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1

### 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)

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

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

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

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

$$R = 60\,\Omega$$
2

### 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
3

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

A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil:
A
decreases by a factor of $$9\sqrt 3$$
B
increases by a factor of 27
C
decreases by a factor of 9
D
increases by a factor of 3

## Explanation

Total length L will remain constant

L = (3a) N        (N = total turns)

and length of winding = (d) N

(d = diameter of wire)

self inductance = $$\mu$$0n2A$$\ell$$

= $$\mu$$0n2$$\left( {{{\sqrt 3 {a^2}} \over 4}} \right)$$ dN

$$\propto$$ a2 N $$\propto$$ a

So self inductance will become 3 times
4

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

A galvanometer having a resistance of 20 $$\Omega$$ and 30 divisions on both sides has figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt, is:
A
120 $$\Omega$$
B
125 $$\Omega$$
C
80 $$\Omega$$
D
100 $$\Omega$$

## Explanation

Rg = 20$$\Omega$$

NL = NR = N = 30

FOM = $${1 \over \phi }$$ = 0.005 A/Div.

Current sentivity = CS = $$\left( {{1 \over {0.005}}} \right)$$ = $${\phi \over {\rm I}}$$

$${\rm I}$$gmax = 0.005 $$\times$$ 30

= 15 $$\times$$ 10$$-$$2 = 0.15

15 = 0.15 [20 + R]

100 = 20 + R

R = 80

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