiences a greater buoyant force (upthrust) from the air, making its apparent weight slightly less than that of iron.

Teacher's Note:
a) Apparent weight equals true weight minus upthrust of air.
b) Since volume of cotton is greater, air upthrust is greater on cotton.

(ii) A body, whose volume is \( 100\text{ cm}^3 \), weighs \( 1\text{ kg} \) in the air. Find its weight in water.
(Density of water = \( 10^3\text{ kg/m}^3 \), \( g = 10\text{ m/s}^2 \)) [2 Marks]

Answer:
1. Volume \( V = 100\text{ cm}^3 = 10^{-4}\text{ m}^3 \).
2. Weight in air \( W_a = m \times g = 1\text{ kg} \times 10\text{ m/s}^2 = 10\text{ N} \).
3. Upthrust in water = Weight of displaced water = \( V \times \rho_w \times g = 10^{-4}\text{ m}^3 \times 10^3\text{ kg/m}^3 \times 10\text{ m/s}^2 = 1\text{ N} \).
4. Weight in water = \( W_a - \text{Upthrust} = 10\text{ N} - 1\text{ N} = 9\text{ N} \).

Teacher's Note:
a) Convert all units carefully to SI units before calculation.
b) Apparent weight in liquid is real weight minus buoyant force.

(iii) Give the order of magnitude of the following
i. Height of Mount Everest = \( 9 \times 10^3\text{ m} \)
ii. Mass of a virus = \( 1 \times 10^{-15}\text{ kg} \)
iii. Life expectancy of a man = \( 2 \times 10^9\text{ s} \)
iv. A speck of dust = \( 7 \times 10^{-10}\text{ kg} \). [3 Marks]

Answer:
i. \( 9 \times 10^3\text{ m} \) (\( 9 > 3.2 \)): Order of magnitude is \( 10^{3+1} = \mathbf{10^4} \).
ii. \( 1 \times 10^{-15}\text{ kg} \) (\( 1 < 3.2 \)): Order of magnitude is \mathbf{\( 10^{-15} \)}.
iii. \( 2 \times 10^9\text{ s} \) (\( 2 < 3.2 \)): Order of magnitude is \mathbf{\( 10^9 \)}.
iv. \( 7 \times 10^{-10}\text{ kg} \) (\( 7 > 3.2 \)): Order of magnitude is \( 10^{-10+1} = \mathbf{10^{-9}} \).

Teacher's Note:
a) Rule for order of magnitude: Express number as \( N \times 10^n \) where \( 1 \le N < 10 \). If \( N > \sqrt{10} \approx 3.22 \), order is \( 10^{n+1} \).
b) Check significant figures and rounding rules carefully.

(iv) Two bodies of masses \( m_1 \) and \( m_2 \) are released from heights \( h_1 \) and \( h_2 \), respectively. Obtain the ratio of the time taken by both the bodies to reach the earth. [2 Marks]

Answer:
1. From the second equation of motion, \( h = ut + \frac{1}{2}gt^2 \). Since initial velocity \( u = 0 \), \( h = \frac{1}{2}gt^2 \).
2. Time \( t = \sqrt{\frac{2h}{g}} \).
3. Since time of fall is independent of mass, for heights \( h_1 \) and \( h_2 \), the ratio of times is \( \frac{t_1}{t_2} = \sqrt{\frac{h_1}{h_2}} \).

Teacher's Note:
a) Emphasize that free-fall acceleration \( g \) is constant and independent of object mass.
b) Keep algebraic steps clean and clear.

(v) Explain why deep-sea divers need special protective wear. [2 Marks]

Answer:
1. As depth increases, hydrostatic pressure increases significantly, which can cause severe injury or damage to human tissues and blood vessels.
2. Special diving suits protect the body from high pressure and also help prevent loss of body heat in cold deep waters.

Teacher's Note:
a) Water pressure increases by about \( 1\text{ atm} \) for every \( 10\text{ m} \) of depth.
b) Protective wear also includes breathing gas mixtures to avoid nitrogen narcosis.

(vi) In summer, why do dogs hang their tongue while lying under trees? [2 Marks]

Answer:
1. Dogs have very few sweat glands on their body. To regulate their body temperature, they pant and hang their tongues out.
2. Saliva on the tongue evaporates, taking away latent heat from the body and providing a cooling effect.

