Question

A ring of mass $m$ and radius $R$ is placed on a rough surface in a uniform electric field. The friction is sufficient to avoide sliding. The charge per unit length on the ring is $\lambda$ such that half ring is positive and half is negatively charged. E = 5 V/m, m = 0.5 kg, $\lambda$ = 2.5 C/m and R = 1m. Column 1 gives a question and in column 2 there are numberical answers in S.I unit.

(A)	Maximum torque on the ring about 0 $\left(kg{m}^2/s^2\right)$	1) 50
(B)	Maximum acceleration of the ring (m/)	2) 25
(C)	Maximum friction force on the ring (N)	3) 10
(D)	Maximum speed of the ring (m/s)	4) 12.5
		5) None

• A. A$\to$ 2, B$\to$ 1, C$\to$ 2, D$\to$ 3
• B. A$\to$ 1, B$\to$ 2, C$\to$ 2, D$\to$ 3
• C. A$\to$ 2, B$\to$ 1, C$\to$ 3, D$\to$ 2
• D. A$\to$ 2, B$\to$ 1, C$\to$ 2, D$\to$ 1

Answer 1

The correct option is A A$\to$ 2, B$\to$ 1, C$\to$ 2, D$\to$ 3


Electric dipole moment of ring is $P=\int dPcos\theta =\int dq2Rcos\theta =2R^2\lambda {\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}cos\theta d\theta$
$P=4\lambda R^2$
${\tau }_{net}=PEsin\theta -fR$

$So,{\tau }_{net,max}=PE-fR$
${\tau }_{max}=4\lambda R^{2}E-fR$
and ${\tau }_{max}=I{\alpha }_{max}$
$4\lambda R^{2}E-fR=M{R}^2{\alpha }_{max}$ ---(1) and f = ma where a = R$\alpha$
So $a_{max}=\frac{\lambda RE}{m}$
and ${\alpha }_{max}=\frac{2\lambda E}{m}$
$a_{max}=\frac{2\times 2.5\times 1\times 5}{0.5}=50m/s^2$
and ${\alpha }_{max}=50rad/s^2$
$f_{max}=m{a}_{max}=0.5\times 50=25N$

By Conservation of Energy:
$PE=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2$ $\Rightarrow u\lambda R^{2}E$
$=\frac{1}{2}m{v}^2+\frac{1}{2}m{R}^2-\frac{v^2}{R^2}$
$v_{max}=10m/s$

${\tau }_{max}=I{\alpha }_{max}=M$$R^2$${\alpha }_{max}$=0.5 x 1 x 1 x 50 = 25 N m