A thin rod of mass ‘m’ and length 2L is made to rotate about a normal axis through centre of rod. If its angular velocity changes from ‘O’ to $‘ ω ’$ in time ‘t’. The torque acting on it is

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${mL}^{2} ω / 12 t$

${mL}^{2} ω / 3 t$

${mL}^{2} ω / t$

$4 {mL}^{2} ω / 3 t$

answer is B.

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Detailed Solution

$τ = \frac{dL}{dt} = = \frac{Iω}{t}; I = \frac{m {(2 L)}^{2}}{12} τ = \frac{m 4 L^{2} ω}{12 t}; τ = \frac{{mL}^{2} ω}{3 t}$

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