JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 11)
Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density
d. The area of the base of both vessels is S but the height of liquid in one vessel is x1
and in the
other, x2
. When both cylinders are connected through a pipe of negligible volume very close to the
bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height.
The change in energy of the system in the process is:
gdS(x2 + x1)2
gdS$$\left( {x_2^2 + x_1^2} \right)$$
$${1 \over 4}gdS{\left( {{x_2} - {x_1}} \right)^2}$$
$${3 \over 4}gdS{\left( {{x_2} - {x_1}} \right)^2}$$
Explanation
$${u_i} = \left[ {dS{x_1}.{{{x_1}} \over 2} + dS{x_2}.{{{x_2}} \over 2}} \right]g\left\{ {dS{x_1} \to m,\,{{{x_1}} \over 2} \to h(C.O.M)} \right\}$$
Here total volume remains same.
$$ \therefore $$ Vi = Vf
$$ \Rightarrow $$ S(x1 + x2) = S(h + h)
$$ \Rightarrow $$ h = $${{{x_1} + {x_2}} \over 2}$$
uf = (dSh)g$${h \over 2} \times 2$$
$$ \Rightarrow $$ $${u_f} = \left[ {dS\left( {{{{x_1} + {x_2}} \over 2}} \right) \times \left( {{{{x_1} + {x_2}} \over 4}} \right) \times 2} \right]g$$
$$ \therefore $$ $${u_i} - {u_f} = dsg\left[ {{{x_1^2} \over 2} + {{x_2^2} \over 2} - {{{{\left( {{x_1} + {x_2}} \right)}^2}} \over 4}} \right]$$
$$ = dsg{{{{\left( {{x_1} - {x_2}} \right)}^2}} \over 4}$$
Here total volume remains same.
$$ \therefore $$ Vi = Vf
$$ \Rightarrow $$ S(x1 + x2) = S(h + h)
$$ \Rightarrow $$ h = $${{{x_1} + {x_2}} \over 2}$$
uf = (dSh)g$${h \over 2} \times 2$$
$$ \Rightarrow $$ $${u_f} = \left[ {dS\left( {{{{x_1} + {x_2}} \over 2}} \right) \times \left( {{{{x_1} + {x_2}} \over 4}} \right) \times 2} \right]g$$
$$ \therefore $$ $${u_i} - {u_f} = dsg\left[ {{{x_1^2} \over 2} + {{x_2^2} \over 2} - {{{{\left( {{x_1} + {x_2}} \right)}^2}} \over 4}} \right]$$
$$ = dsg{{{{\left( {{x_1} - {x_2}} \right)}^2}} \over 4}$$
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