Average torque on a projectile of mass `m` (initial speed `u` and angle of projection `theta`) between initial and final positions `P` and `Q` as shown in figure, about the point of projection is :
.

Options

A. `1(m u^2 sin 2 theta)/(2)`
B. `(m u^2 cos theta)/(2)`
C. `m u^2 sin theta`
D. `m u^2 cos theta`

Answer

A

Explanation

(a) `vec tau. Delta t = Delta vec L`…(i)
Here `Delta t = time of flight = (2 u sin theta)/(g)`
Change in angular about point of projection (initially it is zero)
`|vec(Delta L)| = |vec L_f = vec L_i|= (m u sin theta)` Range
=`((m u sin theta)(u^2 sin 2 theta))/(g) = (m u^2 sin theta sin 2 theta)/(g)`
Now `|vec tau_(av)| = |vec(Delta L)/(Delta t)| = (m u^3 sin theta sin 2 theta)/(g) xx (g)/(2u sin theta)`
=`(m u^2 sin 2 theta)/(2)`.