Search any question & find its solution
Question:
Answered & Verified by Expert
A metal sphere of radius $1 \mathrm{~m}$ is charged with $10^{-2} \mathrm{C}$ in air. Its bulk modulus is
$10^{11} / 4 \pi^{2}$. The volume strain in the sphere is $\left(\epsilon_{0}=\right.$ pemittivity of free space)
Options:
$10^{11} / 4 \pi^{2}$. The volume strain in the sphere is $\left(\epsilon_{0}=\right.$ pemittivity of free space)
Solution:
2647 Upvotes
Verified Answer
The correct answer is:
$\frac{10^{-15}}{8 \epsilon_{0}}$
surface charge density
$\sigma=\frac{q}{4 \pi r^{2}}=\frac{10^{-2}}{4 \pi \times 1}=\frac{10^{-2}}{4 \pi} \mathrm{C} / \mathrm{m}^{2}$
strain $=$
$\frac{F}{A}=\frac{1}{2} \frac{\sigma^{2}}{\varepsilon_{2}}=\frac{1}{2 \varepsilon_{0}} \times\left(\frac{10^{-2}}{4 \pi}\right)^{2}=\frac{10^{-4}}{32 \pi^{2} \varepsilon_{0}}$
strain $=\frac{\text { stress }}{k}=\frac{1}{2} \frac{\sigma^{2}}{\varepsilon_{0} k}$
where $k=\frac{10^{11}}{4 \pi^{2}}$
substituting and calculating we get strain =
$\frac{10^{-15}}{8 \varepsilon_{0}}$
$\sigma=\frac{q}{4 \pi r^{2}}=\frac{10^{-2}}{4 \pi \times 1}=\frac{10^{-2}}{4 \pi} \mathrm{C} / \mathrm{m}^{2}$
strain $=$
$\frac{F}{A}=\frac{1}{2} \frac{\sigma^{2}}{\varepsilon_{2}}=\frac{1}{2 \varepsilon_{0}} \times\left(\frac{10^{-2}}{4 \pi}\right)^{2}=\frac{10^{-4}}{32 \pi^{2} \varepsilon_{0}}$
strain $=\frac{\text { stress }}{k}=\frac{1}{2} \frac{\sigma^{2}}{\varepsilon_{0} k}$
where $k=\frac{10^{11}}{4 \pi^{2}}$
substituting and calculating we get strain =
$\frac{10^{-15}}{8 \varepsilon_{0}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.