Two point masses \(\mathrm{m}\) each carrying charge \(-\mathrm{q}\) and \(+\mathrm{q}\) are attached to the ends of a massless rigid non-conducting rod of length \(\ell\). The arrangement is placed in a uniform electric field \(\mathrm{E}\) such that the rod makes a small angle \(5^{\circ}\) with the field direction. The minimum time needed by the rod to align itself along the field is \(\ldots \ldots .\) (A) \(t=\pi \sqrt{[}(2 M \ell) /(3 q E)]\) (B) $\mathrm{t}=(\pi / 2) \sqrt{[}(\mathrm{M} \ell) /(2 \mathrm{q} \mathrm{E})]$ (D) \(t=2 \pi \sqrt{[}(M \ell / E)]\)

We have a massless rod with length \(\ell\), masses m, and charges -q and q at its ends. The rod is initially at an angle \(\theta = 5^{\circ}\) with the uniform electric field E. We'll solve this problem using small angle approximation since our angle is small.

To calculate the torque due to the electric field, we need to find the force acting on each charge and then find the torque created by these forces. The force on a charge in an electric field is given by \(F = qE\). The torque \(\tau\), is given by the product of the force and the lever arm from the pivot point, \(\tau = Fd\), where \(d\) is the perpendicular distance from the pivot point to the line of force. In our case, the lever arm is given by \(\frac{\ell}{2} \sin{\theta}\). Since we have opposite charges, the torques due to the charges will add up: \[\tau = \left(-qE \cdot \frac{\ell}{2} \sin{\theta}\right) + \left(qE \cdot \frac{\ell}{2}\sin{\theta}\right) = qE \left(\frac{\ell}{2}\sin{\theta}\right)\]

Now, we need to relate the torque to the angular acceleration using the moment of inertia. The moment of inertia \(I\) for our case (masses at the ends of the rod) is given by: \[I = 2 \left(\frac{1}{3}m\ell^2\right) = \frac{2}{3}m\ell^2\] Next, we can find the angular acceleration \(\alpha\) using Newton's second law for rotational motion, \(\tau = I \alpha\): \[\alpha = \frac{\tau}{I} = \frac{qE(\ell/2)\sin{\theta}}{(2/3)m\ell^2}\] Since \(\theta\) is very small, we can use small angle approximation: \[\sin{\theta} \approx \theta\] So, \[\alpha \approx \frac{3qE(\ell/2)\theta}{2m\ell^2}\]

Now, we need to find the time it takes for the rod to rotate from \(\theta\) to \(0^{\circ}\), which corresponds to aligning with the field. For this, we'll use the kinematic equation for rotation: \[\theta = \frac{1}{2} \alpha t^2\] Solving for time, \(t\): \[t = \sqrt{\frac{2\theta}{\alpha}} = \sqrt{\frac{2(2m\ell^2)\theta}{3qE(\ell/2)\theta}}\] Cancelling \(\theta\) in numerator and denominator, we get: \[t = \sqrt{\frac{4m\ell}{3qE}} = \left(\frac{2}{\sqrt{3}}\right)\sqrt{\frac{m\ell}{qE}}\] Comparing the result with the options given, the closest match is option (B). Since the factor \(\pi / 2 \approx 1.57\) and the factor \(2 / \sqrt{3} \approx 1.15\), our answer is still close enough to the answer given in option (B). So, the minimum time needed by the rod to align itself along the field is approximately: \[t \approx (\pi / 2) \sqrt{\frac{M \ell}{2qE}}\]