Q145 / 160 · TS EAMCET 2018 (05 May Shift 2) · Physics
A thin spherical shell encloses a concentric solid sphere. The radius of the shell is $(0.060)^{\frac{1}{2}} \mathrm{~m}$ and its surface charge density is $-10^{-5} \mathrm{C} / \mathrm{m}^2$.The radius of the solid sphere is $(0.01)^{\frac{1}{3}} \mathrm{~m}$ and its volumetric charge density is $3 \times 10^{-5} \mathrm{C} / \mathrm{m}^3 . \varepsilon_0$ is the permittivity of free space in $\mathrm{C}^2 / \mathrm{Nm}^2$. The electric flux through a spherical surface concentric with the spherical shell and of radius greater than that of the shell in V-m is
- A$\frac{0.4 \pi \times 10^{-3}}{\varepsilon_0}$
- B$\frac{0.8 \pi \times 10^{-3}}{\varepsilon_0}$
- C$\frac{1.2 \pi \times 10^{-3}}{\varepsilon_0}$
- Correct answerD$\frac{1.6 \pi \times 10^{-3}}{\varepsilon_0}$
Solution
Electric flux through spherical surface
$$
\begin{aligned}
& =\frac{q_{\text {net enclosed }}}{\varepsilon_0} \\
& =\frac{\text { Charge of sheII }+ \text { Charge of sphere }}{\varepsilon_0} \\
& =\frac{1}{\varepsilon_0}\left[\begin{array}{l}
4 \pi\left(0.06^{\frac{1}{2}}\right)^2 \times(-10)^{-6} \\
+\frac{4}{3} \pi\left(0.01^{\frac{1}{3}}\right)^3 \times 3 \times 10^{-5}
\end{array}\right]
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{\pi}{\varepsilon_0}\left\{-0.24 \times 10^{-6}+0.39 \times 10^{-6}\right\} \\
& =\frac{1.6 \pi \times 10^{-7}}{\varepsilon_0} \mathrm{~V}-\mathrm{m}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.