1

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

A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is :
A
508 pJ
B
692 pJ
C
560 pJ
D
600 pJ

## Explanation

Initial energy of capacitor

Ui = ${1 \over 2}$ ${{{v^2}} \over c}$

= ${1 \over 2}$ $\times$ ${{120 \times 120} \over {12}}$ = 600 J

Since battery is disconnected so charge remain same.

Final energy of capacitor

Uf = ${1 \over 2}{{{v^2}} \over c}$

= ${1 \over 2} \times {{120 \times 120} \over {12 \times 6.5}}$ = 92

W + Uf = Ui

W = 508 J
2

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

The actual value of resistance R, shown in the figure is 30$\Omega$. This is measuered in an experiment as shown using the standard formula R = ${V \over {\rm I}}$, where V and I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is 5% less, then the internal resistance of the voltmeter is - A
570 $\Omega$
B
600 $\Omega$
C
350 $\Omega$
D
35 $\Omega$

## Explanation

0.95 R = ${{R{R_v}} \over {R + {R_v}}}$

0.95 $\times$ 30 = 0.05 Rv

Rv = 19 $\times$ 30 = 570 $\Omega$
3

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

Two equal resistances when connected in series to a battery, consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be :
A
240 W
B
60 W
C
30 W
D
120 W

## Explanation

In series condition, equivalent resistance is 2R

thus power consumed is 60W = ${{{\varepsilon ^2}} \over {2R}}$

In parallel condition, equivalent resistance is R/2 thus new power is

P' = ${{{\varepsilon ^2}} \over {\left( {R/2} \right)}}$

or   P' = 4P = 240W
4

### 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