A steel wire of length $$3.2 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{s}}=2.0 \times 10^{11} \,\mathrm{Nm}^{-2}\right)$$ and a copper wire of length $$4.4 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{c}}=1.1 \times 10^{11} \,\mathrm{Nm}^{-2}\right)$$, both of radius $$1.4 \mathrm{~mm}$$ are connected end to end. When stretched by a load, the net elongation is found to be $$1.4 \mathrm{~mm}$$. The load applied, in Newton, will be: $$\quad\left(\right.$$ Given $$\pi=\frac{22}{7}$$)

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 $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$ ] :

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)