Q78 / 120 · NEET 2011 (Mains) · Physics

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 Figure

  • A
    remain constant
  • B
    increase by a factor of 2
  • Correct answer
    C
    increase by a factor of 4
  • D
    decrease by a factor of 2

Solution

From the law of conservation of angular momentum So, $$ \begin{aligned} m v r & =m v^{\prime} \frac{r}{2} \\ y^{\prime} & =2 y \\ \frac{k_1}{k_2} & =\frac{\frac{1}{2} m v^2}{\frac{1}{2} m v^2} \\ \frac{k_1}{k_2} & =\frac{y^2}{v^{\prime 2}} \\ & =\frac{y^2}{(2 v)^2} \\ \frac{k_1}{k_2} & =\frac{1}{4} \\ k_1 & =\frac{1}{4} k_2 \\ k_2 & =4 k_1 \end{aligned} $$

Hence, the correct answer is option: 'C'.

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