Exemplar MCQs

Hint

(c) In fluids, when the pebble is dropped from the top of a tall cylinder filled with viscous oil, a variable force called viscous force will act which increases with increase in speed. And at equilibrium this velocity becomes constant, that constant velocity is called terminal velocity.
When the pebble is falling through the viscous oil, the viscous force is \(F=6 \pi n r v\) where \(r\) is the radius of the pebble, \(v\) is instantaneous speed, \(\eta\) is coefficient of viscosity. As the force is variable, hence acceleration is also variable so \(\mathrm{v}\) – \(\mathrm{t}\) graph will not be a straight line. First velocity increases and then becomes constant known as terminal velocity.

Hint

In a streamlined flow at any given point. the velocity of each passing fluid particle remains constant. If we consider a cross-sectional area. then a point in the area cannot have different velocities at the same time, hence two streamlines of flow cannot cross each other.

Hint

(b) As discussed above for a streamline flow of a liquid velocity of each particle at a particular cross-section is constant, because \(Av\) = constant (law of continuity) between two cross-section of a tube of flow. So we can say that along a streamline, the velocity of every fluid particle while crossing a given position is the same.

Hint

(a) The situation is shown in the diagram below in which an ideal fluid is flowing through a pipe of circular cross sections.
\(
\begin{aligned}
& \mathrm{a} 1 \mathrm{v} 1=\mathrm{a} 2 \mathrm{v} 2 \\
& \frac{v 1}{v 2}=\frac{a 2}{a 1} \\
& \frac{v 1}{v 2}=\frac{d 2^2}{d 1^2}
\end{aligned}
\)
putting values,
\(
\frac{v 1}{v 2}=\frac{(3.75 \times 3.75)}{2.5 \times 2.5}=\frac{9}{4}
\)
so, \(v1: v2 =9: 4\). Hence option a is correct.

Hint

(c) Key concept: Shape of Liquid Meniscus:
(i) Force of adhesion \(F_a\) (acts outwards at right angle to the wall of the tube).
(ii) Force of cohesion \(F_c\) (acts at an angle \(45^{\circ}\) to the vertical). Resultant force \(F_N\) depends upon the value of \(F_a\) and \(F_c\). If resultant force \(F_N\) makes an angle \(\alpha\) with \(F_a\).
Then \(\tan \alpha=\frac{F_c \sin 135^{\circ}}{F_a+F_c \cos 135^{\circ}}=\frac{F_c}{\sqrt{2} F_a-F_c}\)
By knowing the direction of resultant force we can find out the shape of meniscus because the free surface of the liquid adjust itself at right angle to this resultant force.
(1) If \(F_c=\sqrt{2} F_a\), \(\tan \alpha=\infty \quad \therefore \alpha=90^{\circ}\) i.e., the resultant force acts vertically downwards. Hence the liquid meniscus must be horizontal.
(2) If \(F_c<\sqrt{2} F_a\), \(\tan \alpha=\) positive \(\quad \therefore \alpha\) is acute angle
i.e., the resultant force directed outside the liquid. Hence the liquid meniscus must be concave upward.
(3) \(F_c>\sqrt{2} F_a\) \(\tan \alpha=\) negative
\(\therefore \alpha\) is obtuse angle. i.e., the resultant force directed inside the liquid. Hence the liquid meniscus must be convex upward.

According to the question, the observed meniscus of liquid in a capillary tube is of convex upward which is only possible when angle of contact is obtuse. It is so when one end of glass capillary tube is immersed in a trough of mercury. Hence, the combination will be of mercury-glass \(\left(140^{\circ}\right)\) as shown in the figure.

