Question

can you plz answer fast​

Answer 1

Answer

(D) Q/8

Explanation

$\rm Given,\ q_A=Q\\\\\rm 1.When\ A\ touched\ B:\ \\\\\to\ \rm q_1=\dfrac{q_A+q_B}{2}\\\\\to\ \rm q_1=\dfrac{Q+0}{2}\\\\\to\ \rm q_1=\dfrac{Q}{2}\ \; \bigstar\\\\\rm When\ A\ touched\ C:\\\\\to\ \rm q_2=\dfrac{q_1+q_c}{2}\\\\\to\ \rm q_2=\dfrac{\frac{Q}{2}+0}{2}\\\\\to\ \rm q_2=\dfrac{Q}{4}\ \; \bigstar\\\\\rm When\ A\ touched\ D:\\\\\to\ \rm q_{3}=\dfrac{q_2+q_D}{2}\\\\\to\ \rm q_{3}=\dfrac{\frac{Q}{4}+0}{2}\\\\\to\ \rm q_3=\dfrac{Q}{8}\ \; \bigstar$

$\rm So\ final\ charge\ on\ A\ is\ \dfrac{Q}{8}$

So , Option (D) is correct

Answer 2

(D) Q/8

Explanation

\begin{gathered}\rm Given,\ q_A=Q\\\\\rm 1.When\ A\ touched\ B:\ \\\\\to\ \rm q_1=\dfrac{q_A+q_B}{2}\\\\\to\ \rm q_1=\dfrac{Q+0}{2}\\\\\to\ \rm q_1=\dfrac{Q}{2}\ \; \bigstar\\\\\rm When\ A\ touched\ C:\\\\\to\ \rm q_2=\dfrac{q_1+q_c}{2}\\\\\to\ \rm q_2=\dfrac{\frac{Q}{2}+0}{2}\\\\\to\ \rm q_2=\dfrac{Q}{4}\ \; \bigstar\\\\\rm When\ A\ touched\ D:\\\\\to\ \rm q_{3}=\dfrac{q_2+q_D}{2}\\\\\to\ \rm q_{3}=\dfrac{\frac{Q}{4}+0}{2}\\\\\to\ \rm q_3=\dfrac{Q}{8}\ \; \bigstar\\\\\end{gathered}

Given, q

A

=Q

1.When A touched B:

→ q

1

=

2

q

A

+q

B

→ q

1

=

2

Q+0

→ q

1

=

2

Q

★

When A touched C:

→ q

2

=

2

q

1

+q

c

→ q

2

=

2

2

Q

+0

→ q

2

=

4

Q

★

When A touched D:

→ q

3

=

2

q

2

+q

D

→ q

3

=

2

4

Q

+0

→ q

3

=

8

Q

★

\rm So\ final\ charge\ on\ A\ is\ \dfrac{Q}{8}So final charge on A is

8

Q

So , Option (D) is correct