Solved Sample Paper 1

Section A

Question 1(i)

A certain mass of solid substance is heated for a period of 50 min. Its heating curve is shown below :

From the graph, which one of the following statement is correct?

1. The boiling point of the given substance is 40°C.
2. The specific latent heat of fusion of substance is the same as the specific latent heat of vaporisation.
3. The substance has a higher specific heat capacity in solid state than that in liquid state.
4. The substance has a lower specific heat capacity in solid state than that in liquid state.

Answer

The substance has a higher specific heat capacity in solid state than that in liquid state

Reason — From graph, 40°C is the melting point of the substance. It takes 10 mins for melting for temperature change of 20°C (i.e.,40°-20°) and takes 10 mins for boiling for temperature change of 40°C (i.e., 80°-40°).

Hence, the latent heat of fusion is less than the latent heat of vapourisation

From relation,

Q = m x c x △T

C ∝ $\dfrac{1}{\Delta\text{T}}$

So, from graph, the specific heat capacity in solid state is greater than that in liquid state.

Question 1(ii)

The figure given below shows the cooling curve of pure wax material after heating. It cools from A and B and solidifies along BD. If L and c represents the value of latent heat and the specific heat of liquid wax, then the ratio of $\dfrac{\text{L}}{\text{c}}$ is

1. 20
2. 40
3. 60
4. 100

Answer

20

Reason — Curve AB represents the cooling of wax material, the specific heat of the cooling of wax material is given by C, and temperature change Δt = 90°C − 80°C

Heat supplied in 2 min = m x c x △T

= m x c x (90-80) = 10mc    .....[1]

Curve BD represents the latent heat of fusion during the solidification of wax material for the next 4 minutes. It is represented by

Q = mL    .....[2]

As process of solidification takes twice as long as the cooling process,

∴ 2[10mc] = mL    [From Eq 1 & 2]
20mc = mL
20c = L

Hence, $\dfrac{\text{L}}{\text{c}}$ = 20

Question 1(iii)

In the circuit shown, lamps L₁ and L₂, are identical. The sliding contact at O moved towards Q, then which of the statements about the effect on the brightness in L₂ is correct?

1. Initially it has no light, then its brightness increases gradually and becomes equal to that of L₁
2. Initially it is same as that of L₁, and finally become half as that of L₁
3. Initially it is half as that of L₁ and finally becomes equal to that of L₁.
4. Initially it is half as that of L₁ and finally becomes zero.

Answer

Initially it is half as that of L₁ and finally becomes equal to that of L₁.

Reason — Initially the pointer of sliding contact is at mid point of potential divider. So, the voltage across L₂ is half as that across L₁ and hence its brightness will also be half.

When the contact is moved to Q, the voltage across both lamps become equal and hence their brightness will also be equal.

Question 1(iv)

In which of the following figures shown, will the sound of a tuning fork hit with rubber pad at end A be heard maximum at end B?

1. I
2. II
3. Both I and II
4. Cannot be decided

Answer

I

Reason — For reflection of sound, angle of incidence should be equal to angle of reflection. So, maximum sound will be heard at end B for Fig. I.

Question 1(v)

The force of 20 N is applied at distance of 40 cm from the axis. The moment of force is

1. 8 N-m
2. 0.4 N-m
3. 8.5 N-m
4. 20 N-m

Answer

8 N-m

Reason — Given,

Force (f) = 20 N

Distance (d) = 40 cm = 0.4 m

Moment of force = ?

Moment of force = force (f) x distance (d)

= 20 x 0.4

= 8 N m

So, the moment of force = 8 N-m.

Question 1(vi)

When the displacement is along the opposite direction of the force, then the work is said to be

1. zero
2. negative
3. positive
4. infinite

Answer

negative

Reason — When the displacement of the body is in opposite direction i.e., θ = 180°

W = F × S cos 180° and cos 180° = -1

Hence, W = -F × S

Therefore, work is said to be negative.

