NSEP PHYSICS

DAILY PHYSICS TEST QUESTIONS

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A phenomenon of super conductivity has been observed at 95 K. This temperature is nearly equal to:
1.178�C
2. - 288�F
3. - 146�C
4. - 288�C

Option 1

Option 2 Solution:

Option 3

Option 4

2 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

3 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 $$\lambda = \dfrac{hc}{E (in \ Joule)}$$
$$\displaystyle \lambda=\frac { hc }{ E (in \ Joule)} =\frac { hc }{ e\times E\left( in \ eV \right) } $$
$$\displaystyle \Rightarrow \quad E\left(in \ eV \right) =\frac { hc }{ e\lambda } =\frac { 6.63\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 1.6\times { 10 }^{ -19 }\times 2.1\times { 10 }^{ -11 } } eV$$
$$\displaystyle E(in \ eV)=5.9\times { 10 }^{ 4 }eV=59keV$$

4 Kinematics

A Projectile is given an initial velocity of ( i+2j) m/s, + where i is along the ground and j is along the vertical. If g = 10 m/s², the equation of its trajectory is
1) y = x − 5x² (2) (3) (4)
2) y = 2x − 5x²
3) 4y = 2x − 5x²
4) 4y = 2x − 25x²

Option 1

Option 2 Solution:

Solution

Equation of trajectory

y= 2x -5x²

Option 3

Option 4

5 Gravitation

At what height from the surface of earth thegravitation potential and the value of g are –5.4 × 10⁷ J kg^–2 and 6.0 ms^–2 respectively ?Take the radius of earth as 6400 km :

1) 2600 km
2) 1600 km
3) 1400 km
4) 2000 km

Option 1 Solution:

Solution

r= 0.9×10⁷ m =9000 Km
Now height from surface (h) = height (r) - radius (R)
Height = 8000 – 6400 =2600 km

Option 2

Option 3

Option 4

6 Magnetic Effects of Current and Magnetism

The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by 36%. By doing this the periodic time of the magnetometer will:

1.Increase by 12.5%
2.Increase by 25%
3.Decrease by 12.5%
4.Decrease by 25%

Option 1

Option 2 Solution:

SOLUTION

$$\displaystyle{T}={2}\pi\sqrt{{\frac{I}{{{M}{B}_{{{H}}}}}}}\Rightarrow{T}\propto\frac{1}{\sqrt{{{M}}}}\Rightarrow\frac{{T}_{{{1}}}}{{T}_{{{2}}}}=\sqrt{{\frac{{M}_{{{2}}}}{{M}_{{{1}}}}}}$$

If M1=100 then M2=(100−36)=64

So $$\displaystyle\frac{{T}_{{{1}}}}{{T}_{{{2}}}}=\sqrt{{\frac{64}{{100}}}}=\frac{8}{{4}}\Rightarrow{T}_{{{2}}}=\frac{10}{{8}}{T}_{{{1}}}={1.25}{T}_{{{1}}}$$
So % increase in time period =25%

Option 3

Option 4

In space charge limited region the plate voltage is 50 Volt and plate current is 100 mA. If the plate voltage is changed to 20 Volt, the plate current will become :

(1) 10 mA (2) 19 mA

(3) 25 mA (4) 40 mA

Option 1

Option 2

Option 3 Solution:

Option 4

8 Gravitation

The orbital velocity of an artificial satellite in a circular orbit of a height R above earth's surface is v (R is radius of earth). The orbit velocity of a satellite orbiting at an altitude 2R above the earth's surface will be:

1) v/2
2) $$\dfrac {3}{2}v$$
3) $$\dfrac {2}{3}v$$
4) $$\sqrt {\dfrac {2}{3}}v$$

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Orbital velocity $$=\sqrt { \dfrac { GM }{ r } } $$

Where $$r$$ is the distance of satellite from the centre of earth.

$$V=\sqrt { \dfrac { GM }{ 2R } } $$

$${ V }^{ 1 }=\sqrt { \dfrac { GM }{ 3R } \times \dfrac { 2 }{ 2 } } $$

$${ V }^{ 1 }=\sqrt { \dfrac { 2 }{ 3 } } \sqrt { \dfrac { GM }{ 2R } } $$

$${ V }^{ 1 }=\sqrt { \dfrac { 2 }{ 3 } } V$$

9 Kinematics

A projectile is fired into the air and 12.6 seconds later hits the ground. With what speed was the projectile fired into the air?
1) 97.5 m/s
2) 14.6 m/s
3) 61.7 m/s
4) 38.6 m/s

Option 1

Option 2

Option 3 Solution:

Option 4

10 Electromagnetic Induction

A helicopter rises vertically with a speed of 100 m/s. If helicopter has length 10 m and horizontal component of earth's magnetic field is 5 × 10⁻³Wb/m²then the induced emf between the tip of nose and tail of helicopter is:

1) 50 V
2) 0.5 V
3) 5 V
4) 25 V

Option 1

Option 2

Option 3 Solution:

Solution

In case of motional emf, the motion of the conductor in the field exerts a force on the free charge in the conductor, so that one end of the conductor becomes positive, while the other negative resulting in a potential difference across its ends due to which a non-conservative electric field is set up in the conductor. In steady state the magnetic force on the free charge is balanced by the electric force due to induced field.
$$qE = qvB$$
or $$q\left(\frac{V}{l}\right)=qvB$$
ie, $$V=Bvl$$
So, the induced emf between tip of nose and tail of helicopter is given by
$$e=Bvl$$
$$=5\times 10^{-3}\times 10\times 100$$
$$=5V.$$

Option 4

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