Sample Exercises

Hint

\(
\begin{array}{rlrl}
\text { Mass of girl, } m =50 \mathrm{~kg} . \\
\therefore \text { Force on the heel, } F  =m g=50 \times 9.8=490 \mathrm{~N} \\
\text { Diameter, } D  =1.0 \mathrm{~cm}=1 \times 10^{-2} \mathrm{~m} \\
\therefore \quad \text { Area, } A  =\frac{\pi D^2}{4}=\frac{3.14 \times\left(1 \times 10^{-2}\right)^2}{4}=7.85 \times 10^{-5} \mathrm{~m}^2 \\
\therefore \quad \text { Pressure, } P  =\frac{F}{A}=\frac{490}{7.85 \times 10^{-5}}=6.24 \times 10^6 \mathrm{~Pa} .
\end{array}
\)

Hint

We know that atmospheric pressure, \(P=1.01 \times 10^5 \mathrm{~Pa}\).
If we use French wine of density, \(\rho=984 \mathrm{~kg} \mathrm{~m}^{-3}\), then the height of the wine column should be \(h_{m^{\prime}}\) such that \(P=h \rho g\)
\(
\Rightarrow \quad h_m=\frac{P}{\rho g}=\frac{1.01 \times 10^5}{984 \times 9.8}=10.47 \mathrm{~m} \simeq 10.5 \mathrm{~m}
\)

Hint

Here, Maximum stress \(=10^9 \mathrm{~Pa}, \mathrm{~h}=3 \mathrm{~km}=3 \times 10^3 \mathrm{~m}\);
\(p\) (water) \(=10^3 \mathrm{~kg} / \mathrm{m}^3\) and \(g=9.8 \mathrm{~m} / \mathrm{s}^2\).
The structure will be suitable for putting upon top of an oil well provided the pressure exerted by seawater is less than the maximum stress it can bear.
Pressure due to sea water, \(P=h \rho g=3 \times 10^3 \times 10^3 \times 9.8 \mathrm{~Pa}=2.94 \times 10^7 \mathrm{~Pa}\)
Since the pressure of seawater is less than the maximum

Hint

Pressure on the piston due to car
\(
\begin{aligned}
& =\frac{\text { Weight of car }}{\text { Area of piston }} \\
P & =\frac{3000 \times 9.8}{425 \times 10^{-4}} \mathrm{Nm}^{-2}=6.92 \times 10^5 \mathrm{~Pa}
\end{aligned}
\)
This is also the maximum pressure that the smaller piston would have to bear.

Hint

For water column in one arm of \(U\) tube, \(h_1=10.0 \mathrm{~cm} ; \rho_1\) (density) \(=1 \mathrm{~g} \mathrm{~cm}^{-3}\)
For spirit column in other arm of U tube, \(h_2=12.5 \mathrm{~cm} ; \rho_2=\) ?
As the mercury columns in the two arms of \(U\) tube are in level, therefore pressure exerted by each is equal.
Hence \(h_1 \rho_1 g=h_2 \rho_2 g\) or \(\rho_2=h_1 \rho_1 / h_2=10 \times 1 / 12.5=0.8 \mathrm{~g} \mathrm{~cm}^{-3}\)
Therefore, relative density of spirit \(=\rho_2 / \rho_1=0.8 / 1=0.8\)

Hint

Height of the water column, \(h_1=10+15=25 \mathrm{~cm}\)
Height of the spirit column, \(h_2=12.5+15=27.5 \mathrm{~cm}\)
Density of water, \(\rho_1=1 \mathrm{~g} \mathrm{~cm}^{-3}\)
Density of spirit, \(\rho_2=0.8 \mathrm{~g} \mathrm{~cm}^{-3}\)
Density of mercury \(=13.6 \mathrm{~g} \mathrm{~cm}^{-3}\)
Let \(h\) be the difference between the levels of mercury in the two arms.
Pressure exerted by height \(h\), of the mercury column:
\(=h p \mathrm{~g}\)
\(=h \times 13.6 g \ldots\) (i)
Difference between the pressures exerted by water and spirit:
\(
\begin{aligned}
& =\rho_1 h_1 g-\rho_2 h_2 g \\
& =g(25 \times 1-27.5 \times 0.8) \\
& =3 g \ldots \text { (ii) }
\end{aligned}
\)
Equating equations (i) and (ii), we get:
\(
\begin{aligned}
& 13.6 \mathrm{hg}=3 \mathrm{~g} \\
& h=0.220588 \approx 0.221 \mathrm{~cm}
\end{aligned}
\)
Hence, the difference between the levels of mercury in the two arms is \(0.221 \mathrm{~cm}\).

Hint

Bernoulli’s equation cannot be used to describe the flow of water through a rapid in a river because of the turbulent flow of water. This principle can only be applied to a streamlined flow. Bernoulli’s equation itself is derived under the condition it must be a streamline flow.

Hint

No, it does not matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation, provided the atmospheric pressure at the two points where Bernoulli’s equation is applied are significantly different.

Hint

\(
\begin{aligned}
l & =1.5 \mathrm{~m}, \quad r=1 \times 10^{-2} \mathrm{~m}, \\
\text { Volume/s, } \quad V & =\frac{\text { Mass } / \mathrm{s}}{\text { Density }}=\frac{4 \times 10^{-3}}{1.3 \times 10^3} \mathrm{~m}^3 \mathrm{~s}^{-1} \\
& =\frac{4}{1.3} \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1} \\
\eta & =0.83 \mathrm{~Pa} \mathrm{~s} \\
\text { Now, } \quad V & =\frac{\pi p r^4}{8 \eta l},
\end{aligned}
\)
where \(p\) is the pressure difference across the capillary.
or
\(
p=\frac{8 V \eta l}{\pi r^4}
\)
Substituting values,
\(
\begin{aligned}
p & =\& \times \frac{4}{1.3} \times 10^{-6} \times 0.83 \times 1.5 \times \frac{7}{22} \times \frac{1}{10^{-8}} \mathrm{~Pa} \\
& =9.75 \times 10^2 \mathrm{~Pa}
\end{aligned}
\)
The Reynolds number is \(0.3\). So, the flow is laminar.

Hint

Let \(v_1, v_2\) be the speeds on the upper and lower surfaces of the wing of aeroplane, and \(P_1\) and \(P_2\) be the pressures on upper and lower surfaces of the wing respectively. Then \(v_1=70 \mathrm{~ms}^{-1} ; \quad v_2=63 \mathrm{~ms}^{-1} ; \rho=1.3 \mathrm{~kg} \mathrm{~m}^{-3}\).
From Bernoulli’s theorem
\(
\begin{array}{rlrl}
\frac{P_1}{\rho}+g h+\frac{1}{2} v_1^2 & =\frac{P_2}{\rho}+g h+\frac{1}{2} v_2^2 \\
\therefore & \frac{P_1}{\rho}-\frac{P_2}{\rho} & =\frac{1}{2}\left(v_2^2-v_1^2\right)
\end{array}
\)
or
\(
P_1-P_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right)=\frac{1}{2} \times 1.3\left[(70)^2-(63)^2\right] \mathrm{Pa}=605.15 \mathrm{~Pa} .
\)
This difference of pressure provides the lift to the aeroplane. So, lift on the aeroplane \(=\) pressure difference \(\times\) area of wings
\(
\begin{aligned}
& =605.15 \times 2.5 \mathrm{~N}=1512.875 \mathrm{~N} \\
& =1.51 \times 10^3 \mathrm{~N} .
\end{aligned}
\)

Hint

Figure (a) is incorrect. It is because of the fact that at the kink, the velocity of the flow of liquid is large, and hence using Bernoulli’s theorem the pressure is less. As a result, the water should not rise higher in the tube where there is a kink (i.e., where the area of the cross-section is small).

Hint

Total cross-sectional area of 40 holes, \(a_2\)
\(
\begin{aligned}
& =40 \times \frac{22}{7} \times \frac{\left(1 \times 10^{-3}\right)^2}{4} \mathrm{~m}^2 \\
& =\frac{22}{7} \times 10^{-5} \mathrm{~m}^2
\end{aligned}
\)
Cross-sectional area of tube, \(a_1=8 \times 10^{-4} \mathrm{~m}^2\)
Speed inside the tube, \(v_1=1.5 \mathrm{~m} \mathrm{~min}^{-1}=\frac{1.5}{60} \mathrm{~ms}^{-1}\);
Speed of ejection, \(\quad v_2=\)?
Using \(a_2 v_2=a_1 v_1\)
we get
\(
v_2=\frac{a_1 v_1}{a_2}=\frac{8 \times 10^{-4} \times \frac{1.5}{60} \times 7}{22 \times 10^{-5}} \mathrm{~ms}^{-1}=0.64 \mathrm{~ms}^{-1} .
\)

Hint

In present case force of surface tension is balancing the weight of \(1.5 \times 10^{-2} \mathrm{~N}\), hence force of surface tension, \(F=1.5 \times 10^{-2} \mathrm{~N}\).
Total length of liquid film, \(l=2 \times 30 \mathrm{~cm}=60 \mathrm{~cm}=0.6 \mathrm{~m}\) because the liquid film has two surfaces.
Surface tension, \(T=F / l=1.5 \times 10^{-2} \mathrm{~N} / 0.6 \mathrm{~m}=2.5 \times 10^{-2} \mathrm{Nm}^{-1}\)

Hint

Take case (a):
The length of the liquid film supported by the weight, \(l=40 \mathrm{~cm}=0.4 \mathrm{~cm}\)
The weight supported by the film, \(W=4.5 \times 10^{-2} \mathrm{~N}\)
A liquid film has two free surfaces.
\(\therefore\) Surface tension \(=W / 2l\)
\(
\begin{aligned}
& =4.5 \times 10^{-2} /(2 \times 0.4) \\
& =5.625 \times 10^{-2} \mathrm{~N} / \mathrm{m}
\end{aligned}
\)
In all the three figures, the liquid is the same.
Temperature is also the same for each case.
Hence, the surface tension in figure (b) and figure (c) is the same as in figure (a).
i.e., \(5.625 \times 10^{-2} \mathrm{~N} \mathrm{~m}^{-1}\).
Since the length of the film in all the cases is \(40 \mathrm{~cm}\), the weight supported in each case is \(4.5 \times 10^{-2} \mathrm{~N}\).