Question 1(vii)

In a radioactive decay, the atomic number of parent nuclei is not changed. The radiation emitted will be

1. β-rays
2. γ-rays
3. α-rays
4. the atomic number is affected by all forms of radioactive decay

Answer

γ-rays

Reason — Gamma decay involves the emission of gamma rays, which are high-energy electromagnetic waves. Gamma decay doesn't change the atomic number or mass number of the nucleus. It is often emitted after other types of decay (such as alpha or beta decay) to release excess energy.

Question 1(viii)

A submarine produces an ultrasonic wave of velocity 1500 $\dfrac{\text{m}}{\text{s}}$ in water. The officer receives signal after 50 s of emission of ultrasonic waves. Find the distance of object which is present in the bottom of sea.

1. 25.7 km
2. 37.5 km
3. 37.5 m
4. 50 km

Answer

37.5 km

Reason — As we know,

$\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \\[0.5em]$

Total distance travelled by the sound in going and then coming back = 2d

Given,

t = 50 s

V = 1500 ms^-1

Substituting the values in the formula above, we get,

$1500 = \dfrac{2d}{50} \\[0.5em] \Rightarrow d = \dfrac{1500 \times 50}{2} \\[0.5em] \Rightarrow d = 1500 \times 25 \\[0.5em] \Rightarrow d = 37500 \text{ m}$

Converting m to km

1000 m = 1 km

∴ 37500 m = $\dfrac{37500}{1000}$ km = 37.5 km

Hence, distance of the submarine from the object = 37.5 km

Question 1(ix)

Assertion A ray of light passing through the centre of curvature of a concave mirror is reflected back along the same path.

Reason The incident ray, the reflected ray and the normal at the point of incidence lie on the same plane.

1. Both Assertion and Reason are true.
2. Both Assertion and Reason are false.
3. Assertion is false but Reason is true.
4. Assertion is true but Reason is false.

Answer

Both Assertion and Reason are true

Reason — According to the laws of reflection, angle of incidence = angle of reflection and the incident ray, the reflected ray and the normal at the point of incidence lie in the same plane hence, when a ray of light passes through the centre of curvature of a concave mirror, it strikes the concave mirror at right angles to its surface due to which the angle of incidence and angle of reflection both are zero degree and the ray is reflected back along the same path.

Question 1(x)

If temperature of medium increases, then critical angle

1. decreases
2. increases
3. zero
4. first increases then decreases

Answer

increases

Reason — On increasing the temperature of medium, its refractive index decreases. So, the critical angle for the pair of medium increases with increase in temperature.

Question 1(xi)

The focal length does not depend on

1. refractive index
2. radius of curvature
3. direction of light
4. None of the above

Answer

direction of light

Reason — The focal length does not depend on direction of light, but depends on refractive index and radius of curvature.

Question 1(xii)

If current through a resistance is doubled and its resistance made one-fourth, then the value of power dissipation will

1. remain same
2. double
3. become 4 times
4. become 1⁄4 time

Answer

remain same

Reason — Power if given by the formula,

$P = I^2R$

New Current (I') = 2I

New Resistance (R') = $\dfrac{1}{4}$R

Substituting we get

$\dfrac{P_2}{P_1}$ = $\dfrac{(2I)^2 \times \dfrac{R}{4}}{I^2 \times R}$

Hence, P₂ = P₁

∴ Power dissipation will remain same.

Question 1(xiii)

A bar magnet is used to pick up an iron nail. At which parts of the nail it is easiest for the magnet to pick it up?

1. At end X
2. At mid-point Y
3. At end Z
4. Same at all points

Answer

At end Z

Reason — End Z is the broadest and has more area of material. More the area of material more is the number of magnetic field lines around it. Hence, end Z has maximum strength of magnetization and will be easily picked up.

Question 1(xiv)

For which position of the object in front of convex lens, the image distance will be negative ?

