The dimensional formula moment of couple is - M - 1 - L - 2 - T - -1 - M - 1 - L - 2 - T - -2 - M - -1 - L - 2 - T - -2 - M - 1 - L - 1 - T - -2 -

Question

Toppr Admin · Accepted Answer

Moment-xA0-of a-xA0-couple-xA0-is calculated by multiplying the size of one of the force -F- by the perpendicular distance between the two forces-Thus the units are that of Nm and dimensions are-xA0-M1LT-x2212-2-L-M1L2T-x2212-2-

List-I	List-II
A.Spring constant	1. $M^{1} L^{2} T^{- 2}$
B.pascal	2. $M^{0} L^{0} T^{- 1}$
C.hertz	3. $M^{1} L^{0} T^{- 2}$
D.joule	4. $M^{1} L^{- 1} T^{- 2}$

List - I	List - II
A) Spring constant	I) [ $M^{1} L^{2} T^{- 2}$ ]
B) Pascal	II) [ $M^{0} L^{0} T^{- 1}$ ]
C) Hertz	III)[ $M^{1} L^{0} T^{- 2}$ ]
D) Joule	IV) [ $M^{1} L^{- 1} T^{- 2}$ ]

The dimensional formula for moment of couple is :
$[M^{1} L^{2} T^{- 1}]$
$[M^{1} L^{2} T^{- 2}]$
$[M^{- 1} L^{2} T^{- 2}]$
$[M^{1} L^{1} T^{- 2}]$

Moment of a couple is calculated by multiplying the size of one of the force (F) by the perpendicular distance between the two forces.
Thus the units are that of $N m$ and dimensions are $[(M^{1} L T^{- 2}) (L)] = [M^{1} L^{2} T^{- 2}]$

The dimensional formula for moment of couple is :[M1L2T−1][M1L2T−2][M−1L2T−2][M1L1T−2]

Moment of a couple is calculated by multiplying the size of one of the force (F) by the perpendicular distance between the two forces.Thus the units are that of Nm and dimensions are [(M1LT−2)(L)]=[M1L2T−2]

The dimensional formula for moment of couple is :
$[M^{1} L^{2} T^{- 1}]$
$[M^{1} L^{2} T^{- 2}]$
$[M^{- 1} L^{2} T^{- 2}]$
$[M^{1} L^{1} T^{- 2}]$

Moment of a couple is calculated by multiplying the size of one of the force (F) by the perpendicular distance between the two forces.
Thus the units are that of $N m$ and dimensions are $[(M^{1} L T^{- 2}) (L)] = [M^{1} L^{2} T^{- 2}]$