Hint

Key concept: To understand the concept of tension acting on the free
surface of a liquid, let us consider four liquid molecules like A, B, C and D. Their sphere of influence are shown in the figure.
(1) Molecule A is well within the liquid, so it is attracted equally in all directions. Hence the net force on this molecule is zero and it moves freely inside the liquid.
(2) Molecule B is little below the free surface of the liquid and it is atso attracted equally in all directions. Hence the resultant force acts on it is also zero
(3) Molecule C is just below the upper surface of the liquid film and the part of its sphere of influence is outside the free liquid surface. So the number of molecules in the upper half (attracting the molecules upward) is less than the number of molecule in the lower half (attracting the molecule downward). Thus the molecule C experiences a net downward force.
(4) Molecule D is just on the free surface of the liquid. The upper half of the sphere of influence has no liquid molecule. Hence the molecule D experiences a maximum downward force.
Thus all molecules lying on surface film experiences a net downward force. Therefore, free surface of the liquid behaves like a stretched membrane.
From the key concept, it is clear from point (4) that molecules on the surface experiences a net downward force. As shown in figure above, molecule D experiences a net downward force. Because on the above side of this molecule there is no liquid molecule. So, the potential energy is more than that of a molecule inside. Hence option (b) and (d) are correct.

Hint

\((b, c)\) Pressure is defined as the ratio of magnitude of component of the force normal to the area and the area under consideration.
i.e \(P=F / A\)
Pressure is a scalar quantity. Pressure acts normal to a surface and it is always compressive in nature, therefore, only its magnitude is required for its complete description.

Hint

Key concept: When a body of density \(\rho\) and volume \(V\) is immersed in a liquid of density \(\sigma\), the forces acting on the body are:
\(\text { 1. Weight of body } W=m g=V \rho g \text {, acting vertically downwards through centre of gravity of the body }\)
2. Upthrust force \(=V \sigma g\) acting vertically upwards through the centre of gravity of the displaced liquid i.e., centre of buoyancy.
\(
\text { Case 1: If density of body is greater than that of liquid } \rho>\sigma \text {. }
\)
\(
\text { Case 2: If density of body is lesser than that of liquid } \rho<\sigma \text {. }
\)

When coin is in water, volume of water displaced by coin is equal to the volume of coin \(\mathrm{V}_1\) (say).
When the coin falls into the water, weight of the (block + coin) system decreases, which was balanced by the upthrust force earlier. As weight of the system decreases, block moves up. Hence Idecreases.
When coin is at the top of wooden block, it displaces a volume of water \(V_2\), which is more than \(V_1\) Because,
Weight of coin = Weight of volume of water displaced by coin (when coin is at .the top of wooden block)
\(
\begin{aligned}
& \Rightarrow \quad \rho_c V_1 g=\rho_l V_2 g \\
& \Rightarrow \quad \frac{V_2}{V_1}=\frac{\rho_c}{\rho_l} \\
& \text { Since, } \rho_c>\rho_l \\
& \text { So, } V_2>V_1
\end{aligned}
\)
Hence, upthrust force will also decrease. As volume of water displaced by the block decreases, hence h decreases.

Hint

\((c, d)\) The viscosity of gases increases with increase of temperature, because on increasing temperature the rate of diffusion increases.
The viscosity of liquid decreases with increase of temperature, because the cohesive force between the liquid molecules decreases with increase of temperature.
Relation between coefficient of viscosity and temperature (Andrade formula)
\(
\eta=\frac{A e^{C \rho / T}}{\rho^{-1 / 3}}
\)
where \(T\) = Absolute temperature of liquid, \(p=\) density of liquid, \(A\) and \(C\) are constants.
Important point: With increase in temperature, the coefficient of viscosity of liquids decreases but that of gases increases. The reason is that as temperature rises, the atoms of the liquid become more mobile, whereas in case of a gas, the collision frequency of atoms increases as their motion becomes more random.

Hint

(b, c) Streamline flow is more likely for liquids having low density. We know that greater the coefficient of viscosity of a liquid more will be the velocity gradient, hence each line of flow can be easily differentiated. Streamline flow is related with critical velocity. The critical velocity is that velocity of liquid flow up to which its flow is streamlined and above which its flow becomes turbulent. As the critical velocity is related to viscosity \((\eta)\) and density \((\rho)\) of the liquid as:
\(
\left(V_c\right) \alpha \eta / \rho
\)
Hence if the density will be low and viscosity will be high, the value of critical velocity will be more. So, option (b) and (c) are correct.

Hint

Viscosity is not a vector quantity. It is a scalar quantity because viscosity is a property of liquid as it does not have any direction.