Answer

Between optical center and focus of convex lens:

Reason — For a convex lens, the image distance is considered negative when the image is formed on the same side as the incident light. This typically occurs when the object is placed within the focal length (between the lens and its focal point). So, if the object is positioned anywhere within the focal length of a convex lens, the image distance will be negative.

Question 1(xv)

A ray of light suffers refraction through an equilateral prism. The deviation produced by the prism does not depend on the

1. angle of incidence
2. colour of light
3. material of prism
4. size of prism

Answer

size of prism

Reason — The deviation produced by the prism does not depend on the size of prism.

Question 2(i)

(a) Give two reasons as to why the efficiency of a single removable pulley system is always less than 100%.

(b) Why a thick lead container is required to keep a radioactive substance?

(c) What is the principle of moments?

Answer

(a) The efficiency of a single movable pulley system is not 100% because of the following reasons:

1. There is friction in the pulley bearings or at the axle.
2. The weight of the pulley and string is not zero.

(b) The radioactive substances must be kept in a thick lead container with a very narrow opening, so that the walls of the container absorb the radiations which strike on them from the inside and thus radiations only come out through the narrow opening.

(c) According to the principle of moments, if the algebraic sum of moments of all the forces, acting on the body, about the axis of rotation is zero, the body is in equilibrium. Therefore, as per the principle of moments,
sum of anticlockwise moments = sum of clockwise moments

Question 2(ii)

Three students perform the experiments on series and parallel combinations of two given resistors R₁ and R₂ and the following V-I graphs were obtained.

Which of the graphs is/are correctly formed? Justify your answer.

Answer

According to Ω's law:

V = IR,

V ∝ I hence, the V-I slope will be a straight line.

In series combination, the resistance is more than that in parallel hence slope of V-I graph for series will be more than the parallel and so the correct answer will be : fig. b

Question 2(iii)

Two waves of the same pitch have their amplitudes in the ratio 2:3.

(a) What will be the ratio of their loudness?

(b) What will be the ratio of their frequencies?

Answer

(a) Given, amplitude ratio, a₁ : a₂ = 2 : 3

Loudness is directly proportional to square of amplitude

∴ L ∝ a²

$\phantom{\Rightarrow} \dfrac{\text{L}_1}{\text{L}_2} = \dfrac{\text{a}_1^2}{\text{a}_2^2} \\[1em] \Rightarrow \dfrac{\text{L}_1}{\text{L}_2} = \dfrac{2^2}{3^2} \\[1em] \Rightarrow \dfrac{\text{L}_1}{\text{L}_2} = \dfrac{4}{9}$

Hence, the ratio of their loudness will be 4 : 9.

(b) Waves with same pitch have the same frequency.

Hence, the ratio of their frequencies will be 1 : 1.

Question 2(iv)

(a) What is meant by potential energy of body?

(b) State whether the potential energy will be maximum or minimum at the surface of earth.

Answer

(a) The energy possessed by a body at rest due to its position or size and shape is called the potential energy.

(b) Potential energy will be minimum at the surface of earth because potential energy at any point = mgh, and at the surface of earth h = 0, hence, potential energy is minimum.

Question 2(v)

A uniform hall meter rule balances horizontally on a knife edge at 29 cm mark when a weight of 20 gf is suspended from one end.

(a) Draw a diagram of the arrangement

(b) What is the weight of the half meter rule?

Answer

(a) Diagram of the arrangement is shown below:

(b) If W gf is the weight of half meter rule, by the principle of moments,

$W \times (29 - 25) = 20 \times (50 - 29) \\[0.5em] \Rightarrow W \times 4 = 20 \times 21 \\[0.5em] \Rightarrow W = \dfrac{20 \times 21}{4} \\[0.5em] \Rightarrow W = 105 \text{ gf}$

Question 2(vi)

(a) Derive the relation between CGS and SI unit of power.

