Question 614: Centre of Mass and Rotational Motion

Question

The dimensions of torque are same as that of

Accepted Answer

Torque is expressed as,
$	au=$ force $(F) 	imes$ perpendicular distance $(r)$
So,
$$\begin{aligned}
{[	au] } & =[F] 	imes[r]=\left[\mathrm{MLT}^{-2}ight][\mathrm{L}] \
& =\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]
\end{aligned}$$
Now, the dimension of moment of force $=$ force $	imes$ distance $=\left[\mathrm{MLT}^{-2}ight] 	imes\left[\mathrm{L}^{\mathbf{1}}ight]=\left[\mathrm{ML}^2 \mathbf{T}^{-2}ight]$
(b) Dimension of pressure $=$ Force $	imes[	ext { Area }]^{-2}$
$$=\left[M L T^{-2}ight] 	imes\left[\mathrm{L}^{-2}ight]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}ight]$$
(c) Dimension of acceleration $=\left[M^0 L^1 T^{-2}ight]$
(d) Dimension of impulse $=[$ Force $] 	imes[$ time $]$
$$=\left[\mathrm{MLT}^{-2}ight]\left[\mathrm{T}^1ight]=	ext { MLT }^{-1}$$
As, the torque has dimension $\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]$, which is correctly matched by the dimensions of moment of force. So, option (a) is correct.

Answer

moment of force

Answer

pressure.

Answer

acceleration

Answer

impulse

Centre of Mass and Rotational Motion

The dimensions of torque are same as that of

Choose the correct answer: