Solved Sample Paper 2

Section A

Question 1(i)

Shimmering effect around a camp fire is due to

1. reflection of light
2. dispersion of light
3. scattering of light
4. refraction of light

Answer

refraction of light

Reason — The shimmering effect around a campfire is primarily caused by the refraction of light through varying temperature and density layers in the air.

Question 1(ii)

In the sound wave produced by a vibrating tuning fork shown below, the half wavelength is given by:

1. PQ
2. QR
3. QS
4. PT

Answer

QS

Reason — The distance between two successive troughs or crests is defined as one wavelength and so QS represents half of wavelength.

Question 1(iii)

The work done by machine to load luggage in 60s is 1800 J. Calculate the power spent by machine

1. 50 W
2. 30 W
3. 80 W
4. 70 W

Answer

30 W

Reason — Given,

work done = 1800 J

time = 60 s

Power spent = $\dfrac{\text{work done}}{\text{time taken}}$ = $\dfrac{1800}{60}$ = 30 W

Question 1(iv)

The number of beta particles emitted by radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter nuclei is as

1. isobar of parent
2. isomer of parent
3. isotone of parent
4. isotope of parent

Answer

isotope of parent

Reason — Given, number of beta particles emitted by radioactive substance is twice the number of alpha particles emitted by it.

Hence,

After alpha decay : $_{Z}^{A}\text{X} \longrightarrow _{Z-2}^{A-4}\text{Y} + _{2}^{4}\text{He}$

After beta decay : $_{Z-2}^{A-4}\text{Y} \longrightarrow _{Z-1}^{A-4}\text{Z} \longrightarrow _{Z}^{A-4}\text{W}$

• The daughter nucleus after alpha decay has its mass number reduced by 4.

• There will be no change in the reduced mass number after beta decay.

• The obtained daughter nucleus has a different mass number = (A-4) and the same atomic number = Z as the parent nucleus. So, the resulting daughter nuclei is an isotope of the parent.

Question 1(v)

Which of the following forces cannot create a moment of force?

1. Action-reaction forces
2. Gravitational forces
3. Electric forces
4. Frictional forces

Answer

Action-reaction forces

Reason — Action-reaction forces cannot create a moment of force as they act on different bodies.

Question 1(vi)

A ray of light incident at an angle of 48° on a prism of refracting angle 60°, suffers minimum deviation. The angle of minimum deviation would be

1. 12°
2. 24°
3. 36°
4. 48°

Answer

36°

Reason — As we know,
$δ_ {min} = 2i - A$

Given,

i = 48°

A = 60°

So, substituting the values in the formula above we get,

$δ_ {min} = (2 \times 48°) - 60° \\[0.5em] δ_ {min} = 96° - 60° \\[0.5em] \Rightarrow δ_ {min} = 36° \\[0.5em]$

Hence, the angle of minimum deviation is equal to 36°

Question 1(vii)

Assertion Bending of wire does not affect electrical resistance.

Reason Resistance of a wire is proportional to resistivity of material.

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 — Resistance of a wire depends on the resistivity, length and area of cross section. Bending a wire does not affect any of these, hence it does not affect electrical resistance.

Question 1(viii)

A moving ship continuously sends in ultrasonic pulses down to the sed-bed and records the time delays of the reflected pulses. The variation of time delays recorded versus the positions of the ship is plotted in the following graph. Which of the position indicates the deepest sea bed?

1. A
2. B
3. C
4. D

Answer

B

Reason — The depth of the ocean floor is determined by measuring the time it takes for a signal to be received back. The greatest time delay, observed at position B, indicates the deepest sea-bed. This is because sound must travel a greater distance, resulting in a longer time delay.

Question 1(ix)

Mechanical energy is equal to

1. the sum of total energy and potential energy
2. the sum of potential energy and kinetic energy
3. the energy acquired by machines
4. All of the above

Answer

the sum of potential energy and kinetic energy

Reason — Mechanical energy is equal to the sum of potential energy and kinetic energy.

