JEE Physics Mechanical Properties Of Fluids MCQs Set 21

Answer: (d) 20 km

Explanation:
1. The stress at the top is \( \dfrac{\rho L A g}{A} = \rho Lg \).
2. \( L = \dfrac{15.8\times10^8}{7.9\times10^3\times10} = 2\times10^4 \) m = 20 km.

Teacher's Note:
1. The weight of the wire per unit area grows with its length, so the maximum stress is at the top of the wire.

Answer: (d) 1.5 mm

Explanation:
1. At the lowest point \( v^2 = 2g\ell = 100 \), so the tension is \( T = mg + \dfrac{mv^2}{\ell} = 40 + \dfrac{4\times100}{5} = 120 \) N.
2. \( \Delta\ell = \dfrac{T\ell}{AY} = \dfrac{120\times5}{2\times10^{-6}\times2\times10^{11}} = 1.5\times10^{-3} \) m = 1.5 mm.

Teacher's Note:
1. The extension is greatest when the tension is greatest, which is at the bottom of the swing. Tension there is more than the weight.

Answer: (b) 4 μJ

Explanation:
1. The compression is \( x = \dfrac{FL}{AY} = \dfrac{40\times0.2}{4\times10^{-4}\times10^{11}} = 2\times10^{-7} \) m.
2. Energy stored \( U = \tfrac12Fx = \tfrac12\times40\times2\times10^{-7} = 4\times10^{-6} \) J = 4 μJ.

Teacher's Note:
1. The elastic energy is the area under the force-extension graph, half of force times extension.

Answer: (b) \( \ell^2 \)

Explanation:
1. With volume V fixed, \( A = V/\ell \).
2. \( \Delta\ell = \dfrac{F\ell}{AY} = \dfrac{F\ell^2}{VY} \propto \ell^2 \).

Teacher's Note:
1. When the wire is drawn longer, it also becomes thinner, so the extension grows as the square of the length.

Answer: (b) \( 1\times10^9 \) N/m²

Explanation:
1. Change in stress \( = 7000\times10^3 = 7\times10^6 \) N/m².
2. Change in strain \( = \dfrac{7\times10^{-3}}{1} = 7\times10^{-3} \).
3. \( Y = \dfrac{7\times10^6}{7\times10^{-3}} = 10^9 \) N/m².

Teacher's Note:
1. Young's modulus is the slope of the stress-strain graph. Divide the extension by the original length to get strain.

Answer: (a) \( \dfrac{W_1+W/4}{s} \)

Explanation:
1. The part of the wire below this point has length L/4 and weight W/4.
2. The tension at that section supports W1 and this weight: \( T = W_1 + \dfrac W4 \).
3. Stress \( = \dfrac{W_1+W/4}{s} \).

Teacher's Note:
1. The tension at a point equals the weight of everything hanging below that point.

Answer: (b) \( Y_A > Y_B \)

Explanation:
1. The slope of the load-extension graph is \( \dfrac{F}{\Delta\ell} = \dfrac{YA}{L} \).
2. With the same A and L, the steeper line has the larger Y: \( Y_A > Y_B \).

Teacher's Note:
1. The slope of the load-extension graph is YA/L. The slope of the stress-strain graph is Y itself.

Answer: (a) 16 : 1

Explanation:
1. Energy per unit volume \( = \dfrac12\dfrac{\text{stress}^2}{Y} \), and stress \( = F/A \propto 1/d^2 \).
2. So the energy per unit volume \( \propto \dfrac1{d^4} \). For diameters 1 : 2 the ratio is \( 2^4 : 1 = 16 : 1 \).

Teacher's Note:
1. Stress goes as 1/d², and the energy density goes as the square of stress, giving 1/d⁴.

Answer: (c) OA

Explanation:
1. The slope is \( \dfrac{\text{load}}{\text{elongation}} = \dfrac{YA}{L} \).
2. With the same Y and L, the thinnest wire (smallest A) has the smallest slope, so it is OA.

Teacher's Note:
1. A thin wire stretches more for the same load, so its graph is the flattest.

Answer: (a) \( \Delta\ell \) versus \( \ell^2 \)

Explanation:
1. With V fixed, \( A = V/\ell \), so \( \Delta\ell = \dfrac{F\ell^2}{VY} \).
2. Hence \( \Delta\ell \) is proportional to \( \ell^2 \), and the graph of \( \Delta\ell \) against \( \ell^2 \) is a straight line through the origin.

Teacher's Note:
1. A straight-line graph needs a linear relation between the plotted quantities.

Answer: (c) 8F

Explanation:
1. Same volume: the second wire has length \( \ell/2 \) when the first has \( \ell \).
2. \( \Delta\ell = \dfrac{F'\ell'}{A'Y} \). First: \( \Delta\ell = \dfrac{2F\ell}{AY} \). Second: \( \Delta\ell = \dfrac{F'(\ell/2)}{2AY} = \dfrac{F'\ell}{4AY} \).
3. Equating: \( F' = 8F \).

Teacher's Note:
1. Equal volumes mean that doubling the area halves the length. Both changes make the second wire stiffer.

Answer: (c) 84 kPa

Explanation:
1. \( \Delta P = B\dfrac{\Delta V}{V} = 2100\times10^6\times4\times10^{-5} \).
2. \( = 8.4\times10^4 \) Pa = 84 kPa.

