JEE PHYSICS question bank, MCQ , Notes, Lectures

1 Dual Nature Of Radiation And Matter

A monochromatic light of frequency $$v = \frac{1}{6.63} \times 10^{16}$$ Hz is produced by a laser. The power emitted is $$P = 10^{-2}$$ W. The average number of photos per second emitted by the source is

1) $$\frac{1}{(6.63)^2} \times 10^{16}$$
2) $$(6.63)^2 \times 10^{16}$$
3) $$10^{20}$$
4) $$10^{16}$$

Option 1

Option 2

Option 3

Option 4 Solution:

2 Work Power Energy

Consider an elliptically shaped rail PQ in the vertical plane with OP = 3 m and OQ = 4 m. A block of mass 1 kg is pulled along the rail from P to Q with a force of 18 N, which is always parallel to line PQ (see the figure given). Assuming no frictional losses, the kinetic energy of the block when it reaches Q is (n × 10) Joules. The value of n is (take acceleration due to gravity = 10 ms⁻²)

1) 5
2) 2
3) 3
4) 6

Option 1 Solution:

SOLUTION:−

Work done by external force against gravity
w = 18 × 5 = -90 joules,
Potential energy at height U = −1 × 10 × 4 = −40 joules
W =ΔU+ ΔK
ΔK = W -ΔU
ΔK = 90 − 40 = 50J
∴ n × 10 = 50J
∴ n = 5

Option 2

Option 3

Option 4

3 Kinematics

A lift is moving down with acceleration a . A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively:

1) g, g2) g — a, g — a 3) g — a, g 4) a, g

Option 1

Option 2

Option 3 Solution:

Apparent weight of ball
w' = w - R
R = ma (acts upward)
w' = mg - ma = m(g - a)
Hence, apparent acceleration in the lift is g - a . Now if the man is standing stationary on the ground, then the apparent acceleration of the falling ball is g.

Option 4

4 Magnetism

The horizontal component of earths magnetic field at a certain place is $$3.0 \times 10^{-5}\, T$$ and having a direction from the geographic south to geographic north. The force per unit length on a very long straight conductor carrying a steady current of $$1.2\, A$$ in east to west direction is

1) $$3.0 \times 10^{-5} \, N\, m^{-1}$$
2) $$3.2 \times 10^{-5}\, N\, m^{-1}$$
3) $$3.6 \times 10^{-5}\, N\, m^{-1}$$
4) $$3.8 \times 10^{-5}\, N\, m^{-1}$$

Option 1

Option 2

Option 3 Solution:

Solution:

Force per unit length $$f=\frac{F}{l}=IB\, sin\,\theta $$

When the current is flowing from east to west then $$\theta=90^{\circ}$$, hence

$$f=IBsin90^{\circ}=1.2\times3\times10^{-5}\times1$$

$$=3.6\times10^{-5}\,N \,m^{-1}$$

Option 4

5 Thermal Properties Of Matter

Four rods with different radii r and length / are used to connect two heat reservoirs at different temperature. Which one will conduct most heat ?

1) $$r = 1 \, cm, l = 1 \, m $$
2) $$r = 1 \, cm , l = \frac{1}{2} m$$
3) $$r = 2 \, cm, l = 2 \, m $$
4) $$r = 2 \, cm , l = \frac{1}{2} m$$

Option 1

Option 2

Option 3

Option 4 Solution:

Solution:

$$H = \frac{KAd\theta}{L} \propto \frac{A}{L } = \frac{\pi r^{2}}{L} $$

$$ \left(\frac{A}{L}\right)_{1} = \frac{\pi\left(1\right)^{2}}{1} = 2\pi\,\, units \hspace10mm \left(\frac{A}{L}\right)_{2} = \frac{\pi\left(2\right)^{2}}{\frac{1}{2}} = 8\pi \,\, units.$$

$$ \left(\frac{A}{L}\right)_{3} = \frac{\pi\left(1\right)^{2}}{1} = \pi \,\, units \hspace10mm \left(\frac{A}{L}\right)_{_4 }= \frac{\pi\left(2\right)^{2}}{2} = 2\pi\,\, units$$

So rod (2) will conduct most heat per second.|

6 Kinematics

A small particle of mass m is projected at an angle θ with the X–axis with an initial velocity v_o in the X–Y plane as shown in the figure. At a time

Option 1

Option 2

Option 3 Solution:

SOLUTION

Option 4

7 Magnetic Effects of Current

A solenoid 60 mm long has 50 turns on it and is wound on an iron rod of 7.5 mm radius. Find the flux through the solenoid when the current in it, is 3A. The relative permeability of iron is 600.

