Question

If the unit of mass is $1\text{ kg}$, the unit of length is $1\text{ m}$ and the unit of time is $1\text{ minute}$, the unit of pressure in $\text{Nm}^{-2}$ is

• A. $1/60$
• B. $60$
• C. $1/3600$
• D. $3600$

Hint:

When converting units using dimensions, always remember that $n_1u_1 = n_2u_2$. If the unit of time is in the denominator (like $T^{-2}$), increasing the size of the time unit (from seconds to minutes) will decrease the numerical value of the physical quantity.

Answer 1

The Correct Option is C

1. Dimensional Formula of Pressure: Pressure ($P$) = $\frac{\text{Force}}{\text{Area}} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.

2. Unit Conversion Relationship: Let the value in the SI system be $n_1u_1$ and the value in the new system be $n_2u_2$. $$n_1 [M_1L_1^{-1}T_1^{-2}] = n_2 [M_2L_2^{-1}T_2^{-2}]$$ We want to find the value of $1$ unit of pressure in the new system relative to the SI system ($\text{Nm}^{-2}$).

3. Substituting Given Units: In the new system: $M_2 = 1\text{ kg}$, $L_2 = 1\text{ m}$, $T_2 = 1\text{ minute} = 60\text{ s}$. In the SI system: $M_1 = 1\text{ kg}$, $L_1 = 1\text{ m}$, $T_1 = 1\text{ s}$. $$1 \text{ New Unit} = (1\text{ kg})(1\text{ m})^{-1}(60\text{ s})^{-2}$$ $$1 \text{ New Unit} = 1 \times 1 \times \frac{1}{60^2} \text{ kg m}^{-1} \text{s}^{-2}$$ $$1 \text{ New Unit} = \frac{1}{3600} \text{ Nm}^{-2}$$ Therefore, the unit of pressure in this system is $1/3600$ of the SI unit.