Question

Correct option
(d) ( 2 times 10^{-2} mathrm{C} )
Explanation:
Radii of sphere ( left(mathrm{R}_{1}right)=1 mathrm{cm}=1 times 10^{-2} mathrm{m}: )
( left(R_{2}right)=2 mathrm{cm}=2 times 10^{-2} mathrm{m} ) and charges on sphere;
( left(Q_{1}right)=10^{-2} mathrm{C} ) and ( left(Q_{2}right)=5 times 10^{-2} mathrm{C} )
Common potential ( (V)=frac{text { Total charge }}{text { Total capacity }}=frac{Q_{1}+Q_{2}}{C_{1}+C_{2}} )
[
=frac{left(1 times 10^{-2}right)+left(5 times 10^{-2}right)}{4 pi varepsilon_{0} 10^{-2}+4 pi varepsilon_{0}left(2 times 10^{-2}right)}=frac{6 times 10^{-2}}{4 pi varepsilon_{0}left(3 times 10^{-2}right)}
]
Therefore final charge on smaller sphere ( left(C_{1} Vright) )
[
=4 pi varepsilon_{0} times 10^{-2} times frac{6 times 10^{-2}}{4 pi varepsilon_{0} times 3 times 10^{-2}}=2 times 10^{-2} mathrm{C}
]

# (39)Two metallic spheres of radii 1 cm and 2 cm are given charges 10-2 C and 5 * 10-2 C respectively. If they are connected by a conducting wire, the final charge on the smaller sphere is (1) 3 × 10-2 C (2) 2* 10-2 C (3) 1 * 10-2 C (4) 6 × 10-20

Solution