Hint

\(
\begin{aligned}
\text { Excess pressure } & =\frac{2 \sigma}{R}=\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}=310 \mathrm{~Pa} \\
\text { Total pressure } & =1.01 \times 10^5+\frac{2 \sigma}{R} \\
& =1.01 \times 10^5+310=1.0131 \times 10^5 \mathrm{~Pa}
\end{aligned}
\)
Since data is correct up to three significant figures, we should write total pressure inside the drop as \(1.01 \mathrm{x}\) \(10^5 \mathrm{~Pa}\).

Hint

Here surface tension of soap solution at room temperature
\(\mathrm{T}=2.50 \times 10^{-2} \mathrm{Nm}^{-1}\), radius of soap bubble, \(r=5.00 \mathrm{~mm}=5.00 \times 10^{-3} \mathrm{~m}\).
\(\therefore\) Excess pressure inside soap bubble, \(P=P_i-P_0=\frac{4 T}{r}\)
\(
=\frac{4 \times 2.50 \times 10^{-2}}{5.00 \times 10^{-3}}=20.0 \mathrm{~Pa}
\)
When an air bubble of radius \(r=5.00 \times 10^{-3} \mathrm{~m}\) is formed at a depth \(h=40.0 \mathrm{~cm}=\) \(0.4 \mathrm{~m}\) inside a container containing a soap solution of relative density \(1.20\) or density \(\rho=1.20 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\), then excess pressure
\(
\begin{aligned}
P & =P_i-P_0=\frac{2 T}{r} \\
\therefore \quad P_i & =P_0+\frac{2 T}{r}=\left(\mathrm{P}_a+h \rho g\right)+\frac{2 T}{r} \\
& =\left[1.01 \times 10^5+0.4 \times 1.2 \times 10^3 \times 9.8+\frac{2 \times 2.50 \times 10^{-2}}{5.00 \times 10^{-3}}\right] \mathrm{Pa} \\
& =\left(1.01 \times 10^5+4.7 \times 10^3+10.0\right) \mathrm{Pa} \\
& \simeq 1.06 \times 10^5 \mathrm{~Pa} .
\end{aligned}
\)

Hint

Pressure difference across the door
\(
\begin{aligned}
= & (4 \times 1700 \times 9.8-4 \times 1000 \times 9.8) \mathrm{Pa} \\
\left(6.664 \times 10^4-3.92 \times 10^4\right) & \mathrm{Pa}=2.774 \times 10^4 \mathrm{~Pa} \\
\text { Force on the door } & =\text { Pressure difference } \times \text { Area of door } \\
& =2.774 \times 10^4 \times 20 \times 10^{-4} \mathrm{~N} \\
& =54.88 \mathrm{~N}=55 \mathrm{~N} .
\end{aligned}
\)
Note. The base area does not affect the answer.

Hint

The atmospheric pressure, \(P=76 \mathrm{~cm}\) of mercury
(a) From figure \((a)\),
Pressure head, \(\quad h=20 \mathrm{~cm}\) of mercury
\(\therefore\) Absolute pressure \(=p+h=76+20=96 \mathrm{~cm}\) of mercury
Also, Gauge pressure \(=h=20 \mathrm{~cm}\) of mercury
From figure \((b)\),
pressure head, \(\quad h=-18 \mathrm{~cm}\) of mercury
\(\therefore\) Absolute pressure \(=p+h=76+(-18)=58 \mathrm{~cm}\) of mercury
Also, Gauge pressure \(=h=-18 \mathrm{~cm}\) of mercury

(b) When \(13.6 \mathrm{~cm}\) of water is poured into the right limb of the manometer of figure \((b)\), then, using the relation:
\(
\text { Pressure }=\rho g h=\rho^{\prime} g^{\prime} h^{\prime}
\)
We get \(h^{\prime}=\frac{\rho h}{\rho^{\prime}}=\frac{1 \times 13.6}{13.6}=1 \mathrm{~cm}\) of mercury
\(\left[\rho^{\prime}=\right.\) density of mercury \(]\)
Therefore, pressure at the point \(B\),
\(
p_B=P+h^{\prime}=76+1=77 \mathrm{~cm} \text { of mercury }
\)
If \(h^{\prime \prime}\) is the difference in the mercury levels in the two limbs, then taking \(P_A=P_B\) \(\Rightarrow \quad 58+h^{\prime \prime}=77 \Rightarrow h^{\prime \prime}=77-58=19 \mathrm{~cm}\) of mercury.

Hint

 Pressure (and therefore force) on the two equal base areas are identical. But force is exerted by water on the sides of the vessels also, which has a non-zero vertical component when sides of the vessel are not perfectly normal to the base. This net vertical component of force by water on the sides of the vessel is greater for the first vessel than the second. Hence, the vessels weigh different even when the force on the base is the same in the two cases.

Hint

\(h=P / \rho g=200 /\left(1.06 \times 10^3 \times 9.8\right)=0.1925 \mathrm{~m}\)
The blood may just enter the vein if the height at which the blood container be kept must be slightly greater than \(0.1925 \mathrm{~m} = 0.2 \mathrm{~m}\).

Hint

(a) If dissipative forces are present, then some forces in liquid flow due to pressure difference is spent against dissipative forces, due to which the pressure drop becomes large.
(b) The dissipative forces become more important with increasing flow velocity, because of turbulence.