Hint

Surface tension of a liquid is measured by the force acting per unit length on either side of an imaginary line drawn on the free surface of liquid, the direction of this force being perpendicular to the line and tangential to the free surface of liquid. So if \(\mathrm{F}\) is the force acting on one side of imaginary line of length \(L\), then \(\mathrm{T}=(\mathrm{F} / \mathrm{L})\). It depends only on the nature of liquid and is independent of the area of surface or length of line considered. It is a scalar quantity as it has a unique direction which is not to be specified.

Hint

According to the problem, density of ice \(\left(\rho_{\text {ice }}\right)=0.917 \mathrm{~g}^{\prime} / \mathrm{cm}^3\), Density of water \(\left(\rho_w\right)=1 \mathrm{~g} / \mathrm{cm}^3\)
Let \(V_i=\) Volume of iceberg,
\(V_w=\) Volume of water displaced by iceberg,
Weight of iceberg, \(W=\rho_i V_i g\),
Upthrust, \(F_B=\rho_w V_w g\)
At equilibrium, Weight of the iceberg \(=\) Weight of the water displaced by the submerged part by ice
\(
\begin{array}{ll}
\Rightarrow & \rho_w V_w g=\rho_i V_i g \\
\Rightarrow & \frac{V_w}{V_i}=\frac{\rho_i}{\rho_w}=\frac{0.917}{1}=0.917
\end{array}
\)

Hint

Let \(x\) be the compression on the spring. As the block is in equilibrium
\(
M g-\left(k x+\rho_w V g\right)=0
\)
where \(\rho_w\) is the density of water and \(V\) is the volume of the block. The reading in the pan is the force applied by the water on the pan i.e.,
\(
m_{\text {vessel }}+m_{\text {water }}+\rho_w V g .
\)
Since the scale has been adjusted to zero without the block, the new reading is \(\rho_{\mathrm{w}} \mathrm{Vg}\).

Hint

As of block is accelerating upward, so pseudo force on the container as well as on the block must be downward.
The situation is shown in the diagram above.
According to the problem,
Given, density of block \(=\rho\), Height of block \(=L\),
Density of water \(=\rho_w\), Volume of the block \((V)=L^3\),
Mass of the block \((m)=V \rho=L^3 \rho\),
Weight of the block \(=m g=L^3 \rho g\)
Let \(x\) be the height of cube submerged in water.
1st case:
Volume of part of cube submerged in water \(=x L^2\)
\(\therefore \quad\) Weight of water displaced by block \(=x L^2 \times \rho_w g\)
As block is floating, so
\(\therefore \quad\) weight of block \(=\) weight of water displaced by block
\(\Rightarrow \quad L^3 \rho g=l L^2 \rho_w g\)
\(\Rightarrow \quad \frac{l}{L}=\frac{\rho}{\rho_w}=x\)

2nd case:
When vessel is placed in an elevator moving upward with acceleration \(a\), then effective acceleration \(=(g+a)\)
( \(\because\) Pseudo force is downward)
Then, weight of the block \(=m(g+a)\)
\(
=L^3 \rho(g+a)
\)
\(\therefore\) Effective weight of block \(=m(g+a)\)
Let \(x_1=\) new fraction of block submerged in water. Since block is still floating in water, then
\(
\begin{aligned}
& m(g+a)=\left(x_1 L^3\right) \rho_w(g+a) \\
& x_1=\frac{m}{L^3 \rho_w}=\frac{L^3 \rho}{L^3 \rho_w}=\frac{\rho}{\rho_w}=x
\end{aligned}
\)
Hence, the fraction of the block submerged is independent of any type of acceleration.

Hint

According to the problem, radius \((r)=2.5 \times 10^{-5} \mathrm{~m}\) Surface tension \((S)=7.28 \times 10^{-2} \mathrm{~N} / \mathrm{m}\)
Angle of contact \((\theta)=0^{\circ}\)
Density \((\rho)=10^3 \mathrm{~kg} \mathrm{~m}^{-3}\)
If \(h\) is the maximum height to which sap can rise in trees through capillarity action, is given by
\(
h=\frac{2 S \cos \theta}{r \rho g}
\)
where \(S\) = surface tension, \(\rho=\) density and \(r\) = radius
\(
=\frac{2 \times 7.28 \times 10^{-2} \times \cos 0^{\circ}}{2.5 \times 10^{-5} \times 1 \times 10^{-3} \times 9.8}=0.6 \mathrm{~m}
\)
This is the maximum height to which the sap can rise due to surface tension. Many trees have heights much greater than \(0.6 \mathrm{~m}\), so only this action is not sufficient for supply of water to the top of such long tree.

