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Standard XII

Physics

Average and Instaneous Power

For a project...

Question

For a projectile of mass $m$ projected with velocity $v_{0}$ at an angle $θ_{0}$ above horizontal as shown in the diagram, magnitude of average torque of its weight about origin over its time of flight would be:

\frac{m v_{0}^{2} sin 2 θ_{0}}{2 g}

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m v_{0}^{2} sin 2 θ_{0}

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m g

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None of these

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Solution

The correct option is B $m v_{0}^{2} sin 2 θ_{0}$
Average torque= Force $\times$ RangeSin $θ$ ------- (1)
Here, the force $F = m g$
At maximum height, the range is half. That is $R / 2$
Substituting the range $R / 2$ in (1),
=> $τ = \frac{m g R}{2}$ ------ (2)
where, $R = \frac{v^{2} S i n 2 θ}{g}$ ------- (3)

From (2) and (3),
$τ = \frac{m {g v}^{2} S i n 2 θ}{g}$

$τ = m v^{2} S i n 2 θ$

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For a projectile of mass m projected with velocity v0 at an angle θ0 above horizontal as shown in the diagram, magnitude of average torque of its weight about origin over its time of flight would be:

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For a projectile of mass $m$ projected with velocity $v_{0}$ at an angle $θ_{0}$ above horizontal as shown in the diagram, magnitude of average torque of its weight about origin over its time of flight would be: