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JEE MAIN 2021
27-07-21 S1
Question
Two identical tennis balls each having mass ' $m$ ' and charge ' $q$ ' are suspended from a fixed point by threads of length ' $I$ '. What is the equilibrium separation when each thread makes a small angle ' $\theta$ ' with the vertical ?
Select the correct option:
A
$x=\left(\frac{q^2 \mid}{2 \pi \varepsilon_0 m g}\right)^{\frac{1}{2}}$
B
$x=\left(\frac{q^2 \mid}{2 \pi \varepsilon_0 m g}\right)^{\frac{1}{3}}$
C
$x=\left(\frac{q^2 l^2}{2 \pi \varepsilon_0 m^2 g}\right)^{\frac{1}{3}}$
D
$x=\left(\frac{q^2 l^2}{2 \pi \varepsilon_0 m^2 g^2}\right)^{\frac{1}{3}}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
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