ICSE Class 9 Physics Sample Paper with Solutions Set 03

ICSE EXAMINATION

Sample Question Paper - 3

Physics

Time: 2 Hours.                                                                               Total Marks: 80

SECTION A (40 Marks)

Attempt all Questions from this Section

Question 1

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

(i) The radius of moon is \( 3.84 \times 10^6 \) m. It's order of magnitude is [1 Mark]
(a) \( 10^5 \)
(b) \( 10^6 \)
(c) \( 10^7 \)
(d) \( 10^8 \)

Answer: (c) \( 10^7 \)

Since \( 3.84 \) is greater than \( 3.2 \), rounding to the nearest power of 10 gives \( 10^1 \times 10^6 = 10^7 \).

Teacher's Note:
a) To find the order of magnitude, express the number in standard form \( M \times 10^n \), where \( 1 \le M \lt 10 \). If \( M \gt 3.2 \) (more precisely \( \sqrt{10} \)), add 1 to the exponent \( n \).
b) Students often forget to check the coefficient \( M \) and simply copy the exponent from the given value.

(ii) The velocity-time equation is represented by [1 Mark]
(a) \( v - u = at \)
(b) \( v = at - u \)
(c) \( v^2 - u^2 = 2as \)
(d) \( s = ut + \frac{1}{2}at^2 \)

Answer: (a) \( v - u = at \)

Rearranging \( v = u + at \) gives \( v - u = at \), which is the first equation of motion relating velocity and time.

Teacher's Note:
a) The first equation of motion is fundamentally \( v = u + at \), representing velocity as a linear function of time.
b) Ensure proper algebraic signs are maintained when transposing terms.

(iii) In case of accelerated motion along a straight line, the distance travelled by the body is: [1 Mark]
(a) directly proportional to square of time
(b) directly proportional to time
(c) inversely proportional to square of time
(d) inversely proportional to time

Answer: (a) directly proportional to square of time

From \( s = ut + \frac{1}{2}at^2 \), for initial rest (\( u = 0 \)) and constant acceleration, \( s \propto t^2 \).

Teacher's Note:
a) For uniformly accelerated motion starting from rest, distance is proportional to \( t^2 \).
b) Do not confuse distance-time relation with velocity-time relation.

(iv) A body weighs \( 48 \) N on the surface of the Earth, its weight at the centre of the Earth will be [1 Mark]
(a) \( 24 \) N
(b) \( 48 \) N
(c) \( 96 \) N
(d) Zero

Answer: (d) Zero

Weight \( W = mg \). At the centre of the Earth, acceleration due to gravity \( g = 0 \), hence weight is zero.

Teacher's Note:
a) Acceleration due to gravity varies inversely with distance squared above/below surface and becomes zero at the geometric center.
b) Mass remains constant everywhere, but weight depends on local gravity \( g \).

(v) Assertion (A): Thrust is a scalar quantity.
Reason (R): Thrust is a force acting normally on a surface. [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: (c) Assertion is false but reason is true.

Thrust is a force (vector quantity), so assertion is false. Reason correctly defines thrust as normal force.

Teacher's Note:
a) All forces, including thrust and pressure force, possess both magnitude and a specific direction (normal to the surface), making them vectors.
b) Read assertion statements carefully to distinguish between scalar pressure and vector thrust.

(vi) In which direction does the buoyant force act on a body immersed in a liquid? [1 Mark]
(a) Vertically upward
(b) Sideways towards the walls of the container
(c) Vertically downward
(d) None of the above

Answer: (a) Vertically upward

Buoyant force acts vertically upwards through the centre of buoyancy.

Teacher's Note:
a) Upthrust originates from the pressure difference between the lower and upper surfaces of a submerged body.
b) It always opposes the force of gravity acting downwards.

(vii) Which of the following pair is the one which shows abiotic: biotic component combination? [1 Mark]
(a) Microorganism: Plant
(b) Microorganism: Human being
(c) Soil: Temperature
(d) Temperature: Microorganism

Answer: (d) Temperature: Microorganism

Temperature is an abiotic (non-living) factor, while microorganism is a biotic (living) factor.

