Solved Sample Paper 4

W_A = W_B

Reason — Gravitational potential energy (PE) does not depend on the path but on initial and final position of the body. Hence, W_A = W_B

The below wave represents damped oscillation:

Reason — A damped oscillation is one which fades away with time, hence, its amplitude decreases with time.

The direction of induced current changes alternatively.

Reason — Due to rotation the magnetic polarity of coil keeps on changing. Hence, the direction of induced current changes alternatively. Strength of induced emf decreases as the rotation speed is decreased and the resistance of coil does not affect the strength of induced emf.

I²Rt

Reason — According to Joule's Law of heating, the heat produced in a conductor is directly proportional to the square of current, the resistance offered and the duration of time for which the current flows. Hence, H = I²Rt

Both Assertion and Reason are true.

Reason — Current is the rate of flow of charge and it flows form a higher potential to lower potential, hence, electric current will not flow between two charged bodies when connected, if they are at the same potential.

$^{23}_{11}\text{Na}$ and $^{23}_{12}\text{Mg}$

Reason — Isobars are atoms having the same mass number but different atomic number. Hence, $^{23}_{11}\text{Na}$ and $^{23}_{12}\text{Mg}$ are isobars.

Rolling

Reason — Rolling is a type of motion and not an equilibrium state.

microwaves > infrared rays > UV rays > γ-rays

Reason — The order of electro magnetic waves in terms of decreasing wavelengths is : microwaves > infrared rays > UV rays > γ-rays

Both (a) and (b)

Reason — Both joule and erg are absolute units of work in SI and CGS respectively.

aperture of lens

Reason — The effective diameter of the circular outline of a spherical lens is known as aperture of lens.

same as that source

Reason — The frequency does not depend on medium and depends on the frequency of the source. Hence, the frequency of the sound in air and water will be the same.

$π$

Reason — The maximum deviation produced by a prism is $π$ or 180°, when the ray of light enters the prism at an angle of 90° as shown below.

None of the above

Reason — All of the above uses of water are due to its high specific heat.

it is a toxic metal

Reason — Lead is not used in calorimeter because it is a toxic metal and can be harmful.

$\dfrac{2}{3}$

Reason — The refractive index of glass w.r.t. air is _aμ_g = $\dfrac{3}{2}$

We know that:

$_{\text{g}}\text{μ}_{\text{a}} = \dfrac{1}{_a\mu_g} \\[1em]$

So, substituting we get,

$_{\text{g}}\text{μ}_{\text{a}} = \dfrac{1}{\dfrac{3}{2}} = \dfrac{2}{3} \\[1em]$

Hence, _gμ_a = $\dfrac{2}{3}$

(a) Given,

mass (m) = 5.5 kg

height (h) = 15 m

force acting on it during its fall = ?

Force acting on the body during its fall = weight of the body.

And F = mg

Substituting we get,

F = 5.5 x 10 = 55 N

Hence, Force acting on the body during its fall = 55 N

(b) The atoms of different elements which have the same mass number A but different atomic number Z are called isobars.

Example — $^{23}_{11}\text{Na}$ and $^{23}_{12}\text{Mg}$ are isobars.

(c) The number of pulleys required in the movable block of a block and a tackle system with velocity ratio 5 will be 5.

(a) The motion of the blades of the helicopter is a rotatory motion.

(b) The particles on the axis of rotation are stationary in pure rotational motion.

Centre of gravity for cylinder is the mid point on the axis of cylinder and for triangle it is the point of intersection of medians:

(a) In the figure below, the force on the ball is the weight W acting vertically downwards and the force on the thread is the tension T acting upwards.

(b) No, the couple acting on a rigid body cannot produce translatory motion, it can cause only rotatory motion as the resultant force is zero.

Yes, it is possible.

For example, when an object is moving in a circular path due to centripetal force acting on it.

At any instant of time, a constant acceleration due to the centripetal force acts on the object along the radius towards the centre while the direction of displacement is tangential to the circle i.e., force and displacement are perpendicular to each other and hence, no work is being done.

Yes, SONAR is used for determining the depth of the sea.

To find the depth of the sea, the time interval (t) between the instant when waves are sent and the instant when waves are received, after reflection from the depth is measured.

The depth d of the sea is then —

$\text{d} = \dfrac{Vt}{2} \\[0.5em]$

where, V is the speed of ultrasonic wave in water.