Teacher's Note:
a) Evaporation causes cooling due to the extraction of latent heat.
b) This is a physiological mechanism of thermoregulation in canines.

(vii) Define retardation. Which of the following graph represents uniform retardation? [2 Marks]

[Figure: Four graphs labeled (a) Distance-time horizontal line, (b) Speed-time sloping up, (c) Speed-time curved, (d) Speed-time sloping down to zero]

Answer:
Retardation (or deceleration) is defined as negative acceleration, meaning the rate of decrease of velocity with respect to time.
Graph (d) represents uniform retardation.

Teacher's Note:
a) A speed-time graph with a constant negative slope indicates uniform retardation.
b) State the definition precisely in terms of velocity decrease per unit time.

SECTION B (40 Marks)

Attempt any four Questions from this Section

Question 4

(i) What is the function of ratchet in a screw gauge? A screw gauge has positive error of 7 divisions such that its main scale is marked in half mm and circular scale has 100 divisions. The spindle of the screw advances by 1 division on one complete revolution. If the screw gauge reading is 9 divisions on the main scale and 67 divisions on the circular scale for the diameter of a wire, calculate
a) Pitch
b) Least count and
c) Correct diameter. [4 Marks]

Answer:
Ratchet prevents undue pressure on the object by slipping when optimal contact is made.
a) Pitch = \( \frac{\text{Distance moved}}{\text{Number of rotations}} = \frac{0.5\text{ mm}}{1} = 0.5\text{ mm} = 0.05\text{ cm} \).
b) Least Count (\(\text{LC}\)) = \( \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{0.05\text{ cm}}{100} = 0.0005\text{ cm} \).
Observed Reading = Main Scale Reading + (Circular Scale Reading \(\times\) LC)
\(= 9 \times 0.05\text{ cm} + 67 \times 0.0005\text{ cm} = 0.45\text{ cm} + 0.0335\text{ cm} = 0.4835\text{ cm} \).
c) Correction = \( -(\text{Positive error} \times \text{LC}) = -(7 \times 0.0005\text{ cm}) = -0.0035\text{ cm} \).
Correct Diameter = Observed Reading - Error = \( 0.4835\text{ cm} - 0.0035\text{ cm} = 0.4800\text{ cm} \).

Teacher's Note:
a) Always subtract positive zero error and add negative zero error to find the correct reading.
b) Maintain proper unit conversions between millimeters and centimeters.

(ii) A girl standing on an oscillating swing sits down. How does the time period of a swing get affected? What happens to its frequency of oscillation? [3 Marks]

Answer:
1. When the girl sits down, her centre of gravity is lowered, effectively shortening the length of the pendulum (\( l \)).
2. Since time period \( T = 2\pi\sqrt{l/g} \), a decrease in effective length decreases the time period.
3. Frequency \( f = 1/T \), so the frequency of oscillation increases.

Teacher's Note:
a) Link effective length changes to the time period formula of a simple pendulum.
b) Time period and frequency are inversely related.

(iii) Define the term volume. Mass of a sphere of radius \( 1.4\text{ m} \) is \( 500\text{ kg} \). Calculate the density of the material of the sphere. [3 Marks]

Answer:
Volume is defined as the three-dimensional space occupied by an object or substance.
Given: Radius \( r = 1.4\text{ m} \), Mass \( m = 500\text{ kg} \).
Volume of sphere \( V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (1.4\text{ m})^3 = \frac{4}{3} \times \frac{22}{7} \times 2.744 = 11.498\text{ m}^3 \).
Density \( \rho = \frac{\text{Mass}}{\text{Volume}} = \frac{500\text{ kg}}{11.498\text{ m}^3} = 43.48\text{ kg/m}^3 \).

Teacher's Note:
a) Memorize standard geometric volume formulas like that of a sphere.
b) Double-check decimal arithmetic when cubing radii.

Question 5

(i) Given figure shows a velocity-time graph for two cars A and B starting from the same point in the same direction. [4 Marks]
Calculate the following:
a) Acceleration of car A.
b) Acceleration of car B between \( 2\text{ s} - 5\text{ s} \).
c) At what time intervals, both cars have same velocity?
d) Which car is ahead after \( 10\text{ s} \) and how much?