Teacher's Note:
a) Abiotic components include physical and chemical factors like light, temperature, soil, and water.
b) Biotic components consist of all living organisms in the ecosystem.

(viii) If the magnification has a positive sign, the image formed by the concave mirror must be [1 Mark]
(a) Real and erect
(b) Real and inverted
(c) Virtual and erect
(d) Virtual and inverted

Answer: (c) Virtual and erect

Positive magnification indicates that the image height is positive, meaning the image is erect and virtual.

Teacher's Note:
a) Real images are always inverted (negative magnification), while virtual images are always erect (positive magnification).
b) Concave mirrors produce virtual and erect images only when the object is placed between the pole and the focus.

(ix) How is the current flowing in a conductor changed if the resistance of a conductor is doubled, keeping the potential difference across it the same? [1 Mark]
(a) doubled
(b) one-fourth
(c) halved
(d) four times

Answer: (c) halved

According to Ohm's law, \( I = \frac{V}{R} \). If \( R \) becomes \( 2R \), current becomes \( I' = \frac{V}{2R} = \frac{1}{2}I \).

Teacher's Note:
a) Current is inversely proportional to resistance at constant potential difference.
b) Double the resistance results in exactly half the current flow.

(x) A bulb is connected to a cell. How is the resistance of circuit is affected if another identical bulb is connected in series with the first bulb? [1 Mark]
(a) Resistance is doubled
(b) resistance is halved
(c) Resistance is four times.
(d) resistance is one-fourth.

Answer: (a) Resistance is doubled

In series combination, equivalent resistance \( R_s = R_1 + R_2 \). For two identical bulbs, \( R_s = R + R = 2R \).

Teacher's Note:
a) Series connection adds up individual resistances directly.
b) Adding resistors in series decreases total current in the circuit.

(xi) The magnetism acquired by a magnetic material when it is kept near (or in contact with) a magnet, is called [1 Mark]
(a) temporary magnetism
(b) induced magnetism
(c) permanent magnetism
(d) Induction

Answer: (b) induced magnetism

Temporary magnetism acquired by induction when placed near a magnet is termed induced magnetism.

Teacher's Note:
a) Magnetic induction is a temporary phenomenon occurring without direct physical contact.
b) Soft iron acquires induced magnetism easily but loses it quickly.

(xii) State the positions of neutral points when a magnet is placed with its axis in the magnetic meridian and with its north pole pointing towards geographic south. [1 Mark]
(a) in east-west direction
(b) in west-east direction
(c) in north-south direction
(d) in south-north direction

Answer: (a) in east-west direction

When a magnet's north pole points south, neutral points lie on the broadside-on position, which is along the east-west direction.

Teacher's Note:
a) Neutral points are locations where the magnetic field of the bar magnet equals and opposes the Earth's horizontal magnetic field.
b) Orientation of the magnet determines whether neutral points lie along the axial (north-south) or equatorial (east-west) line.

(xiii) A solid of mass \( 22 \) kg is immersed in water. If it loses half of its weight in water, its R.D. will be [1 Mark]
(a) \( 1/2 \)
(b) \( 1/4 \)
(c) \( 4 \)
(d) \( 2 \)

Answer: (d) 2

Relative Density = Weight in air / Loss of weight in water = \( W / (W/2) = 2 \).

Teacher's Note:
a) Relative density of a solid is defined as the ratio of its weight in air to the buoyant force (loss of weight) in water.
b) Since weight loss equals half the weight in air, the R.D. evaluates to 2.

(xiv) Population growth can be controlled through [1 Mark]
(a) Economic measure
(b) Technological measures
(c) Women empowerment and family planning
(d) Economic growth

Answer: (c) Women empowerment and family planning

Family planning and women empowerment directly educate and regulate population growth rates effectively.