(b) A man is holding a suitcase in his hand at rest. What is the work done by him?

Answer

(a) SI unit of power = watt = 1 J/s

∴ 1 W = 1 J/s = 10⁵ dyne x 10² cm/s [∵ J = N-m]

= 10⁷ erg/s [∵ dyne x cm = erg]

(b) Work done by the man is zero as his displacement is zero.

Question 2(vii)

A meter scale is balanced in horizontal position as shown in figure below. Find the value of w.

Answer

From the given figure we can see that,

Anticlockwise moment = 6 kg × 60 cm
Clockwise moment = w × 40 cm

As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment.

$6 \times 60 = \text{w} \times 40 \\[0.5em] \Rightarrow \text{w} = \dfrac{6 \times 60}{40}\\[0.5em] \Rightarrow \text{w} = 9 \text{ kg} \\[0.5em]$

Hence, the value of w = 9 kg

Question 3(i)

10125 J of heat energy boils off 4.5 g of water at 100°C to steam at 100°C. Find the specific latent heat of steam.

Answer

Given,

Mass (m) = 4.5 g = $\dfrac{4.5}{1000}$ = 4.5 x 10^-3 kg

Heat energy given out (Q) = 10,125 J

Specific latent heat of steam = ?

From relation Q = m x L

Substituting the values in the formula we get,

$10125 = 4.5 \times 10^{-3} \times L \\[0.5em] \Rightarrow L = \dfrac{10125}{4.5 \times 10^{-3}} \\[0.5em] \Rightarrow L = 2.25 \times 10^{6} \text{ J kg}^{-1} \\[0.5em]$

Hence, the specific latent heat of steam = 2.25 x 10⁶ J kg^-1

Question 3(ii)

It is established that an electric current through a metallic conductor produces a magnetic field around it. Is there a similar magnetic field produced around a thin beam of moving

(a) α-particles and

(b) neutrons? Justify your answer.

Answer

When movement of a charged particle takes place, a magnetic field is produced around the path on which the charged particle moves.

(a) Yes, a thin beam of α-particles (positively charged) is like a straight current carrying conductor in the direction of motion.

(b) No, as neutrons carry no charge, so no magnetic field would be created around its path.

Question 3(iii)

Write the uses of radio isotopes in

(a) agriculture

(b) industries.

Answer

(a) Agriculture — Radioisotopes are used to trace and study the uptake of nutrients by plants. For example, the use of radioactive phosphorous ³²P in fertilizers has revealed how phosphorous is absorbed by plants.

(b) Industries — Radio isotopes such as $^{235}_{\space \space 92}\text{U}$ are used as fuel for atomic energy reactors.

Question 3(iv)

Two lenses have power of

(a) +2 D and

(b) -4 D

What is the nature and focal length of each lens?

Answer

(a) As we know,

$\text{Power} = \dfrac{1}{\text{focal length}}$

Given,

Power of the lens = +2.0 D

As the given power is positive,

Therefore, we can say that the lens used is convex in nature.

Substituting the value of power in formula we get,

$\text{2.0} = \dfrac{1}{\text{focal length}} \\[0.5em] \Rightarrow \text{focal length} = \dfrac{1}{2} \\[0.5em] \Rightarrow \text{focal length} = \text{0.5 m} \\[0.5em] \Rightarrow \text{focal length} = \text{50 cm} \\[0.5em]$

Therefore, the focal length is 50 cm and the lens used is convex in nature.

(b) As we know,

$\text{Power} = \dfrac{1}{\text{focal length}}$

Given,

Power of the lens = -4.0 D

As the given power is negative hence, we can say that the lens used is concave in nature.

Substituting the value of power in formula we get,

$-\text{4.0} = \dfrac{1}{\text{focal length}} \\[0.5em] \Rightarrow \text{focal length} = -\dfrac{1}{4} \\[0.5em] \Rightarrow \text{focal length} = -\text{0.25 m} \\[0.5em] \Rightarrow \text{focal length} = -\text{25 cm} \\[0.5em]$

Therefore, the focal length is 25 cm and the lens used is concave in nature.