[Figure: Velocity-time graph showing two lines for Car A and Car B intersecting across time axis from 0 to 10 s]

Answer:
a) Acceleration of car A = Slope of line A = \( \frac{(35 - 0)\text{ m/s}}{10\text{ s}} = 3.5\text{ m/s}^2 \).
b) Acceleration of car B between \( 2\text{ s} \) and \( 5\text{ s} \) = \( \frac{(25 - 0)\text{ m/s}}{3\text{ s}} = 8.33\text{ m/s}^2 \).
c) Both cars have the same velocity at \( t = 3\text{ s} \) and \( t = 7\text{ s} \) (points of intersection).
d) Distance by Car A = Area under graph A = \( \frac{1}{2} \times 35 \times 10 = 175\text{ m} \).
Distance by Car B = Area under graph B = trapezium area = \( \frac{1}{2} \times (8 + 5) \times 25 = 162.5\text{ m} \).
Car A is ahead by \( 175 - 162.5 = 12.5\text{ m} \).

Teacher's Note:
a) Slope of velocity-time graph gives acceleration; area gives displacement.
b) Read intersection points and graph coordinates carefully.

(ii) Derive the first equation of motion. [3 Marks]

Answer:
1. Let a body move with initial velocity \( u \). After time \( t \), its velocity becomes \( v \) under uniform acceleration \( a \).
2. By definition, acceleration is the rate of change of velocity: \( a = \frac{v - u}{t} \).
3. Rearranging the terms gives \( at = v - u \), which yields the first equation of motion: \( v = u + at \).

Teacher's Note:
a) Define all symbols (\( u, v, a, t \)) clearly at the beginning of the derivation.
b) This is a fundamental kinematic relation used frequently in numerical problems.

(iii) A stone dropped from the top of a cliff reaches ground level in \( 4\text{ s} \) and buries itself \( 0.8\text{ m} \) into the mud. Calculate,
a) height of cliff
b) final velocity of stone on reaching level and
c) retardation produced by the mud. Take \( g = 9.8\text{ m/s}^2 \). [3 Marks]

Answer:
a) For free fall: \( u = 0 \), \( t = 4\text{ s} \), \( g = 9.8\text{ m/s}^2 \).
Height \( h = ut + \frac{1}{2}gt^2 = 0 + \frac{1}{2}(9.8)(4)^2 = 78.4\text{ m} \).
b) Final velocity before hitting ground \( v = u + gt = 0 + (9.8 \times 4) = 39.2\text{ m/s} \).
c) In the mud: initial velocity \( u = 39.2\text{ m/s} \), final velocity \( v = 0 \), distance \( s = 0.8\text{ m} \).
Using \( v^2 = u^2 + 2as \): \( 0 = (39.2)^2 + 2 \times a \times 0.8 \)
\( 1.6a = -1536.64 \implies a = -960.4\text{ m/s}^2 \). Retardation is \( 960.4\text{ m/s}^2 \).

Teacher's Note:
a) Split the problem into two distinct stages: free fall in air and motion inside the mud.
b) Retardation is negative acceleration, but when asked for magnitude or term retardation, specify the positive value or state negative sign clearly.

Question 6

(i) It is experienced that in our daily life Newton's first law is contradicted. Comment. [3 Marks]

Answer:
1. In daily life, a rolling ball stops on its own without any visible agency acting on it, which seemingly contradicts Newton's first law.
2. However, this happens because of external frictional forces exerted by the ground and air resistance.
3. In a hypothetical space free from all external forces, an object continues its state of uniform motion indefinitely, validating Newton's first law.

Teacher's Note:
a) Friction is an ever-present force in terrestrial experiments that masks the law of inertia.
b) Newton's first law defines force qualitatively and introduces the concept of inertia.

(ii) Sound takes \( 3\text{ s} \) to reach a certain distance from the source placed in air. How much time will it take to reach the same distance when the source is in water?
Take speed of sound in air as \( 330\text{ m/s} \) and in water as \( 1650\text{ m/s} \). [3 Marks]

Answer:
Given: Speed in air \( v_a = 330\text{ m/s} \), time \( t_a = 3\text{ s} \).
Distance \( d = v_a \times t_a = 330 \times 3 = 990\text{ m} \).
Time taken in water \( t_w = \frac{d}{v_w} = \frac{990\text{ m}}{1650\text{ m/s}} = 0.6\text{ s} \).

