Question

In a system of units in which the unit of mass is $a$ kg, unit of length is $b$ metre and the unit of time is $c$ second, the magnitude of a calorie is :
A. $\dfrac{4.2c}{a{{b}^{2}}}$
B. $\dfrac{4.2c}{ab}$
C. $\dfrac{abc}{4.2}$
D. $\dfrac{4.2}{abc}$

Answer 1

Hint: Kilogram (kg), metre (m) and second (s) are the units of the MKS system for mass, length and time. The SI unit of energy is joules (J) and $1J=1kg{{m}^{2}}{{s}^{-2}}$. Then use the given relation between the unit of MKS system and the units of the new system to find the value of one calorie in the units of the new system.

Complete step by step answer:
Let the first the value of 1 calorie in terms of the system in the units for mass, length and time are kg (kilogram), m (metre) and s (second). This system of units is called the MKS system or SI system.Calorie (cal) is a unit for energy. In the SI system the unit for energy is joules (J). The relation between joules and calories is given as $1cal=4.2J$ …. (i)
The unit of energy in the MKS system is $kg{{m}^{2}}{{s}^{-2}}$.
This means that $1J=1kg{{m}^{2}}{{s}^{-2}}$.
Substitute the value of 1 J in (i).
$1cal=4.2kg{{m}^{2}}{{s}^{-2}}$ …. (ii).
It is given that in another system of units, the unit mass is a kg, the unit of length is b metre and the unit of time is c second.
In that system , let the unit of mass, length and time be x, y and z, respectively.
This means that $x='a'kg$, $y='b'm$ and $z='c's$.
Then,
$1kg=\dfrac{1}{a}x$, $1m=\dfrac{1}{b}y$ and $1s=\dfrac{1}{c}z$.
Substitute these values in (ii).
$1cal=4.2\left( \dfrac{1}{a}x \right){{\left( \dfrac{1}{b}y \right)}^{2}}{{\left( \dfrac{1}{c}z \right)}^{-2}}$
$\Rightarrow 1cal=\left( \dfrac{4.2{{c}^{-2}}}{a{{b}^{2}}} \right)x{{y}^{2}}{{z}^{-2}}$.
Therefore, the magnitude of a calorie in terms of the new units x,y and z is equal to $\dfrac{4.2{{c}^{-2}}}{a{{b}^{2}}}$.

Hence, the correct option is A.

Note: x,y and z are units of mass, length and time just like kg, m and s. Therefore, x,y,z are numbers. Energy is equivalent to work done on a body. Work done is equal to the product of applied force and the displacement. Therefore, if you do not know the units of energy in the MKS system, then you can find the units of energy by the formula for work done.