Q: A glass rod of radius r_{1} is inserted symmetrically into a vertical capillary tube of radius r_{2} such that their lower ends are at same level. The arrangement is now dipped in water the height to which water will enter into the tube (σ = surface tension , ρ = density of water)

(a) $ \displaystyle \frac{2 \sigma }{(r_2 -r_1)\rho g} $

(b) $ \displaystyle \frac{ \sigma }{(r_2 -r_1)\rho g} $

(c) $ \displaystyle \frac{2 \sigma }{(r_2 +r_1)\rho g} $

(d) $ \displaystyle \frac{ 2\sigma }{(r_2^2 -r_1^2)\rho g} $

Ans: (a)

Sol: Let h = height up to which water rises

Upward force = Weight of water

$\large 2\pi r_1 \sigma + 2\pi r_2 \sigma = \pi (r_2^2 – r_1^2)h \rho g $

$\large 2\pi ( r_1 + r_2 ) \sigma = \pi (r_2^2 – r_1^2)h \rho g $

$ \displaystyle h = \frac{2 \sigma }{(r_2 -r_1)\rho g} $