Teacher's Note:
a) Calculate the constant distance first using air conditions.
b) Sound travels faster in water, hence time taken is much less.

(iii) Define coefficient of linear expansion. Give its SI unit. State the factors on which the thermal expansion of a solid depends. [4 Marks]

Answer:
Coefficient of linear expansion is defined as the fractional change in length per degree rise in temperature.
SI unit: \( \text{K}^{-1} \) or \( ^\circ\text{C}^{-1} \).
Factors on which thermal expansion depends:
1. Nature of the material.
2. Initial dimensions of the solid.
3. Rise in temperature.

Teacher's Note:
a) State the formula \( \alpha = \frac{\Delta l}{l_0 \Delta t} \) to support the definition.
b) Different materials expand by different amounts for the same temperature change.

Question 7

(i) Describe an experiment to demonstrate thermal expansion in gases. [4 Marks]

[Figure: Round bottom flask fitted with one-holed rubber cork and a capillary tube dipping into a beaker of water]

Answer:
1. Take a round-bottom flask filled with air, fitted with a one-holed rubber cork and a narrow capillary tube dipping into a beaker of colored water.
2. Gently heat the flask with a spirit lamp or by cupping it with your hands.
3. Observation: Air inside expands and bubbles emerge out of the capillary tube into the water.
4. Conclusion: Gases expand on heating.

Teacher's Note:
a) Gases show maximum thermal expansion compared to liquids and solids due to weak intermolecular forces.
b) Mention apparatus setup clearly in experimental questions.

(ii) The mercury falls by \( 8/15 \) parts between two standard points on a Celsius thermometer, when the boiling water at \( 100^\circ\text{C} \) is allowed to cool to room temperature. Calculate room temperature in
(i) Celsius scale (ii) Fahrenheit scale. [3 Marks]

Answer:
Total divisions on Celsius scale between lower fixed point and upper fixed point = \( 100 \).
Fall in temperature = \( \frac{8}{15} \times 100^\circ\text{C} = \frac{160}{3}^\circ\text{C} = 53.33^\circ\text{C} \).
(i) Celsius temperature = \( 100^\circ\text{C} - 53.33^\circ\text{C} = 46.67^\circ\text{C} \).
(ii) Fahrenheit temperature: \( F = \frac{9}{5}C + 32 = \frac{9}{5}(46.67) + 32 = 84.006 + 32 = 116.01^\circ\text{F} \).

Teacher's Note:
a) Use the standard temperature conversion formula \( \frac{C}{5} = \frac{F - 32}{9} \).
b) Pay close attention to fractional scale drops.

(iii)
i. What is the difference in the wavelength of infrared rays emitted from the sun and that radiated from the earth?
ii. How is the size of a degree defined on the Celsius scale and on the Fahrenheit scale? [3 Marks]

Answer:
i. Infrared rays emitted from the sun have shorter wavelengths because the sun has a much higher surface temperature than the earth. Terrestrial radiation has longer wavelengths.
ii. A degree on the Celsius scale is \( 1/100^\text{th} \) part of the interval between ice point and steam point. A degree on the Fahrenheit scale is \( 1/180^\text{th} \) part of the same interval.

Teacher's Note:
a) Wien's displacement law relates temperature inversely to peak emission wavelength.
b) Fundamental intervals are \( 100 \) divisions for Celsius and \( 180 \) divisions for Fahrenheit.

Question 8

(i) Given figure shows the image I of a point object O. How will you differentiate between point object and its image? [3 Marks]

[Figure: Ray diagram showing curved mirror forming an image from a point object]

Answer:
1. A point object is a source from which rays actually diverge, whereas an image is a point where reflected or refracted rays actually meet or appear to diverge from.
2. If you place your eye at position \( E_1 \), both object and image can be seen, whereas at position \( E_2 \), only the object can be seen.
3. Objects emit or reflect light independently, while images are formed as an optical effect by mirrors or lenses.

Teacher's Note:
a) Real images can be caught on a screen, virtual images cannot.
b) Clear ray diagrams are essential for optical distinction.