Question 3(v)

Study the following electric circuit. Find the reading of

(a) the ammeter and

(b) the voltmeter

Answer

(a) In the given circuit, resistances are connected in series, hence, equivalent resistances (R_e) = R₁ + R₂ = 4 + 2 = 6 Ω

Potential difference (V) = 3 V

Current (I) = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

$3 = I \times 6 \\[0.5em] \Rightarrow I = \dfrac{3}{6} \\[0.5em] \Rightarrow I = 0.5 A$

As in series combination, current flowing through each component is same and is equal to the total current flowing in the circuit hence, the reading of ammeter = 0.5 A

(b) Reading of voltmeter = Potential difference across 2 Ω resistor.

Current flowing through bulb = 0.5 A

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

V = 0.5 x 2 = 1 V

Hence, the reading of voltmeter = 1 V

Section B

Question 4(i)

The diagram below shows a point source P inside a water container. Four rays A, B, C and D starting from the source P are shown upto the water surface.

(a) Show in the diagram, the path of these rays after striking the water surface. The critical angle for water-air surface is 48°.

(b) Name the phenomenon which the rays B and D exhibit.

(c) Can total internal reflection occur when rays travel from rarer medium to denser medium?

Answer

(a) Below diagram shows the path of the rays after striking the water surface:

(b) Ray B shows the phenomenon of 'refraction of light' and ray D shows the phenomenon of 'total internal reflection'.

(c) No, for total internal reflection to occur, the light ray must travel from denser medium to rarer medium.

Question 4(ii)

Name three factors on which the deviation produced by a prism depends and state how does it depend on the factors stated by you.

Answer

The three factors on which the deviation produced by a prism depends are as follows —

1. The angle of incidence (i) — As the angle of incidence increases, the angle of deviation first decreases, reaches to a minimum value for a certain angle of incidence and then on further increasing the angle of incidence, the angle of deviation begins to increase.
2. The angle of prism (A) — Angle of deviation increases with increase in the angle of prism (A).
3. Refractive index of the material of prism — For a given angle of incidence, the prism with a higher refractive index produces a greater deviation than the prism which has a lower refractive index.
   For example — A flint glass prism produces more deviation than a crown glass prism for same refracting angle since μ_flint > μ_crown

Question 4(iii)

(a) A boy uses blue colour of light to find the refractive index of glass, then he 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.

(b) A ray of light incident on a glass slab and follow the path as shown below. If a mirror is placed at S, such that emergent ray falls normally on it, then

I. Draw diagram to show the path of ray reflected from the mirror.

II. On the basis of which principle this refraction of ray is done?

Answer

(a) The speed of blue light in glass is less that of red light.

i.e. c_b < c_r

We know that,

$μ = \dfrac{\text{Speed of light in vacuum or air (c) }}{\text {Speed of light in glass (V)}} \\[0.5em]$

Hence, the refractive index of blue light is greater than that of red light.

i.e. μ_b > μ_r.

Hence, the refractive index of red and blue light in glass will be different.

b.

I. Below diagram shows the path of the ray reflected from the mirror:

II. Refraction of ray is done on the basis of principle of reversibility of light.

Question 5(i)

A ray of light PQ falls normally on the hypotenuse of a right angled prism ABC as shown below

(a) Copy the diagram and complete the path of ray PQ till it emerges from the prism.

(b) What is the value of the angle of deviation of the ray?

(c) Name an instrument where this action of prism is used.

Answer

(a) Below diagram shows the path of the ray PQ till it emerges from the prism with all the angles labelled:

(b) The angle of deviation of the ray PQ is 180°.

(c) This action of the prism is used in Binoculars.

