1

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

In the figure shown below, the charge on the left plate of the 10$\mu$F capacitor is –30$\mu$C. The charge on the right plate of the 6 $\mu$F capacitor is :

A
+ 12 $\mu$C
B
+ 18 $\mu$C
C
$-$ 18 $\mu$C
D
$-$ 12 $\mu$C

## Explanation

6$\mu$F & 4$\mu$F are in parallel & total charge on this combination is 30 $\mu$C

$\therefore$  Charge on 6$\mu$F capacitor = ${6 \over {6 + 4}} \times 30$

= 18 $\mu$C

Since charge is asked on right plate therefore is +18$\mu$C
2

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

The given graph shows variation (with distance r form centre) of :

A
Electric field of a uniformly charged sphere
B
Electric field of a uniformly charged spherical shell
C
Potential of a uniformly charged sphere
D
Potential of a uniformly charged spherical shell

## Explanation

As the field inside the uniformly charged hollow sphere or spherical shell is zero, so the potential inside it is constant, whereas outside it varies inversely with distance.
3

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

In the circuit shown,

the switch S1 is closed at time t = 0 and the switch S2 is kept open. At some later time(t0), the switch S1 is opened and S2 is closed. The behavior of the current I as a function of time 't' is given by :
A
B
C
D

## Explanation

From time t = 0 to t = t0, growth of current takes place and after that decay of current takes place.

most appropriate is (A)
4

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

In a Wheatstone bridge(see fig.), Resistances P and Q are approximately equal. When R = 400 $\Omega$, the bridge is balanced. On interchanging P and Q, the value of R, for balance, is 405 $\Omega$. The value of X is close to :

A
402.5 ohm
B
401.5 ohm
C
403.5 ohm
D
404.5 ohm

## Explanation

For a balanced bridge,

${P \over Q} = {{400} \over X}$ ......(i)

After interchanging P and Q,

${Q \over P} = {{405} \over X}$ ......(ii)

Multiplying equation (i) and (ii), we get

X2 = 400$\times$405

$\Rightarrow$ X = $\sqrt {400 \times 405}$ = 402.5 $\Omega$