1

### JEE Main 2019 (Online) 10th January Morning Slot

In the given circuit the cells have zero internal resistance. The currents (in Amperes) passing through resistance R1 and R2 respectively, are -

A
0.5, 0
B
0, 1
C
1, 2
D
2, 2

## Explanation

i1 = ${{10} \over {20}}$ = 0.5A

i2 = 0
2

### JEE Main 2019 (Online) 10th January Morning Slot

A potentiometer wire AB having length L and resistance 12 r is joined to a cell D of emf $\varepsilon$ and internal resistance r. A cell C having emf $\varepsilon$/2 and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in figure shows no deflection is –

A
${{11} \over {12}}L$
B
${{13} \over {24}}L$
C
${{5} \over {12}}L$
D
${{11} \over {24}}L$

## Explanation

$i = {\varepsilon \over {13r}}$

$i\left( {{x \over L}12r} \right) = {\varepsilon \over 2}$

${\varepsilon \over {13r}}\left[ {{x \over L}.12r} \right] = {\varepsilon \over 2}$ $\Rightarrow \,\,\,\,x = {{13L} \over {24}}$
3

### JEE Main 2019 (Online) 10th January Evening Slot

Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, –2) and (0, –2). The work required to put a fifth charge Q at the origin of the coordinate system will be -
A
${{{Q_2}} \over {4\pi {\varepsilon _0}}}$
B
${{{Q^2}} \over {2\sqrt 2 \pi {\varepsilon _0}}}$
C
${{{Q_2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 3 }}} \right)$
D
${{{Q_2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 5 }}} \right)$

## Explanation

Potential at origin = ${{KQ} \over 2} + {{KQ} \over 2} + {{KQ} \over {\sqrt {20} }} + {{KQ} \over {\sqrt {20} }}$

(Potential at $\infty$ = 0)

= KQ$\left( {1 + {1 \over {\sqrt 5 }}} \right)$

$\therefore$  Work required to put a fifth charge Q

at origin is equal to ${{{Q^2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 5 }}} \right)$
4

### JEE Main 2019 (Online) 10th January Evening Slot

A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is -
A
11 $\times$ 10–5 W
B
11 $\times$ 10–3 W
C
11 $\times$ 105 W
D
11 $\times$ 10–4 W

## Explanation

P = I2R

4.4 = 4 $\times$ 10$-$6 R

R = 1.1 $\times$ 106 $\Omega$

P' = ${{{{11}^2}} \over R}$ = ${{{{11}^2}} \over {1.1}}$ $\times$ 10$-$6 = 11 $\times$ 10$-$5 W