Q163 / 225 · JEE Main 2003 · Physics

A particle performing uniform circular motion has angular frequency is doubled \& its kinetic energy halved, then the new angular momentum is

  • Correct answer
    A
    $\frac{\mathrm{L}}{4}$
  • B
    $2 \mathrm{~L}$
  • C
    $4 \mathrm{~L}$
  • D
    $\frac{\mathrm{L}}{2}$

Solution

Angular momentum $\propto \frac{1}{\text { Angular frequency }} \propto$ Kinetic energy $\Rightarrow \vec{L}=\frac{\mathrm{K} . \mathrm{E} .}{\mathrm{w}}$ $\frac{\mathrm{L}_1}{\mathrm{~L}_2}=\left(\frac{\mathrm{K} . \mathrm{E}_1}{\mathrm{w}_1}\right) \times \frac{\mathrm{w}_2}{\mathrm{KE}_2}=4 \Rightarrow \mathrm{L}_2=\frac{\mathrm{L}}{4}$

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

Relevant Questions