1
MCQ (Single Correct Answer)

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 $${{{Q^2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 5 }}} \right)$$
2
MCQ (Single Correct Answer)

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
3
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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 $$\Omega $$

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