Properties of Bulk Matter Questions and Answers

Answer: C

The forces are in the ratio 1 : 25.

Force equals pressure times area, and the area of a circular cross-section is \(\pi r^{2}\). With the same pressure, the forces are in the ratio of the squares of the radii.

\[ \begin{aligned} \frac{F_1}{F_2} &= \frac{r_1^{2}}{r_2^{2}} \\ &= \frac{1^{2}}{5^{2}} = \frac{1}{25} \end{aligned} \]

1 : 5 compares radii instead of areas, 25 : 1 reverses the ratio, and 1 : 10 doubles the radius ratio instead of squaring it.

Answer: A

The highest modulus of elasticity belongs to steel.

Young's modulus of steel is about 200 GPa (2 × 10¹¹ N m^-2). Copper is about 110 to 120 GPa, brass about 100 GPa, and aluminium and glass about 70 GPa.

That is why steel stretches less than these materials under the same stress and is preferred in structures such as bridges and cables. It is a common misconception that rubber or soft metals are 'more elastic'; in physics the stiffest material has the highest modulus.

Answer: E

The ratio of heights is 2:1.

By Jurin's law, the rise in a capillary is

\[ h = \frac{2T\cos\theta}{r\rho g} \propto \frac{1}{r} \]

for the same liquid. With radii in the ratio 1:2, the heights are in the inverse ratio, 2:1: the narrower tube draws the liquid higher.

1:2 takes the height as proportional to the radius, and 4:1 or 1:4 treat it as depending on the square of the radius.

Answer: C

The relative viscosity of blood is constant between 0 °C and 37 °C.

The viscosities of both blood and water fall as temperature rises, and over the range from 0 °C up to normal body temperature, 37 °C, they fall in step, so their ratio \(\frac{\eta}{\eta_{water}}\) stays nearly constant. This is the range stated in the NCERT Class 11 physics text in its discussion of viscosity.

The other ranges extend beyond body temperature to 47 °C, or begin at 10, 20 or 30 °C, and do not match that statement.

Answer: C

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

At terminal velocity, weight minus buoyancy equals Stokes' viscous drag \(6\pi\eta r v\). The weight and buoyancy both depend on the volume, \(\frac{4}{3}\pi r^{3}\).

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

The \(r^{3}\) from the volume is divided by the \(r\) in the drag, leaving \(r^{2}\). So \(r^{3}\) ignores the drag's dependence on radius, and the inverse options have the trend backwards: bigger balls fall faster.

Answer: D

It depends on the density of the liquid and the depth.

The pressure due to a liquid at depth h is p = hρg. It depends only on the depth below the surface, the density of the liquid and g.

It does not depend on the shape of the container, the area of its base, or the total volume or mass of liquid. This is the hydrostatic paradox: vessels of very different shapes filled to the same height have the same pressure at the bottom.

Answer: A

The capillary rise is 2.5 cm.

For the same tube (same radius, and taking the same angle of contact), \(h = \frac{2T\cos\theta}{r\rho g}\), so \(h \propto \frac{T}{\rho}\):

\[ \begin{aligned} \frac{h'}{4} &= \frac{T/2}{T} \times \frac{1000}{800} = \frac{5}{8} \\ h' &= 2.5\ \text{cm} \end{aligned} \]

Halving the surface tension alone would give 2 cm; the lower density raises it by a factor of 1.25 to 2.5 cm. Using the density ratio the wrong way up gives 1.6 cm, and 5.0 cm ignores the halved surface tension.

Answer: D

The force required is 1100 N.

From \(Y = \frac{F/A}{\Delta L/L}\), the force is \(F = YA\frac{\Delta L}{L}\). Convert the area: 0.1 cm² = \(0.1 \times 10^{-4} = 10^{-5}\) m², and a 0.1 % increase is a strain of \(10^{-3}\).

\[ \begin{aligned} F &= 11 \times 10^{10} \times 10^{-5} \times 10^{-3} \\ &= 11 \times 10^{2} = 1100 \text{ N} \end{aligned} \]

550 N is half the value, and \(11 \times 10^{4}\) N comes from taking the strain as 0.1 instead of 0.001.

Answer: B

The velocity is \(2\sqrt{10}\) m s^-1.

By Torricelli's theorem, water leaves a small hole at depth \(h\) below the free surface with speed \(v = \sqrt{2gh}\), the speed of a body falling freely through \(h\). In a huge tank the surface barely moves, so its speed can be ignored.

\[ \begin{aligned} v &= \sqrt{2 \times 10 \times 2} = \sqrt{40} \\ &= 2\sqrt{10} \text{ m s}^{-1} \end{aligned} \]

\(2\sqrt{5} = \sqrt{20}\) is \(\sqrt{gh}\), which drops the factor 2, and \(\sqrt{10}\) is smaller still.

Answer: C

It is due to the maximum density of water at 4° C.

As a lake cools, the surface water becomes denser and sinks until the whole body of water reaches 4 °C, where water is densest. Further cooling makes the surface water less dense, so it stays on top, cools to 0 °C and freezes, while the denser water at about 4 °C remains below. The ice layer then insulates the water underneath.

Pressure differences do not cause this, water is not densest at 0 °C (ice floats because it is less dense than water), and there is no elevation of the freezing point.