1) 1.66 Wb
2) 1.66 nWb
3) 1.66 mWb
4) 1.66 μWb

Option 1

Option 2

Option 3 Solution:

SOLUTION

Option 4

8 Thermodynamics

Consider a spherical shell of radius R at temperature $$T$$. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume $$u=\frac{U}{V} ∝ T^4$$ and pressure $$p=\frac{1}{3}\bigg(\frac{U}{V}\bigg)$$ If the shell now undergoes an adiabatic expansion, the relation between $$T$$ and $$R$$ is

1) $$T ∝ e^{-R}$$
2) $$T ∝ \frac{1}{R}$$
3) $$T ∝ \frac{1}{R}$$
4) $$T ∝ \frac{1}{R^3}$$

Option 1

Option 2 Solution:

Solution:

Given, $$\frac{U}{V} ∝ T^4$$

$$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{U}{V} =\alpha T^4 \, \, \, \, \, \, \, \, \, \, \, \, ...(i)$$

It is also given that, $$p=\frac{1}{3}\bigg(\frac{U}{V}\bigg)$$

$$\Rightarrow \, \, \, \, \frac{nR_0T}{V} =\frac{1}{3}(\alpha T^4) \, \, \, \, \, \, \, \, \, \, \, \, (R_0$$ = Gas constant)

or $$VT^3=\frac{3nR_0}{\alpha}=constant $$

$$\therefore \, \, \, \, \, \, \, \, \, \, \, \bigg(\frac{4}{3}\pi R^3\bigg)T^3 = constant $$

or $$or \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $$ RT = constant

$$\therefore \, \, \, \, \, \, \, \, \, T ∝ \frac{1}{R}$$

Option 3

Option 4

9 Laws of Motion

Two balls of equal masses are thrown upwards along the same vertical direction at an interval of 2 s, with the same initial velocity of 39.2 m/s. The two balls will collide at a height of

1) 39.2 m
2) 73.5 m
3) 73.5 m
4) 117.6 m

Option 1

Option 2 Solution:

Solution:

Let two balls collide at a height s from the ground after $$t$$ second when second ball is thrown upwards.

$$\therefore$$ Time taken by first ball to reach the point of collision = $$\left(t+2\right)$$ s

s = 39.2 $$\left(t+2\right)+\frac{1}{2} \left(-9.8\right)\left(t+2\right)^{2}$$

$$= 39.2\left(t+2\right)-4.9\left(t+2\right)^{2} \ldots\left(i\right)$$

For second ball

s = 39.2t + $$\frac{1}{2}\left(-9.8\right)t^{2}$$

$$= 39.2 t -4.9 t^{2} \ldots\left(ii\right)$$

From eqs. (i) and (ii)

39.2 (t + 2) - 4.9 (t + 2)2=(39.2)t - 4.9t2

On solving we get, t = 3s

From Eq. (ii),

s = 39.2 × 3 - 4.9 × (3)2 = 117.6 - 44.1 = 73.5 m

Option 3

Option 4

10 Magnetic Effects of Current

The electric current in a circular coil of two turns produced a magnetic induction of $$0.2\, T$$ at its centre. The coil is unwound and then rewound into a circular coil of four turns. If same current flows in the coil, the magnetic induction at the centre of the coil now is

1) $$0.2\, T$$
2) $$0.4\, T$$
3) $$0.6\,T$$
4) $$0.8\, T$$

Option 1

Option 2

Option 3

Option 4 Solution:

Solution:

When there are two turns in the coil, then

$$l=2\times2\pi r_{1}$$ or $$r_{1}=\frac{l}{4\pi}$$

then $$B_{1}=\frac{\mu_{0} N_{1}I}{2r_{1}}=\frac{\mu_{0}\times2\times I}{2\times\left(l / 4\pi\right)}=\frac{\mu_{0}\,4\pi I}{l}$$

When there are four turns in the coil, then

$$l=4 \times2\pi r_{2}$$ or $$r_{2}=\frac{l}{8\pi}$$

Then $$B_{2}=\frac{\mu_{0}\, N_{2}I}{2r_{2}} = \frac{\mu_{0}\times4\times I}{2\times\left(l 8\pi\right)}=\frac{\mu_{0}\,16\pi I}{l}$$

$$\frac{B_{1}}{B_{2}}=\frac{4}{16}=\frac{1}{4}$$ or $$B_{2}=4B_{1}=4\times0.2\,T$$

$$=0.8\, T $$

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