Solved Sample Paper 2

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.

QS

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

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

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.

Action-reaction forces

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

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°

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.

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.

the sum of potential energy and kinetic energy

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

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.

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.

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

Reason — According to Snell's law,

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

lead and tin

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

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.

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

(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.

(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

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

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.

(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.

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

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

(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.

(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:

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

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.

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

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.

(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.

(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:

(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.

(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.

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.

(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

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

(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:

(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.

(a) Penetrating power — α < β < γ.

(b) Ionising power — γ < β < α.

(c) Biological effect — α < β < γ.

(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.

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.

(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
$_{11}^{23}\text{X} \longrightarrow \space _{9}^{19}\text{C} + \space _{2}^{4}\text{He}$

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.

(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.

(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.

(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.