Hint

Here,
\(
\begin{aligned}
& r=2 \times 10^{-3} \mathrm{~m} ; \quad D=2 r=2 \times 2 \times 10^{-3}=4 \times 10^{-3} \mathrm{~m} ; \\
& \eta=2.084 \times 10^{-3} \quad \text { Pa-s; } \rho=1.06 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3} .
\end{aligned}
\)
For flow to be laminar, \(N_R=2000\)
(a) Now, \(v_{\mathrm{c}}=\frac{N_R \eta}{\rho D}=\frac{2000 \times\left(2.084 \times 10^{-3}\right)}{\left(1.06 \times 10^3\right) \times\left(4 \times 10^{-3}\right)}=0.98 \mathrm{~m} / \mathrm{s}\).
(b) Volume flowing per second \(=\pi r^2 v_c=\frac{22}{7} \times\left(2 \times 10^{-3}\right)^2 \times 0.98=1.24 \times 10^{-5} \mathbf{m}^3 \mathrm{~s}^{-1}\).

Hint

Here speed of air over lower wing, \(v_1=180 \mathrm{~km} / \mathrm{h}=180 \times \frac{5}{18}=50 \mathrm{~ms}^{-1}\)
Speed over the upper wing, \(v_2=234 \mathrm{~km} / \mathrm{h}=234 \times \frac{5}{18}=65 \mathrm{~ms}^{-1}\)
\(\therefore\) Pressure difference, \(P_1-P_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right)=\frac{1}{2} \times 1\left(65^2-50^2\right)=862.5 \mathrm{~Pa}\)
\(\therefore\) Net upward force, \(F=\left(P_1-P_2\right) A\)
This upward force balances the weight of the plane.
\(
\begin{array}{ll}
\therefore \quad m g & =F=\left(P_1-P_2\right) A \\
\therefore \quad & {\left[A=25 \times 2=50 \mathrm{~m}^2\right]} \\
& m=\frac{\left(P_1-P_2\right) A}{g}=\frac{862.5 \times 50}{9.8}=4400 \mathrm{~Kg} .
\end{array}
\)

Hint

Here radius of drop, \(r=2.0 \times 10^{-5} \mathrm{~m}\), density of drop, \(p=1.2 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\), viscosity of air \(\mathrm{TI}=1.8 \mathrm{x}\) \(10^{-5} \mathrm{~Pa}-\mathrm{s}\)
Neglecting upward thrust due to air, we find that terminal speed is
\(
\begin{aligned}
v_T & =\frac{2}{9} \frac{r^2 \rho g}{\eta}=\frac{2 \times\left(2.0 \times 10^{-5}\right)^2 \times\left(1.2 \times 10^3\right) \times 9.8}{9 \times\left(1.8 \times 10^{-5}\right)} \\
& =5.81 \times 10^{-2} \mathrm{~ms}^{-1} \quad \text { or } 5.81 \mathrm{~cm} \mathrm{~s}^{-1}
\end{aligned}
\)
Viscous force at this speed,
\(
\begin{aligned}
F=6 \pi \eta r v & =6 \times 3.14 \times\left(1.8 \times 10^{-5}\right) \times\left(2.0 \times 10^{-5}\right) \times\left(5.81 \times 10^{-2}\right) \\
& =3.94 \times 10^{-10} \mathrm{~N} .
\end{aligned}
\)

Hint

Radius of tube, \(\quad r=1.00 \mathrm{~mm}=10^{-3} \mathrm{~m}\)
Surface tension of mercury, \(\sigma=0.465 \mathrm{Nm}^{-1}\)
Density of mercury, \(\sigma=13.6 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\)
Angle of contact, \(\quad \theta=140^{\circ}\)
\(
\begin{aligned}
\therefore \quad h & =\frac{2 \sigma \cos \theta}{r \rho g}=\frac{2 \times 0.465 \times \cos 140^{\circ}}{10^{-3} \times 13.6 \times 10^3 \times 9.8} \\
& =\frac{2 \times 0.465 \times(-0.7660)}{10^{-3} \times 13.6 \times 10^3 \times 9.8} \\
& =-5.34 \times 10^{-3} \mathrm{~m}=-5.34 \mathrm{~mm}
\end{aligned}
\)
Negative sign shows that the mercury level is depressed in the tube.

Hint

Let \(r_1\) be the radius of one bore and \(r_2\) be the radius of second bore of the \(U\)-tube. The, if \(h 1\) and \(h_2\) are the heights of water on two sides, then
\(
h_1=\frac{2 S \cos \theta}{r_1 \rho g} \text { and } h_2=\frac{2 S \cos \theta}{r_2 \rho g}
\)
On subtraction, we get
Here,
\(
\begin{aligned}
& h_1-h_2=\frac{2 S \cos \theta}{r_1 \rho g}-\frac{2 S \cos \theta}{r_2 \rho g}=\frac{2 S \cos \theta}{\rho g}\left[\frac{1}{r_1}-\frac{1}{r_2}\right] \\
& S=7.3 \times 10^{-2} \mathrm{Nm}^{-1}, \quad \theta=0, \quad \rho=1.0 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3} \text {, } \\
& g=9.8 \mathrm{~ms}^{-2}, \quad r_1=\frac{3}{2} \mathrm{~min}=1.5 \times 10^{-3} \mathrm{~m} \quad \text { and } \quad r_2=\frac{6}{2} \mathrm{~mm} \\
& =3 \times 10^{-3} \mathrm{~m} \\
& \therefore \quad h_1-h_2=\frac{2 \times 7.3 \times 10^{-2} \times \cos \theta}{1 \times 10^3 \times 9.8}\left[\frac{1}{1.5 \times 10^{-3}}-\frac{1}{3 \times 10^{-3}}\right] \\
& =1.49 \times 10^{-5} \times \frac{1}{3 \times 10^{-3}} \approx 4.97 \times 10^{-3} \mathrm{~m}=4.97 \mathrm{~mm} \\
&
\end{aligned}
\)