Question 1(x)

A light moves from denser to rarer medium, which of the following is correct?

1. Energy increases
2. Frequency increases
3. Velocity increases
4. None of the above

Answer

Velocity increases

Reason — When a ray of light travels from a denser medium to a rarer medium, the speed of light increases in the rarer medium and it bends away from the normal.

Question 1(xi)

Observe the heating curve of water

Which of the following represent boiling?

1. OA
2. AB
3. BC
4. CD

Answer

BC

Reason — In part BC of the graph, the temperature of water does not increase even though more heat is supplied with time. Hence, BC represents boiling.

Question 1(xii)

A ray of light travels from a medium of refractive index n₁ to a medium of refractive index n₂. If angle of incidence is i and angle of refraction is r, then $\dfrac{\text{sin i}}{\text{sin r}}$ is equal to

1. n₁

2. n₂

3. $\dfrac{\text{n}_2}{\text{n}_1}$

4. $\dfrac{\text{n}_1}{\text{n}_2}$

Answer

$\dfrac{\text{n}_2}{\text{n}_1}$

Reason — According to Snell's law,

$\dfrac{\text{sin i}}{\text{sin r}}$ = $\dfrac{\text{n}_2}{\text{n}_1}$

Question 1(xiii)

Fuse is an alloy of

1. lead and copper
2. copper and tin
3. lead and tin
4. None of the above

Answer

lead and tin

Reason — Fuse wire is an alloy of lead and tin having 50% of each metal to have a low melting point.

Question 1(xiv)

For cooking of food, which of the following types of utensil is most suitable?

1. High specific heat and low conductivity
2. High specific heat and high conductivity
3. Low specific heat and low conductivity
4. Low specific heat and high conductivity

Answer

Low specific heat and high conductivity

Reason — Low specific heat allows the utensil to heat up quickly with less amount of heat and high conductivity ensures that heat is distributed throughout the utensil, promoting uniform cooking of the food.

Question 1(xv)

Which of the following is not a representation of electric power?

1. I²R
2. $\dfrac{\text{V}^2}{\text{R}}$
3. VI
4. I²V

Answer

I²R

Reason — One watt is the electric power consumed when a current of 1 ampere flows through a circuit having potential difference of 1 volt. Hence, P = VI = $\dfrac{\text{V}^2}{\text{R}} = \text{I}^2\text{R}$

Question 2(i)

(a) Write the two uses of radioactivity.

(b) A crow-bar of length 80 cm has its fulcrum situated at a distance of 10 cm from the load. Calculate the mechanical advantage of the crow-bar.

(c) Why fission nuclei undergo β-decay?

Answer

(a) Uses of radioactivity :

1. Many diseases such as leukemia, cancer, etc., are cured by radiation therapy.
2. 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) Given,

Total length of a crowbar = 80 cm

Effort arm = 80 - 10 = 70 cm

Load arm = 10 cm

We know that,

$\text{M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em] \text{M.A.}. = \dfrac{70}{10} \\[0.5em] \Rightarrow\text{M.A.} = 7 \\[0.5em]$

Hence, mechanical advantage = 7

(c) Fission reactions primarily release energy in the form of neutrons. Some of the neutrons produced during fission are initially in an excited state. Beta decay occurs when a neutron in this excited state transforms into a proton, emitting a beta particle (an electron) and an antineutrino. This process helps stabilize the nucleus by converting a neutron into a proton, maintaining a balance between protons and neutrons in the nucleus.

Question 2(ii)

A person standing 22 m away from a wall produces a sound and receives the reflected sound.

(a) In what time interval, he receives the reflected sound, if speed of sound in air is 330 ms^-1 ?

(b) At which distance should he stand to receive the sound in 0.12 s?

