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JEE Main Mechanical Properties of Fluids - Detailed Solutions
This blog post features previous year JEE Main questions from the Mechanical Properties of Fluids chapter.
Question 1:
A hollow spherical shell at outer radius \(R\) floats just submerged under the water surface. The inner radius of the shell is \(r\). If the specific gravity of the shell material is \(\frac{27}{8}\) with respect to water, the value of \(r\) is:
Detailed Solution:
The buoyant force equals the weight of the displaced water (volume of the outer sphere, \( \frac{4}{3}\pi R^3 \)). The weight of the shell is given by the mass of the shell material: density \( \left(\frac{27}{8}\rho\right) \) times its volume \( \left(\frac{4}{3}\pi (R^3 - r^3)\right) \). Equate the two:
\( \frac{27}{8}\rho\cdot\frac{4}{3}\pi (R^3 - r^3)=\rho\cdot\frac{4}{3}\pi R^3 \)
Simplify to get: \( \frac{27}{8}(R^3 - r^3)= R^3 \) → \( R^3 - r^3=\frac{8}{27}R^3 \), hence
\( r^3 = R^3\left(1-\frac{8}{27}\right)=\frac{19}{27}R^3 \) and so \( r= R\left(\frac{19}{27}\right)^{\frac{1}{3}}\approx \frac{8}{9}R \).
Question 2:
An air bubble of radius 1 cm in water has an upward acceleration of \(9.8\, \text{cm/s}^2\). The density of water is \(1\, \text{gm/cm}^3\) and water offers negligible drag force. Given \(g = 980\, \text{cm/s}^2\), the mass of the bubble is:
Detailed Solution:
The buoyant force on the bubble is the weight of the displaced water:
\( F_b = \rho_{\text{water}}\,g\,\frac{4}{3}\pi (1)^3 = 980\cdot\frac{4}{3}\pi \).
If the bubble’s mass is \(m\), then by Newton’s second law:
\( F_b - m\,g = m\,a \) → \( m = \frac{F_b}{g+a} \).
Substituting the values gives \( m\approx 4.15\, \text{gm}\).
Question 3:
A leak-proof cylinder (length 1 m) made of a metal with negligible expansion floats vertically in water at \(0^\circ\)C with 20 cm above water. When water is heated to \(4^\circ\)C, the height above water becomes 21 cm. The density of water at \(4^\circ\)C relative to that at \(0^\circ\)C is approximately:
Detailed Solution:
At \(0^\circ\)C, the submerged length is 80 cm. At \(4^\circ\)C, it is 79 cm. Since the weight of the cylinder remains constant, the buoyant force must also remain equal to the cylinder’s weight:
\( \rho_{0}\times0.8 = \rho_{4}\times0.79 \) → \( \frac{\rho_{4}}{\rho_{0}}\approx\frac{0.8}{0.79}\approx1.01 \).
Question 4:
A cubical block of side \(0.5\, \text{m}\) floats on water with 30% of its volume submerged. What is the maximum weight that can be added without fully submerging it? (Density of water = \(10^3\, \text{kg/m}^3\))
Detailed Solution:
The cube’s volume is \(0.5^3=0.125\, \text{m}^3\). Initially, 30% is submerged so the block’s mass is \(1000\times0.125\times0.3=37.5\, \text{kg}\). When fully submerged, it displaces \(125\, \text{kg}\) of water. Thus, the maximum extra weight is \(125-37.5=87.5\, \text{kg}\).
Question 5:
A submarine experiences a pressure of \(5.05\times10^6\, \text{Pa}\) at depth \(d_1\) and \(8.08\times10^6\, \text{Pa}\) at depth \(d_2\). With water density \(10^3\, \text{kg/m}^3\) and \(g=10\, \text{m/s}^2\), find \(d_2-d_1\).
Detailed Solution:
The pressure difference is \(8.08\times10^6 - 5.05\times10^6 = 3.03\times10^6\, \text{Pa}\). Using \(\Delta P=\rho\,g\,(d_2-d_1)\):
\(d_2-d_1=\frac{3.03\times10^6}{1000\times10}\approx303\, \text{m}\), i.e. about 300 m.
Question 6:
A Venturi meter is used to measure the flow rate of water in a horizontal pipe. The pipe has a diameter of 10 cm at the entrance and narrows to 5 cm at the throat. The measured pressure difference between the entrance and the throat is \(30{,}000\, \text{Pa}\). Assuming the density of water is \(1000\, \text{kg/m}^3\) and neglecting friction losses, determine the speed of water in the entrance section.
