JEE Mains Mechanical Properties of Fluids Test
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1. Water from a tap emerges vertically downwards with an initial speed of \(1.0\, \text{m/s}\). The cross-sectional area of the tap is \(10^{-4}\, \text{m}^2\). Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross-sectional area of the stream, \(0.15\, \text{m}\) below the tap would be :
Detailed Solution:
Using the continuity equation, \(A_1v_1=A_2v_2\), where \(A_1=10^{-4}\, \text{m}^2\) and \(v_1=1.0\, \text{m/s}\). The speed at \(0.15\, \text{m}\) below is found by energy conservation:
\(v_2=\sqrt{v_1^2+2gh}=\sqrt{1^2+2\times10\times0.15}=\sqrt{4}=2\, \text{m/s}\).
Thus, \(A_2=\frac{10^{-4}\times1}{2}=5\times10^{-5}\, \text{m}^2\). The correct answer is (b).
2. Water from a pipe is coming at a rate of \(100\, \text{liters/minute}\). If the radius of the pipe is \(5\, \text{cm}\), the Reynolds number for the flow is of the order of: (density of water = \(1000\, \text{kg/m}^3\), coefficient of viscosity of water = \(1\, \text{mPa s}\))
Detailed Solution:
Convert \(100\, \text{liters/minute}\) to \(m^3/s\): \(100/1000=0.1\, \text{m}^3/\text{min}\) and then \(0.1/60\approx1.67\times10^{-3}\, \text{m}^3/s\). The pipe area is \(A=\pi(0.05)^2\approx7.85\times10^{-3}\, \text{m}^2\), so the velocity is \(v=\frac{Q}{A}\approx0.213\, \text{m/s}\). Taking the diameter \(d=0.1\, \text{m}\), the Reynolds number is
\(Re=\frac{\rho v d}{\mu}\approx\frac{1000\times0.213\times0.1}{0.001}\approx2.13\times10^4\), which is of the order of \(10^4\). Hence, the correct answer is (b).
3. Water flows into a large tank with flat bottom at the rate of \(10^{-4}\, \text{m}^3/s\). Water is also leaking out of a hole of area \(1\, \text{cm}^2\) at its bottom. If the height of the water in the tank remains steady, then this height is:
Detailed Solution:
The leak rate is \(Q_{\text{out}}=A\sqrt{2gh}\) with \(A=1\, \text{cm}^2=10^{-4}\, \text{m}^2\) and \(g=10\, \text{m/s}^2\). Setting inflow equal to outflow:
\(10^{-4}=10^{-4}\sqrt{20h}\) ⟹ \(\sqrt{20h}=1\) ⟹ \(20h=1\) ⟹ \(h=0.05\, \text{m}=5\, \text{cm}\). The nearest option is \(5.1\, \text{cm}\); therefore, the answer is (a).
4. The top of a water tank is open to air and its water level is maintained. It is giving out \(0.74\, \text{m}^3\) water per minute through a circular opening of radius \(2\, \text{cm}\) in its wall. The depth of the centre of the opening from the level of water in the tank is close to:
Detailed Solution:
Convert the outflow rate: \(0.74\, \text{m}^3/\text{min}\approx0.01233\, \text{m}^3/s\). The opening’s area is \(A=\pi(0.02)^2\approx1.2566\times10^{-3}\, \text{m}^2\). Hence, the velocity is \(v=\frac{Q}{A}\approx9.812\, \text{m/s}\). Applying Torricelli’s law \(v=\sqrt{2gh}\) yields
\(h=\frac{v^2}{2g}\approx\frac{9.812^2}{20}\approx4.8\, \text{m}\). The correct answer is (b).
