Question

In the figure shown, there is a smooth tube of radius 'R', fixed in the vertical plane. A ball 'B' of mass 'm' is released from the top of the tube. B slides down due to gravity and compresses the spring. The end 'C' of the spring is fixed and the end A is free. Initially the line OA makes an angle of ${60}^{\circ }$ with OC and finally it makes an angle of ${30}^{\circ }$ after compression. Find the spring constant of the spring.

• A. $\frac{12mg(2+\sqrt{3})}{{\pi }^{2}R}$
• B. $\frac{36mg(2+\sqrt{3})}{{\pi }^{2}R}$
• C. $\frac{36mg(2+\sqrt{3})}{{\pi }^{2}R}$
• D. None of these

Answer 1

The correct option is B

$\frac{36mg(2+\sqrt{3})}{{\pi }^{2}R}$

Applying conservation of energy

$K{E}_i+U_i=K{E}_f+U_f$

$U_i=mg2R,K{e}_i=0$

$U_f=mg(R-Rcos{30}^{\circ })+\frac{1}{2}k{x}^2,K.E_f=0$

$X=△\theta \times R$

= $\frac{\pi }{6}\times R$

$U_f=mg(R-Rcos{30}^{\circ })\times \frac{1}{2}K(\frac{R\pi }{6}{)}^2$

$\therefore mg2R=mg(R-Rcos{30}^{\circ })+\frac{1}{2}(\frac{R\pi }{6}{)}^2$

$\Rightarrow 2mgR=mgR(1-\frac{\sqrt{3}}{2})+\frac{1}{2}\frac{k{R}^2{\pi }^2}{36}$

$\Rightarrow K=\frac{36mg(2+\sqrt{3})}{{\pi }^{2}R}$