JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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Analyzing +4 | -1

The velocity of a particle as a function of time is given as

v(t)=a t^2+b t+\frac{c}{t^2+d}

, where

a, b, c

and

d

are constants, The dimensions of

\left[\frac{a b}{c d}\right]

View answer and explanation

Correct option: B

\left[L T^{-8}\right]

Option A is correct because by equating dimensions of each term in the given equation, we can find the dimensions of a, b, c, and d, and then calculate the dimensions of ab/cd.

More Questions from UNITS & MEASUREMENTS

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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