Q57 / 75 · JEE Main 2026 (23 January Shift 1) · Physics
A small bob $A$ of mass $m$ is attached to a massless rigid rod of length 1 m pivoted at point $P$ and kept at an angle of $60^{\circ}$ with vertical as shown in figure. At distance of 1 m below point $P$, an identical bob $B$ is kept at rest on a smooth horizontal surface that extends to a circular track of radius $R$ as shown in figure. If bob $B$ just manages to complete the circular path of radius $R$ upto a point $Q$ after being hit elastically by $\operatorname{bob} A$, then radius $R$ is $\_\_\_\_$ m.
- Correct answerA$\frac{1}{5}$
- B$\frac{2-\sqrt{3}}{5}$
- C$\frac{3}{5}$
- D$\frac{2+\sqrt{3}}{5}$
Solution
Bob A swings from $60°$ to vertical. Height drop $h = l(1-\cos 60°) = 1 \times 0.5 = 0.5$ m.
Velocity of A at bottom: $v_A = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.5} = \sqrt{10}$ m/s.
Elastic collision between equal masses: B gets $v_B = \sqrt{10}$ m/s, A stops.
For B to just complete circular loop: $v_B^2 = 5gR$.
$10 = 5 \times 10 \times R \Rightarrow R = \frac{1}{5}$ m.
Velocity of A at bottom: $v_A = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.5} = \sqrt{10}$ m/s.
Elastic collision between equal masses: B gets $v_B = \sqrt{10}$ m/s, A stops.
For B to just complete circular loop: $v_B^2 = 5gR$.
$10 = 5 \times 10 \times R \Rightarrow R = \frac{1}{5}$ m.
Hence, the correct answer is option: 'A'.