Hint

If the tanker accelerates in the positive \(x\) direction, then the water will bulge at the back of the tanker. The free surface will be such that the tangential force on any fluid parcel is zero.

Consider a parcel at the surface, of unit volume. The forces on the fluid are
\(
-\rho g \hat{\mathbf{y}} \text { and }-\rho a \hat{\mathbf{x}}
\)
The component of the weight along the surface is \(\rho g \sin \theta\) The component of the acceleration force along the surface is \(\rho a \cos \theta\)
\(
\therefore \rho g \sin \theta=\rho a \cos \theta
\)
Hence, \(\tan \theta=a / g\)

Hint

Let \(v_1\) and \(v_2\) be the volume of the droplets and \(v\) of the resulting drop.
Then \(v=v_1+v_2\)
\(
\begin{aligned}
& \Rightarrow r^3=r_1^3+r_2^3=(0.001+0.008) \mathrm{cm}^3=0.009 \mathrm{~cm}^3 \\
& \therefore \quad r=0.21 \mathrm{~cm} \\
& \therefore \Delta U=4 \pi T\left(r^2-\left(r_1^2+r_2^2\right)\right) \\
& \quad=4 \pi \times 435.5 \times 10^{-3}\left(0.21^2-0.05\right) \times 10^{-4} \mathrm{~J} \\
& \quad=-32 \times 10^{-7} \mathrm{~J}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& R^3=N r^3 \\
& \Rightarrow r=\frac{R}{N^{1 / 3}} \\
& \Delta U=4 \pi T\left(R^2-N r^2\right)
\end{aligned}
\)
Suppose all this energy is released at the cost of lowering the temperature. If \(s\) is the specific heat then the change in temperature would be,
\(\Delta T=\frac{\Delta U}{m s}=\frac{4 \pi T\left(R^2-N r^2\right)}{\frac{4}{3} \pi R^3 \rho s}\), where \(\rho\) is the density.
\(
\begin{aligned}
\therefore \Delta T & =\frac{3 T}{\rho s}\left(\frac{1}{R}-\frac{r^2}{R^3} N\right) \\
& =\frac{3 T}{\rho s}\left(\frac{1}{R}-\frac{r^2 R^3}{R^3 r^3}\right)=\frac{3 T}{\rho s}\left(\frac{1}{R}-\frac{1}{r}\right)
\end{aligned}
\)

\(
\because \quad R>r \Rightarrow \frac{1}{R}<\frac{1}{r} \Rightarrow\left(\frac{1}{R}-\frac{1}{r}\right)<0
\)
\(\therefore \quad \Delta T\) will be negative. Hence, temperature of droplet falls.

Hint

The drop will evaporate if the water pressure is more than the vapour pressure. The membrane pressure (water)
\(
\begin{aligned}
& p=\frac{2 T}{r}=2.33 \times 10^3 \mathrm{~Pa} \\
& \therefore r=\frac{2 T}{p}=\frac{2\left(7.28 \times 10^{-2}\right)}{2.33 \times 10^3}=6.25 \times 10^{-5} \mathrm{~m}
\end{aligned}
\)

Hint

Consider a horizontal parcel of air with cross section \(A\) and height \(d h\). Let the pressure on the top surface and bottom surface be \(p\) and \(p+d p\). If the parcel is in equilibrium, then the net upward force must be balanced by the weight.
i.e. \((p+\mathrm{d} p) A-p A=-P g A d h\)
\(
\Rightarrow \mathrm{d} p=-\rho g \mathrm{~d} h
\)

