UPSEE 2018 Solved Paper

As we know, surface tension, $S = \frac{\text{force}}{\text{length}} = \frac{F}{l}$ Hence, its SI unit is $\frac{\mathrm{N}}{\mathrm{m}}$. So, option (a) is unit of surface tension. Again, option (b) $\frac{\text{joule}}{(\text{metre})^{2}} \cong \frac{\text{energy}}{(\text{length})^{2}}$ $= \frac{\text{force} \times \text{length}}{(\text{length})^{2}} \frac{\text{force}}{\text{length}} = \frac{F}{l}$ In terms of SI unit, it will be $\frac{\mathrm{N}}{\mathrm{m}}$, hence it is also equal to surface tension. Again, option (c) $\frac{\mathrm{kg}}{\mathrm{s}^{2}} \approx \frac{\text{mass}}{(\text{time})^{2}}$ But $\frac{\text{energy}}{(\text{length})^{2}} = \frac{F \times s}{s^{2}} = \frac{m \times a \times s}{s^{2}}$ $= m \times \frac{\text{distance}}{(\text{time})^{2}} \times \frac{1}{\text{distance}} = \frac{\mathrm{kg}}{\mathrm{s}^{2}} = \frac{F}{1}$ Hence, option (c) also equivalent to surface tension. But, option $(d) = \frac{\text{watt}}{\text{metre}} = \frac{\text{power}}{\text{mass}}$ $= \frac{\text{force} \times \text{distance}}{\text{mass}} = \frac{\mathrm{N} - \mathrm{m}}{\mathrm{kg}}$ cannot be equal to surface tension. Hence, option (d) is correct.