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