Hint

Let the density of atr on the earth’s surface be \(\rho_0\), then
\(
\begin{aligned}
& \frac{p}{p_o}=\frac{\rho}{\rho_o} \\
& \Rightarrow \rho=\frac{\rho_o}{p_o} p \\
& \therefore d p=-\frac{\rho_o g}{p_o} p d h \\
& \Rightarrow \frac{d p}{p}=-\frac{\rho_o g}{p_o} d h \\
& \Rightarrow \int_{p_o}^p \frac{d p}{p}=-\frac{\rho_o g}{p_o} \int_o^h d h \\
& \Rightarrow \ln \frac{p}{p_o}=-\frac{\rho_o g}{p_o} h \\
& \Rightarrow p=p_o \exp \left(-\frac{\rho_o g}{p_o} h\right)
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \ln \frac{1}{10}=-\frac{\rho_o g}{p_o} h_o \\
& \therefore h_o=-\frac{p_o}{\rho_o g} \ln \frac{1}{10} \\
& =\frac{p_o}{\rho_o g} \times 2.303 \\
& =\frac{1.013 \times 10^5}{1.29 \times 9.8} \times 2.303=0.16 \times 10^5 \mathrm{~m}=16 \times 10^3 \mathrm{~m}
\end{aligned}
\)

Hint

The assumption \(p \propto \rho\) is valid only for the isothermal case which is only valid for small distances.

Hint

According to the problem, latent heat of vaporisation for water
\(
\left(L_v\right)=540 \mathrm{kcal}=540 \times 10^3 \times 4.2 \mathrm{~J}=2268 \times 10^3 \mathrm{~J}
\)
Therefore, energy required to evaporate \(1 \mathrm{kmol}(18 \mathrm{~kg})\) of water
\(
\begin{aligned}
& =\left(2268 \times 10^3 \mathrm{~J}\right)(18) \\
& =40824 \times 10^3 \mathrm{~J}=4.0824 \times 10^7 \mathrm{~J}
\end{aligned}
\)
Since, there are \(N_A\) molecules in \(M_A \mathrm{~kg}\) of water, the energy required for 1 molecule to evaporate is
\(
U=\frac{M_A L_v}{N_A} \mathrm{~J}
\)
[where \(N_A=6 \times 10^{26}=\) Avogadro number]
\(
\begin{aligned}
U & =\frac{4.0824 \times 10^7}{6 \times 10^{26}} \mathrm{~J}=0.68 \times 10^{-19} \mathrm{~J} \\
& =6.8 \times 10^{-20} \mathrm{~J}
\end{aligned}
\)

Hint

Let the water molecules to be at points and are placed at a distance \(d\) from each other,
Volume of \(N_A\) molecule of water \(=\frac{M_A}{\rho_w}\)
Thus, the volume around one molecule is
\(
=\frac{\text { volume of } 1 \mathrm{kmol}}{\text { number of molecules } / \mathrm{kmol}}=\frac{M_A}{N_A \rho_w}
\)
And also volume around one molecule \(=d^3\)
Thus, by equating these, we get
\(
\begin{aligned}
d^3 & =\frac{M_A}{N_A \rho_w} \\
\therefore \quad d & =\left(\frac{M_A}{N_A \rho_w}\right)^{1 / 3}=\left(\frac{18}{6 \times 10^{26} \times 10^3}\right)^{1 / 3} \\
& =\left(30 \times 10^{-30}\right)^{1 / 3} \mathrm{~m} \approx 3.1 \times 10^{-10} \mathrm{~m}
\end{aligned}
\)

Hint

Volume occupied by \(1 \mathrm{kmol}(18 \mathrm{~kg}\) ) of water molecules
\(
\begin{aligned}
& =\frac{1601 \times 10^{-6} \mathrm{~m}^3}{\mathrm{~g}}\left(18 \times 10^3 \mathrm{~g}\right) \\
& =28818 \times 10^{-3} \mathrm{~m}^3
\end{aligned}
\)
Since \(6 \times 10^{26}\) molecules occupies \(18 \times 1601 \times 10^{-3} \mathrm{~m}^3\)
\(\therefore\) Volume occupied by 1 molecule
\(
=\frac{28818 \times 10^{-3} \mathrm{~m}^3}{6 \times 10^{26}}=48030 \times 10^{-30} \mathrm{~m}^3
\)
If \(d^{\prime}\) is the intermolecular distance, then
\(
\left(d^{\prime}\right)^3=48030 \times 10^{-30} \mathrm{~m}^3
\)
So, \(\quad d^{\prime}=36.3 \times 10^{-10} \mathrm{~m}=36.3 \times 10^{-10} \mathrm{~m}\)

