Solved Sample Paper 3

refraction of light

Reason — As light passes from water to air, it bends away from the normal. The refracted rays appear to come from a point above the actual position of the coin, hence the coin appears raised.

tripled

Reason — Let velocity be v then

Kinetic energy (KE) = $\dfrac{1}{2}mv^2$ and

momentum (ρ) = mv

$\dfrac{KE}{ρ} = \dfrac{\dfrac{1}{2}mv^2}{mv} = \dfrac{v}{2}$    [Equation 1]

Now, when velocity is tripled then

KE' = $\dfrac{1}{2}m(3v)^2$ = $\dfrac{9}{2}m(v)^2$

and ρ' = 3mv

$\dfrac{KE'}{ρ'} = \dfrac{\dfrac{9}{2}mv^2}{3mv} = \dfrac{3v}{2}$    [Equation 2]

On comparing equation 1 and 2, we can see that the ratio gets tripled.

solar energy to electrical energy

Reason — Solar cells convert solar energy to electrical energy.

centre of gravity

Reason — The point at which entire weight of the body can be supposed to be concentrated is called centre of gravity.

3.36 x 10⁵ Jkg^-1

Reason — Latent heat of fusion of ice is 3.36 x 10⁵ Jkg^-1

get reversed

Reason — On reversing the direction of current in a wire, the magnetic field produced by it get reversed according to right hand thumb rule.

0 K

Reason — The zero on the kelvin scale (called absolute zero or 0 K) is the temperature at which molecular motion completely ceases.

8.4 V

Reason — Given,

Emf of battery (V) = 9 V

Internal resistance = 12 Ω

Current (I) = 50 mA = 50 x 10^-3 A

From formula

V = E - Ir

V = 9 - (50 x 10^-3 x 12) = 9 - 0.6 = 8.4 V

Both Assertion and Reason are true

Reason — The current is the rate of flow of charges and it flows from a body at higher potential to a body at lower potential, in a direction opposite to the direction of flow of electrons. When the potential is same at both the ends there will not be any flow of charges due to absence of driving force, hence no current will flow.

0.1 s

Reason — The sensation of sound persists in our brain for about 0.1 s.

intensity

Reason — The intensity of the sound wave at a point of the medium is the amount of sound energy passing per second normally through a unit area at that point.

2

Reason — As we know, the lens formula is —

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

Given,

f = 10 cm

u = -15 cm

v = ?

Substituting the values in the formula, we get,

$\dfrac{1}{10} = \dfrac{1}{v} – \dfrac{1}{-15} \\[1em] \dfrac{1}{10} = \dfrac{1}{v} + \dfrac{1}{15} \\[1em] \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15} \\[1em] \dfrac{1}{v} = \dfrac{3-2}{30} \\[1em] \dfrac{1}{v} = \dfrac{1}{30} \\[1em] \Rightarrow v = 30 \text{ cm} \\[1em]$

∴ Image distance is 30 cm.

As we know, the formula for magnification of a lens is —

$m = \dfrac{v}{u}$

Given,

v = 30 cm

u = -15 cm

Substituting the values in the formula, we get,

$m = \dfrac{30}{- 15} \\[1em] m = -2$

∴ The magnification is 2, negative sign implies image is inverted.

lateral displacement is more

Reason — The perpendicular distance between the incident ray and the emergent ray is defined as lateral displacement. This shift depends upon the angle of incidence, the angle of refraction and the thickness of the medium. When angle of incidence is more, lateral displacement is also more.

The sum of its kinetic and potential energy remains constant throughout the motion.

Reason — The kinetic energy decreases and the potential energy becomes maximum at B.

After a moment the the to and fro movement starts again.

So, from B to A, again the potential energy changes into kinetic energy and this process repeat again and again.

So, when the bob in its state of to and from movement it has potential energy at the extreme position B or C and kinetic energy at resting position A.

It has both the kinetic energy and potential energy at an intermediate position.

Hence, the sum of kinetic and potential energy remain same at every point of movement.

Co-60

Reason — Many diseases such as leukemia, cancer, etc., are cured by radiation therapy. γ radiations from cobalt-60 are used to treat cancer by killing the cells in the malignant tumor of the patient.

(a) As some effort is wasted in overcoming the friction between the strings and the grooves of the pulley as a result the efficiency of a movable pulley is always less than 100%.

(b) 1 horse power = 746 watt.

(c) 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.

Let W gf be the weight of metre rule which acts downwards from 50 cm mark.

In equilibrium position, taking moments about fulcrum.

