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1

### 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

2

### 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
3

### 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
4

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

A parallel plate capacitor is of area 6 cm2 and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants K1 = 10, K2 = 12 and K3 = 14. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be -

A
12
B
36
C
14
D
4

## Explanation

Let dielectric constant of material used be K.

$$\therefore$$  $${{10{ \in _0}\,A/3} \over d} + {{12{ \in _0}\,A/3} \over d} + {{14{ \in _0}\,A/3} \over d} = {{K{ \in _0}A} \over d}$$

$$\Rightarrow$$    K $$=$$ 12

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