Answer

(a) 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,

d = 22 m

V = 330 ms^-1

Substituting the values in the formula above, we get,

$330 = \dfrac{2 \times 22}{t} \\[0.5em] \Rightarrow t = \dfrac{44}{330} \\[0.5em] \Rightarrow t = \dfrac{4}{30} \\[0.5em] \Rightarrow t = 0.13 \text{ s} \\[0.5em]$

Hence, the time after which he receives the reflected sound = 0.13 seconds

(b) As we know,

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

t = 0.12 s

V = 330 ms^-1

d = ?

Substituting the values in the formula above, we get,

$330 = \dfrac{2d}{0.12} \\[0.5em] \Rightarrow d = \dfrac{330 \times 0.12}{2} \\[0.5em] \Rightarrow d = 19.8 \text{ m} \\[0.5em]$

Hence, distance to receive the sound in 0.12 s = 19.8 m

Question 2(iii)

The moment of a force of 20 N about a point O is 18 N-m. Find the distance of point of application of the force from the point O.

Answer

Given,

moment of a force = 18 N-m

force = 20 N

distance = ?

Moment of force = force × distance

Substituting the values in the formula above, we get,

$18 = 20 \times \text{d} \\[0.5em] \Rightarrow d = \dfrac{18}{20} \\[0.5em] \Rightarrow d = 0.9 \text{ m} \\[0.5em]$

Hence, distance of point of application of the force from the point O = 0.9 m

Question 2(iv)

Define the term 'centre of gravity of a body'.

Answer

The centre of gravity of a body is defined as the point about which the algebraic sum of moments of weights of particles constituting the body is zero and the entire weight of the body is to act at this point.

Question 2(v)

The diagram below shows a lever in use:

(a) To which class of levers does it belong?

(b) Without changing the dimensions of the lever, if the load is shifted towards the fulcrum, what happens to the mechanical advantage of the lever?

Answer

(a) It belongs to second class lever as load is situated between fulcrum and effort.

(b) The mechanical advantage of the lever will increase when load is shifted towards the fulcrum.

Question 2(vi)

A beam of metal is acted upon by two forces (parallel) of 15 N and 20 N at a distance of 35 cm from each other. At what point the beam will be balanced ?

Answer

Let the balance be at x cm from 20 N force.

Anticlockwise moment = 20(35-x)

Clockwise moment = 15x

According to principle of moment :

Anticlockwise moment = Clockwise moment

Substituting the values in the formula above, we get,

$20(35-\text{x}) = 15\text{x} \\[0.5em] \Rightarrow (20 \times 35) - 20 \text{x} = 15\text{x} \\[0.5em] \Rightarrow (20 \times 35) = (15 + 20)\text{x} \\[0.5em] \Rightarrow (20 \times 35) = 35 \text{x} \\[0.5em] \Rightarrow \text{x} = \dfrac{(20 \times 35)}{35} = 20 \text{ cm} \\[0.5em]$

Hence, the beam will be balanced at 20 cm

Question 2(vii)

Calculate the equivalent resistance between P and Q from the following diagram:

Answer

In the circuit, there are two parts. In the first part, two resistors of 10 Ω each are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (10 + 10) Ω = 20 Ω

In the second part, resistance R'_s = 20 Ω and 5 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{20} + \dfrac{1}{5} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1 + 4}{20} \\[0.5em] \dfrac{1}{R_p} = \dfrac{5}{20} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1}{4} \\[0.5em] \Rightarrow R_p = 4 Ω \\[0.5em]$

∴ Equivalent resistance of second part = 4 Ω

In the last part, 3 Ω, R_p = 4 Ω and 2 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = (3 + 4 + 2) Ω = 9 Ω

∴ Equivalent resistance between the points P and Q = 9 Ω

Question 3(i)

(a) The diagram shows a bar magnet surrounded by four plotting compasses. Copy the diagram and mark the direction of the compass needle for each of the cases B, C and D.

(b) Which is the North pole, X or Y?