Hint

(a) We know that rate of decrease of density \(p\) of air is directly proportional to the height \(y\). It is given as \(d \rho / d y=-\rho / y_0\)
where \(y\) is a constant of proportionality and -ve sign signifies that density is decreasing with an increase in height. On integration, we get
\(
\begin{aligned}
\int_{\rho_0}^\rho \frac{d \rho}{\rho}=-\int_0^y \frac{1}{y_0} d y \\
\Rightarrow \quad[\log \rho]_{\rho_0}^\rho=-\left[\frac{y}{y_0}\right]_0^y, \text { where, } \rho_0=\text { density of air at sea level } \text { i.e., } y=0 \\
\text { or } \log _e \frac{\rho}{\rho_0}=-\frac{y}{y_0} \text { or } \rho=\rho_0 e^{-\frac{y}{y_0}}
\end{aligned}
\)
Here dimensions and units of constant \(y_0\) are same as of \(y\).
(b) Here volume of He balloon, \(V=1425 \mathrm{~m}^3\), mass of payload, \(m=400 \mathrm{~kg}\)
\(
y_0=8000 \mathrm{~m} \text {, density of } \mathrm{He} \rho_{\mathrm{He}}=0.18 \mathrm{kgm}^{-3}
\)
Mean density of balloon, \(\rho=\frac{\text { Total mass of balloon }}{\text { Volume }}=\frac{m+V \cdot \rho_{\mathrm{He}}}{\mathrm{Va}} \mathrm{Pa}\)
\(
=\frac{400+1425 \times 0.18}{1425}=0.4608=0.46 \mathrm{kgm}^{-3}
\)
As density of air at sea level \(\rho_0=1.25 \mathrm{~kg} \mathrm{~m}^{-3}\). The balloon will rise up to a height \(y\) where density of air \(=\) density of balloon \(\rho=0.46 \mathrm{kgm}^{-3}\)
As \(\quad \rho=\rho_0 e^{-\frac{y}{y_0}}\) or \(\frac{\rho_0}{\rho}=e^{\frac{y_0}{y}}\)
\(
\begin{aligned}
\therefore \quad \log _e\left(\frac{\rho_0}{\rho}\right) & =\frac{y_0}{y} \quad \text { or } \quad y=\frac{y_0}{\log _e\left(\frac{\rho_0}{\rho}\right)}=\frac{8000}{\log _e\left(\frac{1.25}{0.46}\right)} \\
& =8002 \mathrm{~m} \text { or } 8.0 \mathrm{~km} .
\end{aligned}
\)

Hint

Length of thread \(l=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}\)
Surface tension of water \(\mathrm{T}=0.76 \mathrm{~N} / \mathrm{m}\)
We know that:
\(
\begin{aligned}
F & =T \times l=0.76 \times 5 \times 10^{-2} \\
& =3.8 \times 10^{-3} \mathrm{~N}
\end{aligned}
\)
Therefore, the water surface on one side of the thread pulls it with a force of \(3.8 \times 10^{-3} \mathrm{~N}\)

Hint

Radius of mercury drop \(\mathrm{r}=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
Radius of soap bubble \(r=4 \mathrm{~mm}=4 \times 10^{-3} \mathrm{~m}\)
Radius of air bubble \(r=4 \mathrm{~mm}=4 \times 10^{-3} \mathrm{~m}\)
Surface tension of mercury \(\mathrm{T}_{H g}=0.465 \mathrm{~N} / \mathrm{m}\)
Surface tension of soap solution \(\mathrm{T}_s=0.03 \mathrm{~N} / \mathrm{m}\)
Surface tension of water \(\mathrm{T}_w=0.076 \mathrm{~N} / \mathrm{m}\)

(a) Excess pressure inside mercury drop:
\(
\begin{aligned}
& P=\frac{2 T_{H g}}{r} \\
& =\frac{0.465 \times 2}{2 \times 10^{-3}}=465 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)

(b) Excess pressure inside the soap bubble:
\(
\begin{aligned}
& P=\frac{4 T_s}{r} \\
& =\frac{4 \times 0.03}{4 \times 10^{-3}}=30 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)

(c) Excess pressure inside the air bubble:
\(
\begin{aligned}
& P=\frac{2 T_w}{r} \\
& =\frac{2 \times 0.076}{4 \times 10^{-3}}=38 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)

