NEET 2025 Physics Circular Motion MCQ Question

To solve the problem of finding the ratio of the speed $v$ of the bob at point $P$ to its initial speed $v_0$, we can use two key principles: the condition for the string to go slack and the conservation of mechanical energy.

Step 1: Condition for the String to Go Slack

At point $P$, the string becomes slack. This means that the tension $T$ in the string is zero. The only force providing the necessary centripetal force is the component of the bob's weight acting toward the center of the circular path.

1. The weight of the bob is given by $mg$, where $m$ is the mass of the bob and $g$ is the acceleration due to gravity.

2. The component of weight acting toward the center at angle $\theta$ is $mg \sin \theta$.

3. The centripetal force required to keep the bob moving in a circular path is provided by this component:

   $\frac{mv^2}{l} = mg \sin \theta$

4. Rearranging gives:

   $v^2 = gl \sin \theta$

Step 2: Conservation of Mechanical Energy

Next, we apply the principle of conservation of mechanical energy between the initial position (bottom of the swing) and point $P$.

1. Initial Energy (at the bottom): The potential energy at the bottom can be considered zero, and the kinetic energy is:

   $E_i = \frac{1}{2} mv_0^2$

2. Energy at Point $P$: At point $P$, the bob has kinetic energy and potential energy. The potential energy is determined by the height $h$ of the bob above the reference point (the bottom):

   • The height $h$ at point $P$ is given by:

     $h = l(1 + \sin \theta)$

   • Therefore, the potential energy at point $P$ is:

     $U_P = mgh = mg l(1 + \sin \theta)$

   • The kinetic energy at point $P$ is:

     $K_P = \frac{1}{2} mv^2$

   Thus, the total energy at point $P$ is:

   $E_f = \frac{1}{2} mv^2 + mg l (1 + \sin \theta)$

3. Setting Initial Energy Equal to Final Energy:

   From conservation of energy, we have:

   $E_i = E_f$

   So,

   $\frac{1}{2} mv_0^2 = \frac{1}{2} mv^2 + mg l (1 + \sin \theta)$

   Dividing through by $m$ and multiplying by 2 gives:

   $v_0^2 = v^2 + 2g l (1 + \sin \theta)$

Step 3: Finding the Ratio

Now we substitute $gl = \frac{v^2}{\sin \theta}$ (from Step 1) into the energy equation:

1. Substituting gives:

   $v_0^2 = v^2 + 2 \left(\frac{v^2}{\sin \theta}\right)(1 + \sin \theta)$

   Simplifying this:

   $v_0^2 = v^2 + \frac{2v^2(1 + \sin \theta)}{\sin \theta}$

   $v_0^2 = v^2 \left( 1 + \frac{2 + 2 \sin \theta}{\sin \theta} \right)$

   $v_0^2 = v^2 \left( \frac{3 \sin \theta + 2}{\sin \theta} \right)$

2. Now, to find the ratio $\frac{v^2}{v_0^2}$:

   $\frac{v^2}{v_0^2} = \frac{\sin \theta}{2 + 3 \sin \theta}$

3. Taking the square root gives:

   $\frac{v}{v_0} = \left( \frac{\sin \theta}{2 + 3 \sin \theta} \right)^{1/2}$

Conclusion

The correct answer is:

Option D: (sin θ / (2 + 3 sin θ))¹/²