Answer

(a) Below diagram shows the direction of the compass needle for cases B, C and D:

(b) X is the north pole as the magnetic field lines are coming out of the bar magnet at point X.

Question 3(ii)

An object is placed in front of a converging lens. Then, draw the ray diagrams for following cases

(a) When object distance (u) > 2f.

(b) When image formed has the same size as that of object.

Answer

(a) Ray diagram for the formation of image when object distance (u) > 2f for a convex lens is shown below:

(b) Ray diagram for the formation of image when image formed has the same size as that of object for a convex lens is shown below:

Question 3(iii)

(a) Ring system is a better wiring system, why?

(b) State the relation between electrical power, resistance and potential difference in an electrical circuit.

Answer

(a) Advantages of a ring system are —

1. Each appliance has a separate fuse. Therefore, if due to some fault, the fuse of one appliance burns, it does not affect the operation of the other appliances.
2. In this system all the plugs and sockets used can be of same size, but each socket should have it's own fuse of rating suitable for the appliances to be connected with it.

(b) The relation between electrical power, resistance and potential difference in an electrical circuit is given by:

P = $\dfrac{\text{V}^2}{\text{R}}$

Question 3(iv)

Explain any two factors on which the internal resistance of a cell depends.

Answer

Two factors on which the internal resistance of a cell depends are:

1. The surface area of the electrodes — Larger the surface area of the electrodes, less is the internal resistance.
2. The distance between the electrodes — More the distance between the electrodes, greater is the internal resistance.

Question 3(v)

When 2240 J of heat is supplied to a metal piece of mass 450 g, its temperature rises from 25°C to 65°C. Then, find

(a) heat capacity of metal piece and

(b) specific heat capacity of metal.

Answer

Given,

Amount of heat supplied (Q)= 2240 J

Rise in temperature (△t) = (65 – 25)°C = 40°C

Heat capacity = ?

(a) Heat capacity (C') = $\dfrac{\text{\text{Q}}}{\text{△t}}$

Substituting the values in the formula above we get,

$= \dfrac{2240}{40} = 56 \text{ J°C}^{-1}$

Hence, heat capacity of metal piece = 56 J per °C

(b) Specific heat capacity (c) = ?

mass (m) = 450 g

We know that,

$\text{c} = \dfrac{\text{C}'}{\text{m}}$

Substituting the values in the formula above we get,

$\text{c} = \dfrac{56}{450} = 0.12\text{ J g}^{-1} \text{°C}^{-1}$

Hence, Specific heat capacity = 0.12 J g^-1 °C^-1

Section B

Question 4(i)

An object of height 5 cm is placed perpendicular to the principal axis of a concave lens of focal length 10 cm. If the distance of the object from the optical centre of the lens is 20 cm, determine the position, nature and size of the image formed using the lens formula.

Answer

Given,

f = -10 cm

u = -20 cm

v = ?

Lens formula is —

$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\[0.5em]$

Substituting the values in the formula, we get,

$\dfrac{1}{v} - \dfrac{1}{-20} = \dfrac{1}{-10} \\[0.5em] \dfrac{1}{v} + \dfrac{1}{20} = - \dfrac{1}{10} \\[0.5em] \dfrac{1}{v} = -\dfrac{1}{10} - \dfrac{1}{20} \\[0.5em] \dfrac{1}{v} = \dfrac{-2-1}{20} \\[0.5em] \dfrac{1}{v} = -\dfrac{3}{20} \\[0.5em] \Rightarrow v = -\dfrac{20}{3} \\[0.5em] \Rightarrow v = -6.67$

Therefore, the image in formed 6.67 cm in front of the lens.

As we know,

the formula for magnification of a lens is —

$m = \dfrac{v}{u} = \dfrac{h_i}{h_o} \\[0.5em]$ Given,

u = -20 cm

v = -6.67 cm

height of object (h_o) = 5 cm

height of image (h_i) = ?

