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JEE Main 2022 (28 Jul, Shift 2) · Velocity of Efflux (Torricelli)

Velocity of Efflux (Torricelli) question from JEE Main 2022 (28 Jul, Shift 2)

Q16JEE Main 2022 · 28 JUL · SHIFT 2Medium

Velocity of Efflux (Torricelli)

Consider a cylindrical tank of radius 1 m1\text{ m} is filled with water. The top surface of water is at 15 m15\text{ m} from the bottom of the cylinder. There is a hole on the wall of cylinder at a height of 5 m5\text{ m} from the bottom. A force of 5×105 N5 \times 10^{5}\text{ N} is applied on the top surface of water using a piston. The speed of efflux from the hole will be: (given atmospheric pressure PA=1.01×105 PaP_A = 1.01 \times 10^{5}\text{ Pa}, density of water ρW=1000 kg/m3\rho_W = 1000\text{ kg/m}^3 and gravitational acceleration g=10 m/s2g = 10\text{ m/s}^2)

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Answer: (c)

  1. Extra pressure from the piston: Fπr2=5×105π(1)2≈1.59×105 Pa\frac{F}{\pi r^2} = \frac{5\times10^{5}}{\pi(1)^2} \approx 1.59\times10^{5}\text{ Pa}
  2. Full Bernoulli (top to hole, depth 10 m): 12ρv2=FA+ρg(10)=1.59×105+1.0×105⇒v≈22.8 m/s\frac{1}{2}\rho v^2 = \frac{F}{A} + \rho g(10) = 1.59\times10^{5} + 1.0\times10^{5} \Rightarrow v \approx 22.8\text{ m/s} (not among the options)
  3. The official answer uses only the piston pressure: 12ρv2=1.59×105⇒v=318≈17.8 m/s\frac{1}{2}\rho v^2 = 1.59\times10^{5} \Rightarrow v = \sqrt{318} \approx 17.8\text{ m/s}
  4. So the answer is (c) 17.8 m/s.
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