Anticlockwise moment = 40 x (30 - 5) = 40 x 25 = 1000 gf cm

Clockwise moment = W (50 - 30) = 20 W gf cm

∴ Using principle of moments:

Clockwise moment = anticlockwise moment

So, 20 W = 1000

W = $\dfrac{1000}{20}$ = 50 gf

∴ Weight of the metre scale = 50 gf

As we know,

F = G $\dfrac{\text{m}_1\text{m}_2}{\text{r}^2}$

If separation between two bodies is decreased by a factor of 3 then,

F = G $\dfrac{\text{m}_1\text{m}_2}{{(\dfrac{\text{r}}{3}})^2}$ = 9G $\dfrac{\text{m}_1\text{m}_2}{\text{r}^2}$

Therefore, the gravitational force will be increased by a factor of 9.

Given,

Force (F) = 20 N

and

Displacement (s) = 4 m

(a) When displacement is in the direction of force, θ = 0°

W = Fs cos θ°

= Fs cos 0°

= 20 x 4 x 1 [∵ cos 0° = 1]

= 80 J

(b) When displacement is at an angle 60° with the direction of force,

W = Fs cos θ°

= Fs cos 60°

= 20 x 4 x $\dfrac{1}{2}$ [∵ cos 60° = $\dfrac{1}{2}$]

= 40 J

(a) The two parts of SONAR are

1. Transmitter (for emitting ultrasonic waves)
2. Receiver (for detecting ultrasonic waves)

(b) Free vibrations

(a) Parallelogram — point of intersection of diagonals

(b) Hollow cone — at a height $\dfrac{h}{3}$ from the base, on its axis (if h = height of cone).

(a) Green

(b) Metallic body of an electrical appliance is earthed.

Given,

Mass of water (m) = 600 g

Initial temperature = 50°C

Final temperature = 12°C

∴ fall in temperature = 50 - 12 = 38 °C

Heat lost by water = m x c x Δt

= 600 x 4.2 x 38 = 95,760 J

If m' g ice is added, heat gained by it to melt to 0 °C = m'L = m' x 336 J

By the principle of method of mixture,

heat lost by water = heat gained by ice

∴ 95,760 = m' x 366

m' = $\dfrac{95,760}{366}$ = 261.639 g

Hence, mass of ice required = 261.64 g

(a) Ray diagram showing the image formation by a convex lens when an object is placed between optical centre and focus of the lens is given below:

(b) Ray diagram showing the image formation by a convex lens when an object is placed between focus and twice the focal length of the lens is given below:

(a) ₁₁Na²⁴ ⟶ ₁₂Mg²⁴ + _-1β⁰

(b) ₉₂U²³⁸ ⟶ ₉₀Th²³⁴ + ₂He⁴ + energy

In the circuit, there are three parts. In the first part, resistors of R₁ and R₂ Ω are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = R₁ + R₂

In the second part, resistors of R₃ and R₄ are connected in series. If the equivalent resistance of this part is R''_s then

R''_s = R₃ + R₄

In the third part, the two parts of resistance R'_s and R''_s are connected in parallel. If the equivalent resistance between points A and B is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{R'_s} + \dfrac{1}{R''_s} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1}{R_1 + R_2} + \dfrac{1}{R_3 + R_4} \\[0.5em] \dfrac{1}{R_p} = \dfrac{R_3 + R_4 + R_1 + R_2}{(R_1 + R_2)(R_3 + R_4)} \\[0.5em] R_p = \dfrac{(R_1 + R_2)(R_3 + R_4)}{R_1 + R_2 + R_3 + R_4} \\[0.5em]$

Hence, equivalent resistance = $\dfrac{(R_1 + R_2)(R_3 + R_4)}{R_1 + R_2 + R_3 + R_4}$

(a) In the North-South direction, the wire is moving parallel to the Earth's magnetic field, hence, there would be minimal or no induced emf.

(b) In the East-West direction, the wire is moving perpendicular to the Earth's magnetic field, hence, there would be an induced emf due to the motion of the wire across the magnetic field lines.

(a) Angle of deviation is the angle between the direction of incident ray and the emergent ray. It is represented by the greek alphabet δ (delta).

(b) The completed ray diagram showing the complete path of the light ray till it emerges out of the prism is given below:

(a) A prism having an angle of 90° between its two refracting surfaces and the other two angles each equal to 45° is called a total reflecting prism. In such a prism, the light incident normally on any of its faces suffers total internal reflection inside the prism.

(b) Three actions that it can produce —

1. To deviate a ray of light through 90°.
2. To deviate a ray of light through 180°.
3. To erect the inverted image without producing deviation in its path.

(c) Below diagram shows the deviation of a ray of light through 90° by a total reflecting prism:

(a) When a girl puts her pencil into an empty trough, and later pours water in the same trough, the part of the pencil which is immersed in water will look bent or broken.

Below is the diagram showing how the girl's eye sees the pencil through water:

(b) The above observation is based on the phenomenon of refraction of light.

