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1

### JEE Main 2019 (Online) 12th January Morning Slot

In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the energy dissipated in the circuit in terms of capacitance 'C' and total charge 'Q' is :

A
$${1 \over 8}{{{Q^2}} \over C}$$
B
$${5 \over 8}{{{Q^2}} \over C}$$
C
$${3 \over 4}{{{Q^2}} \over C}$$
D
$${3 \over 8}{{{Q^2}} \over C}$$

## Explanation

Vi = $${1 \over 2}$$CE2

Vf = $${{{{\left( {CE} \right)}^2}} \over {2 \times 4c}}$$ = $${1 \over 2}{{C{E^2}} \over 4}$$

$$\Delta$$E = $${1 \over 2}$$CE2 $$\times$$ $${3 \over 4}$$ = $${3 \over 8}$$ CE2
2

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

Seven capacitors, each of capacitance 2 $$\mu$$F, are to be connected in a configuration to obtain an effective capacitance of $$\left( {{6 \over {13}}} \right)\mu F.$$ Which of the combinations, shown in figures below, will achieve the desired value
A
B
C
D

## Explanation

Ceq = $${6 \over {13}}$$$$\mu$$F

Therefore three capacitors most be in parallel to get 6 in

$${1 \over {{C_{eq}}}} = {1 \over {3C}} + {1 \over C} + {1 \over C} + {1 \over C} + {1 \over C}$$

Ceq = $${{3C} \over {13}}$$ = $${6 \over {13}}$$$$\mu$$F

3

### 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
4

### 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

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