Hint

Work done to change the distance from \(d\) to \(d^{\prime}\) is \(U=F\left(d^{\prime}-d\right)\), This work done is equal to energy required to evaporate 1 molecule.
\(
\begin{aligned}
& \therefore \quad F\left(d^{\prime}-d\right)=6.8 \times 10^{-20} \\
& \text { or } \\
& F=\frac{6.8 \times 10^{-20}}{d^{\prime}-d} \\
& =\frac{6.8 \times 10^{-20}}{\left(36.3 \times 10^{-10}-3.1 \times 10^{-10}\right)} \\
& =2.048 \times 10^{-11} \mathrm{~N} \\
\end{aligned}
\)

Hint

\(
\text { Surface tension }=\frac{F}{d}=\frac{2.05 \times 10^{-11}}{3.1 \times 10^{-10}}=6.6 \times 10^{-2} \mathrm{~N} / \mathrm{m}
\)

Hint

Let the pressure inside the balloon be \(P_t\) and the outside pressure be \(P_0\)
\(
P_{\mathrm{t}}-P_o=\frac{2 \gamma}{r}
\)
Considering the air to be an ideal gas
\(P_t V=n_t R T_t\) where \(V\) is the volume of the air inside the balloon, \(n_1\) is the number of moles inside and \(T_t\) is the temperature inside, and \(P_o V=n_o R T_o\) where \(V\) is the volume of the atr displaced and \(n_0\) is the number of moles displaced and \(T_0\) is the temperature outside. \(n_1=\frac{P_i V}{R T_i}=\frac{M_i}{M_A}\) where \(M_i\) is the mass of air inside and \(\mathrm{M}_{\mathrm{A}}\) is the molar mass of air and \(n_0=\frac{P_o V}{R T_o}=\frac{M_O}{M_A}\) where \(M_0\) is the mass of air outside that has been displaced. If \(W\) is the load it can raise, then
\(
\begin{aligned}
& W+M_t g=M_o g \\
& \Rightarrow W=M_o g-M_t g
\end{aligned}
\)
Alr is \(21 \% \mathrm{O}_2\) and \(79 \% \mathrm{~N}_2\)
\(\therefore\) Molar mass of air \(M_{\mathrm{A}}=0.21 \times 32+0.79 \times 28=28.84 \mathrm{~g}\).
\(\Rightarrow W=\frac{M_A V}{R}\left(\frac{P_o}{T_o}-\frac{P_i}{T_i}\right) g\)
\(=\frac{0.02884 \times \frac{4}{3} \pi \times 8^3 \times 9.8}{8.314}\left(\frac{1.013 \times 10^5}{293}-\frac{1.013 \times 10^5}{333}-\frac{2 \times 5}{8 \times 313}\right) \mathrm{N}\) \(=\frac{0.02884 \times \frac{4}{3} \pi \times 8^3}{8.314} \times 1.013 \times 10^5\left(\frac{1}{293}-\frac{1}{333}\right) \times 9.8 \mathrm{~N}\) \(=3044.2 \mathrm{~N}\).

Hint

No, the water will neither rise nor fall in the silver capillary.
According to Jurin’s law, the level of water inside a capillary tube is given by \(h=\frac{2 T \cos \theta}{r \rho g}\)
Here, \(\theta=90^{\circ}\)
\(
\begin{aligned}
& \Rightarrow h=\frac{2 T \cos 90^0}{r \rho g} \\
& \Rightarrow h=0
\end{aligned}
\)
Thus, the water level neither rises nor falls.

