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Units & Dimensions

JEE Main level, single correct, medium

128 of 149

Question 128 of 149

Time $\left(T\right)$ , velocity $\left(C\right)$ and angular momentum $\left(h\right)$ are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be:

Show answer and solution

Answer: (A) $\left[M\right]=\left[{T}^{-1}{C}^{-2}h\right]$

M\propto {T}^{x}{C}^{y}{h}^{z}

{M}^{1}{L}^{0}{T}^{0}={\left({T}^{1}\right)}^{x}{\left({L}^{1}{T}^{-1}\right)}^{y}{\left({M}^{1}{L}^{2}{T}^{-1}\right)}^{z}

[{M}^{1}{L}^{0}{T}^{0}]=[{M}^{z}{L}^{y+2z}{T}^{x-y-z}]

]

⟹z=1

y+2z=0,x-y-z=0

solving

y=-2

and

x=-1

M\Rightarrow \left[{T}^{-1}{C}^{-2}{h}^{1}\right]

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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More Physics practice

JEE Main Physics practice question on Units & Dimensions (Dimensions), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

Time (T)\left(T\right)(T), velocity (C)\left(C\right)(C) and angular momentum (h)\left(h\right)(h) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be:

JEE Main question paper PDF

More Physics practice

Time $\left(T\right)$ , velocity $\left(C\right)$ and angular momentum $\left(h\right)$ are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be: