JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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Application +4 | -1

The surface tension is

T = \frac{F}{l},

then the dimensions of surface tension are

View answer and explanation

Correct option: B

\left[ {\mathrm{M}{\mathrm{T}^{ - 2}}} \right]

Option B (

{\mathrm{M}{\mathrm{T}^{ - 2}}}

) is correct because surface tension is force per unit length, and its dimensions are mass times time to the power of -2.

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\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}