This process is called echo depth sounding.

(a) Neutral and earth wires

(b) Live wire

Convex lens

(a) In the circuit, there are two parts. In the first part, resistors of R₁ (6 Ω) and R₂ (2 Ω) are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{{R}_P} = \dfrac{1}{6} + \dfrac{1}{2} \\[0.5em] \dfrac{1}{{R}_P} = \dfrac{1 + 3}{6} \\[0.5em] \dfrac{1}{{R}_P} = \dfrac{4}{6} \\[0.5em] \dfrac{1}{{R}_P} = \dfrac{2}{3} \\[0.5em] \Rightarrow {R}_P = \dfrac{3}{2} = 1.5 Ω$

Hence, R_P = 1.5 Ω

In the second part, R_P (1.5 Ω) and R₃ (6 Ω) are connected in series. If the equivalent resistance of this part is R_s then

R_s = (1.5 + 6) Ω = 7.5 Ω

Hence, combined resistance = 7.5 Ω

(b) The earth wire is connected to the metallic body of the appliance.

(a) South pole

Reason — A moving south pole of a magnet near a coil induces a current. The induced current creates a magnetic field opposing the change, following Lenz's law. This leads to the coil's side facing the south pole developing a south pole as well.

(b) North pole

Reason — Moving a magnet's south pole away from a coil induces a current. The induced current creates a magnetic field opposing the change, following Lenz's law. This leads to the coil's side facing the south pole developing a north pole.

When common salt is added to ice, it takes heat from ice and mixes with the water droplets present on the outer layer of ice. Hence, it absorbs heat from ice and temperature of the mixture falls below 0°C.

The nuclear waste obtained from laboratories, hospitals, scientific establishments or power plants must be first kept in thick casks and then they must be buried in the specially constructed deep underground stores. These underground stores must be far away from the populated areas.

(a) Refractive index of the liquid is given by Snell's law and is shown as below,

_airμ_liquid = $\dfrac{\text{sin i}}{\text{sin r}}$

_airμ_liquid = $\dfrac{\text{sin 45}\degree}{\text{sin 30}\degree}$

$= \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{2}} \\[1em] = \dfrac{2}{\sqrt{2}} \\[1em] = \sqrt{2} \\[1em] = 1.414 \\[1em]$

Hence, the refractive index of the lens is 1.4

(b) Below labelled diagram shows the path of the ray after it strikes the mirror and re-enters in air:

(c) Below labelled diagram shows the path of the ray when it strikes the plane mirror normally inside the liquid:

The principle of reversibility is used.

(a) Difference between reflection of light from a plane mirror and total internal reflection of light from a prism are:

(b) Total internal reflection occurs when light traveling through a medium with a higher refractive index (such as water) encounters a boundary with a medium of lower refractive index (such as air) at an angle greater than the critical angle. In this case, the water acts as a more optically dense medium than air.

As a result, the light reflects completely back into the water instead of passing into the air above the test tube. This reflection creates a mirror-like effect on the surface of the water inside the test tube when viewed from above, giving the appearance of a mirror.

(a) The glass piece is not seen when the refractive index of the liquid becomes equal to the refractive index of glass.

(b) The light of a single colour is used as the refractive index of a medium (glass or liquid) is different for the light of different colours.

(c) The refractive index of a medium is defined as the ratio of the speed of light in vacuum (or air) to the speed of light in that medium.

$μ = \dfrac{\text{Speed of light in vacuum or air(c)}}{\text{Speed of light in that medium (V)}}$

The refractive index of a transparent medium is always greater than 1 (it cannot be less than 1) because the speed of light in any medium is always less than that in vacuum ( i.e. V < c ) and μ = 1 for air or vacuum.

(a) The deviation of the incident light rays produced by a lens on refraction through it is a measure of its power. It can be defined as a reciprocal of focal length:

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

(b) As the given power is negative hence, we can say that the lens used is concave in nature and the child is suffering from myopia or short sightedness (i.e., unable to see the far objects distinctly).

(c) As we know,

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

Given,

Power of the lens = -2.5 D

Substituting the value of power in formula we get,

$-\text{2.5} = \dfrac{1}{\text{focal length}} \\[0.5em] \Rightarrow \text{focal length} = -\dfrac{1}{2.5} \\[0.5em] \Rightarrow \text{focal length} = -\text{0.4 m} \\[0.5em] \Rightarrow \text{focal length} = -\text{40 cm} \\[0.5em]$

∴ Focal length = - 40 cm.

