JEE PHYSICS MCQ: UNITS & MEASUREMENTS

1 Medium

1 of 1

Application +4 | -1

If velocity (v), time (T) and mass (M) are taken as fundamental quantities, then dimensional formula for gravitational potential would be.

View answer and explanation

Correct option: C

\left[ {{M^0}{v^2}{T^0}} \right]

Option C (

[{M^0}{v^2}{T^0}]

) is correct because gravitational potential is energy per unit mass, which has dimensions of velocity squared when velocity and mass are fundamental.

More Questions from UNITS & MEASUREMENTS

Practice All Chapter MCQs →

Assertion :

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

has the unit

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

Reason: Acceleration has the dimensions of

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

The method of dimensional analysis can be used to derive which of the following relations? Statement-I: Dimensions of constant of proportionality of constants can be derived from dimensional method
Statement-II: Numerical value of constant of proportionality can be found from experiments only The dimensions of torque are The physical quantity having the dimensions

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

is A quantity X is given by

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

, where

{\varepsilon _0}

is the permittivity of free space, L is a length,

{\Delta V}

is a potential difference and

{\Delta t}

is a time interval. The dimensional formula for X is the same as that of