Teacher's Note:
a) Sustainable population control relies on social awareness, education, and family planning.
b) Empowering women leads to smaller, healthier families and better resource management.

(xv) An inverted image can be seen by using a convex mirror under what circumstances? [1 Mark]
(a) Under no circumstances
(b) When the object is placed at infinity
(c) When the object is at a distance equal to the radius of curvature of the mirror
(d) When the distance of the object from the mirror is equal to the focal length of the mirror.

Answer: (a) Under no circumstances

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

Teacher's Note:
a) Convex mirrors diverge light rays, preventing real image formation for real objects.
b) Consequently, inverted (real) images are never formed by a convex mirror.

Question 3

(i)
(a) There are ____ [One/two/three] types of inertia. [1 Mark]
(b) The slope of a straight-line graph gives the value of ____ [constant of proportionality/ Distance between x-axis and y axis/Angle of inclination]. [1 Mark]
(c) The ratio of buoyant forces experienced by a solid body when immersed in two liquids whose relative densities are 1 and 0.5 respectively is ____ [2:1/1:2/1:1]. [1 Mark]
(d) 1 ml = ____ [\( 10^{-6} \)/\( 10^{-3} \)/1000] \( \text{m}^3 \). [1 Mark]
(e) The magnetic field lines of the earth are ____ [Normal/Diagonal/Vertical] to earth's surface near the magnetic poles and ____ [Perpendicular/Parallel/Horizontal] to earth's surface near the magnetic equator. [1 Mark]

Answer:
(a) three
(b) constant of proportionality
(c) 2:1
(d) \( 10^{-6} \)
(e) Vertical, Horizontal

Teacher's Note:
a) Inertia is classified into inertia of rest, motion, and direction.
b) Buoyant force equals weight of displaced liquid (\( V \rho g \)), hence proportional to liquid density.

(ii)
[Figure: A simple pendulum with a rigid support, a string of length l, a bob of mass m oscillating between extreme positions A and B, passing through mean position O, with vertical reference line Y and displacement angles \( \theta \) shown.]
Above diagram represents a simple pendulum. A simple pendulum is a heavy point mass suspended from a rigid support by a massles and inextensible string. Answer the following questions: [2 Marks]
(a) What is the length of given simple pendulum?
(b) What is the value of amplitude of this simple pendulum?

Answer:
(a) The length of a simple pendulum is the distance from the point of suspension to the centre of gravity of the bob.
(b) Amplitude is the maximum displacement of the bob from its mean position (represented by distance OB or OA).

Teacher's Note:
a) Effective length is measured from the point of support to the center of the spherical bob.
b) Amplitude measures the maximum angular or linear displacement from equilibrium.

(iii) Raghu tied a rope to a tree and started shaking it such that oscillation produce periodic waves of time period 10 seconds.
[Figure: A person shaking a rope tied to a tree, showing transverse wave lobes with peak-to-trough height marked as 2 cm and horizontal wavelength marker as 10 cm.]
Answer the following question based on the given information. [2 Marks]
(a) Define the term amplitude of a wave. What will the maximum amplitude of wave for the given case? State it in S.I system.
(b) How is the frequency of a wave related to its time period?

Answer:
(a) Amplitude is the maximum displacement of a particle from its mean position. From the figure, peak-to-trough distance is 2 cm, so amplitude is \( 2 / 2 = 1 \) cm \( = 0.01 \) m.
(b) Frequency (\( f \)) is inversely proportional to time period (\( T \)), given by \( f = 1/T \).

Teacher's Note:
a) Always convert units to S.I. standards (meters for length, seconds for time).
b) Frequency is measured in Hertz (Hz) or \( \text{s}^{-1} \).

