NSEP PHYSICS

DAILY PHYSICS TEST QUESTIONS

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1 Wave Optics

When light passes from one medium to another, there is a change in:

(1) Wavelength and velocity
(2) Frequency, wavelength and velocity
(3) Frequency and velocity
(4) Frequency and wavelength.

Option 1 Solution:

Solution:

Frequency remains same.

Option 2

Option 3

Option 4

2 Current Electricity

Just as electricity is supplied at 220 V for domestic use in India, It is supplied at 110 V in U.S. A. If resistance of 60 W bulb for use in India is R, that of a 60 W bulb for use in U.S.A. will be:

(1) R
(2) 2R
(3) R/4
(4) R/2

Option 1

Option 2

Option 3 Solution:

Option 4

A wave of frequency 400 Hz has a velocity 360 m/s. The distance between two nearest points which are 60� out of phase, is:
1.12 cm
2. 14 cm
3.10 cm
4.15 cm

Option 1

Option 2

Option 3

Option 4 Solution:

4 Laws of Motion

What force is required to give a projectile weighing 450. N an acceleration of 500. m/s²?
1) 5.6 E 4 N
2) 2.3 E 4 N
3) 1.6 E 4 N
4) 8.7 E 4 N

Option 1

Option 2

Option 3

Option 4

5 Dual Nature of Matter and Radiation

The Kα X-ray emission line of tungsten occurs at λ = 0.021nm. The energy difference between K and L levels in this atom is about

1) 0.51MeV
2) 1.2MeV
3) 59keV
4) 13.6eV

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Line $\displaystyle { K }_{ \alpha }$ corresponds to transfer of electron from L-shell to K-shell.
Here, $\displaystyle \lambda =0.021nm = 2.1\times 10^{-11}\ m$

We know that

$\lambda = \dfrac{hc}{E (in \ Joule)}$

$\displaystyle \lambda=\frac { hc }{ E (in \ Joule)} =\frac { hc }{ e\times E\left( in \ eV \right) }$

$\displaystyle \Rightarrow \quad E\left(in \ eV \right) =\frac { hc }{ e\lambda } =\frac { 6.63\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 1.6\times { 10 }^{ -19 }\times 2.1\times { 10 }^{ -11 } } eV$

$\displaystyle E(in \ eV)=5.9\times { 10 }^{ 4 }eV=59keV$

6 Wave Optics

Which of the following is correct ?

(1) Light and solid are both transverse waves.
(2) Coherent source of light passing through two slits causes interference.
(3) Interference of light is evidence for transverse character of light.
(4) Rainbow appears because of scattering of light.

Option 1

Option 2 Solution:

Option 3

Option 4

7 Dual Nature of Matter and Radiation

Electromagnetic radiations described by the equation: $E = 100 \left[ \sin \left( 2 \times 10 ^ { 15 } \right) t + \sin \left( 6.28 \times 10 ^ { 15 } \right) t \right] V / m$ are incident on a photosensitive surface having work function $1.5\ eV$ . The maximum kinetic energy of photoelectrons will be

1) 0.125 eV
2) 2.625 eV
3) 4.125 eV
4) 6.125 eV

Option 1

Option 2 Solution:

Solution

Standard form will be like $E=E_0[sin(\omega _1t) +sin (\omega _2 t)]$

so we have

$\omega _1 =2\times 10^{15}rad/s$ and

$\omega _2 =6.28\times 10^{15}rad/s$

corresponding frequency will be

$\nu _1=\omega _1/2\pi=\dfrac{2\times 10^{15}}{6.28}$

and

$\nu _2=\omega _2/2\pi=\dfrac{6.28\times 10^{15}}{6.28}=10^{15}Hz$

We can see that

$\nu _2$ is

$higher$ so it will provide

$maximum$ kinetic energy.

Energy of the

$\nu_2$ photon will be

$E=h\nu _2=6.6\times 10^{-34}Js\times 10^{15} Hz=6.6\times 10^{-19} Joule=\dfrac{6.6\times 10^{-19}Joule}{1.6\times 10^{-19}J/eV}$

$E=4.125eV$

Consumed energy in work function will be

$W=1.5eV$

So remaining energy that is maximum kinetic energy will be

$KE=E-W=(4.125-1.5)eV=2.625eV$

Option 3

Option 4

8 Dual Nature of Matter and Radiation

An electric field of intensity $6\times 10^4 volts/m$ is applied perpendicular to the direction of motion of the electron. A magnetic field of intensity $8\times 10^{-2} wb/m^2$ is applied perpendicular to both the electric field and direction of motion of the electron. What is the velocity of the electron if it passes undeflected?

1) $7.5\times 10^5 m/s$
2) $7.5\times 10^{-5} m/s$
3) $6\times 10^4\times 8\times 10^{-2} m/s$
4) It is never possible

Option 1 Solution:

Solution

In case of no deviation, $eE=evB$
$\therefore v=\dfrac {E}{B}=\dfrac {6\times 10^4}{8\times 10^{-2}}$

$=7.5\times 10^5 m/sec.$

Option 2

Option 3

Option 4

9 Laws of Motion

A man of mass 70 kg stands on a weighing scale in a lift. Which is moving upwards with a uniform acceleration of 5 m/s^-2?

The reading on the scale is

1) 35 kg

2) 70 kg

3) 105 kg

4) 120 kg

Option 1

Option 2

Option 3 Solution:

Solution
N - mg = ma
N = m(g — a) = (70kg)(10— 5)m / s^-2
N=l050kgm^-2
The scale reads the normal force in kg
.

Option 4

10 Electrostatics

Two charges, each equal to q, aer kept at x=−a and x=a on the x-axis. A particle of mass m and charge q₀=q / 2 is placed at the origin. If charge q₀ is given a small displacement (y<< a) along the y-axis, the net force acting on the particle is proportional to

1) y
2) −y
3) 1 / y
4) −1 / y

Option 1 Solution:

Solution

⇒Fnet = 2Fcosθ

$\displaystyle{F}_{{{n}{e}{t}}}=\frac{{{2}{k}{q}{\left(\frac{q}{{2}}\right)}}}{{{\left(\sqrt{{{y}^{2}+{a}^{2}}}\right)}^{2}}}\cdot\frac{{{y}}}{{\sqrt{{{y}^{2}+{a}^{2}}}}}$

$\displaystyle{F}_{{{n}{e}{t}}}=\frac{{{2}{k}{q}{\left(\frac{q}{{2}}\right)}{y}}}{{{\left({y}^{2}+{a}^{2}\right)}^{{{3}\text{/}{2}}}}}\Rightarrow\frac{{{k}{q}^{2}{y}}}{{{a}^{3}}}$

So, F∝y

Option 2

Option 3

Option 4

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