MHTCET 11 Apr 2026 Shift 2 Paper

Concept:Water rises in the annular space between the capillary wall and the wire, and surface tension acts along both circular contact lines.Explanation:For water in a capillary, the angle of contact is $0^\circ$, so $\cos 0^\circ = 1$.The upward force due to surface tension acts along the inner circumference of the capillary, $2\pi R$, and the outer circumference of the wire, $2\pi r$.Therefore, total upward force is:$F = T(2\pi R + 2\pi r) = 2\pi T(R+r)$.The water column occupies the annular area between the capillary and the wire:$A = \pi R^2 - \pi r^2 = \pi(R^2-r^2)$.If the height of rise is $h$, the weight of the water column is:$W = \rho g \pi(R^2-r^2)h$.At equilibrium, the upward force equals the weight of the water column:$2\pi T(R+r) = \rho g \pi(R^2-r^2)h$.Canceling $\pi$ from both sides:$2T(R+r) = \rho g(R^2-r^2)h$.Using $R^2-r^2=(R-r)(R+r)$, we obtain:$h = \frac{2T(R+r)}{\rho g(R-r)(R+r)} = \frac{2T}{(R-r)\rho g}$.Answer:The rise of water in the capillary is $\frac{2T}{(R-r)\rho g}$.Correct option: B.