TS EAMCET 20-July-2022 Shift 2 Solved Paper

Given, ratio of radii of two metal sphere,
R1:R2‌=4:7
‌

R1

R2

‌=‌

4

7

If q1 and q2 be the charges on both the spheres after being contact with charge of 8.8×10−7C. then q1+q2=8.8×10−7C=0.88µC
⇒q1+q2=0.88µC
If V be the common potential on both the spheres after connection, then
‌q1=C1V‌ and ‌q2=C2V
∴‌‌

q1

q2

=‌

C1V

C2V

=‌

C1

C2

=‌

4πε0R1

4πε0R2

⇒‌

q1

q2

=‌

R1

R2

=‌

4

7

⇒‌‌q1=‌

4

7

q2 ......(ii)
From Eqs. (i) and (ii), we get
‌‌

4

7

q2+q2=0.88
⇒‌‌q2=‌

7

11

×0.88=0.56µC
‌∴q1=0.88−q1=0.88−0.56=0.32µC
i.e charge on smaller sphere, q1=0.32×10−6C
∴ Potential due to smaller sphere at r=60m