Question 2

(i)
(a) Express one day in milliseconds.
(b) Why are the passengers' cabins in an aeroplane pressurised?
(c) What does a straight-line graph signify? [3 Marks]

Answer:
(a) 1 day = \( 24 \times 60 \times 60 \times 1000 = 8,64,00,000 \) ms \( = 8.64 \times 10^7 \) ms.
(b) At high altitudes, atmospheric pressure drops significantly, causing oxygen starvation and health risks. Cabins are pressurised to maintain normal physiological pressure.
(c) A straight-line graph signifies that the two quantities are directly proportional to each other with a constant slope.

Teacher's Note:
a) Conversion factors: 1 day = 24 h, 1 h = 3600 s, 1 s = 1000 ms.
b) Direct proportionality is graphically represented by a straight line passing through the origin.

(ii) The following table gives the distance travelled by a particle at different times.

Draw a distance-time graph representing the motion of the particle. [2 Marks]

Answer:
Plot Time (s) along the x-axis and Distance (cm) along the y-axis, using the given data points to sketch a smooth upward-curving parabola indicating non-uniform acceleration.

Teacher's Note:
a) Non-linear distance-time data results in a curve, confirming non-uniform speed.
b) Proper labeling of axes with units is mandatory for full credit.

(iii) What will be the magnitude and direction of the reaction force acting on a coin of 10 g lying on the surface of the floor? Take g = 9.8 m/\( \text{s}^2 \) [2 Marks]

Answer:
Weight of coin \( W = mg = 0.01 \text{ kg} \times 9.8 \text{ m/s}^2 = 0.098 \) N.
Magnitude of reaction force = \( 0.098 \) N, acting vertically upwards.

Teacher's Note:
a) By Newton's third law, the normal reaction equals the downward weight of the object.
b) Convert mass from grams to kilograms before calculating force.

(iv) Distinguish between fundamental units and derived units. (Any 2 points) [2 Marks]

Answer:

Teacher's Note:
a) Fundamental units form the base of system of units (like SI).
b) Derived units depend directly on fundamental units through physical formulas.

(v) The earth attracts an apple. Does the apple also attract the earth? If it does, why does the earth not move towards the apple? [2 Marks]

Answer:
Yes, by Newton's third law, the apple also attracts the Earth with an equal and opposite gravitational force. The Earth does not move towards the apple because its mass is enormously large, resulting in an infinitesimally small acceleration according to \( a = F/m \).

Teacher's Note:
a) Gravitational force is mutual and acts equally on both bodies.
b) Acceleration is inversely proportional to mass, making Earth's motion unnoticeable.

(vi) An empty truck and a loaded truck are moving with the same velocity. On applying brakes, which truck will stop first and why? [2 Marks]

Answer:
The empty truck will stop first. Since it has lesser mass and hence lesser inertia of motion, the same braking force produces a higher deceleration, bringing it to rest sooner.

Teacher's Note:
a) Inertia depends directly on mass.
b) A heavier (loaded) truck requires a larger opposing force or longer stopping distance to come to rest.

(vii) In cold countries, ponds freeze only at the surface. Why? [2 Marks]

Answer:
Water has a maximum density at \( 4^{\circ}\text{C} \). As surface water cools below \( 4^{\circ}\text{C} \), it becomes lighter and stays at the top, freezing into ice, while water at \( 4^{\circ}\text{C} \) remains at the bottom, sustaining aquatic life.

Teacher's Note:
a) This anomalous expansion of water prevents lakes from freezing solid from the bottom up.
b) Ice is a poor conductor of heat, acting as an insulating layer.

SECTION B (40 Marks)

Attempt any four Questions from this Section

Question 4

(i) A Vernier scale has 40 divisions and its main scale is divided in millimeters. It has an error of +0.0125 cm. While measuring the length of a cylinder, the reading on the main scale is 75 mm and the 12th Vernier scale division coincides with the main scale. Calculate the corrected length. [4 Marks]

Answer:
1. Least Count (LC) = \( 1 \text{ mm} / 40 = 0.025 \text{ mm} = 0.0025 \text{ cm} \).
2. Main Scale Reading (MSR) = \( 75 \text{ mm} = 7.5 \text{ cm} \).
3. Vernier Coincidence (VC) = 12.
Vernier Reading = \( 12 \times 0.0025 \text{ cm} = 0.03 \text{ cm} \).
Observed Length = \( 7.5 \text{ cm} + 0.03 \text{ cm} = 7.53 \text{ cm} \).
Corrected Length = Observed Length - Error = \( 7.53 \text{ cm} - (+0.0125 \text{ cm}) = 7.5175 \text{ cm} \).

