JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 22)
Consider a water tank as shown in the figure. It's cross-sectional area is 0.4 m2. The tank has an opening B near the bottom whose cross-section area is 1 cm2. A load of 24 kg is applied on the water at the top when the height of the water level is 40 cm above the bottom, the velocity of water coming out the opening B is v ms-1.
The value of v, to the nearest integer, is ___________. [Take value of g to be 10 ms-2]
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The value of v, to the nearest integer, is ___________. [Take value of g to be 10 ms-2]
Answer
3
Explanation
_18th_March_Evening_Shift_en_22_3.png)
Force at point A,
mg + p0A = pA
$$ \Rightarrow $$ p = p0 + $${{mg} \over A}$$
Given, area of A = 0.4 m2 = 0.4 $$\times$$ 104 cm2 and area of B = 1 cm2
applying continuity equation
AV1 = av [V1 = velocity at point A]
$$ \Rightarrow $$ V1 = $${a \over A}v$$
As A >>> a so $${a \over A}$$ very small
$$ \therefore $$ V1 <<< v
So we can neglect V1 by assuming V1 = 0
On applying Bernoulli's theorem at points A and B,
$${p_0} + {{mg} \over A} + {{\rho V_1^2} \over 2} + \rho gh = {p_0} + {{\rho {v^2}} \over 2} + 0$$
$$ \Rightarrow {p_0} + {{mg} \over A} + 0 + \rho gh = {p_0} + {{\rho {v^2}} \over 2}$$
$$ \Rightarrow {{mg} \over A} + \rho gh = {{\rho {v^2}} \over 2}$$
$$ \Rightarrow {{24 \times 10} \over {0.4}} + 1000 \times 10 \times 0.4 = {{1000 \times {v^2}} \over 2}$$
$$ \Rightarrow $$ v $$ \simeq $$ 3 m/s
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