(ii) Give three applications of plane mirrors. [3 Marks]

Answer:
1. Used as looking glasses in dressing rooms.
2. Used in optical instruments like periscopes and kaleidoscopes.
3. Used in solar cookers and scientific instruments to reflect light rays in desired directions.

Teacher's Note:
a) Plane mirrors form virtual, erect, and laterally inverted images of same size as the object.
b) List practical daily-life and laboratory uses.

(iii) In a physics lab there were two charged spheres A and B, initially both were kept at some distance from each other in an enclosed box as shown in figure below.
After some time, the teacher brought both the sphere close to each other such that they are now in contact with each other.
Answer the following questions based on the given information.
a) Which sphere is at higher potential before electrical contact based on convention?
b) Which sphere is at lower potential before electrical contact based on convention?
c) In which direction conventional current flows?
d) In which direction electronic current flows? [4 Marks]

[Figure: Two charged spheres A (orange, positive signs) and B (blue/white, neutral/negative) inside a box]

Answer:
a) Sphere A (positively charged) is at higher potential.
b) Sphere B is at lower potential.
c) Conventional current flows from sphere A to sphere B.
d) Electronic current flows from sphere B to sphere A.

Teacher's Note:
a) Conventional current flows from higher potential to lower potential (direction of positive charge flow).
b) Electronic current flows in the opposite direction, from lower to higher potential (flow of electrons).

Question 9

(i) An engineer wrapped copper coils around a soft iron bar XY. Which is then connected to a switch, battery and rheostat as shown below.
Answer the following questions based on the given information.
a) State the polarity at the ends X and Y as the switch is pressed.
b) What is an electromagnet? Name the material used for preparing an electromagnet.
c) Suggest two ways of increasing the strength of electromagnet so formed.
d) State two differences between an electromagnet and a permanent magnet. [4 Marks]

[Figure: Solenoid wound around an iron bar XY connected to a battery circuit with switch and rheostat]

Answer:
a) Polarity at X is North and at Y is South (based on current direction and clock rule).
b) An electromagnet is a temporary strong magnet made by winding an insulated wire around a soft iron core carrying electric current. Material used is soft iron.
c) Ways to increase strength: (1) Increase the magnitude of current, (2) Increase the number of turns in the coil.
d) Differences:

Teacher's Note:
a) Use the clock face rule to determine north and south poles of a current-carrying solenoid.
b) Soft iron acquires high permeability and low retentivity, making it ideal for electromagnets.

(ii) In the given circuit diagram two cells are connected in parallel. Answer the following: [3 Marks]
a) What is the potential difference between points A and B if e.m.f. of each cell is \( 1.5\text{ V} \)?
b) How does the effective voltage of the two cells connected in series compare to their arrangement in parallel?
c) Are the cells in a conventional flash light connected in series or in parallel?

[Figure: Circuit diagram showing two cells connected in parallel across terminals A and B with a resistor]

Answer:
a) The potential difference between points A and B is \( 1.5\text{ V} \) because the voltage across parallel combination of identical cells remains equal to the e.m.f. of a single cell.
b) In series combination, e.m.f. is additive (\( 3\text{ V} \)), which is twice the effective voltage of the parallel arrangement (\( 1.5\text{ V} \)).
c) The cells in a conventional flashlight are connected in series.

Teacher's Note:
a) Parallel connection increases current capacity rather than voltage.
b) Series connection increases total voltage output.

(iii) You are given following three bars exactly similar in size and shape:
a) A permanent magnet
b) A bar of soft iron
c) A bar of non-magnetic substance.
Describe how you will identify each of the bars if only a piece of thread is supplied to you as the extra piece of apparatus. [3 Marks]

Answer:
1. Suspend each bar freely by a thread. The permanent magnet will always rest in the geographic north-south direction due to Earth's magnetic field, while the other two bars can rest in any random direction.
2. Take the permanent magnet and bring its ends near the middle of the remaining two bars. The soft iron bar will experience strong attraction, whereas the non-magnetic bar will show no attraction.
3. This allows clear identification of all three bars.

Teacher's Note:
a) Test for a magnet is repulsion, but suspension test identifies directive property.
b) Soft iron is attracted by both poles of a magnet, while non-magnetic substances experience no magnetic force.