Question

In a particular system of unit, if the unit of mass becomes twice and that of time becomes half, then 8 joules will be written as ______ units of work.

• A. 16
• B. 1
• C. 4
• D. 64

Answer 1

Answer: (B)

1

Answer 2

The dimension of work is $[M^1 L^2 T^{-2}]$. Let the original units of mass, length, and time be $u_m, u_l, u_t$. So, 1 Joule $= 1 \ u_m \cdot u_l^2 \cdot u_t^{-2}$. The given work is $W = 8$ Joules, which is $8 \ u_m \cdot u_l^2 \cdot u_t^{-2}$.

Let the new units be $u_m', u_l', u_t'$. According to the problem statement:

1. The unit of mass becomes twice: $u_m' = 2 u_m$, which implies $u_m = \frac{1}{2} u_m'$.
2. The unit of time becomes half: $u_t' = \frac{1}{2} u_t$, which implies $u_t = 2 u_t'$.
3. The unit of length is assumed to remain unchanged: $u_l' = u_l$, which implies $u_l = u_l'$.

We want to express the work $W$ in the new system of units. Let the value be $N$ in the new system. So, $W = N \ u_m' \cdot u_l'^2 \cdot u_t'^{-2}$.

Equating the two expressions for work: $8 \ u_m \cdot u_l^2 \cdot u_t^{-2} = N \ u_m' \cdot u_l'^2 \cdot u_t'^{-2}$

Substitute the relations for the old units in terms of the new units: $8 \left(\frac{1}{2} u_m'\right) \cdot (u_l')^2 \cdot (2 u_t')^{-2} = N \ u_m' \cdot u_l'^2 \cdot u_t'^{-2}$

Simplify the left side: $8 \cdot \frac{1}{2} \cdot u_m' \cdot u_l'^2 \cdot \left(\frac{1}{4} u_t'^{-2}\right) = N \ u_m' \cdot u_l'^2 \cdot u_t'^{-2}$ $8 \cdot \frac{1}{8} \ u_m' \cdot u_l'^2 \cdot u_t'^{-2} = N \ u_m' \cdot u_l'^2 \cdot u_t'^{-2}$ $1 \ u_m' \cdot u_l'^2 \cdot u_t'^{-2} = N \ u_m' \cdot u_l'^2 \cdot u_t'^{-2}$

By comparing both sides, we find $N=1$. Therefore, 8 Joules will be written as 1 unit of work in the new system.