Substituting the values in the formula, we get,

$\dfrac{-6.67}{-20} = \dfrac{\text{h}_i}{5} \\[0.5em] \text{h}_i = 5 \times \dfrac{-6.67}{-20} \\[0.5em] \text{h}_i = 1.67 \text{ cm} \\[0.5em]$

Hence, the image is formed 6.67 cm in front of the lens, it is of size 1.67 cm and it is virtual (as the magnification is positive), erect and diminished.

Question 4(ii)

(a) Figs. I and II show the incident and transmitted rays through lens kept in a box in each case. Draw the lens and complete the path of rays through it.

(b) What is the difference between double convex and bi-convex lens?

(c) Does convex lens always form real images?

Answer

(a) In Fig I, the rays are diverging, hence a concave lens is placed inside the box:

In Fig II, the rays are converging, hence a convex lens is placed inside the box:

(b) In case of double convex lens the two surfaces may have any values for the two radii of curvature of the two surfaces while in bi-convex lens the radii of curvature of the two surfaces, have the same value. So, all double convex lenses may not be bi-convex but every bi-convex lens is a double convex lens.

(c) No, when the distance of the object from the lens is less than the focal length of the lens, then convex lens will form virtual image.

Question 4(iii)

(a) The velocity of light in diamond is 121000 km/s. What is its refractive index?

(b) With the help of a labelled diagram, show that the apparent depth of an object such as a coin in water is less than its real depth.

Answer

(a) Given,

velocity of light in diamond = 121000 kms^-1

To convert kms^-1 into ms^-1, we multiply by 1000

121000 x 1000 ms^-1 = 121 x 10⁶ ms^-1

Speed of light in air, c = 3 × 10⁸ ms^-1

As we know,

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

Substituting the values in the formula above we get,

$μ = \dfrac{ 3 \times 10^8 } {121 \times 10^6} \\[0.5em] μ = \dfrac{ 3 \times 10^8 } {1.21 \times 10^8} \\[0.5em] μ = 2.48 \\[0.5em]$

Hence, refractive index = 2.48

(b) Below labelled diagram, shows that the apparent depth of a coin in water is less than its real depth:

Question 5(i)

(a) A stick partly immersed in water appears to be bent. Draw a ray diagram to show the bending of the stick when placed in water and viewed obliquely from above. On which phenomenon, it is based?

(b) Two parallel rays of orange and green light in air, meet an air-water interface as shown below

I. Complete the diagram for path inside water.

II. What conclusion will you draw for this?

Answer

(a) Ray diagram showing bending of the stick when placed in water and viewed obliquely from above is given below:

The apparent bending of the stick is due to the refraction of light when it passes from water into air.

(b)

I. The complete path of the rays is shown in the below diagram:

II. The two refracted rays are not parallel in glass as orange light with faster speed bends less while green light with slower speed bends more.

Question 5(ii)

(a) What are the necessary conditions for TIR?

(b) Draw a diagram to show how rays of light are totally reflected when fall normally on the face of a right angled prism. Where is this action of prism used?

Answer

(a) The two necessary conditions for total internal reflection are :

1. The light must travel from a denser medium to a rarer medium.
2. The angle of incidence must be greater than the critical angle for the pair of media.

(b) Below ray diagram shows the total internal reflection of light rays falling normally on the face of a right angled prism:

This action of prism is used in a periscope where a total reflecting prism is preferred over a plane mirror.

Question 5(iii)

Which type of rays exist beyond X-rays of the electro magnetic spectrum? State one property and one use of these rays.

Answer

Beyond X-rays, ultraviolet rays exist.

Property — Ultraviolet radiations travel in a straight line with a speed of 3 x 10⁸ ms^-1 in air or vacuum.

Use — Ultraviolet radiations are used for sterilizing purposes.