5. When an air bubble of radius \(r\) rises from the bottom to the surface of a lake, its radius becomes \(\frac{5r}{4}\). Taking the atmospheric pressure to be equal to a \(10\, \text{m}\) water column, the depth of the lake would approximately be (ignore surface tension and temperature effects):
Detailed Solution:
Assume an isothermal process so that \(P_1V_1=P_2V_2\). Initially, the bubble at depth \(H\) has volume \(V_1\propto r^3\) and pressure \(P_1=P_{\text{atm}}+\rho gH\) (with \(P_{\text{atm}}\) equivalent to a \(10\, \text{m}\) water column). At the surface, the bubble expands to volume \(V_2\propto \left(\frac{5r}{4}\right)^3=\frac{125}{64}r^3\) at pressure \(P_2=P_{\text{atm}}\). Equating, we have:
\((10+H)r^3=10\left(\frac{125}{64}r^3\right)\) ⟹ \(H\approx9.53\, \text{m}\). The closest option is (d) \(9.5\, \text{m}\).
6. An open glass tube is immersed in mercury so that \(8\, \text{cm}\) of its length extends above the mercury level. The open end is then closed and sealed, and the tube is raised vertically by an additional \(46\, \text{cm}\). What is the length of the air column above the mercury now? (Atmospheric pressure = \(76\, \text{cm Hg}\))
Detailed Solution:
Initially, the trapped air occupies \(8\, \text{cm}\) at atmospheric pressure (\(76\, \text{cm Hg}\)). After raising the tube by \(46\, \text{cm}\), the top is \(54\, \text{cm}\) above the external mercury level. Using Boyle’s law, the product \(P \times L\) remains constant, and the new equilibrium yields a mercury rise of \(38\, \text{cm}\) inside the tube. Thus, the air column length becomes \(54-38=16\, \text{cm}\), making the correct answer (a).
7. Water flows at a speed of \(1.5\, \text{m/s}\) through a horizontal tube of cross-sectional area \(10^{-2}\, \text{m}^2\). To stop the flow by placing your palm, assuming the water stops immediately, the minimum force required is (density of water = \(10^3\, \text{kg/m}^3\)):
Detailed Solution:
The mass flow rate is \(\dot{m}=\rho Av=1000\times10^{-2}\times1.5=15\, \text{kg/s}\). The force required is the rate of change of momentum: \(F=\dot{m}\times v=15\times1.5=22.5\, \text{N}\). Hence, the answer is (b).
8. Air of density \(1.2\, \text{kg/m}^3\) flows across an aeroplane’s horizontal wings so that the speeds above and below are \(150\, \text{m/s}\) and \(100\, \text{m/s}\), respectively. The pressure difference between the upper and lower sides is:
Detailed Solution:
Using Bernoulli’s principle, the pressure difference is given by \(\Delta P=\frac{1}{2}\rho\left(v_{\text{below}}^2-v_{\text{above}}^2\right)\). Substituting, we get \(\Delta P=0.5\times1.2\times(150^2-100^2)=7500\, \text{Nm}^{-2}\). Therefore, the answer is (c).
9. Water flows through a horizontal tube whose two ends have cross-sectional areas \(A\) and \(A'\) with \(\frac{A}{A'}=5\). If the pressure difference between the two ends is \(3\times10^5\, \text{N/m}^2\), the velocity of water as it enters the tube is (neglecting gravity):
Detailed Solution:
Let the entry velocity be \(v\) and at the smaller end \(v'\). By continuity, \(Av=A'v'\) with \(\frac{A}{A'}=5\) so that \(v'=5v\). Bernoulli’s equation gives \(\Delta P=\frac{1}{2}\rho\left((5v)^2-v^2\right)=12\,000\,v^2\). Setting \(12\,000\,v^2=3\times10^5\) yields \(v^2=25\) and \(v=5\, \text{m/s}\). The correct answer is (a).
10. A cylinder of height \(20\, \text{m}\) is completely filled with water. The velocity of efflux through a small hole on its side near the bottom is:
Detailed Solution:
By Torricelli’s law, \(v=\sqrt{2gh}\) with \(h=20\, \text{m}\) and \(g=10\, \text{m/s}^2\), so \(v=\sqrt{400}=20\, \text{m/s}\). Thus, the answer is (b).
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