The dimensional formula for angular momentum | vedclass

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Question

(B) Angular momentum $(L)$ is defined as the product of linear momentum $(p)$ and the perpendicular distance $(r)$ from the axis of rotation.Mathemati

Vedclass · Accepted Answer

$M{L^2}{T^{ - 1}}$

Vedclass · Answer

(B) Angular momentum $(L)$ is defined as the product of linear momentum $(p)$ and the perpendicular distance $(r)$ from the axis of rotation.Mathematically,$L = p 	imes r$.Since linear momentum $p = m 	imes v$,where $m$ is mass and $v$ is velocity.The dimensional formula for mass $m$ is $[M]$.The dimensional formula for velocity $v$ is $[L{T^{ - 1}}]$.The dimensional formula for distance $r$ is $[L]$.Therefore,the dimensional formula for angular momentum is $[M] 	imes [L{T^{ - 1}}] 	imes [L] = [M{L^2}{T^{ - 1}}]$.

Column-$I$	Column-$II$
$(a)$ Viscous force	$(i)$ $[M^1 L^1 T^{-2}]$
$(b)$ Coefficient of viscosity	$(ii)$ $[M^1 L^{-1} T^{-1}]$
	$(iii)$ $[M^1 L^{-1} T^{-2}]$

The dimensional formula for angular momentum is

Similar Questions

If the dimensions of a physical quantity are given by $[L^{a} M^{b} T^{c}]$,then the physical quantity is:

The dimensional formula $M{L^2}{T^{ - 3}}$ represents:

Different physical quantities are given in Column-$I$ and their dimensional formulas are given in Column-$II$. Match them appropriately.

Column-$I$ Column-$II$

$(a)$ Viscous force $(i)$ $[M^1 L^1 T^{-2}]$

$(b)$ Coefficient of viscosity $(ii)$ $[M^1 L^{-1} T^{-1}]$

$(iii)$ $[M^1 L^{-1} T^{-2}]$

$A$ function $f(\theta)$ is defined as $f(\theta) = 1 - \theta + \frac{\theta^2}{2!} - \frac{\theta^3}{3!} + \frac{\theta^4}{4!} - \dots$. Why is it necessary for $f(\theta)$ to be a dimensionless quantity?

The quantity having dimension $-1$ in the length is:

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