Question 6(i)

Monu pulls a toy car through a distance of 20 m on smooth and horizontal floor. The string field in Monu's hand makes an angle 60° with horizontal surface. If the force applied by the Monu be 10N,

(a) calculate the work done by Monu is pulling the car.

(b) If the string makes an angle 30° with the horizontal surface, then calculate work done.

Answer

(a) Given,

Distance moved (d) = 20 m

Force applied (F) = 10 N

Angle = 60°

Work done (W) = F d cos θ

Substituting the values in the formula above we get,

W = 10 x 20 cos 60 = 10 x 20 x 0.5 = 100 J

Hence, work done = 100 J

(b) If angle = 30° then work done = ?

Substituting the values in the formula above we get,

W = 10 x 20 cos 30 = 10 x 20 x $\dfrac{\sqrt{3}}{2}$ = 100 $\sqrt{3}$ J

Hence, work done = 100 $\sqrt{3}$ J

Question 6(ii)

A coolie of mass 55 kg move a load of 25 kg on his head upto 50 stairs, each of height 20 cm in 50 s. If g is 10 m/s², then find

(a) gravitational force acting on coolie,

(b) work done by him

(c) his power in watt

(d) the gravitational force is increased by two times, then calculate the power.

Answer

Given,

Mass (m_c) = 55 kg

Load = 25 kg

Total load

50 stairs of height 20 cm,

so distance = 50 x 20 = 1000 cm

Converting cm into m

100 cm = 1m

so, 1000 cm = $\dfrac{1000}{100}$ = 10 m

Time = 50 s

(a) Gravitational force (F) = (m_c + load) x g

Substituting the values in the formula above we get,

F = (55 + 25) x 10 = 800 N

Hence, gravitational force acting on coolie = 800 N

(b) F = 800 N

distance (S) = 10 m

work done = ?

work done = F x S = (m_c + load) x g x S

Substituting the values in the formula above we get,

W = (55 + 25) x 10 x 10 = 8000 J

(c) work done (W) = 8000 J

time taken (t) = 50

Power = $\dfrac{\text{work done}}{\text{time taken}}$

Substituting the values in the formula above we get,

Power = $\dfrac{8000}{50}$ = 160 W

Hence, power in watt = 160 W

(d) If gravitational force is increased two times i.e., F = 2 x 800 = 1600 N

Substituting the values in the formula above we get,

W = 1600 x 10 = 16000 J

Power = $\dfrac{16000}{50}$ = 320 W

Hence, power in watt = 320 W

Question 6(iii)

(a) What is a simple machine? State the principle of an ideal machine.

(b) A block and tackle pulley system has a velocity ratio 3. Draw a labelled diagram of this system. In your diagram, indicate clearly the points of application and the directions of the load and effort.

Answer

(a) A machine is a device by which we can either overcome a large resistive force (or load) at some point by applying a small force (or effort) at a convenient point and in a desired direction or by which we can obtain a gain in speed.
An ideal machine is that in which there is no loss of energy in any manner. Here, the work output is equal to the work input, i.e., its efficiency is 100%.

(b) Given,

V.R = n (number of pulleys) = 3

Hence, number of pulleys = 3

Below is a labelled diagram of the system indicating the points of application and the directions of load L and effort E:

Question 7(i)

(a) A wave has a frequency of 800 MHz and a wave length of 30 cm. Calculate velocity.

(b) A radio can tune in to any station in 17.5 MHz to 13.5 MHz band, what is the corresponding wavelength band?

Answer

(a) As we know,

Velocity of wave (c) = frequency (f) x wavelength (λ)

Given,

f = 800 MHz

Hence, f = 800 x 10⁶ Hz or

f = 8 x 10⁸ Hz

λ = 30 cm

As 100 cm = 1m

Therefore,

30 cm = $\dfrac{30}{100}$ m

Hence, λ = 0.3 m

Substituting the values in the formula above we get,

c = 8 x 10⁸ x 0.3 = 2.4 x 10⁸ ms^-1

Hence,

Velocity of wave = 2.4 x 10⁸ ms^-1

(b) As we know,

Velocity of wave (c) = frequency (f) x wavelength (λ)

Given,

f = 17.5 MHz = 17.5 x 10⁶ Hz

Speed of light in air/vacuum = 3 x 10⁸ ms^-1.