Teacher's Note:
a) Always convert all units to the same system (centimeters or millimeters) before calculating.
b) Positive error is subtracted from the observed reading, while negative error is added.

(ii) Draw a graph between effective length 'l' and square of time period '\( T^2 \)' of a simple pendulum. How will you obtain the value of acceleration due to gravity from the graph? [3 Marks]

Answer:
[Figure: A straight line graph showing \( T^2 \) on the y-axis and length \( l \) on the x-axis, passing through the origin with a positive slope.]
1. Plot \( l \) on the x-axis and \( T^2 \) on the y-axis to obtain a straight line passing through the origin.
2. The formula is \( T = 2\pi\sqrt{l/g} \), which gives \( T^2 = (4\pi^2 / g)l \).
3. Therefore, slope = \( T^2 / l = 4\pi^2 / g \), and acceleration due to gravity \( g = 4\pi^2 / \text{slope} \).

Teacher's Note:
a) The straight line confirms that \( T^2 \) is directly proportional to \( l \).
b) Calculating the reciprocal of the slope helps determine \( g \) accurately.

(iii) The mass of a block is 1.35 kg and its volume is \( 1.5 \times 10^{-3}\text{ m}^3 \). Find the density of the block. Will this block float in water? Give reason. [3 Marks]

Answer:
1. Density = Mass / Volume = \( 1.35 \text{ kg} / (1.5 \times 10^{-3}\text{ m}^3) = 900 \text{ kg/m}^3 \).
2. Yes, the block will float in water.
3. Reason: The density of the block (\( 900 \text{ kg/m}^3 \)) is less than the density of water (\( 1000 \text{ kg/m}^3 \)).

Teacher's Note:
a) Density formula is \( d = M/V \) expressed in SI units (\( \text{kg/m}^3 \)).
b) An object floats if its density is lower than that of the surrounding fluid.

Question 5

(i) An athlete runs around a circular track of circumference 360 m in \( (1/60) \) h and reaches the starting point. Calculate,
a) The distance covered by athlete
b) The displacement
c) The average speed and
d) The average velocity. [4 Marks]

Answer:
a) Distance = Circumference = \( 360 \text{ m} = 0.36 \text{ km} \).
b) Displacement = 0 m (since the athlete returns to the starting point).
c) Average Speed = Total Distance / Time = \( 0.36 \text{ km} / (1/60 \text{ h}) = 21.6 \text{ km/h} \) (or \( 6 \text{ m/s} \)).
d) Average Velocity = Total Displacement / Time = \( 0 / (1/60) = 0 \text{ km/h} \).

Teacher's Note:
a) Displacement is zero for any complete round trip in a closed circular path.
b) Average velocity depends on displacement, whereas average speed depends on total distance.

(ii) The table shows the velocity of a two-wheeler at various intervals of time.

a) Plot the velocity-time graph.
b) Calculate the rate of change of velocity between 5s - 7s, 7s - 10 s and 10s - 15 s. [3 Marks]

Answer:
a) [Figure: Velocity-time graph with time on x-axis and velocity on y-axis showing horizontal and sloped line segments connecting data points (0,10), (5,10), (7,7), (10,10), (15,0).]
b) Rate of change of velocity (acceleration):
- For 5s to 7s: \( a = (7 - 10) / (7 - 5) = -3 / 2 = -1.5 \text{ m/s}^2 \).
- For 7s to 10s: \( a = (10 - 7) / (10 - 7) = 3 / 3 = 1 \text{ m/s}^2 \).
- For 10s to 15s: \( a = (0 - 10) / (15 - 10) = -10 / 5 = -2 \text{ m/s}^2 \).

