JEE PHYSICS question bank, MCQ , Notes, Lectures

1 Magnetism

The magnitude of torque experienced by a square coil of side $$12 \,cm$$ which consists of $$25$$ turns and carries a current $$10\, A $$ suspended vertically and the normal to the plane of coil

1) $$1.6 \,N\,m$$
2) $$1.2 \,N\,m$$
3) $$1.4 \,N\,m$$
4) $$1.8 \,N\,m$$

Option 1 Solution:

Solution

$$\tau =NIA\, B\, sin \, \theta$$

Here, $$N=25$$,

$$I=10\, A$$,

$$B=0.9 \, T$$,

$$\theta =30^{\circ}$$

$$A=a^{2}=12 \times 10^{-2} \times 12\times 10^{-2}$$

$$144 \times 10^{-4}\, m^{2}$$

$$\therefore \tau=25 \times 10 \times = 144\times 10^{-4}\times 0.9\times sin \, 30^{\circ}$$

$$=16.\, N\,m$$

Option 2

Option 3

Option 4

2 Units and Dimensions

The energy [E ] , angular momentum ( L) and universal gravitational constant (G) are chosen as fundamental quantities. The dimensions of universal gravitational constant in the dimensional formula of planks constant (h) is.
1) 0
2) -1
3) 5 / 3
4) 1

Option 1 Solution:

Sol : From the given problem

h=E^aL^bG^c Where a, b, c are constants

From the principle of homogeneity of dimensions
Dimensions of h [ML²T-1]
Dimensions of E= [ML² T^-2 ]
Dimensions of L= [ML ²T ^-1 ]
Dimensions of G= [M^-1L³ T^-2 ]

∴ [ML²T-1] = [ML² T^-2 ]^a [ML ²T ^-1 ]^b[M^-1L³ T^-2 ] ^c

Solving we get c = 0

Option 2

Option 3

Option 4

3 Gravitation

A particle hanging from a spring stretches it by 1cm at earth's surface. Radius of earth is 6400 km. At a place 800 km above the earths surface, the same particle will stretch the spring by

1) 0.79 cm
2) 1.2 cm
3) 4 cm
4) 17 cm

Option 1 Solution:

Solution

as, $$mg=kx$$

where, $$x=1cm, k=$$spring constant

for second case $$mg'=kx'$$

Hence $$ \dfrac{g}{g'}=\dfrac{x}{x'}$$

$$x'=xg'/g$$

$$x'=\dfrac{xgR^2}{(R+h)^2 g}$$

Earth's radius, $$R=6400$$

At height, $$h=800$$

$$x'=\dfrac{1\times (6400 \times 10^3)^2}{(7200 \times 10^3)^2}$$

$$x'=0.79 cm$$

Option 2

Option 3

Option 4

4 Kinetic Theory of Gases

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as V$$^q$$ , where $$V$$ is the volume of the gas. The value of $$q$$ is $$\bigg(\gamma =\frac{C_p}{C_v}\bigg)$$

1) $$\frac{3\gamma +5 }{6}$$
2) $$\frac{\gamma +1}{2}$$
3) $$\frac{3 \gamma-5}{6}$$
4) $$\frac{\gamma-1}{2}$$

Option 1

Option 2 Solution:

Solution:

Average time between two collisions is given by $$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \tau =\frac{1}{\sqrt{2 \pi n v_{rms}d^2}} \, \, \, \, \, \, \, \, \, \, \, \, $$ .....(i) Here, n = number of molecules per unit volume = $$\frac{N}{V}$$ and $$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, C_{rms} =\sqrt{\frac{3RT}{M}}$$ Substituting these values in Eq.(i) we have, $$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \tau = \frac{V}{\sqrt T} \, \, \, \, \, \, \, \, \, \, \, \, $$ ....(ii) For adiabatic process $$TV^{\gamma-1}=constant$$ substituting in Eq. (ii), we have $$\tau = \frac{V}{\sqrt{\bigg(\frac{1}{V^{\gamma-1}}\bigg)}}$$ or $$\tau =V^{1+\bigg(\frac{\gamma -1}{2}\bigg)}$$ or $$\tau = V^{\bigg(\frac{1+\gamma}{2}\bigg)}$$

Option 3

Option 4

5 Current Electricity

A coil having magnetic moment $$15 \,A\,m^{2}$$ placed in a uniform magnetic field of $$4 \, T$$ in the horizontal direction exists such that initially the axis of coil is in the direction of the field. If the coil is rotated by $$45^{\circ}$$ and the moment of inertia of the coil is $$0.5 \,kg\, m^{2}$$ then the angular speed acquired by the coil is

1) $$20 \, rad \, s^{-1}$$
2) $$10 \, rad \, s^{-1}$$
3) $$8.34\, rad \, s^{-1}$$
4) $$4.5\, rad \, s^{-1}$$

Option 1

Option 2

Option 3 Solution:

Solution:

Since $$\tau =I {\frac {d\omega} {dt}}= mB\, sin\, \theta$$ $$\ldots\left(i\right)$$

NOW, $$\frac{d \omega}{dt}=\frac{d\omega}{d\theta} \frac{d\theta}{dt}$$

$$=\omega \frac{d\omega}{d\theta} \ldots\left(ii\right)$$

then, from $$\left(i\right)$$ and$$ \left(ii\right)$$

$$I\omega \frac{d\omega}{d\theta}=mB\,sin\,\theta ; I\omega d\omega=mBsin\theta d\theta$$

Integrating both sides

$$I \int\limits_{0}^{\omega_{f}} \omega\, d\omega=mB \int\limits_{0}^{45^{\circ}} sin\, \theta\,d\theta$$

$$I \left[ \frac{\omega^{2}}{2}\right]_{0}^{\omega_{f}} =mB\left[-cos \,\theta\right]_{0^{\circ}}^{45^{\circ}}$$

$$\frac{1}{2}I \omega_{f}^{2}=-mB\left[cos\,45^{\circ}-cos\,0^{\circ}\right]$$

$$=-mB\left[\frac{1}{\sqrt{2}}-1\right]=0.29\, mB$$

$$\omega_{f}^{2}=\frac{2\times0.29mB}{f}$$

$$=\frac{2\times0.29\times15\times4}{0.50}$$

$$\omega_{f}^{2}=69.6 ; \omega_{f} =\sqrt{69.6}$$

$$=8.34\, rad \, s^{-1}$$solution

Option 4

6 Nuclei

The ionigation energy of hydrogen atom is 13.6 eV. The ionigation energy of helium atom would be

1)   27.2 eV
2)   13.6 eV
3)   54.4 eV
4)   6.8 eV

Option 1

Option 2

Option 3 Solution:

SOLUTION

Option 4

7 Electrostatics

Two equal point charges are fixed at x = −a and x = +a on the x−axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x−axis, is approximately proportional to:

1) x
2) x²
3) x³
4) 1 / x

Option 1

Option 2 Solution:

Solution

$$\displaystyle{U}_{{{i}}}={\left(\frac{{{2}{Q}{q}}}{{{4}\pi\epsilon_{{{0}}}{\left({a}\right)}}}\right)},{U}_{{{f}}}=\frac{{{Q}{q}}}{{{4}\pi\epsilon_{{{0}}}}}{\left[\frac{{{1}}}{{{a}+{x}}}+\frac{{{1}}}{{{a}-{x}}}\right]}$$ ?=πr²Eo = 3.14 × 0.0004 × 100 = 0.125Nm²/C

$$\displaystyle{U}_{{{i}}}-{U}_{{{f}}}=\frac{{{\left({Q}{q}{x}^{{{2}}}\right)}}}{{{2}\pi{\epsilon_{{{0}}}^{{{3}}}}}}$$

[for x<<a,x² can be neglected in comparison to a² in denominator]

Option 3

Option 4

8 Mechanical Properties Of Fluids

On heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r < < R, and the surface tension of water is T value of r just before bubbles detach is (density of water is p_w)

Option 1

Option 2

Option 3

Option 4 Solution:

SOLUTION

When the bubble gets detached

Buyount force = force due to surface tension

9 Current Electricity

A 100-W bulb and 25-W bulbs are designed for the same voltage . They have filaments of the same length and material. The ratio of the diameter of the 100-W bulb to the that of the 25- W bulbs is:

1)�4:1
2) 2:1
3) √2:1
4)�1:2

�

Option 1

Option 2 Solution:

Solution

Given that,

Power of first bulb,$${{P}_{1}}=100\,W$$

Power of second bulb, $${{P}_{2}}=25\,W$$

Power is given by,

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

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

$$\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{\left( \dfrac{{{V}^{2}}}{{{P}_{1}}} \right)}{\left( \dfrac{{{V}^{2}}}{{{P}_{2}}} \right)}=\dfrac{{{P}_{2}}}{{{P}_{1}}}........(1)$$

And the resistance in terms of resistivity is given by:

$$ \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{\dfrac{\rho l}{\pi {{({{d}_{1}}/2)}^{2}}}}{\dfrac{\rho l}{\pi {{({{d}_{2}}/2)}^{2}}}}=\dfrac{d_{2}^{2}}{d_{1}^{2}} $$

$$ so, $$

$$ \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{d_{2}^{2}}{d_{1}^{2}} $$

$$ \dfrac{{{d}_{2}}}{{{d}_{1}}}=\sqrt{\dfrac{{{P}_{2}}}{{{P}_{1}}}} $$

$$ \dfrac{{{d}_{2}}}{{{d}_{1}}}=\sqrt{\dfrac{25}{100}} $$

$$ \dfrac{{{d}_{2}}}{{{d}_{1}}}=\dfrac{1}{2} $$

The ratio of the diameter of the 100 W bulb to the that of the 25 W bulbs is�2:1

Option 3

Option 4

10 Current Electricity

When $$5\,V$$ potential difference is applied across a wire of length $$0.1\,m$$, the drift speed of electrons is $$25\times10^{-4}$$ $$ms^{-1}$$. If the electron density in the wire is $$8\times10^{28} m^{-3}$$ the resistivity of the material is close to

1) $$1.6\times10^{-8}\Omega\,m$$
2) $$1.6\times10^{-7}\Omega\,m$$
3) $$1.6\times10^{-5}\Omega\,m$$
4) $$1.6\times10^{-6}\Omega\,m$$

Option 1

Option 2

Option 3 Solution:

Solution:

$$i=neAv_d$$ or $$\frac{V}{R}=neAv_d$$

or $$\frac{V}{\bigg(\frac{\rho l}{A}\bigg)}=neAv_d$$

$$\therefore$$ $$\rho=\frac{V}{nelv_d}$$=resistivity of wire

Substituting the given values we have

$$\rho=\frac{5}{(8\times10^{28})(16\times10^{-19})(0.1)(2.5\times10^{-4})}$$

$$\approx1.6\times10^{-5}\Omega-m$$

Option 4

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