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(D) Linear density $m$ has dimensions of $\frac{	ext{Mass}}{	ext{Length}} = [ML^{-1}]$.Force $A$ has dimensions $[MLT^{-2}]$.Given the relation $A/B

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Latent heat

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(D) Linear density $m$ has dimensions of $\frac{	ext{Mass}}{	ext{Length}} = [ML^{-1}]$.Force $A$ has dimensions $[MLT^{-2}]$.Given the relation $A/B = m$,we can write $B = A/m$.Substituting the dimensions: $B = \frac{[MLT^{-2}]}{[ML^{-1}]} = [L^2T^{-2}]$.Latent heat is defined as energy per unit mass: $\frac{[ML^2T^{-2}]}{[M]} = [L^2T^{-2}]$.Thus,the dimensions of $B$ are the same as those of latent heat.

Column-$I$	Column-$II$
$(1)$ Torque	$(a)$ $M^1L^1T^{-1}$
$(2)$ Angular momentum	$(b)$ $M^1L^2T^{-1}$
$(3)$ Linear momentum	$(c)$ $M^1L^2T^{-2}$

Two quantities $A$ and $B$ are related by the equation $A/B = m$,where $m$ is linear density and $A$ is force. The dimensions of $B$ are equivalent to which of the following?

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Column-$I$ Column-$II$

$(1)$ Torque $(a)$ $M^1L^1T^{-1}$

$(2)$ Angular momentum $(b)$ $M^1L^2T^{-1}$

$(3)$ Linear momentum $(c)$ $M^1L^2T^{-2}$

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