Hint

Given:
Surface area of mercury drop, \(A=1 \mathrm{~mm}^2=10^{-6} \mathrm{~m}^2\)
Radius of mercury drop, \(r=4 \mathrm{~mm}=4 \times 10^{-3} \mathrm{~m}\)
Atmospheric pressure, \(\mathrm{P}_0=1.0 \times 10^5 \mathrm{P}_{\mathrm{a}}\)
Surface tension of mercury, \(\mathrm{T}=0.465 \mathrm{~N} / \mathrm{m}\)
(a) Force exerted by air on the surface area:
\(
\begin{aligned}
& F=P_0 A \\
& \Rightarrow F=1.0 \times 10^5 \times 10^{-6}=0.1 \mathrm{~N}
\end{aligned}
\)
(b) Force exerted by mercury below the surface area :
Pressure \(\mathrm{P}^{\prime}=\mathrm{P}_0+\frac{2 T}{\mathrm{r}}\)
\(
\begin{aligned}
& F=P^{\prime} A=\left(P_0+\frac{2 T}{r}\right) A \\
& =\left(0.1+\frac{2 \times 0.465}{4 \times 10^{-3}}\right) \times 10^{-6} \\
& =0.1+0.00023=0.10023 \mathrm{~N}
\end{aligned}
\)
(c) Force exerted by mercury surface in contact with it:
\(
\begin{aligned}
& P=\frac{2 T}{r} \\
& F=P A=\frac{2 T}{r} A \\
& =\frac{2 \times 0.465}{4 \times 10^{-3}} \times 10^{-6}=0.00023 \mathrm{~N}
\end{aligned}
\)

Hint

Given:
Surface tension of water \(T=7.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) Taking \(\cos \theta=1\) :
Radius of capillary \(A\left(r_A\right)=0.5 \mathrm{~mm}=0.5 \times 10^{-3} \mathrm{~m}\)
Height of water level in capillary A:
\(
\begin{aligned}
& h_{\mathrm{A}}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r}_{\mathrm{A}} \rho \mathrm{g}} \\
& =\frac{2 \times 7.5 \times 10^{-2}}{0.5 \times 10^{-3} \times 1000 \times 10} \\
& =3 \times 10^{-2} \mathrm{~m}=3 \mathrm{~cm}
\end{aligned}
\)
Radius of capillary \(B\left(r_B\right)=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}\)

Height of water level in capillary B:
\(
\begin{aligned}
& h_B=\frac{2 \mathrm{~T} \cos \theta}{r_B \rho g} \\
& =\frac{2 \times 7.5 \times 10^{-2}}{1 \times 10^{-3} \times 10^3 \times 10} \\
& =15 \times 10^{-3} \mathrm{~m}=1.5 \mathrm{~cm}
\end{aligned}
\)
Radius of capillary \(C\left(r_C\right)=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m}\)

Height of water level in capillary C:
\(
\begin{aligned}
& \mathrm{h}_{\mathrm{C}}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r}_{\mathrm{C}} \rho \mathrm{g}} \\
& =\frac{2 \times 7.5 \times 10^{-2}}{1.5 \times 10^{-3} \times 10^3 \times 10} \\
& =\frac{15}{1.5} \times 10^{-3} \mathrm{~m}=1 \mathrm{~cm}
\end{aligned}
\)

Hint

Rise \(/\) depression \(=2\) S \(\cos \theta / r \rho g\)
For mercury, \(-0.02=2 S^{\prime} \cdot \cos \theta / r \rho^{\prime} \text { g } \dots(i)
\)
For water rise, \(\mathrm{h}=2 \mathrm{~S} \cdot \cos 0^{\circ} / r \rho g \dots(ii)\)
Dividing (ii) by (i)
\(
\begin{aligned}
& \mathrm{h} / 0.02=-\mathrm{S} \rho^{\prime} /\left(\mathrm{S}^{\prime} \times \cos \theta \times \rho\right) \\
& \rightarrow \mathrm{h}=-0.02 \times \mathrm{~S} \rho^{\prime} /\left(1000 \mathrm{~S}^{\prime} \times \cos \theta\right) \\
& \mathrm{S}^{\prime}=7.5 \times 10^{-2} \mathrm{~N} / \mathrm{m} . \mathrm{S}=0.465 \mathrm{~N} / \mathrm{m} \\
& \rho^{\prime}=13600 \mathrm{~kg} / \mathrm{m}^3 \\
& \theta=140^{\circ} \\
& \text { So, } h=-0.02 \times 7.5 \times 10^{-2} \times 13600 /\left(1000 \times 0.465 \times \cos 140^{\circ}\right) \\
& \rightarrow \mathrm{h}=0.0573 \mathrm{~m}=\mathbf{5 . 7 3} \mathrm{cm}
\end{aligned}
\)

Hint

Given:
Radius of tube \(r=1.0 \mathrm{~mm}\)
Atmospheric pressure \(=76 \mathrm{~cm}\) of \(\mathrm{Hg}\)
Contact angle of mercury with glass \(\theta=135^{\circ}\)
Surface tension of mercury \(T=0.465 \mathrm{~N} / \mathrm{m}\)
Density of mercury \(=13600 \mathrm{~kg} \mathrm{~m}^{-3}\)
Let \(\mathrm{h}\) be the rise in level in the baromet
\(
\begin{aligned}
& \mathrm{h}=\frac{2 T \cos \theta}{r \rho g} \\
& =\frac{2 \times 465 \times(1 / \sqrt{2})}{10^{-3} \times 13600 \times 10}=0.0048 \mathrm{~m} \\
& =0.48 \mathrm{~cm} \\
& \therefore \text { Net rise in level in the barometer tube }=\mathrm{H}-h \\
& =76-0.48 \\
& =75.52 \mathrm{~cm} \\
&
\end{aligned}
\)

