NSEP PHYSICS

DAILY PHYSICS TEST QUESTIONS

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1 Magnetic Effects of Current and Magnetism

At the mid-point along the length of a long solenoid, the magnetic field is equal to x, If the length of the solenoid is doubled and the current is reduced to half, the field at the new mid-point will be nearest to:

(1) x/2
(2) x/4
(3) x
(4) 2x

Option 1

Option 2 Solution:

Solution

x/4

Option 3

Option 4

2 Gravitation

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R ?

$$\begin{array}{1 1} (1)\;\large\frac{5GmM}{6R}\\ (2)\;\large\frac{2GmM}{3R} \\ (3)\;\large\frac{GmM}{2R}\\ (4)\;\large\frac{GmM}{3R}\end {array}$$

Option 1 Solution:

Solution

The kinetic energy at altitude $$2R= \large\frac{GMm}{6R}$$

The gravitational potential energy at altitude $$2R= \large\frac{-GMm}{3R}$$

Total energy =$$=k \in +PE=-\large\frac{GMm}{6R}$$

Potential energy at the surface is $$\large\frac{-GMm}{R}$$

∴Req. kinetic energy = $$\large\frac{GMm}{R}-\frac{GMm}{6R};=$$

$$\qquad=\large\frac{5GMm}{6R}$$

Option 2

Option 3

Option 4

3 Wave Optics

In a double slit experiment, the two slits are 1 mm apart and the screen is place 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

1) 0.5mm
2) 0.02mm
3) 0.2mm
4) 0.1mm

Option 1

Option 2

Option 3 Solution:

Solution

$$d = 10^{-3}\:m$$
$$D=1\:m$$
$$ \lambda = 500 \times 10^{-9}\:m$$
The width of central maxima in a single slit diffraction: $$\dfrac{2 \lambda D}{a}$$
Fringe width in double slit pattern: $$\beta = \dfrac{\lambda D}{d}$$
Given,
$$10 \beta = \dfrac{2 \lambda D}{a}$$
$$ \Rightarrow 10 \dfrac{\lambda D}{d} = \dfrac{2 \lambda D}{a}$$
$$ \Rightarrow a =\dfrac{d}{5}\:mm= 0.2\:mm$$

Option 4

4 Electromagnetic Waves

The electric field for an electromagnetic wave in free space is $$\vec{E}=\vec{i}30\cos(kz-5\times 10^8t)$$, where magnitude of $$\vec E$$ is in $$V/m$$. The magnitude of wave vector, $$k$$ is:

(velocity of em wave in free space $$=3\times 10^8m/s$$)

1) $$0.46$$ $$rad\ m^{-1}$$
2) $$3$$ $$rad\ m^{-1}$$
3) $$1.66$$ $$rad\ m^{-1}$$
4) $$0.83$$ $$rad\ m^{-1}$$

Option 1

Option 2

Option 3 Solution:

Solution

Electric field of an EM wave $$\vec{E} =30 (kz - 5\times 10^8 t) \hat{i}$$

$$\implies$$ $$\omega = 5\times 10^8$$ rad/s

Given : Velocity of EM wave $$v = 3\times 10^8$$ $$m/s$$

Using $$v = \dfrac{w}{k}$$ $$\implies k = \dfrac{w}{v}$$

$$\therefore$$ $$k = \dfrac{5\times 10^8}{3\times 10^8} = 1.66$$ rad $$m^{-1}$$

Option 4

5 Properties of Bulk Matter

A small ball of density ρ is immersed in a liquid of density (σ>ρ)(σ>ρ) to a depth hh and released. The height above the surface of water up to which the ball will jump, is:

1) $$\left ( \dfrac{\rho }{\sigma }-1 \right )h$$
2) $$\left ( \dfrac{\rho }{\sigma }+1 \right )h$$
3) $$\left ( \dfrac{\sigma }{\rho }-1 \right )h$$
4) $$\left ( \dfrac{\sigma }{\rho }+1 \right )h$$

Option 1

Option 2

Option 3 Solution:

