Two cylindrical vessels of equal cross-sectional area $$16 \mathrm{~cm}^{2}$$ contain water upto heights $$100 \mathrm{~cm}$$ and $$150 \mathrm{~cm}$$ respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, is [Take, density of water $$=10^{3} \mathrm{~kg} / \mathrm{m}^{3}$$ and $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$ ] :

The area of cross section of the rope used to lift a load by a crane is $$2.5 \times 10^{-4} \mathrm{~m}^{2}$$. The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be :

(take $$g=10 \,m s^{-2}$$ )

A water drop of radius $$1 \mathrm{~cm}$$ is broken into 729 equal droplets. If surface tension of water is 75 dyne/ $$\mathrm{cm}$$, then the gain in surface energy upto first decimal place will be :

(Given $$\pi=3.14$$ )

A drop of liquid of density $$\rho$$ is floating half immersed in a liquid of density $${\sigma}$$ and surface tension $$7.5 \times 10^{-4}$$ Ncm^{$$-$$1}. The radius of drop in $$\mathrm{cm}$$ will be :

(g = 10 ms^{$$-$$2})