In the circuit diagram shown in figure, R = 10 \(\Omega\), L = 5 H, E = 20 V and i = 2 A. This current is decreasing at a rate of 1.0 A/s. $V_{ab}$ at this instant will be:

1.	40 V	2.	35 V
3.	30 V	4.	45 V

When the current in the portion of the circuit shown in the figure is 2 A and increases at the rate of 1 A/s, the measured potential difference $V_{ab}$ $=$ 8 V. However, when the current is 2 A and decreases at the rate of 1 A/s, the measured potential difference $V_{ab}$ $=$ 4 V. The value of R and L is:

Some magnetic flux is changed from a coil of resistance 10 $\mathrm{\Omega }$. As a result, an induced current is developed in it, which varies with time as shown in the figure. The magnitude of change in flux through the coil in Wb is:

A 10 H inductor carries a current of 20 A. How much ice at 0°C could be melted by the energy stored in the magnetic field of the inductor?
Latent heat of ice is  2.26 × ${10}^3$ J/kg .

Two coils of 10 turns each are arranged such that the mutual inductance between them is 150 mH. The magnetic flux linked through one coil when 2 amperes current will flow in another coil, will be:

1.  $1\times {10}^{-\mathrm{3 }}Wb$ $$

2.  $10\times {10}^{-\mathrm{3 }}Wb$ $$

3.  $20\times {10}^{-\mathrm{3 }}Wb$ $$

4.  $30\times {10}^{-\mathrm{3 }}Wb$ $$

A square coil ACDE with its plane vertical is released from rest in a horizontal uniform magnetic field $\overrightarrow{\mathrm{B}}$ of length 2L (figure). The acceleration of the coil is:

A copper rod of mass m slides under gravity on two smooth parallel rails l distance apart and set at an angle $\theta$ to the horizontal as shown in fig. At the bottom, the rails are joined by a resistance R. There is a uniform magnetic field perpendicular to the plane of the rails. The terminal velocity of the rod is:

1. $\frac{mgR\cos \theta }{B^2{l}^2}$

2. $\frac{mgR\sin \theta }{B^2{l}^2}$

3. $\frac{mgR\tan \theta }{B^2{l}^2}$

4. $\frac{mgR\mathrm{co}t\theta }{B^2{l}^2}$ 

A straight solenoid has 50 turns per cm in primary coil and 200 turns in the secondary coil. The area of cross-section of the solenoid is 4 cm². Calculate the mutual inductance.

1. 5.0 H

2. $5.0\times {10}^{-4} H$

3. 2.5 H

4. $2.5\times {10}^{-4} H$

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R):
 

In the light of the above statements choose the correct answer from the options given below:
 

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R):