question_answer

A bead is free to slide down a smooth wire tightly stretched between two points over a vertical wire loop (as shown below).

Now, choose the correct option.

A) Time to slide down along \[{{P}_{1}}{{P}_{4}}\]is maximum

B) Time to slide down along \[{{P}_{1}}{{P}_{3}}\]is minimum

C) Time to slide down along\[{{P}_{1}}{{P}_{2}}\] is more than time to slide along \[{{P}_{1}}{{P}_{3}}\]

D) All are incorrect

Correct Answer: D

Solution :

Acceleration of the bead down the wire is \[g\,\cos \,\,\theta \]

Also, length \[{{p}_{1}}{{p}_{2}}\] is \[2R\,\cos \,\theta \]

So,       \[{{\upsilon }^{2}}={{u}^{2}}+2as\]

Gives,   \[{{\upsilon }^{2}}=2g\,\cos \,\theta \times 2R\cos \theta \]\[=2\sqrt{gR\,\,\cos \theta }\]

Time, \[t\,=\frac{\upsilon }{a}=\frac{2\sqrt{gR}\,\cos \theta }{g\,\cos \theta }=2\sqrt{\frac{R}{g}}\]

which is same regardless of where \[{{P}_{2}}\] is located.

Time of free fall,

\[2R=\frac{1}{2}g{{t}^{2}}\Rightarrow {{t}^{2}}=\frac{4R}{g}\Rightarrow t=2\sqrt{\frac{R}{g}}\]