λ = ?

Substituting the values in the formula above we get,

$3 \times 10^8 = (17.5 \times 10^6) \times λ \\[0.5em] \Rightarrow λ = \dfrac{ 3\times 10^8}{17.5 \times 10^6 } \\[0.5em] \Rightarrow λ = 17.1 \text{ m}$

Hence, wavelength of wave = 17.1 m

When,

f = 13.5 MHz = 13.5 x 10⁶ Hz

Speed of light in air/vacuum = 3 x 10⁸ ms^-1.

λ = ?

Substituting the values in the formula above we get,

$3 \times 10^8 = (13.5 \times 10^6) \times λ \\[0.5em] \Rightarrow λ = \dfrac{ 3\times 10^8}{13.5 \times 10^6 } \\[0.5em] \Rightarrow λ = 22.2 \text{ m}$

wavelength of wave = 22.2 m

∴ The corresponding wavelength band is 17.1 m - 22.2 m.

Question 7(ii)

Arrange α, β and γ-rays in ascending order with respect to their

(a) penetrating power

(b) ionising power

(c) biological effect

Answer

(a) Penetrating power — α < β < γ.

(b) Ionising power — γ < β < α.

(c) Biological effect — α < β < γ.

Question 7(iii)

(a) A man standing between two cliffs produces a sound and hears two successive echoes at intervals of 2 s and 5 s respectively. Calculate the distance between the two cliffs. The speed of sounding in the air is 330 ms^-1.

(b) Explain, why strings of different thicknesses are provided on a stringed instrument.

Answer

(a) 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,

t₁ = 2 s

t₂ = 5 s

V = 330 ms^-1 and

First echo will be heard from the nearer cliff and the second echo from the farther cliff.

Hence, Distance between cliffs = d₁ + d₂

Therefore, substituting the values in the formula above, we get,

For t₁

$330 = \dfrac{2d_1 }{2} \\[1em] \Rightarrow d_1 = \dfrac{330 \times 2}{2} \\[1em] \Rightarrow d_1 = 330 \text{ m}$

For t₂

$330 = \dfrac{2d_2}{5} \\[1em] \Rightarrow d_2 = \dfrac{330 \times 5}{2} \\[1em] \Rightarrow d_2 = 825 \text{ m}$

Distance between cliffs = d₁ + d₂

= 330 + 825

= 1155 m

Hence, distance between cliffs = 1155 m

(b) As we know,

frequency (f) = $\dfrac{1}{2l} \sqrt\dfrac{T}{πr^2d}$

So, we can say that the natural frequency of vibration of a stretched string is inversely proportional to the radius of the string.

Hence, in order to produce sound waves of different frequencies, strings of different thickness are provided on a stringed instrument.

Question 8(i)

Shrey's room have two electric bulbs, these bulbs are connected in series and power supplied by 220 V.

(a) Calculate the total resistance of the circuit.

(b) Calculate the current in the circuit.

(c) Calculate the voltage drop across each bulb.

Answer

Given, bulbs of 40 Ω and 70 Ω are connected in series.

Let their equivalent resistance be R_s then

R_s = 40 + 70 = 110 Ω

Hence, resistance in the circuit = 110 Ω

(b) According to Ohm's law: V = IR

Substituting the values in the formula above, we get

220 = I x 110

I = $\dfrac{220}{110}$ = 2 A

Hence, current in the circuit = 2 A

(c) Voltage drop across the 40 Ω bulb (B₁) = ?

I = 2A

Applying Ohm's Law

V = 2 x 40 = 80 V

Voltage drop across the 70 Ω bulb (B₂) = ?

