Question 22: Properties of Matter

Question

Two cylindrical vessels of equal cross sectional area of $2 \mathrm{~m}^2$ contain water upto heights 10 m and 6 m , respectively. If the vessels are connected at their bottom then the work done by the force of gravity is
(Density of water is $10^3 \mathrm{~kg} / \mathrm{m}^3$ and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

First, when the two vessels are connected, the water levels equalize at
$$H = \frac{H_1 + H_2}{2} = \frac{10 + 6}{2} = 8	ext{ m}.$$
We can get the work done by gravity from the loss of gravitational potential energy:
Initial potential energy (taking the bottom as zero)
For a column of height $H$ and cross-section $A$,
$$U = ho g A\int_0^H z\,dz = \frac{ho g A H^2}{2}.$$
So
$$U_i = \frac{ho g A}{2}\bigl(H_1^2 + H_2^2\bigr)           = \frac{10^3\cdot 10\cdot 2}{2}(10^2 + 6^2)          = 10^4\,(100 + 36)          = 1.36	imes10^6	ext{ J}.$$
Final potential energy (both at height 8Ã¢ÂÂm)
$$U_f = 2\;\frac{ho g A H^2}{2}\;\Big|_{H=8}          = ho g A\,(8^2)          = 10^3\cdot10\cdot2\cdot64          = 1.28	imes10^6	ext{ J}.$$
Work done by gravity = drop in potential energy
$$W = U_i - U_f = 1.36	imes10^6 - 1.28	imes10^6                    = 0.08	imes10^6                    = 8	imes10^4	ext{ J}.$$
Answer: 8ÃÂ10^4Ã¢ÂÂJ (Option C).

Answer

${ }1 	imes 10^5 \mathrm{~J}$

Answer

$4 	imes 10^4 \mathrm{~J}$

Answer

$8 	imes 10^4 \mathrm{~J}$

Answer

$6 	imes 10^4 \mathrm{~J}$

Properties of Matter

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