NSEP PHYSICS

DAILY PHYSICS TEST QUESTIONS

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1 Kinematics

An aeroplane is travelling horizontally at a height of 2000 m from the ground. The aeroplane, when at a point P, drops a bomb
to hit a stationary target Q on the ground. In order that the bomb hits the target, what angle θ must the line PQ make with the vertical ? [g = 10ms^–2]

1) 15°
2) 30°
3) 90°
4) 45°

Option 1

Option 2

Option 3

Option 4 Solution:

Solution
Let t be the time taken by bomb to hit the target.

h = 2000 = 1/2gt² => t = 20 sec

R = ut = (100) (20) = 2000 m

2 Kinematics

A stone falls under gravity. It covers distances h1, h2, h3 in the first 5 seconds, next 5 seconds and the next 5 seconds respectively. The relation between h₁, h₂ ,h₃ is.
1) h₁ =2h₂ =3h₃
2) h₁ = h₂/3 = h₃/5
3) h₂ =3h₁ and h₃ = 3h₂
4) h₁ =h₂ =h₃

Option 1

Option 2 Solution:

Solution

Option 3

Option 4

3 Electrostatics

Four charges q1=1μC, q2=2μC, q3=3μC and q4=4μC are placed at (0,0,0), (1m,0,0), (0,1m,0), (0,0,1m) respectively. Let $\vec { { F }_{ i } }$ be the net force acting on $i$ th charge of the given charges then $\sum { \vec { { F }_{ i } } } =$

1) 0.018
2) 0.02N
3) 0.036N
4) zero

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Let

$F_{ij}$ be the force on jth charge by ith charge.

we have to find

$F_u=\Sigma_{j=1}^{u}{\Sigma_{i=1}^{u}{F_{ij}}}$

And By

$F_{12}=F_{21}$ or say generally

$F_{ij}=F_{ji}, \; Thus, \; F_{Total}=0$

4 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 $L=\cfrac{{ V }_{ 0 }^{ 2 }}{{ i }_{ 0 }^{ 2 }}\cdot C$
$=\cfrac { { \left( 150 \right) }^{ 2 } }{ { \left( 0.2 \right) }^{ 2 } } \times 0.5\times { 10 }^{ -6 }= 0.28 H$

5 Oscillations

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum 20 m/s² at a distance of 5 m from the moving position. The time period of oscillation is?

1) 2s
2) 2πs
3) 1s
4) πs

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Since in SHM, magnitude of acceleration is given as

a=ω²x

20=ω²×5

ω=2ω=2

Now, Time Period (T),

T=2π / ω

T=π s

6 Electromagnetic Induction

A coil has an area of 0.05m² and it has 800 turns. It is placed perpendicular in a magnetic field of strength 4×10⁻⁵Wb/m², it is rotated through 90^o in 0.1sec. The average emf induced in the coil is

1) 0.056V
2) 0.046V
3) 0.026V
4) 0.016V

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

$\phi_1 = NBA$

$\phi_2 = O$

$E_avg = \dfrac{NBA}{Ot}$

$= \dfrac{800 \times 4 \times 10^{-5} \times 0.05}{0.1}$

= 0.016V

7 Electromagnetic Induction

The magnetic flux through a coil perpendicular to its plane and directed into paper is varying according to the relation ?=(2t²+4t+6) mWb. The emf induced in the loop at t = 4 s is:

1) 0.12 V
2) 2.4 V
3) 0.02 V
4) 1.2 V

Option 1

Option 2

Option 3 Solution:

Solution

Given, $\phi= (2t^2 + 4t + 6) mWb$
As, $\displaystyle \epsilon = \frac{d\phi}{dt} =\frac {d}{dt} \,(2t^2 + 4t + 6) \times 10^{-3} Wb\,s^{-1}$
$= (4t + 4) \times 10^{-3} V$
At $t=4 \,s$
$\epsilon$ =(4 x 4 + 4) x 10 $^{-3}$ V = 20 x 10 $^{-3}$ V = 0.02V

Option 4

8 Gravitation

An object weighs 100. newtons on Earth’s surface. When it is moved to a point one Earth radius above Earth’s surface, it will weigh

1) 25.0 N
2) 50.0 N
3) 100. N
4) 400. N

Option 1 Solution:

Solution

At a point one Earth’s radius above the Earth’s surface, the acceleration due to gravity would be

The weight of an object is the mass times the acceleration due to gravity.

On the surface of the Earth, the weight of the body is 100 N.

⇒mg₀ = 100

⇒

At a point one Earth’s radius above the Earth’s surface, the weight would be 2525 N

Option 2

Option 3

Option 4

9 Waves

A vertical smoked plate falls vertically under gravity and traces found by transverse vibrations of a fork show 10 complete vibrations in each of two consecutive lengths which differ by 1.8cm. The frequency of the fork is

1) 233.3 Hz
2) 333.3 Hz
3) 444.3 Hz
4) 555.3 Hz

Option 1 Solution:

Solution

Let the length of $10$ waves in two sets be $l_1$ and $l_2$ .
Then, $l_1=ut+\dfrac{1}{2}gt^2$ and $l_2=u_1t+\dfrac{1}{2}gt^2$
But $u_1=u+gt$

Hence,

$l_2-l_1=gt^2$ or

$t=\sqrt{\dfrac{l_2-l_1}{g}}$

Now frequency of the fork is given by:

$\nu=N/t=10 \times \sqrt{\dfrac{g}{l_2-l_1}}=10\times\sqrt{\dfrac{9.8}{0.018}}=233.3\ Hz$

Option 2

Option 3

Option 4

10 Properties of Bulk Matter

The hydrostatic pressure on the dam wall at the bottom of a deep reservoir depends upon the

(1) Depth of the water
(2) Surface area of the water
(3) Length of the reservoir
(4) Thickness of the wall

Option 1

Option 2

Option 3

Option 4 Solution:

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