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Q.
The dimensions of torque are same as that of
MHT CETMHT CET 2019Physical World, Units and Measurements
Solution:
Torque is expressed as,
$\tau=$ force $(F) \times$ perpendicular distance $(r)$
So, $[\tau] =[F] \times[r]=\left[ MLT ^{-2}\right][ L]$
$=\left[ ML ^{2} T ^{-2}\right]$
Now, the dimension of moment of force $+$ force $\times$ distance
$= \left[M L T^{-2}\right] \times\left[L^{1}\right]=\left[M L^{2} T^{-2}\right].$
(b) Dimension of pressure $=$ Force $\times[\text { Area }]^{-2}$
$=\left[ MLT ^{-2}\right] \times\left[ L ^{-2}\right]=\left[ ML ^{-1} T ^{-2}\right]$
(c) Dimension of acceleration $=\left[ M ^{0} L ^{1} T ^{-2}\right]$
(d) Dimension of impulse $=$ [Force] $x$ [time]
$=\left[M L T^{-2}\right]\left[T^{1}\right]=M L T^{-1}$
As, the torque has dimension $\left[ ML ^{2} T ^{-2}\right],$
which is correctly matched by the dimensions of moment of force.
So, option (a) is correct.