Solution

Neglecting the viscous effect of the fluid,
upward force = ma
$$\Rightarrow V\sigma g - \rho Vg=\rho V a$$
$$\Rightarrow \left ( \dfrac{\sigma-\rho }{\rho} \right )g=a$$
So, Let velocity of ball after moving $$h$$ is $$V$$,
$$V^{2}= U^{2}+2ah$$ as $$U=0$$
$$\Rightarrow V = \sqrt{2\left ( \dfrac{\sigma}{\rho}-1 \right )gh}$$
Now,
After emerging from liquid surface, ball will travel a distance $$s$$,
$$\Rightarrow V_1^{2}=V^{2}-2as$$, as $$V_1=0$$
$$\Rightarrow 2\left ( \dfrac{\sigma }{\rho}-1 \right )gh = 2 \times g \times s$$
$$\Rightarrow \left ( \dfrac{\sigma }{\rho}-1 \right )h=s$$

Option 4

6 Properties of Bulk Matter

If the pressure of gas is increased from 1.01×10 ⁵ pa to 1.165×10⁵ pa and volume is decreases by 10% at constant temperature then the Bulk modulus of the gas is

(1) 15.5 × 10⁵ pa
(2) 1.4 × 10⁵ pa
(3) 1.55 × 10⁵ pa
(4) 0.0155 × 10⁵ pa

Option 1

Option 2

Option 3 Solution:

Solution

Bulk modulus is defined as:
ΔP=(1.165−1.010)∗105=0.155Pa ΔV /v ?=−0.1

Option 4

7 Thermodynamics

Absorption coefficient of an ideal blackbody is :

(1) less then 1
(2) 1
(3) zero
(4) infinity

Option 1

Option 2 Solution:

Option 3

Option 4

8 Physical world and measurement

If the frequency (F), velocity (V) and density (D) are taken as fundamental quantities then the dimensional formula for momentum in this new system will be

1) D^-1V^-4F3
2) D¹V⁴F^-3
3) D¹V¹F^-2
4) D²V²F²

Option 1

Option 2 Solution:

Solution

F = T^-1
V = L¹T^-1
L = VF^-1
D = M¹L^-3
M = D¹ L³
Momentum = M¹L¹T^-1
Momentum = [D¹ (VF^-1)³][ VF^-1] [F]
Momentum = D¹V⁴F^-3

Option 3

Option 4

9 Alternating Currents

In the circuit shown in the figure, E=50.0V,R=250Ω and C=0.500μF. The switch S is closed for a long time, and no voltage is measured across the capacitor. After the switch is opened, the voltage across the capacitor reaches a maximum value of 150V. What is the inductance L?

1) 0.98H
2) 0.78H
3) 0.38H
4) 0.28H

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

In steady state when switch was closed,
$${i}_{0}=E/R=(1/5)A=0.2A$$
After switch is opened, it becomes $$L-C$$ circuit in which peak value current is $$0.2A$$
$$\therefore$$ $$\cfrac { 1 }{ 2 } L{ i }_{ 0 }^{ 2 }\quad =\cfrac { 1 }{ 2 } L{ V }_{ 0 }^{ 2 }\quad $$
or $$L=\cfrac{{ V }_{ 0 }^{ 2 }}{{ i }_{ 0 }^{ 2 }}\cdot C$$
$$=\cfrac { { \left( 150 \right) }^{ 2 } }{ { \left( 0.2 \right) }^{ 2 } } \times 0.5\times { 10 }^{ -6 }= 0.28 H$$

10 Gravitation

Radius of the Earth were increased by a factor of 3 and its mass remained the same, then the acceleration due to gravity on the Earth would

1) reduce by a factor of 3
2) reduce by a factor of 9
3) increase by a factor of 3
4) increase by a factor of 9

Option 1

Option 2 Solution:

The acceleration due to gravity can be expressed as

g = GM/R²
If the mass Mof the Earth remains constant and the radius R were increased by a factor of 3, then the acceleration g would reduce by a factor of 9 .

Option 3

Option 4

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