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1

### JEE Main 2019 (Online) 12th January Morning Slot

In the figure shown, a circuit contains two identical resistors with resistance R = 5$$\Omega$$ and an inductance with L = 2mH. An ideal battery of 15 V is connected in the circuit. What will be the current through the battery long after the switch is closed ?

A
6 A
B
7.5 A
C
3 A
D
5.5 A

## Explanation

Ideal inductor will behave like zero resistance long time after switch is closed

I = $${{2\varepsilon } \over R}$$ = $${{2 \times 15} \over 5} = 6A$$
2

### JEE Main 2019 (Online) 12th January Morning Slot

The galvanometer deflection, when key K1 is closed but K2 is open, equals $$\theta$$0 (see figure). On closing K2 also and adjusting R2 to 5$$\Omega$$, the deflection in galvanometer becomes $${{{\theta _0}} \over 5}.$$ . The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :

A
5 $$\Omega$$
B
25 $$\Omega$$
C
12 $$\Omega$$
D
22 $$\Omega$$

## Explanation

case I :

ig = $${E \over {220 + {R_g}}}$$ = C$$\theta$$0     . . .(i)

Case II :

ig = $$\left( {{E \over {220 + {{5{R_g}} \over {5 + {R_g}}}}}} \right)$$ $$\times$$ $${5 \over {\left( {{R_g} + 5} \right)}}$$ = $${{C{\theta _0}} \over 5}$$       . . .(ii)

$$\Rightarrow$$  $${{5E} \over {225{R_g} + 1100}}$$ = $${{C{\theta _0}} \over 5}$$       . . .(ii)

$${E \over {220 + {R_g}}} = C\theta$$       . . .(i)

$$\Rightarrow$$  $${{225{R_g} + 1100} \over {1100 + 5{R_g}}} = 5$$

$$\Rightarrow$$  5500 + 25Rg = 225Rg + 1100

200Rg = 4400

Rg = 22$$\Omega$$
3

### JEE Main 2019 (Online) 11th January Evening Slot

In the experimental set up of metre bridge shown in the figure, the null point is obtained at a distance of 40 cm from A. If a 10 $$\Omega$$ resistor is connected in series with R1, the null point shifts by 10 cm. The resistance that should be connected in parallel with (R1 + 10) $$\Omega$$ such that the null point shifts back to its initial position is :

A
40 $$\Omega$$
B
30 $$\Omega$$
C
20 $$\Omega$$
D
60 $$\Omega$$

## Explanation

$${{{R_1}} \over {{R_2}}} = {2 \over 3}\,\,$$                                      . . .(i)

$${{{R_1} + 10} \over {{R_2}}} = 1 \Rightarrow {R_1} + 10 = {R_2}$$     . . .(ii)

$${{2{R_2}} \over 3} + 10 = {R_2}$$

$$10 = {{{R_2}} \over 3} \Rightarrow {R_2} = 30\Omega$$

& $${R_1} = 20\Omega$$

$${{{{30 \times R} \over {30 + R}}} \over {30}} = {2 \over 3}$$

$$R = 60\,\Omega$$
4

### JEE Main 2019 (Online) 11th January Evening Slot

In the circuit shown, the potential difference between A and B is :

A
6 V
B
3 V
C
2 V
D
1 V

## Explanation

Potential difference across AB will be equal to battery equivalent across CD

VAB $$=$$ VCD $$=$$ $${{{{{E_1}} \over {{r_1}}} + {{{E_2}} \over {{r_2}}} + {{{E_3}} \over {{r_3}}}} \over {{1 \over {{r_1}}} + {1 \over {{r_2}}} + {1 \over {{r_3}}}}} = {{{1 \over 1} + {2 \over 1} + {3 \over 1}} \over {{1 \over 1} + {1 \over 1} + {1 \over 1}}}$$

$$=$$ $${6 \over 3}$$ $$=$$ 2V

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