(c) The cause of above phenomenon is refraction of light. Rays starting from the dipped portion of the pencil suffer refraction at the water-air surface. Air is a rarer medium than water. So, the refracted rays bend away from the normal and appear to come from a point slightly above the actual position of the pencil. Hence, the pencil appears bent or broken.

(a) The orderly arrangement of electromagnetic waves in increasing or decreasing order of wavelength or frequency is called electromagnetic spectrum.

(b) Two uses of ultraviolet rays are:

1. For sterilising air, surgical equipments, etc.
2. For detecting the purity of gems, eggs, ghee etc.

(c) Given,

f = 4 x 10¹⁴ Hz

λ = 500 nm = 500 x 10^-9 m = 5 x 10^-7 m

As we know,

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

Substituting the values in the formula above we get,

c = (4 x 10¹⁴) x (5 x 10^-7) = 2 x 10⁸

Hence,

Speed of the electromagnetic wave = 2 x 10⁸ ms^-1

(a) Concave lens

(b) Ray diagram showing the formation of the image is given below:

(a) Convex lens

(b) Completed ray diagram showing the formation of the image is shown below:

(c) Magnifying glass or a simple microscope

Let, D be the point from where the scale is hanged so that the meter scale and the various masses are balanced.

Clockwise moments = 80 x GD

Anticlockwise moments = (20 x FD) + (30 x ED) + (60 x CD)

As the meter scale is balanced.

∴ Sum of clockwise moments = sum of anticlockwise moments

80 x GD = (20 x FD) + (30 x ED) + (60 x CD)

[80 x {90 - (50 + x)}] = [20 x (50 + x - 10)] + [30 x (50 + x - 20)] + 60x

[80 x (90 - 50 - x)] = [20 x (50 + x - 10)] + [30 x (50 + x - 20)] + 60x

[80(40 - x)] = [20(40 + x)] + [30(30 + x)] + 60x

3200 - 80x = 800 + 20x + 900 + 30x + 60x

3200 - 80x = 800 + 20x + 900 + 30x + 60x

3200 - 80x = 1700 + 110x

3200 - 1700 = 110x + 80x

1500 = 190x

x = $\dfrac{1500}{190}$ = 7.89 ≈ 7.9 cm

Hence, the string should be hanged at 50 + 7.9 = 57.9 cm mark.

(a) When a body remains in a state of rest under the influence of several forces, the body is said to be in static equilibrium.
For example, if a book is lying on a table, the weight of the book exerted on the table vertically downwards is balanced by an equal and opposite force of reaction exerted by the table on the book vertically upwards. Thus, the book is in static equilibrium.

(b) Given,

Efficiency (η) = 95%

Number of pulleys = 4

Velocity ratio = 4 (as the system has 4 pulleys)

MA = ?

We know, $\eta = \dfrac{\text {MA}}{\text {VR}}$

Substituting in the formula above we get,

$\dfrac{95}{100} = \dfrac{\text {MA}}{\text {4}}$

MA = 0.95 x 4 = 3.8

Hence, MA = 3.8

II. From formula : MA = $\dfrac{\text{Load}}{\text{Effort}}$

Substituting in the formula above we get,

3.8 = $\dfrac{1000}{\text{effort}}$

effort = $\dfrac{1000}{3.8}$ = 263.15 N

Hence, effort required = 263.15 N

(a) Work done = force x displacement

Hence, work done is same for both men as both carry 50 kg weight to a height of 2m.

Let the work done by both men be W.

As W = mgh = 50 x 10 x 2 = 1000 J

Hence,

Work done by first man : Work done by second man = 1 : 1

(b) Time taken by first man = 2 min = 2 x 60 secs = 120 secs

Time taken by second man = 5 min = 5 x 60 secs = 300 secs

Power = $\dfrac{\text{Work done}}{\text{time}}$

Power developed by first man (P₁) = $\dfrac{1000}{120}$ = $\dfrac{25}{3}$

Power developed by second man (P₂) = $\dfrac{1000}{300}$ = $\dfrac{10}{3}$

Ratio of power developed by the two men :

$\dfrac{P_1}{P_2} = \dfrac{\dfrac{25}{3}}{\dfrac{10}{3}} = \dfrac{25}{3} \times \dfrac{3}{10} = \dfrac{5}{2}$

Hence, Ratio of power developed by the two men = 5 : 2

(a) ₁₁Na²⁴ ⟶ ₁₂Mg²⁴ + _-1β⁰

(b) 24 is the mass number (i.e., sum of neutrons and protons) and 11 is the atomic number (i.e., number of electrons).

(c) Isobars

(a) Given,

D₁ = 330 m

D₂ = 660 m

V = 330 ms^-1

t₁ = ?

t₂ = ?

