Q: A glass rod of radius r1 is inserted symmetrically into a vertical capillary tube of radius r2 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} $