UPSEE Solved Model Paper 7

Dimension of density is [d] = $[\mathrm{M}\mathrm{L}^{-3}]$ Dimension of radius is [r] = [L] Dimension of surface tension is [S] = $\left[\frac{\text{force}}{\text{length}}\right]$ = $[\mathrm{M}\mathrm{T}^{-2}]$ The time period of the small liquid drop can be represented as T α $s^x r^y S^z$ where x. y and z have some numerical values, to be evaluated Thus T = $K d^x r^y S^z$ ... (i) where, K is a dimensionless constant. applying the principle of homogeneity of dimensions in eq. (i) we get, [T] = $[\mathrm{M}\mathrm{L}^{-3}]^x [\mathrm{L}]^y [\mathrm{M}\mathrm{T}^{-2}]^z$ ⇒ $[\mathrm{M}^0\mathrm{L}^0\mathrm{T}^{-1}]$ = $[\mathrm{M}]^{x+z} [\mathrm{L}]^{-3x+y} [\mathrm{T}]^{-2z}$ Equating the powers of M L and T on both sides of the above equation, we get x + z = 0 - 3x + y = 0 - 2z = 1 Solving the above equations, we obtain x = $\frac{1}{2}$ y = $\frac{3}{2}$ z = $-\frac{1}{2}$ Using above values of x, y and z in eq (I), we get T α $d^{\frac{1}{2}} r^{\frac{3}{2}} S^{-\frac{1}{2}}$ ⇒ T = $K\sqrt{\frac{d r^3}{S}}$ Hence, option 'B' is correct