1
MCQ (Single Correct Answer)

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Two identical conducting spheres A and B, carry equal charge. They are separated by a distance much larger than their diameters, and the force between theis F. A third identical conducting sphere, C, is uncharged. Sphere C is first touhed to A, then to B, and then removed. As a result, the force between A and B would be equal to :
A
F
B
$${{3F} \over 4}$$
C
$${{3F} \over 8}$$
D
$${{F} \over 2}$$

Explanation

Let, change of A and B = q

$$\therefore\,\,\,$$ Force between them, F = $${{k \times q \times q} \over {{r^2}}} = {{k{q^2}} \over {{r^2}}}$$

When C touched with A then charge of A. Will fl;ow to C and divide into half parts.

$$\therefore\,\,\,$$ charge of A and C ,

qA = qB = $${q \over 2}$$

Then C touched with B, then charge on B,

qB = $${{{q \over 2} + q} \over 2} = {{3q} \over 4}$$

$$\therefore\,\,\,$$ Force between A and B,

F' = $${{k \times {q \over 2} \times {{3q} \over 4}} \over {{r^2}}}$$

= $${{k \times 3{q^2}} \over {8{r^2}}}$$

= $${3 \over 8} \times {{k{q^2}} \over {{r^2}}}$$

= $${3 \over 8}\,F$$
2
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

In the following circuit, the switch S is closed at t = 0. The charge on the capacitor C1 as a function of time will be given by $$\left( {{C_{eq}} = {{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)$$

A
$${C_1}E\left[ {1 - \exp \left( { - tR/{C_1}} \right)} \right]$$
B
$${C_2}E\left[ {1 - \exp \left( { - t/R{C_2}} \right)} \right]$$
C
$${C_{eq}}E\left[ {1 - \exp \left( { - t/R{C_{eq}}} \right)} \right]$$
D
$${C_{eq}}E\,\,\exp \left( { - t/R{C_{eq}}} \right)$$

Explanation

Charge on capacitor at time t,

q = q0 [ 1 $$-$$ e$$-$$t/RCeq]

= Ceq E [ 1 $$-$$ e$$-$$t/RCeq]

[ as $$\,\,\,\,\,\,$$ q0 = Ceq E]

this charge q will be on both capacitor C1 and C2, as both are in series.
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

Three charges + Q, q, + Q are placed respectively, at distance, 0, d/2 and d from the origin, on the x-axis. If the net force experienced by + Q, placed at x = 0, is zero, then value of q is :
A
$$-$$ $${Q \over 4}$$
B
+ $${Q \over 2}$$
C
+ $${Q \over 4}$$
D
$$-$$ $${Q \over 2}$$

Explanation



Force on + Q charge at x = 0 due to q charge, F1 = $${{KQq} \over {{{\left( {{d \over 2}} \right)}^2}}}$$

Force on +Q charge at x = 0 due to + Q charge at x = d is,

      F2 = $${{KQQ} \over {{d^2}}}$$

According to the question,

F1 + F2 = 0

$$ \Rightarrow $$   F1 = $$-$$ F2

$$ \Rightarrow $$   $${{KQq} \over {{{\left( {{d \over 2}} \right)}^2}}}$$ = $$-$$ $${{KQQ} \over {{d^2}}}$$

$$ \Rightarrow $$   4q = $$-$$ Q

$$ \Rightarrow $$   q = $$-$$ $${Q \over 4}$$
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

A resistance is shown in the figure. Its value and tolerance are given respectively by :

A
270 $$\Omega $$, 10 %
B
27 k$$\Omega $$, 10 %
C
27 k$$\Omega $$, 20 %
D
270 $$\Omega $$, 5 %

Explanation

From color code table :

For Red value is 2

For Violet value is 7

For Orange multiplier is 103

For Silver tolarence is 10%

$$ \therefore $$ Resistance and tolerance is

= 27 $$ \times $$ 103 $$ \pm $$ 10%

= 27 k$$\Omega $$ $$ \pm $$ 10%

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