Hint

Given:
Radius of capillary tube \(r=0.5 \mathrm{~mm}=5 \times 10^{-4} \mathrm{~m}\)
Depth (where pressure is to be found) \(\mathrm{h}=5.0 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}\)
Surface tension of water \(\mathrm{T}=0.075 \mathrm{~N} / \mathrm{m}\)
Excess pressure at \(5 \mathrm{~cm}\) before the surface:
\(
P=\rho h g=1000 \times\left(5 \times 10^{-2}\right) \times 9.8=490 \mathrm{~N} / \mathrm{m}^2
\)
Excess pressure at the surface is given by:
\(
\begin{aligned}
& P_0=\frac{2 T}{r}=\frac{2 \times(0.75)}{\left(5 \times 10^{-4}\right)} \\
& =300 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)
Difference in pressure: \(\mathrm{P}_0-\mathrm{P}\)
\(
=490-300=190 \mathrm{~N} / \mathrm{m}^2
\)
Hence, the required difference in pressure is \(190 \mathrm{~N} / \mathrm{m}^2\).

Hint

Given:
Radius of cylindrical vessel, \(r=6.0 \mathrm{~cm}=0.06 \mathrm{~m}\)
Surface tension of water, \(T=0.075 \mathrm{~J} / \mathrm{m}^2\)
Area, \(A=\pi r^2=\pi \times(0.06)^2\)
Surface energy \(=\mathrm{T} \times \mathrm{A}\)
\(
\begin{aligned}
& =(0.075) \times(3.14) \times(0.06)^2 \\
& =8.5 \times 10^{-4} \mathrm{~J}
\end{aligned}
\)
Therefore, the surface energy of water kept in a cylindrical vessel is \(8.5 \times 10^{-4} \mathrm{~J}\).

Hint

Given:
Initial radius of mercury drop \(R=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
Surface tension of mercury \(T=0.465 \mathrm{~J} / \mathrm{m}^2\)
Let the radius of a small drop of mercury be \(r\).
As one big drop is split into 8 identical droplets:
volume of initial drop \(=8 \times\) (volume of a small drop)
\(
\left(\frac{4}{3}\right) \pi R^3=\left(\frac{4}{3}\right) \pi r^3 \times 8
\)
Taking cube root on both sides of the above equation: \(r=\frac{R}{2}=10^{-3} \mathrm{~m}\)
Surface energy \(=\mathrm{T} \times\) surface area
\(
\begin{aligned}
& \therefore \text { Increase in surface energy }=\mathrm{TA}^{\prime}-\mathrm{TA} \\
& =\left(8 \times 4 \pi r^2-4 \pi R^2\right) T \\
& =4 \pi T\left[8 \times\left(\frac{R^2}{4}\right)-R^2\right] \\
& =4 \pi T R^2 \\
& =4 \times(3.14) \times(0.465) \times\left(4 \times 10^{-6}\right) \\
& =23.36 \times 10^{-6} \\
& =23.4 \mu \mathrm{J} \\
&
\end{aligned}
\)
Hence, the required increase in the surface energy of the mercury droplets is \(23.4 \mu \mathrm{J}\).

Hint

(a) Let \(T\) be the surface tension and \(\rho\) be the density of the liquid.
Then, for \(\cos \theta=1\), height \((h)\) of liquid level: \(h=\frac{2 T}{r \rho g} \dots(i)\) where \(g\) is the acceleration due to gravity
\(
\begin{aligned}
& \Rightarrow \mathrm{h}=\frac{2 \times(0.076)}{10^{-3} \times 10 \times 100} \\
& =1.52 \mathrm{~cm} \\
& =1.52 \times 10^{-2} \mathrm{~m} \\
& =1.52 \mathrm{~cm}
\end{aligned}
\)

(b) Let the new length of the tube be \(h\) :
\(
\begin{gathered}
\mathrm{h}^{\prime}=\frac{2 T \cos \theta}{\mathrm{r} \rho \mathrm{g}} \\
\cos \theta=\frac{\mathrm{h}^{\prime} \mathrm{r} \rho g}{2 T}
\end{gathered}
\)
Using equation (i), we get:
\(
\begin{aligned}
& \cos \theta=\frac{h^{\prime}}{h}=\frac{1}{2}\left(\text { Because } h^{\prime}=\frac{h}{2}\right) \\
& \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ}
\end{aligned}
\)
The water surface in the capillary makes an angle of \(60^{\circ}\) with the wall.

Hint

\(
\begin{aligned}
& r=1 \mathrm{~mm}=10^{-3} \mathrm{~m}, \theta=135^{\circ} \\
& h=\frac{2 T \cos \theta}{r \rho g} \theta \\
& =\frac{2 \times 0.465 \times \cos 135^{\circ}}{10^{-3} \times 13600 \times(10)} \\
& =0.0049 \mathrm{~m}=4.9 \mathrm{~mm} \\
& \text { Depression }=4.9 \mathrm{~m} \\
& \mathrm{~h}^{\prime}=\frac{h}{2} \\
& \mathrm{~h}^{\prime}=\frac{2 T \cos \theta}{\operatorname{\rho\times rg}} \\
& \frac{h^{\prime}}{h}=\frac{2 T \cos \theta}{\operatorname{\rho\times rg}} \\
& \frac{h^{\prime}}{h}=\frac{\left(2 \mathrm{~T} \cos \theta^{\prime} \times r 0 \times r g\right)}{\operatorname{\rho\times rg} \times 2 \mathrm{~T} \cos 135}=\frac{\cos \theta^{\prime}}{\cos 135} \\
& \frac{1}{2}=\frac{\left(\cos \theta^{\prime}\right)}{\cos 135} \\
& \theta^{\prime}=111^{\circ}
\end{aligned}
\)

