| A small mass attached to a string rotates on frictionless table top as shown. If the tension is the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic energy of the mass will [AIPMT (M) 2011] |
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A) remain constant
B) increase by a factor of 2
C) increase by a factor of 4
D) decrease by a factor of 2
Correct Answer: C
Solution :
| From the law of conservation of angular momentum |
| \[mvr=mv'\frac{r}{2}\] |
| \[v'=2v\] |
| So, \[\frac{{{k}_{1}}}{{{k}_{2}}}=\frac{\frac{1}{2}m{{v}^{2}}}{\frac{1}{2}m{{v}^{'2}}}\] |
| \[\frac{{{k}_{1}}}{{{k}_{2}}}=\frac{{{v}^{2}}}{v{{'}^{2}}}\] |
| \[=\frac{{{v}^{2}}}{{{(2v)}^{2}}}\] |
| \[\frac{{{k}_{1}}}{{{k}_{2}}}=\frac{1}{4}\] |
| \[{{k}_{1}}=\frac{1}{4}{{k}_{2}}\] |
| \[{{k}_{2}}=4{{k}_{1}}\] |
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