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JEE Advanced 2017 Paper 1 · Physics · Fluid Mechanics, Elasticity and Surface Tension · medium

A cylindrical tank of cross-section 1.0 m^2 stands on level ground and is filled with water to a depth H=1.8 m. Small circular holes of area 100 cm^2…

One or more options correct

A cylindrical tank of cross-section 1.0 m21.0\ \mathrm{m^2} stands on level ground and is filled with water to a depth H=1.8 mH=1.8\ \mathrm{m}. Small circular holes of area 100 cm2100\ \mathrm{cm^2} can be opened in its side wall, one at a time, at different heights above the ground. Take g=10 m s−2g=10\ \mathrm{m\,s^{-2}}, treat the flow as steady at each instant and ignore viscosity. Which of the following statements is/are correct?
  1. A. The largest horizontal range on the ground (from the wall) that any hole can give is about 1.8 m1.8\ \mathrm{m}.
  2. B. Holes at heights 0.3 m0.3\ \mathrm{m} and 1.5 m1.5\ \mathrm{m} above the ground give jets that land at the same distance from the wall.
  3. C. The efflux speed from a hole at height 0.9 m0.9\ \mathrm{m} is about 3.0 m s−13.0\ \mathrm{m\,s^{-1}}.
  4. D. If the hole were at the very bottom of the wall, the tank would drain completely in about 60 s60\ \mathrm{s}.
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Answer: A, B, D

For a hole at height yy above the ground the water above it has depth H−yH-y, so by Bernoulli the efflux speed is v=2g(H−y)v=\sqrt{2g(H-y)} (the surface speed is only 1%1\% of this). The jet falls a height yy in t=2y/gt=\sqrt{2y/g}, so the range is x=v t=2y(H−y)x=v\,t=2\sqrt{y(H-y)}. This is maximal at y=H/2=0.9 my=H/2=0.9\ \mathrm{m}, where x=H=1.8 mx=H=1.8\ \mathrm{m}, so (A) is correct. Since xx is symmetric under y↔H−yy\leftrightarrow H-y and 0.3+1.5=1.80.3+1.5=1.8, both holes give x=20.3×1.5≈1.34 mx=2\sqrt{0.3\times1.5}\approx1.34\ \mathrm{m}; (B) is correct. At y=0.9 my=0.9\ \mathrm{m} the speed is 2×10×0.9=4.24 m s−1\sqrt{2\times10\times0.9}=4.24\ \mathrm{m\,s^{-1}}, not 3.03.0, so (C) is wrong. For a base hole the level hh obeys A h˙=−a2ghA\,\dot h=-a\sqrt{2gh}, giving T=Aa2Hg=100×0.6=60 sT=\dfrac{A}{a}\sqrt{\dfrac{2H}{g}}=100\times0.6=60\ \mathrm{s}, so (D) is correct. Correct options: A, B and D.
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