Explanation
capillarity will not take place when the liquid is at its boiling point, the liquid is at its freezing point and the angle of contact is$$0^0$$
$$h = \dfrac {2T cos \theta}{r \rho g}$$$$h \alpha cos \theta$$for $$\theta = 60^0 cos \theta = \dfrac {1}{2}$$$$\therefore$$ h is double $$\Rightarrow h = 6 cm$$.
Viscosity of gas increase with increase in temperature as randomness in molecules increases and number of collisions also increase, while that of a liquid decrease as molecules move away from each other making it less viscous.If density of sphere equal that of liquid, the two forces itself balance immedeatly leading to zero terminal velocity.
Applying Bernoulli's principle when the pipe was closed and after it was opened,$$\Rightarrow P_{1}=P_{2}+\dfrac{1}{2}\rho V^{2}$$$$\Rightarrow 3.5 \times 10^{5}= 3 \times 10^{5} + \dfrac{1}{2}\rho V^{2}$$$$\Rightarrow \dfrac{0.5 \times 10^{5} \times 2}{10^{3}}=V^{2}$$$$\Rightarrow 10^{2}= V^{2}\Rightarrow V=10 m/s$$
Decreases in pressure$$10^{5}m$$\ apart$$=1500pa$$energy cost per unit volume per unit distance$$=\dfrac{1500\times 10^{5}A}{10^{5}\times A\times 10^{5}}=\dfrac{work done}{lenght\times total volume}$$$$=0.015\ joules$$
Velocity of water from the bottom operture $$=\sqrt{2gh}$$ $$=\sqrt{2 \times 9.8 \times 10}m/s$$So,
$$Q=26.4 m^{3}=\pi r^{2}V$$$$\Rightarrow 26.4 m^{3} = \pi \times r^{2} v$$$$\Rightarrow \dfrac{26.4}{60}m^{3}/s = \pi r^{2} \times \sqrt{2 \times 9.8 \times 10}$$$$\Rightarrow r^{2}=0.010004024$$$$\Rightarrow r =0.1 m$$
So, Diameter of aperture $$=0.2\ m$$
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