Question

The energy of a $100\mu \text{F}$ capacitor charged to $6\text{kV}$ is used to lift a $50\text{kg}$ mass (by powering a motor by energy of capacitor) without incurring any losses. What would be the greatest vertical height through which mass could be raised?
(Take $g=10{\text{m/s}}^2$)

• A. $1.6\text{m}$
• B. $1.2\text{mm}$
• C. $3.6\text{m}$
• D. $0.6\text{mm}$

Answer 1

The correct option is C $3.6\text{m}$

Given:
$C=100\mu \text{F};m=50\text{kg};V=6\text{kV}$

According to question $100\%$ energy of capacitor has been used to lift the mass.

So, applying energy conservation

$\Rightarrow$ Loss of Electrostatic Energy stored in capacitor = Gain in gravitational potential energy of mass

$\Rightarrow \frac{1}{2}C{V}^2=mgh$

$\Rightarrow \frac{1}{2}\times (100\times {10}^{-6})\times (6\times {10}^3{)}^2=50\times 10\times h$

$\Rightarrow 50\times {10}^{-6}\times 36\times {10}^6=500h$

$\Rightarrow h=\frac{36\times 50}{500}=3.6\text{m}$

Hence, option $\left(b\right)$ is correct.