A) 1.5 m/s
B) 2.5 m/s
C) 3.2 m/s
D) 4.7 m/s
Correct Answer: A
Solution :
As shown in figure \[T\sin \theta =\frac{m{{v}^{2}}}{r}\] \[T\cos \theta =mg\] So, \[\tan \theta =\frac{{{v}^{2}}}{rg}=\frac{r}{\sqrt{{{l}^{2}}-{{r}^{2}}}}\] \[\therefore \] \[v={{\left[ \frac{{{r}^{2}}g}{{{({{l}^{2}}-{{r}^{2}})}^{1/2}}} \right]}^{1/2}}\] \[={{\left[ \frac{0.09\times 10}{{{(0.25-0.09)}^{1/2}}} \right]}^{1/2}}=1.5\,m/s\]You need to login to perform this action.
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