Hint

Given:
Surface tension of water \(\mathrm{T}=0.075 \mathrm{~N} / \mathrm{m}\)
Separation between the glass plates \(\mathrm{d}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}\)
Density of water \(\rho=10^3 \mathrm{~kg} / \mathrm{m}^3\)
Applying law of conservation of energy:
\(
\begin{aligned}
& \mathrm{T}(2 \mathrm{~L})=\left[1 \times\left(10^{-3}\right) \times \mathrm{h}\right] \rho \mathrm{g} \\
& \Rightarrow \mathrm{h}=\frac{2 \times(0.075)}{10^{-3} \times 10^3 \times 10} \\
& =0.015 \mathrm{~m}=1.5 \mathrm{~cm}
\end{aligned}
\)
Therefore, the rise of water in the space between the plates is \(1.5 \mathrm{~cm}\).

Hint

Given:
Edge of the ice cube (a) \(=1.0 \mathrm{~cm}\)
The water that is formed due to the melting of ice acquires a spherical surface. In the absence of gravity, let the radius of the spherical surface be \(r\).
Volume of ice cube \(=\) volume of spherical surface of water
\(
\begin{aligned}
& \Rightarrow a^3=\frac{4}{3} \pi r^3 \\
& \Rightarrow r=\left[\frac{3 a^3}{4 \pi}\right]^{1 / 3}
\end{aligned}
\)
Surface area of spherical water surface \(=4 \pi r^2\)
\(
\begin{aligned}
& =4 \pi\left[\frac{3 a^3}{4 \pi}\right]^{2 / 3} \\
& =(36 \pi)^{1 / 3} \mathrm{~cm}^2
\end{aligned}
\)

Hint

Considering a small section on thread subtending an angle of \(2 d \theta\) at center.
Force due to soap solution on this section=2T(2R \(d\theta)\)
This force is balanced by tension in thread \(=2 T^{\prime} \sin d \theta=2 T^{\prime} d \theta\)
By equilibrium of force
\(2 T(2 \mathrm{Rd} \theta)=2 \mathrm{~T}^{\prime} \mathrm{d} \theta\)
\(T^{\prime}=2 T R\)
\(T^{\prime}=2 T\left(\frac{L}{2 \pi}\right)\)

\(T^{\prime}=\frac{T L}{\pi}\)
\(=\frac{(0.03)\left(6.28 \times 10^{-2}\right)}{3.14}\)
\(T^{\prime}=6 \times 10^{-4} \mathrm{~N}\)

Hint

(a)
\(
\text { Viscous force }=6 \pi \eta r v
\)
\(
\begin{aligned}
& =6 \times 3.14 \times 0.8 \times 10^{-3} \times 10^{-2} \\
& \mathrm{~F}=1.5 \times 10^{-4} \mathrm{~N}
\end{aligned}
\)

(b)
\(\text { Hydrostatic force }={ V \rho g }\)
\(
\begin{aligned}
& =\frac{4}{3} \pi^3 \mathrm{\rho g} \\
& =\frac{4}{3} \times 3.14 \times\left(10^{-3}\right)^3 \times 1260 \times 10 \\
& =5.2 \times 10^{-5} \mathrm{~N}
\end{aligned}
\)

(c)
c) At the time of terminal velocity
Download force \(=\) Upward force
\(
\begin{aligned}
& \mathrm{mg}=\mathrm{V \rho g}+6 \pi \eta r v \\
& 50 \times 10^{-3} \times 10=5.2 \times 10^{-5}+6 \times 3.14 \times 0.8 \times 10^{-3} \mathrm{~V} \\
& \mathrm{~V}=2.9 \times 10^{-2} \mathrm{~m} / \mathrm{s} \\
& \mathrm{V}=2.9 \mathrm{~cm} / \mathrm{sec}
\end{aligned}
\)

Hint

Estimate the speed of vertically falling raindrops from the following data. Radius of the drops \(=0.02\) \(\mathrm{cm}\), viscosity of air \(=1.8 \times 10^{-4}\) poise, \(\mathrm{g}=9.9 \mathrm{~m} / \mathrm{s}^2\) and density of water \(=1000 \mathrm{~kg} / \mathrm{m}^3\).
\(
\begin{aligned}
& \mathrm{V}=\frac{2 \mathrm{r}^2(\rho-\sigma) \mathrm{g}}{9 \eta} \\
& \mathrm{V}=\frac{2 \times\left(0.02 \times 10^{-2}\right)^2(1000-0)(9.9)}{9 \times 1.8 \times 10^{-5}} \\
& \mathrm{~V}=5 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \mathrm{V}=6 \times 10^{-2} \mathrm{~m} / \mathrm{s} \\
& \mathrm{R}=1^{\times 10^{-2}} \mathrm{~m} \\
& \eta=0.01 \\
& \text { poise }=10^{-3} \text { poisullie }
\end{aligned}
\)
Reynold’s number
\(
\begin{aligned}
& N=\frac{\rho v D}{\eta} \\
& =\frac{1000 \times 6 \times 10^{-2} \times 2 \times 10^{-2}}{10^{-3}} \\
& N=1200
\end{aligned}
\)
Reynold number is less than 1200 . Therefore, it is a steady Flow.