Teacher's Note:
a) Rate of change of velocity is acceleration, calculated as \( (v - u) / t \).
b) Negative acceleration indicates deceleration or retardation.

(iii) The graph given below shows how three runners A, B and C ran a 200-metre race.

[Figure: Distance-time graph with distance (metres) on y-axis up to 200 and time (seconds) on x-axis up to 14.5, showing three curves for runners A, B, and C with runner B showing a horizontal resting plateau.]

Answer the following questions:
(a) Which runner won the race?
(b) Which runner stopped for a rest and how long did he stop for?
(c) How long did the runner B take to complete the race? [3 Marks]

Answer:
(a) Runner A won the race because he reached 200 m in the shortest time.
(b) Runner C (or B depending on graph label, per official key: Runner C) stopped for a rest at 100 m for \( 10.5 - 6.75 = 3.75 \) seconds.
(c) Runner B took 10.5 seconds to complete the race.

Teacher's Note:
a) In distance-time graphs, steeper slopes represent higher speeds.
b) A horizontal section on a distance-time graph indicates a period of rest (zero speed).

Question 6

(i) Show that Newton's first law of motion can be obtained from the second law. [4 Marks]

Answer:
1. According to Newton's second law, force \( F = m \times a = m \times \frac{v - u}{t} \).
2. If no external force acts on the body, \( F = 0 \).
3. Therefore, \( 0 = m \times \frac{v - u}{t} \), which gives \( v - u = 0 \) or \( v = u \).
4. This proves that an object continues to move with uniform velocity (or remains at rest) unless acted upon by an external force, which is Newton's first law.

Teacher's Note:
a) Newton's first law is a special case of the second law when net external force is zero.
b) State the mathematical steps clearly with proper justification for each line.

(ii)
(a) The velocity of a body is continuously changing. Can its speed remain constant?
(b) If the speed is changing, can the velocity of a body remain constant?
(c) Is it possible for a body to have a constant speed in accelerated motion? [3 Marks]

Answer:
(a) Yes. For example, in uniform circular motion, speed is constant while velocity changes continuously due to changing direction.
(b) No. Velocity includes both magnitude (speed) and direction; if speed changes, velocity must change.
(c) Yes. In uniform circular motion, speed remains constant, but the body is accelerated due to continuous change in direction.

Teacher's Note:
a) Speed is a scalar quantity, whereas velocity is a vector quantity.
b) Acceleration can occur due to change in direction alone, even if speed is constant.

(iii)
(a) The earth attracts a ball with a force of 1 N. If this is the force of action, what would be the force of reaction and who exerts this force?
(b) State two circumstances under which your weight would become zero. [3 Marks]

Answer:
(a) The reaction force is 1 N, exerted by the ball on the Earth.
(b) Weight becomes zero: 1. When a body is in free fall (weightlessness), and 2. At the centre of the Earth.

Teacher's Note:
a) Action and reaction forces are equal in magnitude, opposite in direction, and act on two different bodies.
b) Apparent weight becomes zero in free fall or where gravitational acceleration is zero.

Question 7

(i)
a) What is second's pendulum? What is its approximate effective length?
b) A second's pendulum is set up on the surface of the moon, where acceleration due to gravity is \( 1/6 \)th of that of the earth. How is the time period of the pendulum affected? Give a reason in support of your answer. [4 Marks]

Answer:
a) A second's pendulum is a simple pendulum with a time period of 2 seconds. Its effective length is approximately 1 metre.
b) The time period increases. Since \( T \propto \frac{1}{\sqrt{g}} \), a decrease in gravity (\( g \)) on the moon results in a longer time period.

Teacher's Note:
a) Time period of a pendulum depends directly on square root of length and inversely on square root of gravity.
b) Lower gravity on the moon makes the pendulum swing slower.

