NEET PHYSICS

DAILY PHYSICS TEST QUESTIONS

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Question Bank

1 Atoms and Nuclei

As the electron in Bohr orbit of hydrogen atom passes from state n = 2 to n = 1, the kinetic energy K and the potential energy U change as:

(1) K four-fold, U two- fold
(2) K two-fold, U four-fold
(3) K two-fold, U also two-fold
(4) K four-fold, U also four-fold.

Option 1

Option 2

Option 3

Option 4 Solution:

2 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

A reel of mass less thread rolls itself falling down under gravity. The acceleration of its fall is

(1) g (2) g/2

(3) 2g/3 (4) 4g/3

Option 1

Option 2

Option 3 Solution:

Option 4

4 Laws of Motion

A uniform rope of length l lies on a table. If the coefficient of friction is µ, then the maximum length l₁ of the hanging part of the rope which can overhang from the edge of the table without sliding down is

1) l/µ
2) l/(µ+1)
3) µl/(µ+1)
4) µl/ (l-1)

Option 1

Option 2

Option 3 Solution:

Solution

To prevent sliding of rope
friction due to rope on table = weight of rope hanging
let m be the mass per unit length of rope. l1 be the length hanging
m(l-l₁)gµ = l1mg
lµ - l₁µ = l1
l₁ ( 1+ µ) = lµ
l₁ = lµ / ( 1+ µ)

Option 4

5 Electromagnetic Induction

An alternating voltage is connected in series with a resistance R and an inductance L. If the potential drop across the resistance is 200 volt and across the inductance is 150 volt, the applied voltage is

1) 350 volt
2) 250 volt
3) 500 volt
4) 300 volt

Option 1

Option 2 Solution:

Solution

$$Total\quad PD\quad =\quad \sqrt { PD_{ Inductance }^{ 2 }+PD_{ resistance }^{ 2 } } \\$$

$$ PD\quad =\quad \sqrt { { 150 }^{ 2 }+{ 200 }^{ 2 } } $$

$$=250V$$

Option 3

Option 4

6 Capacitance

A circuit has a self-inductance of 1H and carries a current of 2A. To prevent sparking, when the circuit is switched off, a capacitor which can withstand 400V is used. The least capacitance of capacitor connected across the switch must be equal to

1) 50μF

2) 25μF

3) 100μF

4) 12.5μF

Option 1

Option 2 Solution:

Solution

Energy stored in capacitor = energy stored in inductance

i.e., $$\dfrac { 1 }{ 2 } C{ V }^{ 2 }=\dfrac { 1 }{ 2 } { LI }^{ 2 }$$
$$\Rightarrow C=\dfrac { { LI }^{ 2 } }{ { V }^{ 2 } } =\dfrac { 1\times { \left( 2 \right) }^{ 2 } }{ { \left( 400 \right) }^{ 2 } } =25\mu F$$

Option 3

Option 4

7 Alternating Currents

The potential difference VV and the current II flowing through an instrument in an AC circuit are given by V=5cosωt volt,I=2sinωtAmp . The power dissipated in the circuit is

1) 5 watt
2) 10 watt
3) 2.5 watt
4) 0

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

The voltage can be written as,

$$V = 5\left( {\sin \omega t + \frac{\pi }{2}} \right)$$

The angle between the voltage and the current is $$90^\circ $$.

The power dissipated in the circuit is given as,

$$P = {V_{rms}}{I_{rms}}\cos \phi $$

$$ = {V_{rms}}{I_{rms}}\cos 90^\circ $$

=$$0$$

Thus, the power dissipated in the circuit is $$0$$.

The equation Ip = KVp ^3/2 is called :

(1) Gelusec law (2) Newton�s law

(3) Charles law (4) None

Option 1

Option 2

Option 3

Option 4 Solution:

9 Ray Optics

A student is using a convex lens of focal length 10 cm to study the image formation by a convex lens for the various positions of the object. In one of his observations, he may observe that when the object is placed at a distance of 20 cm from the lens, its image is formed at (select the correct option)

1) 20 cm on the other side of the lens and is of the same size, real and erect.
2) 40 cm on the other side of the lens and is magnified, real and inverted
3) 20 cm on the other side of the lens and is of the same size, real and inverted.
4) 20 cm on the other side of the lens and is of the same size, virtual and erect.

Option 1

Option 2

Option 3 Solution:

Solution

Focal length of the lens, $$f =10 cm$$
Object distance,$$ u = -20 cm$$ (negative sign is used due to sign convention)
Using the lens formula:

$$\dfrac{1}{u} + \dfrac{1}{f} = \dfrac{1}{v}$$
Substituting the values in the above equation we get,
$$\Rightarrow v = 20 cm$$
Hence, the image will be formed on the other side of the lens.
Magnification is given as:
$$ m = \dfrac{image \; distance}{object \; distance} = \dfrac{20}{20} = 1$$
Therefore, the image formed by the convex lens will be of same size, real and inverted.

Option 4

10 Electromagnetic Waves

A plane electromagnetic wave of frequency $$25\ MHz$$ travels in free space along the x-direction. At a particular point in space and time, $$\vec { E } =6.3\hat { j } { V }/{ m }$$. What is $$\vec {B}$$ at this point?

1) 4.1×10−8k
2) 2.1×10⁻⁸k
3) 0.21×10⁻⁸k
4) 20.1×10⁻⁸k

Option 1

Option 2 Solution:

Solution

The magnitude of B is

B = $$\frac{E}{C}$$

= $$\frac{6.3}{3\times10^8}$$

= $$2.1 \times 10^{-8}T$$

To find the direction, we note that E is along y - direction and the wave propagates along x - axis.

Therefore, B should be in a direction perpendicular to both x - and y - axes.

Using vector algebra, E $$\times$$ B should be along x - direction.

since, ( + j ) $$\times ( + k ) = i$$,

so, B is along + k

or the z - direction

so $$\vec B=2.1\times 10^{-8} k$$

Option 3

Option 4

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NEET PHYSICS

DAILY PHYSICS TEST QUESTIONS

Question Bank

Solution

Solution

Solution

i.e., $$\dfrac { 1 }{ 2 } C{ V }^{ 2 }=\dfrac { 1 }{ 2 } { LI }^{ 2 }$$ $$\Rightarrow C=\dfrac { { LI }^{ 2 } }{ { V }^{ 2 } } =\dfrac { 1\times { \left( 2 \right) }^{ 2 } }{ { \left( 400 \right) }^{ 2 } } =25\mu F$$

Solution

Solution

Solution

MOCK TEST

i.e., $$\dfrac { 1 }{ 2 } C{ V }^{ 2 }=\dfrac { 1 }{ 2 } { LI }^{ 2 }$$
$$\Rightarrow C=\dfrac { { LI }^{ 2 } }{ { V }^{ 2 } } =\dfrac { 1\times { \left( 2 \right) }^{ 2 } }{ { \left( 400 \right) }^{ 2 } } =25\mu F$$