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Physics Question on Units and measurement

A calorie is a unit of heat energy and its value is 4.18 J4.18\, J where 1 J=1 kg m2 s−21 \,J = 1 \,kg\, m^2\, s^{-2}. Suppose we use a new system of units in which unit of mass equals α kg\alpha\, kg, the unit of length equals β m\beta\, m and the unit of the time is γ sec\gamma\, sec. Then the value of a calorie in the new system of units is

A

4.18γ2αβ24.18 \frac{\gamma^{2}}{\alpha\beta^{2}}

B

4.18αβ2γ24.18 \frac{\alpha\beta^{2}}{\gamma^{2}}

C

4.18γ2α4.18 \frac{\gamma^{2}}{\alpha}

D

4.18β2αγ24.18 \frac{\beta^{2}}{\alpha\gamma^{2}}

Answer

4.18γ2αβ24.18 \frac{\gamma^{2}}{\alpha\beta^{2}}

Explanation

Solution

1 J=(1 kg)(1 m)2(1 sec)−21\,J=\left(1\,kg\right)\left(1\,m\right)^{2}\left(1\,sec\right)^{-2} 1x=(α kg)(β m)2(γ sec)−21x=\left(\alpha\,kg\right)\left(\beta\,m\right)^{2}\left(\gamma\,sec\right)^{-2} ∴1 J1x=(1α)(1β)2(γ)2=γ2αβ2\therefore \frac{1\,J}{1x}=\left(\frac{1}{\alpha}\right)\left(\frac{1}{\beta}\right)^{2}\left(\gamma\right)^{2}=\frac{\gamma^{2}}{\alpha\beta^{2}} ∴1 J=γ2αβ2x\therefore 1\,J=\frac{\gamma^{2}}{\alpha\beta^{2}}x or 1 cal=4.18γ2αβ21\,cal=4.18 \frac{\gamma^{2}}{\alpha\beta^{2}}