Detailed Solution:
Let the entrance and throat diameters be \(D_1=10\, \text{cm}\) and \(D_2=5\, \text{cm}\) respectively. Their cross-sectional areas are:
\( A_1=\pi\left(\frac{D_1}{2}\right)^2=\pi(0.05)^2\approx0.00785\, \text{m}^2\), and \( A_2=\pi\left(\frac{D_2}{2}\right)^2=\pi(0.025)^2\approx0.00196\, \text{m}^2\).
Using the continuity equation \(A_1v_1=A_2v_2\), we have \(v_2=\frac{A_1}{A_2}v_1\approx4v_1\).
For horizontal flow, Bernoulli's equation gives:
\(P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2\), so the pressure difference is
\(P_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)\).
Substitute \(v_2=4v_1\):
\(30{,}000=\frac{1}{2}\times1000\,(16v_1^2-v_1^2)=500\times15v_1^2=7500v_1^2\).
Solving, \(v_1^2=\frac{30{,}000}{7500}=4\), hence \(v_1=2.0\, \text{m/s}\).
Question 7:
A load of mass \(M\) kg is suspended from a steel wire (length 2 m, radius 1.0 mm) causing an extension of 4.0 mm in Searle’s apparatus. When the load is fully immersed in a liquid of relative density 2—and the load’s material has a relative density of 8—the new extension of the wire is:
Detailed Solution:
The extension is proportional to the tension (apparent weight). In air the tension is \(Mg\) and the extension is 4.0 mm. When immersed, the buoyant force on the load (whose density is 8 times water) in a liquid of density 2 (i.e. 2 times water) is \(\frac{1}{4}Mg\). Thus, the apparent weight becomes \(Mg-\frac{1}{4}Mg=\frac{3}{4}Mg\), reducing the extension to \(\frac{3}{4}\times4.0\, \text{mm}=3.0\, \text{mm}\).
Question 8:
A liquid of density \(\rho\) flows out of a hose (radius \(a\)) with speed \(v\) and strikes a mesh. Here, 50% passes through unchanged, 25% loses all momentum, and 25% rebounds with the same speed. The resultant pressure on the mesh is:
Detailed Solution:
Let the mass flow rate be \(\dot{m}=\rho A v\). The momentum changes per second are:
- 50%: no change.
- 25%: from \(v\) to 0 gives a change of \(0.25\,\dot{m}\,v\).
- 25%: from \(v\) to \(-v\) gives a change of \(0.25\,\dot{m}\,(v-(-v))=0.5\,\dot{m}\,v\).
Total momentum change per second is \(0.25\,\rho A v^2+0.5\,\rho A v^2=0.75\,\rho A v^2\). Dividing by area \(A\) yields \( P=\frac{3}{4}\rho v^2\).
Question 9:
Two identical charged spheres suspended by equal-length strings make an angle of \(30^\circ\) with each other. When immersed in a liquid of density \(0.8\, \text{g/cm}^3\), the angle remains unchanged. Given the spheres’ density is \(1.6\, \text{g/cm}^3\), the dielectric constant of the liquid is:
Detailed Solution:
In air the spheres experience a weight of \(Mg\) and an electrostatic repulsion \(F_e\). In the liquid, the apparent weight is reduced by the buoyant force. For a sphere of density \(1.6\,\rho\) in a liquid of density \(0.8\,\rho\), the effective weight becomes \(\frac{1}{2}Mg\). Since the electrostatic force in the liquid is reduced by the dielectric constant \(K\) (i.e. becomes \(F_e/K\)), the equilibrium condition (which fixes the angle) requires:
\( \frac{Mg}{F_e} = \frac{(Mg/2)}{(F_e/K)} \) → \(K=2\).
Question 10:
An ideal fluid flows through a pipe with non-uniform diameter. The maximum and minimum diameters are 6.4 cm and 4.8 cm, respectively. The ratio of the minimum to maximum velocities is:
Detailed Solution:
For an incompressible fluid, the continuity equation \(A_1v_1=A_2v_2\) applies. The areas are proportional to the square of the diameters. Thus,
\( \frac{v_{\text{min}}}{v_{\text{max}}}=\left(\frac{4.8}{6.4}\right)^2=\left(\frac{3}{4}\right)^2=\frac{9}{16} \).