(a) The lens formed by the combination is concave in nature.

(b) The completed ray diagram when AB is between optical centre and 2F is shown below:

(a) The lens formed by the combination is a convex lens.

(b) The line PQ is called the principal axis.

(c) Below diagram shows the path of the incident ray AB after passing through the convex lens:

(d) The point F is called the focal point.

(a) Let W be the weight of the rod.

For uniform meter rod, its weight acts at 50 cm.

Clockwise moments = [200 x (95-70)] = 200 x 25 = 5000 Nm

Anticlockwise moments = [50 x (70 - 40)] + [W x (70 - 50)]

= [50 x 30] + [W x 20]

= 1500 + 20W

(b) According to principle of moment :

Anticlockwise moment = Clockwise moment

1500 + 20W = 5000

W = $\dfrac{5000 - 1500}{20} = \dfrac{3500}{20}$ = 175 N

Hence, weight of meter rod = 175 N

As W = 175 N, hence, anticlockwise moment = 1500 + 20W = 1500 + (20 x 175) = 1500 + 3500 = 5000

Hence, anticlockwise moment = 5000 Nm

(a) Diagram of a pulley system of velocity ratio 4 is shown below:

Given,

Efficiency (η) = 90%

Velocity ratio = 4

MA = ?

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

Substituting in the formula above we get,

$\dfrac{90}{100} = \dfrac{\text {MA}}{\text {4}}$

MA = 0.90 x 4 = 3.6

Hence, MA = 3.6 N

(b) The pole increases the walker's moment of inertia or resistance to rotating and prevents him from falling off the wire, allowing the walker time to correct his position and stay balanced. When he feels he is falling towards right, he shifts the pole towards left and vice versa.

In addition, holding the pole low helps to keep his center of gravity low, and this further increases his ability to balance.

(a) Let kinetic energy of truck be KE_t and that of car be KE_c and v be their common speed.

Given,

$\dfrac{\text{KE}_\text{t}}{\text{KE}_\text{c}} = \dfrac{25}{12}$

By formula,

KE = $\dfrac{1}{2}$ mv²

$\phantom{\Rightarrow} \dfrac{\text{KE}_\text{t}}{\text{KE}_\text{c}} = \dfrac{25}{12} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2}\text{m}_\text{t}\text{v}^2}{\dfrac{1}{2}\text{m}_\text{c}\text{v}^2} = \dfrac{25}{12} \\[1em] \Rightarrow \dfrac{\text{m}_\text{t}}{\text{m}_\text{c}} = \dfrac{25}{12}$

Hence, ratio of masses = 25 : 12

(b) Mechanical energy is conserved in:

1. Simple pendulum and
2. Motion of a freely falling body.

(a) Two characteristics that a medium should have for propagation of sound are:

1. The medium must be elastic so that it's particles may come back to their initial position after displacement on either side, i.e., the particles are capable of vibrating about their mean position.
2. The medium must have inertia so that it's particles may store mechanical energy.

(b) Echo is heard after 8 vibrations of pendulum. Since frequency of pendulum is 4 vibrations per sec, so the time taken for the echo to be heard is t = $\dfrac{8}{4}$ = 2s

If d is the distance between the cliff and the observer, then velocity of sound V = $\dfrac{2\text{d}}{\text{t}}$

Substituting the values in the equation above:

340 = $\dfrac{2\text{d}}{\text{2}}$

d = 340 m

Hence, distance between the cliff and the observer = 340 m

(a) Any physical change (such as change in pressure and temperature) or chemical change (such as excessive heating, freezing, action of strong electric and magnetic fields, chemical treatment, oxidation etc.) does not change the rate of decay and the nature of radiation emitted by the substance.

This clearly shows that the phenomenon of radioactivity cannot be due to the orbital electrons which could easily be affected by such changes. Therefore radioactivity should be the property of the nucleus. Thus, radioactivity is a nuclear phenomenon.

(b) The radioactive material after its use such as nuclear power generation, nuclear weapons production, and nuclear medicine is known as nuclear waste.

(a) Resonance

Definition — When the frequency of the externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude. This phenomenon is known as resonance.

(b) Frequency of the channel is 93.5 MHz.

