JEE PHYSICS MCQ: UNITS & MEASUREMENTS

1 Easy

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Application +4 | -1

The force F on the sphere of radius ‘a’ moving in a medium with velocity '

v

' is given by

F = 6\pi \eta a\upsilon

. The dimensions of η are

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it correctly derives the dimensions of viscosity (

\eta

) from the given formula using dimensional analysis.

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\sqrt{\frac{\text { Modulus of elasticity }}{\text { Density }}}

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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}