Teacher's Note:
1. 0.004% as a fraction is 4 × 10⁻⁵. The volume of 200 litres itself is not needed.

Answer: (b) \( 39.2\times10^9 \) N/m²

Explanation:
1. Excess pressure at the bottom \( = \rho gh = 1000\times9.8\times400 = 3.92\times10^6 \) Pa.
2. \( B = \dfrac{\Delta P}{\Delta V/V} = \dfrac{3.92\times10^6}{10^{-4}} = 3.92\times10^{10} \) N/m² = \( 39.2\times10^9 \) N/m².

Teacher's Note:
1. 0.01% is 10⁻⁴. The pressure change is the hydrostatic pressure of the water column.

Answer: (a) \( \dfrac{p}{3B} \)

Explanation:
1. \( \dfrac{\Delta V}{V} = \dfrac pB \). Since \( V\propto d^3 \), \( \dfrac{\Delta V}{V} = 3\dfrac{\Delta d}{d} \).
2. \( \dfrac{\Delta d}{d} = \dfrac{p}{3B} \).

Teacher's Note:
1. Volume depends on the cube of the diameter, so the fractional change in volume is three times that in diameter.

Answer: (a) 0.2

Explanation:
1. \( Y = 2\eta(1+\sigma) \).
2. \( 2.4\eta = 2\eta(1+\sigma) \Rightarrow 1+\sigma = 1.2 \), so \( \sigma = 0.2 \).

Teacher's Note:
1. The relation between Young's modulus, rigidity modulus and Poisson's ratio is Y = 2η(1 + σ).

Answer: (b) 1 : 1

Explanation:
1. The thrust on the base is \( F = P\times\text{area} = \rho gh\times\text{area} \).
2. The height and base area are the same, so the thrust is the same: 1 : 1.

Teacher's Note:
1. This is the hydrostatic paradox: the thrust on the base depends on the depth and base area, not on the shape of the vessel.

Answer: (b) Pascal's law

Explanation:
1. In a hydraulic press, pressure applied to the liquid in the small cylinder is transmitted undiminished to the large cylinder.
2. This is Pascal's law, which gives a force multiplication equal to the ratio of the piston areas.

Teacher's Note:
1. Pascal's law: pressure applied to an enclosed fluid is transmitted undiminished through the fluid.

Answer: (c) 915 kg/m³

Explanation:
1. At the level of the interface, the pressures in the two arms are equal: \( \rho_{oil}(\ell+d)g = \rho_w\ell g \).
2. \( \rho_{oil} = \dfrac{1000\times135}{147.5} \approx 915 \) kg/m³.

Teacher's Note:
1. Equal pressure at the same level in the same liquid. The denser liquid stands lower.

Answer: (c) 200 cm³

Explanation:
1. Floating: weight = weight of water displaced, so the displaced volume is \( \dfrac{800}{1} = 800 \) cm³.
2. The total volume is 1000 cm³, so the volume outside water is \( 1000-800 = 200 \) cm³.

Teacher's Note:
1. A floating body displaces its own weight of liquid.

Answer: (c) 4/3 gm/c.c.

Explanation:
1. In water, the submerged fraction is 1/3: \( \rho_s = \tfrac13\times1 \) g/cc.
2. In the liquid, the submerged fraction is 1/4: \( \rho_s = \tfrac14\rho_l \).
3. \( \rho_l = 4\times\tfrac13 = \tfrac43 \) g/cc.

Teacher's Note:
1. Submerged fraction = density of body ÷ density of liquid.

Answer: (d) balance A will read less than 2 kg and B will read more than 5 kg

Explanation:
1. The liquid exerts an upward buoyant force on the block, so the spring balance A reads less than 2 kg.
2. By Newton's third law, the block exerts an equal downward force on the liquid, so balance B reads more than 5 kg.

Teacher's Note:
1. Buoyancy acts on the block and its reaction acts on the liquid. The two readings change by the same amount in opposite directions.

Answer: (d) \( \rho_1 < \rho_3 < \rho_2 \)

Explanation:
1. The liquids do not mix, so the denser one (2) is at the bottom: \( \rho_1 < \rho_2 \).
2. The ball sinks in liquid 1 but floats on liquid 2, so its density lies between the two: \( \rho_1 < \rho_3 < \rho_2 \).

Teacher's Note:
1. A ball floating at the interface is denser than the upper liquid and less dense than the lower liquid.

Answer: (c) ℓ increases and h decreases

Explanation:
1. Without the coin the block carries less load, so it floats higher: ℓ increases.
2. Before, the block and coin displaced water equal to their combined weight. After, the block displaces less, and the coin (denser than water) displaces only its own volume, which is less than before. The total displaced volume decreases, so the water level h falls.

Teacher's Note:
1. A dense object floating on a block displaces more water than when it sinks, because its weight rather than its volume determines the displacement.

Answer: (a) \( W\left(1-\dfrac{\rho_1}{\rho}\right) \)

Explanation:
1. The volume is \( V = \dfrac{W}{\rho g} \). Buoyant force \( = V\rho_1g = W\dfrac{\rho_1}{\rho} \).
2. Apparent weight \( = W - W\dfrac{\rho_1}{\rho} = W\left(1-\dfrac{\rho_1}{\rho}\right) \).

Teacher's Note:
1. Apparent weight = true weight minus the buoyant force.