${10}^{12}\alpha$-particles (nuclei of helium) fall per second on a neutral sphere; calculate the time in which the sphere gets charged by $2\mu C$.

The correct option is B 6.25 s
Total charge required $Q=2\mu C=2\times {10}^{-6}C$

Charge of one $\alpha$ particle $q=2e=2\times 1.6\times {10}^{-19}C$

Charge accumulated per second $Q^{\prime }={10}^{12}\times 2\times 1.6\times {10}^{-19}C$

So, time taken $t=\frac{Q}{Q^{\prime }}=\frac{2\times {10}^{-6}}{{10}^{12}\times 2\times 1.6\times {10}^{-19}}=6.25s$