Back to UNITS & MEASUREMENTS

JEE PHYSICS MCQ: UNITS & MEASUREMENTS

1 Medium

1 of 1

Understanding +4 | -1

Statement-I: Dimensional analysis can give us the numerical value of proportionality constants that may appear in an algebraic expression.
Statement-II: Dimensional analysis make use of the fact that dimensions can be treated as algebraic quantities.

View answer and explanation

Correct option: C

Statement-I is false and statement-II is true

Option C (Statement-I is false and statement-II is true) is correct because dimensional analysis cannot determine the numerical values of dimensionless constants, but it does rely on the fact that dimensions can be treated algebraically.

More Questions from UNITS & MEASUREMENTS

Practice All Chapter MCQs →

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

\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}},\;

{\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