Question 5(ii)

A linear object is placed on the axis of a lens. An image is formed by refraction in the lens. For all positions of the object on the axis of the lens, the positions of the image are always between the lens and the object.

(a) Name the lens.

(b) Draw a ray diagram to show the formation of the image of an object placed infront of the lens at any position of your choice except infinity.

Answer

(a) The lens is concave as the positions of the image are always between the lens and the object.

(b) Ray diagram showing the formation of the image of an object placed infront of the lens at any position except infinity:

Question 5(iii)

Draw ray diagrams showing the image formation by a convex lens when an object is placed

(a) between optical centre and focus of the lens.

(b) between focus and twice the focal length of the lens.

(c) at twice the focal length of the lens.

(d) at infinity.

Answer

(a) between optical centre and focus of the lens:

(b) between focus and twice the focal length of the lens:

(c) at twice the focal length of the lens:

(d) at infinity:

Question 6(i)

(a) A force applied on a

I. rigid body

II. non-rigid body

How does the effect of force differ in the two cases?

(b) Give one example of couple force.

Answer

(a) (i) Rigid body — A force when applied on a rigid object does not change the inter-spacing between it's constituent particles and therefore it does not change the dimensions of the object, but causes only motion in it.

(ii) Non-rigid body — A force when applied on a non-rigid object, changes the inter-spacing between it's constituent particles and therefore causes a change in it's dimensions and can also produce motion in it.

(b) Two equal and opposite parallel forces, not acting along the same line form a couple. E.g. tightening the cap of a bottle.

Question 6(ii)

(a) In a single movable pulley, if the weight of the load is L and the weight of the movable pulley is w, find out the expression for its mechanical advantage.

(b) Which simple machine is used by labourers to load heavy barrels on a truck?

(c) Does a single fixed pulley help us to multiple force? In what way is it useful?

Answer

(a) L + w = 2T and E = T

So, L = 2T - w

Mechanical advantage

= $\dfrac{\text{L}}{\text{E}}$

= $\dfrac{\text{2T - w}}{\text{T}}$

= 2 - $\dfrac{\text{w}}{\text{T}}$

= 2 - $\dfrac{\text{w}}{\text{E}}$

Hence, mechanical advantage = 2 - $\dfrac{\text{w}}{\text{E}}$

(b) An inclined plane

(c) No, a single fixed pulley does not help in multiplying force. It only helps us to change the direction of effort to be applied in a more convenient direction.

Question 6(iii)

(a) A runner of mass 60 kg runs upstairs and reaches the 10 m high first floor in 4 s.

Calculate the

I. force of gravity acting on him

II. work done by him against gravity and

III. power developed by the runner.

(b) Give two examples in which mechanical energy is conserved.

Answer

(a) Given,

Mass of runner = 60 kg

Gravitational force = g = 10ms^-2

I. Force of gravity acting on runner = mg = 60 x 10 = 600 N

Hence, force of gravity acting on him = 600 N

II. Work done by runner against gravity = mgh = 60 x 10 x 10 = 6000 J or 6 kJ

Hence, Work done by runner against gravity = 6 kJ

III. Power developed by the runner,

P = $\dfrac{\text{w}}{\text{t}}$

= $\dfrac{6000}{4}$

= 1500 W

Hence, power developed by the runner = 1500 W

(b) Two examples in which mechanical energy is conserved are:

1. Motion of a simple pendulum.
2. Free fall of a body under gravity.

Question 7(i)

Kunal and Abhimanyu were waiting to go cross a railway crossing. Kunal jumped over the barrier and curiously put his ear on the railway track. Abhimanyu opposed Kunal and pulled him away from the railway track.

(a) Why did Kunal put his ear on the railway track?

(b) Why did Abhimanyu pull Kunal away from the railway track?

(c) Name the waves used for sound ranging. State one reason for their use.

Answer

(a) Kunal put his ear on the railway track to get an idea of the distance of the incoming train as sound travels faster in solid medium.