The SI unit for frequency is the hertz (Hz)

1 MHz = 10⁶ Hz

93.5 MHz = 93.5 x 10⁶ Hz

In the circuit, there are two parts. In the first part, resistors of 2.0 and 4.0 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = 2 + 4 = 6 Ω

In the second part, R_s = 6.0 Ω and resistor of 6.0 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1 + 1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{2}{6} \\[0.5em] R_p = \dfrac{6}{2} \\[0.5em] R_p = 3.0 Ω \\[0.5em]$

Hence, the effective resistance of the circuit = 3 Ω

(b) current drawn = ?

R = 3 Ω

V = 9.0 V

Using Ohm's law,

V = IR

Substituting the values in the formula above we get,

9 = I x 3
⇒ I = $\dfrac{9}{3}$ = 3 A

Hence, the current drawn = 3 A

(a) Background radiations are the radioactive radiations (such as α, β and γ) to which we all are exposed, even in the absence of an actual visible radioactive source. For example, radioactive substances such as potassium (K-40), carbon (C-14) and radium are present inside our body. Cosmic rays also contribute to background radiations.

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

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

(c) The rate of radioactive decay generally depends on the specific properties of the radioactive substance and is not directly proportional to temperature changes. Hence, increasing the temperature will not affect the rate of radioactivity.

(a) Live wire is connected to the switch of an appliance.

(b) Diagrams of a dual control switch when the appliance is switched ON and OFF are shown below:

(c) Given,

Power (P) = 40 W

time (t) = 12.5 h for 30 days

As energy consumed (E) = P × t

$E = 40 \times 12.5 \\[0.5em] \Rightarrow E = 500 \text { Wh}$

Hence, electrical energy consumed in one day = 500 Wh

Electrical energy consumed in 30 days = ?

E = 500 x 30
= 15000 Wh
= 15 kWh

Hence, electrical energy consumed in 30 days = 15 kWh

(a) 1 kg of steam at 100°C contains more heat than 1 kg of water at the same temperature. This is because the heat energy in the steam includes the additional latent heat of vaporization, which is the energy required to change water from liquid to vapor at its boiling point.

(b) Ice contracts on melting and expands on solidification.

(c) The reason for keeping juice bottles underwater is that water due to it's high specific heat capacity can impart a large amount of heat before reaching to it's freezing point. Hence, bottles kept in water remain warm and do not freeze when there is considerable fall in surrounding's temperature.

Given,

m_i = 100 g

Converting g to kg

1000 g = 1 kg

So, 100 g = $\dfrac{100}{1000}$ = 0.1 kg

m_w = 450 g

Converting g to kg

1000 g = 1 kg

So, 450 g = $\dfrac{450}{1000}$ = 0.45 kg

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

specific latent heat of fusion of ice = 336 x 10³ J kg^-1

final temperature of water = ?

Let final temperature = t

Heat energy given out by water when it cools from 45° C to t° C
= m x c x Δt
= 0.45 × 4200 × (45 – t)
= 85050 – 1890t

Heat energy taken by ice when it converts from ice into water at 0° C
= m x L
= 0.1 × 336 x 10³ J
= 33600 J

Heat energy taken by water when it raises it's temperature from 0° to t° C
= m x c x Δt
= 0.1 × 4200 × (t – 0)
= 0.1 × 4200 × t
= 420t

If there is no loss of energy,

Heat energy gained = heat energy lost

Substituting the values we get,

$33600 + 420t = 85050 - 1890t \\[0.5em] 1890t + 420t = 85050 - 33600 \\[0.5em] 2310t = 51450 \\[0.5em] t = \dfrac{51450}{2310} \\[0.5em] t = 22.27° C$

Hence, final temperature = 22.27°C

(a) Below diagram shows an electromagnet made from a soft iron bar with polarity of the ends labelled:

(b) Magnetic field inside a long straight solenoid carrying current is uniform i.e., same at all points.

(c) Direction of magnetic field:
Inside the loop — Going into the table.
Outside the loop — Emerging out of the table.

Explanation:
On applying the right-hand rule to the downward direction of the current flowing in the circular loop, the direction of magnetic field lines will be as if they are emerging out from the table outside the loop and merging into the table inside the loop.
Similarly, for the upward direction of the current flowing in the circular loop, the direction of magnetic field lines will be as if they are emerging out from the table outside the loop and merging into the table inside the loop.