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(B) The average torque $	au_{avg}$ is defined as the rate of change of angular momentum: $	au_{avg} = \frac{\Delta L}{\Delta t}$.At the point of pro

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$50$

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(B) The average torque $	au_{avg}$ is defined as the rate of change of angular momentum: $	au_{avg} = \frac{\Delta L}{\Delta t}$.At the point of projection, the angular momentum $L_i = 0$.At the point of impact, the projectile is at a horizontal distance $R$ (range) from the point of projection. The vertical velocity component is $v_y = -u \sin 	heta$ and the horizontal component is $v_x = u \cos 	heta$.The angular momentum at the point of impact is $L_f = m 	imes v_y 	imes R = m(-u \sin 	heta) 	imes \frac{u^2 \sin 2	heta}{g} = -\frac{m u^3 \sin 	heta \sin 2	heta}{g}$.Alternatively, using the torque due to gravity: $	au = \vec{r} 	imes \vec{F} = \vec{r} 	imes m\vec{g}$.The average torque is $	au_{avg} = \frac{1}{T} \int_0^T 	au dt = \frac{1}{T} \int_0^T (mg \cdot x) dt = \frac{mg}{T} \int_0^T (u \cos 	heta) t dt = \frac{mg u \cos 	heta}{T} [\frac{t^2}{2}]_0^T = \frac{mg u \cos 	heta T}{2}$.Since $T = \frac{2u \sin 	heta}{g}$, we have $	au_{avg} = \frac{mg u \cos 	heta}{2} 	imes \frac{2u \sin 	heta}{g} = m u^2 \sin 	heta \cos 	heta = \frac{1}{2} m u^2 \sin 2	heta$.Substituting the values: $m = 1 \ kg$, $u = 10 \ ms^{-1}$, $	heta = 45^{\circ}$.$	au_{avg} = \frac{1}{2} 	imes 1 	imes (10)^2 	imes \sin(90^{\circ}) = \frac{1}{2} 	imes 100 	imes 1 = 50 \ Nm$.

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