(ii) Indicate on a graph how the density of water at \( 0^{\circ}\text{C} \) changes when it is gradually heated up to \( 10^{\circ}\text{C} \). [3 Marks]

Answer:
[Figure: A graph showing density of water on y-axis and temperature from \( 0^{\circ}\text{C} \) to \( 10^{\circ}\text{C} \) on x-axis, showing density increasing from \( 0^{\circ}\text{C} \) to a peak at \( 4^{\circ}\text{C} \) and then decreasing up to \( 10^{\circ}\text{C} \).]
1. When water at \( 0^{\circ}\text{C} \) is heated, its volume decreases and density increases until it reaches \( 4^{\circ}\text{C} \).
2. At \( 4^{\circ}\text{C} \), density is maximum (\( 1 \text{ g/cm}^3 \)).
3. Upon further heating from \( 4^{\circ}\text{C} \) to \( 10^{\circ}\text{C} \), water expands, causing its density to decrease.

Teacher's Note:
a) This unique behavior between \( 0^{\circ}\text{C} \) and \( 4^{\circ}\text{C} \) is known as anomalous expansion of water.
b) Label axes clearly, marking the maximum density peak at \( 4^{\circ}\text{C} \).

(iii) The below diagram shows that the energy flow in an ecosystem. [3 Marks]

[Figure: Energy flow pyramid diagram showing Sun transferring energy to Producers (20,810 cal), Primary consumers (3,368 cal), Secondary consumers (283 cal), and Tertiary consumers (21 cal), with energy loss marked as decay and respiration.]

Answer the following questions based on the given information.
a) State the law of which governs the energy flow in an ecosystem.
b) State the importance of green plants in ecosystems.
c) What is the difference between primary, secondary, and tertiary consumers? Given 2 examples of each.

Answer:
a) Governed by the laws of thermodynamics (specifically the ten percent law and energy conservation).
b) Green plants act as producers, trapping solar energy and converting it into chemical energy to sustain all trophic levels.
c) Differences:
- Primary consumers: Herbivores feeding directly on producers (e.g., Deer, Rabbit).
- Secondary consumers: Carnivores feeding on primary consumers (e.g., Frog, Fox).
- Tertiary consumers: Top carnivores feeding on secondary consumers (e.g., Snake, Lion).

Teacher's Note:
a) Energy flow in an ecosystem is unidirectional and decreases at each higher trophic level.
b) Give precise examples for each consumer category to secure full marks.

Question 8

(i) The angle between the incident ray and the mirror is \( 30^{\circ} \).
a) What is the angle of incidence?
b) What is the angle of reflection?
c) What is the total angle turned by the ray of light? [3 Marks]

Answer:
a) Angle of incidence \( i = 90^{\circ} - 30^{\circ} = 60^{\circ} \).
b) Angle of reflection \( r = i = 60^{\circ} \).
c) Total angle turned by the ray (deviation) \( \delta = 180^{\circ} - (60^{\circ} + 60^{\circ}) = 60^{\circ} \) or total angular turn = \( i + r = 60^{\circ} + 60^{\circ} = 120^{\circ} \).

Teacher's Note:
a) Angle of incidence is measured with the normal, not the mirror surface.
b) Deviation is the angle between the extended incident ray and the reflected ray.

(ii) Where will the image form if the object is placed at the centre of curvature in front of the concave mirror? Also, state the nature of the image. [3 Marks]

Answer:
[Figure: Ray diagram for a concave mirror showing object at centre of curvature C, ray passing through focus and parallel ray reflecting through C, forming an inverted real image of the same size at C.]
1. The image is formed at the centre of curvature itself.
2. Nature of the image: Real, inverted, and of the same size as the object.

Teacher's Note:
a) When object distance equals radius of curvature (\( u = 2f \)), image distance also equals \( 2f \).
b) This standard ray diagram case is frequently tested in examinations.

(iii) Seema used a laser pointer and kept it normal to a plane mirror at position A as shown below, now after some time she shifted it to position B at some angle \( \theta \) from its initial point. [4 Marks]

[Figure: Laser pointer at position A normal to plane mirror MM, and shifted to position B at angle \( \theta \), showing normal incidence at A and oblique incidence at B with reflected rays.]

Answer the following questions based on the given information.
(a) What is its angle of incidence when the laser pointer is at point A?
(b) Define the reflection of light.
(c) What is the direction of the reflected ray when the laser pointer is at point A? Show it on a diagram.
(d) What will the angle of reflection be if the value of \( \theta \) is \( 60^{\circ} \)? Show it on a diagram.

Answer:
(a) \( 0^{\circ} \).
(b) Reflection of light is the bouncing back of light rays into the same medium after striking a polished surface.
(c) Same as the incident ray (retraces its path normally).
(d) If \( \theta = 60^{\circ} \), the angle of incidence is \( 60^{\circ} \), so by the laws of reflection, the angle of reflection is also \( 60^{\circ} \).

Teacher's Note:
a) Normal incidence results in zero angle of incidence and zero angle of reflection.
b) Always draw arrows on ray diagrams to indicate the direction of light propagation.

Question 9

(i) The figure shows a negatively charged ebonite rod A which is brought near an uncharged metal sphere P touching the other uncharged metal sphere Q. Both the spheres stand on separate insulating stands. [3 Marks]

[Figure: Negatively charged rod A near touching metal spheres P and Q supported on insulating stands, showing induced charges via electrostatic induction.]

a) If keeping the rod A in position, the sphere Q is removed by holding the insulating stand of it, state the kind of charge on and give reason to support your answer.
b) If the rod A is removed first and then the sphere Q is removed by holding the insulating stand of it, what kind of charges will be on the spheres P and Q? Explain.

Answer:
a) Sphere Q will have a negative charge because electrons repel away from rod A into Q due to electrostatic induction.
b) Neither sphere will have any charge because removing rod A first allows redistributed electrons to return to their original neutral state.

Teacher's Note:
a) Electrostatic induction separates charges temporarily within connected conductors.
b) Sequence of removing separating spheres determines whether permanent charging by induction occurs.

(ii)
a) Why will heating the magnet strongly remove its magnetism?
b) What are neutral points? [3 Marks]

Answer:
a) Strong heating agitates molecular magnets, disrupting their aligned order and causing the magnet to lose its magnetism.
b) Neutral points are locations where the magnetic field of a bar magnet is equal in magnitude and opposite in direction to the Earth's horizontal magnetic field, resulting in zero net magnetic field.

Teacher's Note:
a) Thermal agitation disorganises atomic magnetic dipoles.
b) Compasses placed at neutral points show random or erratic orientation due to zero net field.

(iii) Sameer placed a bar magnet placed on the top of a flat table with its north pole pointing towards south. The arrow shows the north-south direction. There are no other magnets or magnetic materials nearby.

[Figure: Bar magnet on table with North pole pointing South, alongside a directional arrow for North-South.]

(i) Insert two magnetic field lines on either side of the magnet using arrowhead to show the direction of each field line.
(ii) Indicate by crosses, the likely positions of the neutral points.
(iii) What is the magnitude of the magnetic field at each neutral point? Give a reason for your answer.
(iv) Explain the method of plotting the magnetic field lines by using a small compass needle. [3 Marks]

Answer:
(i) Field lines drawn originating from North and entering South, looping around the magnet.
(ii) Crosses marked along the broadside-on (east-west) positions.
(iii) Magnitude is zero because the magnet's magnetic field equals and opposes the Earth's horizontal magnetic field.
(iv) Place a small compass near the magnet, mark the positions of its poles with pencil dots, move the compass so its south pole coincides with the previous dot, and repeat to trace continuous field lines.

Teacher's Note:
a) Magnetic field lines always run externally from North to South pole.
b) Compass needle method is standard for mapping invisible magnetic fields experimentally.