(b) Abhimanyu pulled Kunal away from the railway track because it is not safe to be on or very near the railway track when the train is coming.

(c) Ultrasonic waves are used for sound ranging. Ultrasonic waves can travel undeviated through a long distance and so they are used for sound ranging.

Question 7(ii)

A mixture of radio active substance gives off three types of radiations.

I. Name the radiation which travels with the speed of light.

II. Name the radiation which has the highest ionising power.

III. Name the radiation which is negatively charged.

(b) When an alpha particle gains two elections, it becomes neutral and an atom of an element which is a rare gas. Name this rare gas.

Answer

(a) I. The radiation which travels with the speed of light is γ radiation.

II. The radiation which has the highest ionizing power is α radiations.

III. The radiation which is negatively charged is β radiations.

(b) Helium [₂He⁴]

Question 7(iii)

(a) A wire of length 80 cm has a frequency of 256 Hz. Calculate the length of a similar wire under similar tension, which will have frequency of 1024 Hz.

(b) A certain sound has a frequency of 256 Hz and a wavelength of 1.3 m.

I. Calculate the speed with which this sound travels.

II. What difference would be felt by a listener between the above sound and another sound travelling at the same speed, but of wavelength 2.6m?

Answer

(i) Given,

l₁ = 80 cm, f₁ = 256 Hz, f₂ = 1024 Hz, l₂ = ?

Since, f ∝ $\dfrac{1}{l}$

Therefore, f₁l₁ = f₂l₂

or

$f_2 = \dfrac{f_1l_1}{f_2} \\[0.5em] = \dfrac{256 \times 80}{1024} \\[0.5em] = 20 \text{ cm}$

(ii) Given, f = 256 Hz, λ = 1.3 m

(a) V = fλ = 256 x 1.3 = 332.8 m s^-1

(b) Speed of second sound (V') = 332.8 m s^-1

Wavelength of second sound (λ') = 2.6 m

$\text{f} = \dfrac{V'}{λ'} \\[0.5em] = \dfrac{332.8}{2.6} \\[0.5em] = 128 \text{ Hz}$

As frequency of second sound is less as compared to the first sound, hence, to the listener, the first sound of wavelength 1.3 m will appear to be shriller than the second sound of wavelength 2.6 m.

Question 8(i)

(a) The isotope of ₉₂U²³⁸ decays by α-emission to an isotope of thorium(Th). The thorium isotopes decays by β-emission to an isotope of protactinium (Pa). Write down the equations to represent these two nuclear charges.

(b) Calculate the numbers of α and β particles emitted.

Answer

(a) When a radioactive uranium nucleus $_{92}^{238}\text{U}$ emits an α particle, a new nucleus thorium $_{90}^{234}\text{Th}$ is formed and the change is represented as follows —

$^{238}_{92}\text{U} \longrightarrow \space ^{234}_{90}\text{Th} + \space ^{4}_{2}\text{He}$

When thorium $_{90}^{234}\text{Th}$ isotope decays by a β-emission to an isotope of protactinium (Pa), the change is represented as follows —

$^{234}_{90}\text{Th} \longrightarrow \space ^{234}_{91}\text{Pa} + \space ^{\enspace 0}_{-1}\text{e}$

Question 8(ii)

(a) A cell is sending current in an external circuit. How does the terminal voltage compare with the emf of the cell?

(b) At what voltage is the alternating current supplied to our houses?

(c) How should the electric lamps in a building be connected?

Answer

(a) The terminal voltage of the cell is less than its emf due to voltage drop during flow of charge through the electrolyte inside the cell.

(b) 220 V

(c) Parallel

Question 8(iii)

(a) A current of 1 A flows in a series circuit having an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp.

(b) Now, if a resistance of 10 Ω is connected in parallel with this series combination, then what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place? Give reason.

