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11.12 Power of the drilling machine, P = 10 kW= 10 $\times {10}^3$ W

Mass of the Aluminium block, m = 8.0 kg = 8 $\times {10}^3$ gm

Time for which the machine is used, t = 2.5 minute = 2.5 $\times 60$ s = 150 s

Specific heat of Aluminium, c = 0.91 J/gK

Let the rise of temperature in the block after drilling be $\delta$ T

Total energy consumed by the drilling machine= P $\times t$ = 10 $\times {10}^3\times 150$ J = 1.5 $\times {10}^6$ J

It is given 50% of energy is useful.

So useful energy,  $?Q$ = 50% of Pt= 0.5 $\times$ 1.5 $\times {10}^6=$ 7.5 $\times {10}^5$ J

We know,  $?Q$ = mc $?$ T or T = $\frac{?Q}{mc}$ = $\frac{7.5\mathrm{}\times {10}^5}{8\mathrm{}\times {10}^3\times 0.91}$ = 103 $?$

Therefore 2.5 minute drilling will raise the block temperature by 103 $?$

When air is blown under a paper, the velocity of air is more than the upper portion of the paper. As per Bernoulli's principle, atmospheric pressure reduces under the paper and makes it fall. To keep the paper horizontal, the air needs to be blown on the upper surface of the paper.

For a smaller opening, the flow of fluid is more than when it is bigger. When we try to close the tap with our fingers, water gushes through the small openings. Area and velocity are inversely proportional to each other.

Small opening of a syringe needle controls the velocity of the blood oozing out. At the constriction point of the syringe system, the flow rate suddenly increases to a high value for a constant thumb pressure.

When a fluid flows out from a small hole in a vessel, the vessel receives backward thrust, when fluid flows out from a small hole, it attains high velocity. Due to the law of conservation of momentum, the vessel attains backward velocity.

A spinning cricket ball in air has two simultaneous motions – rotary and linear. These two types of motion, opposes each other. The velocity below the ball decreases, the pressure on upper side becomes lesser than the lower side of the ball. An upward force acts on the ball and it takes a curved path, does not follow parabolic path.

The surface tension of a liquid is inversely proportional to temperature. Decreases

Most fluids offer resistance to their motion. It is like internal mechanical friction, known as viscosity. Gas viscosity increases with temperature, whereas liquid viscosity decreases with temperature. Because, intermolecular forces weaken with temperature increase, viscosity decreases.

With reference to the elastic modulus of rigidity for solids, the shearing force is proportional to the shear strain. With reference to elastic modulus of rigidity for fluids, the shearing force is proportional to the rate of shear strain.

For a steady-flowing fluid, an increase in its flow speed at a constriction follows the conservation of mass (Bernouli' principle).

For the model of a plane in wind tunnel, turbulence occurs at a greater speed then it does for an actual plane. This follows from Bernoulli's principle and different Reynolds' numbers associated with the motions of the two planes.

11.9 Initial temperature, $T_1$ = 27°C, Length of the wire $l_1$ at $T_1$ = 1.8 m

Final temperature, $T_2$ = -39°C

Diameter of the wire, d = 2.0 mm = 2 $\times {10}^{-3}$ m

Coefficient of linear expansion of brass, $\alpha =2\times {10}^{-5}$ /K

Youngs’ modulus of brass, Y = 0.91 $\times {10}^{11}$ Pa

Let the tension developed be F

We know Youngs’ modulus = $\frac{Stress}{Strain}$ = $\frac{\frac{F}{A}}{\frac{?L}{L}}=Y$

Y $\times \frac{?L}{L}$ = $\frac{F}{A}$ or F = $\frac{AY?L}{L}$

Here, A = cross-sectional area of the wire = $\frac{\pi }{4}{d}^2$ = $\frac{\pi }{4}{(2\mathrm{}\times {10}^{-3})}^2$ $m^2$ = 3.1416 $\times {10}^{-6}{m}^2$

Now $?L$ can be written as $?L$ = $\alpha L(T_2-\mathrm{}{T}_1)$ = ( $2\times {10}^{-5})\times L\times (-39-27)$ = -1.32 $\times {10}^{-3}$ L

Substituting all values, we get

F = (3.1416 $\times {10}^{-6}\times 0.91\mathrm{}\times {10}^{11}\times (-1.32\times {10}^{-3})$ = -377.37 N

The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called the angle of contact ( $\theta )$ , as shown in the diagram

$S_{la}$ = Interfacial tension between liquid-air interface

$S_{sl}$ = Interfacial tension between solid -liquid interface

$S_{sa}$ = Interfacial tension between solid-air interface

At the line of contact of contact, the surface forces between the three media must be in equilibrium. Hence

$\cos ?\theta$ = $\frac{S_{sa}-S_{la}}{S_{la}}$

The angle of contact $\theta$ is obtuse, if $S_{sa}<S_{la}$ , as in the case of mercury on glass

This angle is acute if $S_{sl}<S_{la}$ , as in the case of water on glass

Mercury molecules (which makes an obtuse angle with glass) have a strong force of attraction between themselves and a weak force of attraction towards solids. Hence, they tend to form drops

Water molecules make acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction towards solids. Hence, they tend to spread out

Surface tension is the force acting per unit length at the interface between the plane of a liquid and any other surface. This force is independent of the area of the liquid surface. Hence, surface tension is independent of the area of the liquid surface.

Water with detergent dissolved in it has a small angle of contact( $\theta ).$ This is because for a small $\theta ,$ there is a fast capillary rise of the detergent in the cloth. The capillary rise of a liquid is directly proportion to the cosine of the angle $\theta .$ If $\theta$ is small, then cos $\theta$ will be large and then the rise of detergent in the cloth will be faster.

A liquid tend to acquire the minimum surface area because of the presence of surface tension. The surface area of a sphere is the minimum for a given volume. Hence, under no external forces, liquid drops always take spherical shape.

The pressure of a liquid is given by the following relation:

P = h $\rho$ g, where P = Pressure, h = height of the liquid column,  $\rho$ = is the density of the liquid and g= acceleration due to gravity

From the above relation, because of the h factor (height of the human body), the pressure is more at the feet and less at the brain

The said phenomenon is due to the $\rho$ factor. Density of air is maximum at the sea level. At height, density decreases and pressure also decreases. At 6 km height, the density of air is nearly half of that of a t sea level

When pressure is applied on the liquid, the pressure is transmitted in all directions in the liquid, Hence in absence of fixed direction, it is a scalar quantity

11.5 (a) For Thermometer A

Triple point of water, T = 273.16 K

At this temperature, the pressure in thermometer A , $P_A$ = 1.250 $*{10}^5$ Pa

Let $T_1$ be the temperature for the normal melting point of sulphur and $P_1$ be the corresponding pressure. It is given, $P_1$ = 1.797 $*{10}^5$ Pa