NSEP PHYSICS

DAILY PHYSICS TEST QUESTIONS

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1 Ray Optics

A microscope is focused on a coin lying at the bottom of a beaker. The microscope is now raised up by 1 cm. to what depth should the water be poured into the beaker so that coin is again in focus? (Refractive index of water is 4 / 3)

1) 4 cm
2) 4 / 3 cm
3) 3 cm
4) 1 cm

Option 1 Solution:

Solution

Water should be poured to a height h cm so that apparent shift in position of coin is 1 cm.
As $$\displaystyle y=\left( 1-\frac { 1 }{ n } \right) h$$
$$\displaystyle \Rightarrow \quad h=\frac { y }{ \left( 1-\frac { 1 }{ n } \right) } =\frac { 1 }{ \left( 1-\frac { 1 }{ { 4 }/{ 3 } } \right) } =\frac { 1 }{ \frac { 1 }{ 4 } } =4cm$$

Option 2

Option 3

Option 4

A shot leaves a gun at the rate of 200 m/s. The greatest distance to which it could be projected is:
1. 200 m
2. 1000 m
3. 2000 m
4. 4000 m

Option 1

Option 2

Option 3 Solution:

Option 4

3 Electrostatics

n identical mercury droplets charged to the same potential V coalesce to form a single bigger drop. The potential of the new drop =

1) nn

2) n^2/3 n

3) n/n

4) n^1/3 n

Option 1

Option 2 Solution:

Solution

Suppose we have n identical drops each having radius r, capacitance C, charge q and potential V. If these drops are combined to form a big drop of radius R, capacitance C', charge Q and potential V will become : Charge on big drop

Q=nq Capacitance of big drop

C′ = n^1/3C

Hence potential of big drop V′=Q / C′ =nq / n1^/3C

⁼n ^2/3C

Option 3

Option 4

4 Electrostatics

The electric field on the axis of a dipole of dipole moment \;\overrightarrow{P}\; at a distance x from the line joining the charges and (x>>d)

$$(1)\;\large\frac{k\overrightarrow{P}}{x^3}\qquad(2)\;-\large\frac{k\overrightarrow{P}}{x^3}\qquad(3)\;\large\frac{2 k\overrightarrow{P}}{x^3}\qquad(4)\;-\large\frac{2 k\overrightarrow{P}}{x^3}$$

Option 1

Option 2 Solution:

Solution

$$E_{D}=\large\frac{2 k\;q }{((\large\frac{d}{2})^{2}+x^2)^{\large\frac{3}{2}}}\times \large\frac{d}{2}$$

This is because only horizontal components will add vertical cancel each other being equal and opposite

$$E_{D} \approx \large\frac{2 \;k \;q\; d}{2x^3}$$

$$E_{D} \approx \large\frac{k \; P}{x^3}$$

From the diagram it is clear the ED & P are in opposite direction
Therefore

$$\overrightarrow{E}_{D}=-\large\frac{k\;\overrightarrow{P}}{x^3}\;.$$

Option 3

Option 4

5 Laws of Motion

A 1,200-kilogram car traveling at 10. meters per second hits a tree and is brought to rest in 0.10 second. What is the magnitude of the average force acting on the car to bring it to rest?

1) 1.2 × 10²N
2) 1.2 × 10³N
3) 1.2 × 10⁴N
4) 1.2 × 10⁵N

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

F = m x a

F = 1200 x -100, ==> F = -120000 N (- sign indicted in the opposite direction)

6 Kinematics

From the top of twer of height 40 m a ball is projected upwards with a speed of 20m/s at an angle of elevation of 30°. Then the ratio of the total time taken by the ball to hit the ground to its time of flight ( time taken to come back to the same elevation ) is

1) 2 : 1
2) 3 : 1
3) 3 : 2
4) 4 : 1

Option 1 Solution:

Solution

Time taken to reach ball on ground
h=-ut + ½ gt ²
40=-(20sin30) t + ½ ×10t²
5t2 - 10t -40=0
thus t=-2 or t=4 seconds time can not be negative thus t=4 sec
time to come to same elevation
T=2usinθ/g=(2×20×sin30 ) /10=2 sec
∴ t/T=4/2=2:1

Option 2

Option 3

Option 4

7 Electronic Devices

Ge at absolute temp is a :

(1) super condutor
(2) conductor
(3) semi conductor
(4) insulator

Option 1

Option 2

Option 3

Option 4 Solution:

8 Work Energy and Power

A force of 7N, making an angle q with the horizantal, acting on an object displaces it by 0.5m along the horizontal direction. If the object gains kinetic energy of 2J, what is the horizontal component of the force?

1) 2N
2) 4N.
3) 1N
4) 14N

Option 1

Option 2 Solution:

Solution

Let the force in horizontal direction be F

By work energy theorem

Gain in KE=Work done

⇒2=F.s = F×0.5

⇒F=4N

Option 3

Option 4

9 Magnetic Effects of Current and Magnetism

Under the influence of a uniform magnetic field a charged particle is moving in a circle of radius R with constant speed v. The time period of the motion

1) depends on R and not v
2) is independent of both R and v
3) depends on both R and v
4) depends on v and not on R

Option 1

Option 2

Option 3 Solution:

Solution

In a uniform magnetic field, a charged particle is moving in a circle of radius R with constant speed v.

$$\therefore \displaystyle \frac{mv^2}{R} = Bqv$$ or $$\displaystyle R = \frac{mv}{Bq}$$ .......... (1)

Time period, $$\displaystyle T = \frac{2 \pi R}{v} = \frac{2 \pi m v}{B qv} = \frac{2 \pi m}{Bq} $$ ......... (2)

Time period T does not depend on both R and v because v is changed, R is also changed proportionately and for period, it is R/v that is taken.

Option 4

10 Thermodynamics

The door of a domestic refrigerator is kept open while the switch is on, Then the room will

(1) Get heated for sometime, cooled for sometime
(2) Get neither heated nor cooled
(3) Get heated
(4) Get cooled

Option 1

Option 2

Option 3

Option 4 Solution:

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