Properties of Bulk Matter Questions and Answers

Answer: B

\(F_{2} = F_{1}\frac{A_{2}}{A_{1}}\).

By Pascal's law, the pressure applied at the small piston is transmitted undiminished through the fluid to the large piston:

\[ \begin{aligned} \frac{F_{1}}{A_{1}} &= \frac{F_{2}}{A_{2}} \\ F_{2} &= F_{1}\frac{A_{2}}{A_{1}} \end{aligned} \]

Because \(A_{2} \gt A_{1}\), the output force is larger, which is why a hydraulic lift can raise a car. \(F_{1}\frac{A_{1}}{A_{2}}\) inverts the ratio and would make the lift reduce the force. The square-root forms have no basis in Pascal's law, and the last option is not even a force.

Answer: A

The constant is called the modulus of elasticity.

Hooke's law says stress is proportional to strain within the elastic limit: stress = k × strain. The constant k is the modulus of elasticity (Young's, bulk or shear modulus according to the type of stress).

Poisson's ratio relates lateral and longitudinal strain; compressibility is the reciprocal of the bulk modulus; 'coefficient of deformation' and 'mechanical strength' are not the names of this constant.

Answer: D

The principle is Bernoulli's principle.

Bernoulli's principle relates the speed and pressure of a flowing fluid. It helps explain blood flow in an artery: where an artery is narrowed, the blood speeds up and the pressure inside drops, which can make the artery collapse under the external pressure (the basis of some arterial disorders).

Pascal's law concerns pressure transmission in a fluid at rest, Archimedes' principle buoyancy, Boyle's law gases, and the Magnus effect the lift on a spinning ball.

Answer: A

The ratio of strains is 1 : 9.

Longitudinal strain is stress divided by Young's modulus: \(\text{strain} = \frac{F}{AY} = \frac{F}{\pi r^{2}Y}\). Both wires are copper (same \(Y\)) under the same force, so the strain depends only on the radius, not on the length.

\[ \begin{aligned} \frac{\text{strain}_{1}}{\text{strain}_{2}} &= \frac{r_{2}^{2}}{r_{1}^{2}} \\ &= \frac{1}{9} \end{aligned} \]

The length ratio 1 : 2 is a distractor: it affects the extension, \(\Delta L = \text{strain} \times L\), but not the strain. Using it gives 1 : 18, and 1 : 3 forgets to square the radius ratio.

Answer: E

Bernoulli's equation holds for non-viscous and incompressible fluids in streamline flow.

Bernoulli's equation, \(P + \frac{1}{2}\rho v^{2} + \rho gh = \text{constant}\) along a streamline, comes from applying energy conservation to an ideal fluid. Its derivation assumes no viscosity (no energy lost to friction), constant density (incompressible), and steady streamline flow, so that each element follows a fixed path.

Viscosity dissipates energy, compressibility changes the density term, and turbulent flow has no steady streamlines, so every other option breaks at least one assumption.

Answer: A

It is called tensile stress.

Stress is the restoring force per unit area set up inside a body by a deforming force. When two equal forces act normal to the cross-section and pull the cylinder outwards, stretching it, the stress is tensile (longitudinal) stress.

If the normal forces pushed inwards, squeezing it, the stress would be compressive. Forces acting parallel to the surface (tangentially) produce shearing or tangential stress, and 'transverse stress' is not the term for a stretching force.

Answer: D

The depth is \(\dfrac{p}{\rho g}\).

The pressure at depth \(h\) is \(P_{\text{abs}} = P + \rho g h\), where \(P\) is atmospheric pressure. Gauge pressure is the pressure above atmospheric, so

\[ \begin{aligned} p &= P_{\text{abs}} - P = \rho g h \\ h &= \frac{p}{\rho g} \end{aligned} \]

Atmospheric pressure has already been taken out of the gauge reading, so options that add or subtract \(P\), or use \(P\) alone, are wrong, and \(\frac{p^{2}}{\rho g}\) does not have the units of length.

Answer: D

The elongation is 1 mm.

From \(Y = \frac{FL}{A\,\Delta L}\):

\[ \begin{aligned} \Delta L &= \frac{FL}{AY} = \frac{150 \times 10^{3} \times 1}{10^{-3} \times 1.5 \times 10^{11}} \\ &= \frac{1.5 \times 10^{5}}{1.5 \times 10^{8}} = 10^{-3}\ \text{m} \end{aligned} \]

\(10^{-3}\) m is 1 mm. Forgetting the 'k' in 150 kN gives \(10^{-6}\) m, and 1 cm is ten times too large.

Answer: D

The terminal velocity is directly proportional to \(r^{2}\).

At terminal velocity the weight minus buoyancy equals Stokes' drag \(6\pi\eta rv\):

\[ \begin{aligned} \tfrac{4}{3}\pi r^{3}(\rho - \sigma)g &= 6\pi\eta r v \\ v &= \frac{2r^{2}(\rho - \sigma)g}{9\eta} \end{aligned} \]

So v grows as \(r^{2}\): a ball twice as large falls four times as fast. It is inversely, not directly, proportional to \(\eta\), and the options in r alone or \(r^{-2}\) give the wrong power.

Answer: D

The incorrect statement is that elastomers have larger Young's modulus than metals.

Elastomers such as rubber stretch a great deal under small stress, which means a very small Young's modulus, of the order of 10⁶ to 10⁷ Pa for rubber, against about 10¹¹ Pa for steel.

The other statements are accepted. Stress is not a vector (it is a tensor quantity), Young's modulus applies only to solids because only solids have a fixed length to stretch, compressibility (the reciprocal of bulk modulus) applies to solids, liquids and gases, and tensile stress is tension per unit area.