Answer

(a) Given,

Current I = 1 A

Voltage V = 10 V

Resistance of conductor = 5 Ω

Resistance of lamp = R_l

Equivalent resistance R_s

= Resistance of conductor + Resistance of lamp

= 5 + R_l

According to Ohm's law V = IR

Substituting we get, 10 = 1(5+R_l)

R_l = 10 - 5 = 5 Ω

Hence, resistance of the electric lamp = 5 Ω

(b) Total resistance

$\dfrac{1}{R_p} = \dfrac{1}{R_1 + 5} + \dfrac{1}{10} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1}{5+5} + \dfrac{1}{10} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1}{10} + \dfrac{1}{10} \\[0.5em] \Rightarrow \dfrac{1}{R_p} = \dfrac{2}{10} \\[0.5em] \Rightarrow R_p = 5 Ω$

Hence, Current flowing through the circuit I = $\dfrac{10}{5}$ = 2A

Thus, 1A current flows through 10 Ω resistor and 1A flows through lamp and conductor, hence, there is no change in current flowing through conductor. Also there is no change in potential difference across the lamp.

Question 9(i)

Farmers fill their fields with water on a cold winter night. Explain in brief along with a reason.

Answer

On a cold winter night, if the atmospheric temperature falls below 0°C, water in the fine capillaries of plants will freeze, so the veins will burst due to the increase in volume of water on freezing. As a result, plants will die and the crop will get destroyed. In order to save crop on such cold nights, farmers fill their fields with water because water has a high specific heat capacity, so it does not allow the temperature in the surrounding area of plants to fall up to 0°C.

Question 9(ii)

(a) How would you demonstrate magnetic effect of current? By whom it was discovered?

(b) For the current carrying solenoid as shown below, draw magnetic field lines. Explain that out of the three points A, B and C at which point, the field strength is maximum and at which point it is minimum.

Answer

(a) If a compass needle is placed in the vicinity of the current carrying wire, the needle is found to deflect in a definite direction. This was discoverd by Hans Oersted.

(b) Below diagram shows the magnetic field lines for the current carrying solenoid:

The magnetic field strength is maximum at point A and minimum at point B because magnetic field is strong where magnetic field lines are crowded and is weak where magnetic field lines are far apart.

Question 9(iii)

In a laboratory experiment to measure specific heat capacity of copper, 0.05 kg of water at 60°C was poured into a copper calorimeter with a stirrer of mass 0.20 kg initially at 10°C. After stirring, the final temperature reached to 25°C.

Specific heat of water is taken as 4200 J/kg°C.

(a) What is the quantity of heat released per kg of water per 1°C fall in temperature?

(b) Calculate the heat energy released by water in the experiment in cooling from 70°C to 45°C.

(c) Assuming that the heat released by water is entirely used to raise the temperature of calorimeter from 10°C to 45°C, calculate the specific heat capacity of copper.

Answer

(a) Quantity of heat released per kg of water per 1°C fall in temperature is given by 4200 J/kg°C.

(b) Given, m_w = 0.05 kg

t_w = 60°C

m_c = 0.20 kg

t_c = 10°C

Specific heat capacity of water = 4200 J kg^-1 K^-1

final temperature of water = 25°C

Heat energy given out by water in lowering it's temperature from 70° C to 45° C
= m x c x ΔT
= 0.05 x 4200 x (70 - 45)
= 0.05 x 4200 x 25
= 5250 J

(c) Heat energy taken by calorimeter when it raises it's temperature from 10°C to 45°C
= m x c x change in temperature
= 0.20 × c × (45 - 10)
= 0.20 × c × 35 = 7c

If there is no loss of energy,

Heat energy gained = heat energy lost

Substituting the values we get,

$5250 = 7c \\[0.5em] \Rightarrow c = \dfrac{5250}{7} \\[0.5em] \Rightarrow c = 750 \text{ J Kg}^{-1}\text{K}^{-1}$

Hence, specific heat capacity of copper = 750 J kg^-1 K^-1