JEE PHYSICS MCQ: UNITS & MEASUREMENTS

1 Hard

1 of 1

Application +4 | -1

Velocity

\left( v \right)

and acceleration

\left( a \right)

in two systems of units 1 and 2 are related as

{v_2} = \frac{n}{{{m^2}}}{v_1}

and

{a_2} = \frac{{{a_1}}}{{mn}}

respectively. Here

m

and

n

are constants. The relations for distance and time in two systems respectively are:

View answer and explanation

Correct option: A

\frac{{{n^3}}}{{{m^3}}}{L_1} = {L_2}\,\,{\text{and}}\,\,\frac{{{n^2}}}{m}{T_1} = {T_2}

Option A (

\frac{{{n^3}}}{{{m^3}}}{L_1} = {L_2}\,\,{\text{and}}\,\,\frac{{{n^2}}}{m}{T_1} = {T_2}

) is correct because it correctly relates the distance and time units between the two systems based on the given velocity and acceleration relationships.

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