NSEP PHYSICS

DAILY PHYSICS TEST QUESTIONS

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1 Dual Nature of Matter and Radiation

A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photo electron is

1) 4.81eV
2) 3.74eV
3) 2.65eV
4) 1.07eV

Option 1

Option 2

Option 3 Solution:

Solution

Energy of incident light

$$E\left( eV \right) =\dfrac { 12375 }{ 3320 } =3.72eV$$

We know that, $$E={ W }_{ 0 }+{ eV }_{ 0 }$$

$${ V }_{ 0 }=\dfrac { \left( E-{ W }_{ 0 } \right) }{ e } =\dfrac { 3.72eV-1.07eV }{ e } $$

$${ V }_{ 0 }=2.65$$ Volt

Option 4

2 Current Electricity

A material 'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of wire made of 'A' then for the two wires to have the same resistance, the ratio, $$\dfrac{I_B}{I_A}$$ of their respective length must be

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

Option 1

Option 2 Solution:

Solution

$$R=\rho\dfrac{L}{A}$$

$$\implies R=\rho\dfrac{4L}{D^2}$$ where $$D=$$ diameter

Given that $$\dfrac{\rho_B}{\rho _A}=2$$

and, $$\dfrac{D_A}{D_B}=\dfrac{1}{2}$$

For same resistance-

$$\rho_A\dfrac{L_A}{D_A^2}=\rho_B\dfrac{L_B}{D_B^2}$$

$$\implies \dfrac{L_B}{L_A}=\dfrac{\rho_A}{\rho_B}\dfrac{D_B^2}{D_A^2}$$

$$\implies \dfrac{L_B}{L_A}=\dfrac{1}{2}\times 2^2$$

$$\implies \dfrac{L_B}{L_A}=2$$

Option 3

Option 4

3 Laws of Motion

A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is

1) Mg(√2 + 1)
2) Mg √2
3) Mg/√2
4) Mg(√2 – 1)

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

4 Gravitation

A planet is moving around the Sun in a circular orbit of circumference C. The work done on the planet by the gravitational force F of the Sun is

1) FC/2
2) FC
3) F/C
4) Zero.

Option 1

Option 2

Option 3

Option 4 Solution:

The work done is given by

W = F S cos ø where F is the force, S is the displacement and ø is the angle between the force and displacement.
Here the angle between the force and the displacement is 90^o. Hence the work done is zero

5 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

6 Electromagnetic Waves

A certain color of light towards the purple end of the visible spectrum has a wavelength of 420 nm in a vacuum.
What is the frequency of this light?
The speed of light in a vacuum is 3.00×10⁸m/s

1) 126Hz
2) 7.1×10¹⁴Hz
3) 1.26×10¹¹Hz
4) 1.4×10⁻¹⁵Hz

Option 1

Option 2 Solution:

Solution

Given : $$v = 3.00 \times 10^8$$ m/s $$\lambda = 420$$ nm $$ = 420 \times 10^{-9}$$ m

$$\therefore$$ Frequency of the light $$\nu = \dfrac{v}{\lambda} = \dfrac{3.00 \times 10^8}{420 \times 10^{-9}} = 7.1 \times 10^{14}$$ $$Hz$$

Option 3

Option 4

7 Laws of Motion

The maximum speed at which a car can turn around a curve of 20m radius on a level road if the coefficient of friction between the tyres and the road is 0.4, is

1) 10.26 m/s
2) 8.85 m/s
3) 9.16 m/s
4) 11.22 m/s

Option 1

Option 2 Solution:

Option 3

Option 4

8 Thermodynamics

Water falls from a height of 300 m assuming that the whole of the energy due to fall is converted into heat, the rise in temperature of water (J = 4.18 J/Cal) is:

1.0.54°C
2.0.62°C
3.0.82°C
4.None of the above

Option 1

Option 2

Option 3

Option 4 Solution:

9 Dual Nature of Matter and Radiation

A proton when accelerated through a potential difference of V volt has a wavelength λ associated with it. An α - particle in order to have the same wavelength λ must be accelerated through a p.d. of

1) V/8 volt
2) V/4 volt
3) V volt
4) 2V volt

Option 1 Solution:

Solution

For proton, $$\lambda_p =\dfrac{0.286\times 10^{-10}}{\sqrt{\nu }}\ m$$
where V is p.d
For $$\alpha -particle,\ \lambda_\alpha =\dfrac{0.101\times 10^{-10}}{\sqrt{\nu }}m$$

so, keeping same $$\alpha $$ for proton and $$\alpha -partside$$

$$1= \dfrac{\lambda _p}{\lambda _{\alpha }} = \dfrac{0.286 \times \sqrt{V_{\alpha }}}{0.101 \times \sqrt{V_{P}}}$$

$$0.101 \times \sqrt{V_{P}} = 0.286 \times \sqrt{V_{\alpha }}$$
$$(0.101)^{2} \times V_{P} = (0.286)^{2}V _{\alpha} $$

$$\dfrac{(0.101)^{2}\nu }{(0.286)^{2}} = V _{\alpha }$$

$$(\because V_{P}=V)$$

$$\Rightarrow V_{\alpha } = \dfrac{V}{8} volt$$

Option 2

Option 3

Option 4

10 Physical world and measurement

Of which the velocity is equal to light velocity :

(1) cathode ray
(2) X-rays
(3) positive ray
(4) all

Option 1

Option 2 Solution:

Option 3

Option 4

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