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Two cylindrical vessels of equal cross-sectional area $\mathrm{ 16 \mathrm{~cm}^{2}}$ contain water upto heights $\mathrm{ 100 \mathrm{~cm}}$ and $\mathrm{ 150 \mathrm{~cm}}$ 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 $\mathrm{ [Take, density \: of \: water =10^{3} \mathrm{~kg} / \mathrm{m}^{3} \: and \: \mathrm{g}=10 \mathrm{~ms}^{-2} ] :}$
Option: 1
$\mathrm{0.25 \mathrm{~J}}$

Option: 2
$\mathrm{1 \mathrm{~J}}$

Option: 3
$\mathrm{8 \mathrm{~J}}$

Option: 4
$\mathrm{12 \mathrm{~J}}$

Answers (1)

best_answer

The final water level is $\mathrm{125 \mathrm{~cm}}$ Rise in the water Level of vessel 1 is $\mathrm{25 \mathrm{~cm}}$ whereas there is corresponding fall in the water level of vessel 2 by $\mathrm{25 \mathrm{~cm}}$ .

Work done by gravity is

$\mathrm{W=-\Delta U}$

$\mathrm{=U_{i}-U_{f}}$

$\mathrm{U_i =\left[\frac{\rho A h_1^2 g}{2}+\frac{\rho A_2^2 g}{2}\right] }$

$\mathrm{U_f =2\left[\frac{\rho A h^2 g}{2}\right] }$

$\mathrm{W =U_i-U_f }$

$\mathrm{\left.=\frac{\rho A g}{2}[\left(h_1^2+h_2^2\right)-2 h^2\right] }$

$\mathrm{=\frac{10^3 \times 16 \times 10^{-4} \times 10}{2}[1+2.25-2 \times 15625] }$

$\mathrm{ W =8[3.25-3.125] }$

$\mathrm{W =1 \mathrm{~J}}$

Hence 2 is correct option.

Posted by

vishal kumar

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