As we know,

$\text {Time interval (t)}_1 = \dfrac{\text{Total distance travelled (2d)}}{\text {Speed of sound (V)}} \\[0.5em]$

$\text{t}_1 = \dfrac{2 \times 330 }{330} \\[0.5em] \Rightarrow \text{t}_1 = 2 \text{s}$

Hence, time taken by first echo to reach the student = 2 s

(b) time taken by second echo to reach the student :

$\text{t}_2 = \dfrac{2 \times 660 }{330} \\[0.5em] \Rightarrow \text{t}_2 = 4 \text{s}$

t₂ - t₁ = 4 - 2 = 2 s

Hence, time interval between two echoes is 2 s

(a) Resonance.

(b) When the frequency of the externally applied periodic force on a body is equal to it's natural frequency, the body begins to vibrate with an increased amplitude. This phenomenon is known as resonance. Resonance is a special case of forced vibration.

(c) Loudness is the characteristic by virtue of which a loud sound can be distinguished from a faint one, both having the same pitch and quality.

(d) The unit of loudness is phon. The level of sound is expressed in decibel (dB)

(a) Schematic diagram of main circuit is shown below:

(b) The reciprocal of specific resistance is known as conductivity. It is represented by symbol σ

σ = $\dfrac{1}{ρ}$

Its SI unit is ohm^-1metre^-1 or siemens metre ^-1.

(c) Neutral and earth are at the same potentials.

(a) In the circuit 5 Ω, 10 Ω and 15 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = 5 + 10 + 15 = 30 Ω

According to Ohm's law: V = IR

Substituting the values in the formula:

12 = I x 30

I = $\dfrac{12}{30}$ = 0.4 A

Hence, combined resistance = 30 Ω and current flowing in the circuit = 0.4 A

As in a series circuit current flowing through each resistor is equal to the total current flowing in the circuit, hence, current flowing through each resistor = 0.4 A

(b) Potential difference across 5 Ω resistor: ?

Applying Ohm's law: V = IR

Substituting the values in the formula:

V = 0.4 x 5 = 2 V

Potential difference across 10 Ω resistor: ?

V = 0.4 x 10 = 4 V

Potential difference across 15 Ω : ?

V = 0.4 x 15 = 6 V

Hence, potential difference across 5 Ω resistor = 2, 10 Ω resistor = 4V, 15 Ω resistor = 6V

(a) Given,

Atomic number = 8, hence, number of protons = number of electrons = 8

Mass number = 16, hence, number of protons + neutrons = 16

So, Number of neutrons = mass number - atomic number = 16 - 8 = 8

Hence, number of neutrons = 8

(b)

1. One proton is emitted:
   As the atomic number is given as 8, the number of protons will be also 8,

$_{8}^{16}\text{P} \longrightarrow \space _{7}^{15}\text{A} + _{+1}^{\enspace 1}\text{p}$

2. One β particle is emitted

$_{8}^{16}\text{P} \longrightarrow \space _{9}^{16}\text{B} + _{-1}^{\enspace 0}\text{e}$

(a) Water has a high specific heat capacity of 4.2 Jg^-1°C^-1, hence, it absorbs more heat from the roads without much rise in its own temperature and the road gets cooled, so it is used for spraying on the roads in the evening of hot summer.

(b) As water has a high specific heat capacity of 4.2 Jg^-1°C^-1, hence, it absorbs more heat from the surrounding without much rise in its own temperature hence, some of its uses are:

1. It is used as an effective coolant,
2. Hot water bottles are used for fomentation
3. It is used as a heat reservoir. In cold countries, wine and juice bottles are kept in water to avoid their freezing.
4. Farmers fill their fields with water to protect the crops from frost.

(a) Temperature-time graph representing the cooling of naphthalene from 90°C to room temperature of 20°C is shown below:

(b) The time taken to boil water involves two main steps: raising the temperature to the boiling point and then converting it to steam. Initially, energy is used to increase the temperature of water (say 25°C to 100°C), and this process takes time. Once the water reaches its boiling point, additional energy is required to undergo the phase change from liquid to vapour (steam), known as the latent heat of vaporization. This extra energy input during the phase transition contributes to the overall time it takes to boil water compared to just reaching the boiling point.

1. When a bar magnet is pushed in to the coil, the magnetic field around the coil will change. Hence, some current will be induced in the coil and the galvanometer will show deflection.
2. When the bar magnet is withdrawn from the coil, the current is induced in opposite direction and so the deflection in the galvanometer will be in opposite direction.
3. When a bar magnet is held stationary inside the coil, no current is induced and there will be no deflection in the galvanometer.

The phenomenon involved is Electromagnetic induction.