I = 2A

Applying Ohm's Law

V = 2 x 70 = 140 V

Hence, voltage drop across B₁ is 80 V and across B₂ is 140 V.

Question 8(ii)

Write reactions of a nucleus ₁₁X²³ when it loses

(a) one proton

(b) one β particle

(c) one α particle

Answer

(a) One proton is emitted:

As the atomic number is given as 11, the number of protons will be also 11,

$_{11}^{23}\text{X} \longrightarrow \space _{10}^{22}\text{A} + _{+1}^{\enspace 1}\text{p}$

(b) one β particle is emitted

$_{11}^{23}\text{X} \longrightarrow \space _{12}^{23}\text{B} + _{-1}^{\enspace 0}\text{e}$

(c) one α particle is emitted
$_{11}^{23}\text{X} \longrightarrow \space _{9}^{19}\text{C} + \space _{2}^{4}\text{He}$

Question 8(iii)

What is meant by electrical resistance of a conductor? State how resistance of a conductor is affected when

(a) a low current passes through it for a short duration and

(b) a heavy current passes through it for about 30 s.

Answer

The obstruction offered to the flow of current by a conductor is called it's electrical resistance.

(a) When a low current is passed through the conductor for a short duration there will be no change in the temperature of the conductor and so the resistance of the conductor will remain same.

(b) When a high amount of current is passed through a conductor for 30 s then it will raise the temperature of the conductor and due to an increase in temperature resistance of the conductor will also increase.

Question 9(i)

The figure below shows a magnetic needle kept just below the conductor AB which is kept in North-South direction.

(a) In which direction will the needle deflect when the key is closed?

(b) Why is the deflection produced?

(c) What will be the change in the deflection, if the magnetic needle is taken just above the conductor AB?

(d) Name one device which works on this principle.

Answer

(a) North pole of needle deflects towards east.

(b) On passing current through the wire AB, a magnetic field is produced around the wire which aligns the magnetic needle in it's direction.

(c) The direction of deflection gets reversed i.e., now it deflects towards west.

(d) An electric bell works on this principle.

Question 9(ii)

(a) Write an expression for the heat energy liberated by a hot body.

(b) Some heat is provided to a body to raise its temperature by 25°C. What will be the corresponding rise in temperature of the body as shown on the Kelvin scale?

(c) What happens to the average kinetic energy of the molecules as ice melts at 0°C?

Answer

(a) The expression for the heat energy liberated by a hot body is:

ΔQ = mcΔT

where

m = mass of substance,

c = specific heat capacity of substance,

ΔT = change in temperature

(b) Rise in temperature will be the same on both Celsius and Kelvin scale as a degree (or temperature difference) is same on both the Celsius and Kelvin scales i.e.,
Δt °C = ΔT K

(c) When ice melts at 0°C, average kinetic energy of the molecules will remain the same. If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance.

Question 9(iii)

(a) Water falls from a height of 150 m. If the specific heat capacity of water is 4.2 kJ kg^-1 °C^-1 and g = 10 ms^-2, then find the rise in temperature of water.

(b) What is determined by the temperature of a body?

Answer

(a) Given,

Height = 150 m

Specific heat capacity of water = 4.2 kJ kg^-1 °C^-1 = 4200 J kg^-1 °C^-1

g = 10 ms^-2

Let, whole of the energy possessed by water at the height is changed to heat energy on striking the bottom then initial energy possessed by water = heat energy gained by water.

mgh = mcΔt

Δt = $\dfrac{\text{gh}}{\text{c}}$ = $\dfrac{10 \times 150}{4200}$ = 0.357 °C ≈ 0.36°C

Hence, rise in temperature = 0.36°C

(b) Temperature is a parameter which tells the thermal state of a body (i.e., the degree of hotness or coolness of the body). It determines the direction of flow of heat when two bodies at different temperatures are placed in contact.