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1

### JEE Main 2021 (Online) 27th August Evening Shift

Numerical
The ratio of the equivalent resistance of the network (shown in figure) between the points a and b when switch is open and switch is closed is x : 8. The value of x is ___________.

## Explanation

$${R_{eq\,open}} = {{3R} \over 2}$$

$${R_{eq\,closed}} = 2 \times {{R \times 2R} \over {3R}} = {{4R} \over 3}$$

$${{{R_{eq\,open}}} \over {{R_{eq\,closed}}}} = {{3R} \over 2} \times {3 \over {4R}} = {9 \over 8}$$

$$\therefore$$ $$x = 9$$
2

### JEE Main 2021 (Online) 27th August Morning Shift

Numerical
First, a set of n equal resistors of 10 $$\Omega$$ each are connected in series to a battery of emf 20V and internal resistance 10$$\Omega$$. A current I is observed to flow. Then, the n resistors are connected in parallel to the same battery. It is observed that the current is increased 20 times, then the value of n is ............... .

## Explanation

In series

$${R_{eq}} = nR = 10n$$

$${i_s} = {{20} \over {10 + 10n}} = {2 \over {1 + n}}$$

In parallel

$${R_{eq}} = {{10} \over n}$$

$${i_p} = {{20} \over {{{10} \over n} + 10}} = {{2n} \over {1 + n}}$$

$${{{i_p}} \over {{i_s}}} = 20$$

$${{\left( {{{2n} \over {1 + n}}} \right)} \over {\left( {{2 \over {1 + n}}} \right)}} = 20$$

$$n = 20$$
3

### JEE Main 2021 (Online) 27th July Evening Shift

Numerical
For the circuit shown, the value of current at time t = 3.2 s will be _________ A.

[Voltage distribution V(t) is shown by Fig. (1) and the circuit is shown in Fig. (2)]

## Explanation

From graph voltage at t = 3.2 sec is 6 volt.

i = $${{6 - 5} \over 1}$$

$$\Rightarrow$$ i = 1 A
4

### JEE Main 2021 (Online) 25th July Morning Shift

Numerical
An electric bulb rated as 200 W at 100 V is used in a circuit having 200 V supply. The resistance 'R' that must be put in series with the bulb so that the bulb delivers the same power is _____________ $$\Omega$$.

## Explanation

Power, $$P = {{{V^2}} \over {{R_B}}}$$

$${R_B} = {{{V^2}} \over P} = {{100 \times 100} \over {200}}$$

$${R_B} = 50\Omega$$

To produce same power, same voltage (i.e. 100 V) should be across the bulb.

Hence, R = RB

R = 50 $$\Omega$$

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