JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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Analyzing +4 | -1

If the velocity

(V)

, acceleration

(A)

, and force

(F)

are taken as fundamental quantities instead of mass

(M)

, length

(L)

, and time

(T)

, the dimensions of Young's modulus

(Y)

would be

View answer and explanation

Correct option: A

F A^2 V^{-4}

Option A (

F A^2 V^{-4}

) is correct because the dimensions of Young's modulus can be expressed as

F A^2 V^{-4}

when velocity, acceleration, and force are fundamental quantities.

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\mathrm{ms}^{-1} \text {. }

\frac{1}{\left(\sqrt{\varepsilon_0 \mu_0}\right) t} \text {. }

\left[ {{\mathrm{M}^{ - 1}}{\mathrm{L}^{ - 3}}{\mathrm{A}^2}{T^3}} \right]

{\varepsilon _0}L\frac{{\Delta V}}{{\Delta t}},\;

{\varepsilon _0}

{\Delta V}

{\Delta t}