JEE PHYSICS MCQ: UNITS & MEASUREMENTS

1 Medium

1 of 1

Application +4 | -1

A calorie is a unit of heat energy and its value is

4.18 \mathrm{~J}

where

1 \mathrm{~J}=1 \mathrm{~kg}-\mathrm{m}^2 / \mathrm{s}^2

. Suppose we use a new system of units in which unit of mass equals

\alpha \,\mathrm{kg}

, the unit of length equals

\beta \,\mathrm{m}

and the unit of the time is

\gamma \,\mathrm{sec}

. Then the value of a calorie in the new system of units is:

View answer and explanation

Correct option: A

4.18 \frac{\gamma^2}{\alpha \beta^2}

Option A (

4.18 \frac{{\gamma^2}}{{\alpha \beta^2}}

) is correct because it correctly converts the units of energy (

\mathrm{{kg \cdot m^2 / s^2}}

) from the old system to the new system.

More Questions from UNITS & MEASUREMENTS

Practice All Chapter MCQs →

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

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

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

\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}