Hint

No, we cannot conclude the surface tension to be zero solely by the fact that the liquid neither rises nor falls in a capillary.
The height of the liquid inside a capillary tube is given by \(h=\frac{2 T \cos \theta}{r \rho g}\). From the equation, we see that the height (h) of the liquid may also be zero if the contact angle \(\theta\) between the liquid and the capillary tube is \(90^0\) or \(270^0\)

Hint

Due to the surface tension, the surface of the liquid acts as a stretched membrane. So even if some more water than the rim level of the glass is poured into it, the stretched water surface which holds the glass rim due to adhesion becomes convex upward and the contact angle between the water and the glass becomes more than \(90^{\circ}\) in this situation.

The zero degrees angle of contact for water and the glass is for the situation when the water level is below the rim level of the glass. In some special situations, the angle of contact may change.

Hint

As the angle of contact is 0 , there is no force between the surface of the tube and the liquid. The diameter of the liquid surface is pulled on both sides by equal and opposite forces of surface tension. This results in no net force remaining on the surface of the liquid. Hence, the liquid stays in equilibrium.

Hint

No, it does not violate the principle of conservation of energy.
There is a force of attraction between glass and water, which is why the liquid rises in the tube. However, when water and glass are not in contact, there exists a potential energy in the system. When they are brought into contact, this potential energy is first converted into kinetic energy, which lets the liquid rush upwards in the tube, and then into gravitational potential energy. Therefore, energy is not created in the process.

Hint

A mosquito thrown into water has its wings wet. Now, wet wing surfaces tend to stick together because of the surface tension of water. This does not let the mosquito fly.

Hint

force \(\left(F_{\mathrm{t}}\right)\) due to the surface tension acts tangentially. The net force \((F)\) is acting downward.

Hint

No. The average intermolecular distances do not increase with an increase in the surface area.
A soap bubble’s layer consists of several thousand layers of molecules. An increase in the surface area causes the surface energy to also increase. This in turn allows more and more molecules from the inner liquid layers of the bubble to attain potential energy, enabling them to enter the outer surface of the bubble. Hence, the surface area increases.

Hint

No. For a liquid at rest, no viscous forces exist.
Viscous forces oppose relative motion between the layers of a liquid. These layers do not exist in a liquid that is at rest. Therefore, it is obvious that viscous forces are non-existent in a static liquid.

Hint

Hint

Castor oil will come to rest more quickly because it has a greater coefficient of viscosity than water.
Castor oil has a higher viscosity than water. It will therefore, lose kinetic energy and come to rest faster than water.

Hint

The surface of a liquid refers to the layer of molecules that have higher potential energy than the bulk of the liquid. This layer is typically 10 to 15 times the diameter of the molecule. Now, the size of an average molecule is around \(1 \mathrm{~nm}=\) \(10^{-9} \mathrm{~m}\), so a diameter of 10 to 15 times would be of order \(10 \times 10^{-9}=10^{-8} \mathrm{~m}\).

Hint

As the ice cube melts completely, the water thus formed will have minimum surface area due to its surface tension. Any state of matter that has a minimum surface area to its volume takes the shape of a sphere. Therefore, as the ice melts, it will take the shape of a sphere.

Hint

As the water droplets merge to form a single droplet, the surface area decreases. With this decrease in surface area, the surface energy of the resulting drop also decreases. Therefore, extra energy must be liberated from the drop in accordance with the conservation of energy.

Hint

Dimension of modulus of elasticity:
\(
\frac{\frac{F}{A}}{\frac{\Delta l}{l}}=\frac{\left[M L T^{-2}\right]}{L^2}=\left[M L^{-1} T^{-2}\right]
\)
Dimension of moment of force:
\(
F L=\left[M L T^{-2}\right][L]=\left[M L^2 T^{-2}\right]
\)
Dimension of surface tension:
\(
\frac{F}{L}=\frac{\left[M L T^{-2}\right]}{L}=\left[M T^{-2}\right]
\)
Dimension of coefficient of viscosity:
\(
\frac{F L}{A v}=\frac{\left[M L T^{-2}\right][L]}{\left[L^2\right]\left[L T^{-1}\right]}=\left[M L^{-1} T^{-1}\right]
\)

Hint

No. of surfaces of a soap bubble \(=2\)
Increase in surface area \(=4 \pi(2 r)^2-4 \pi(r)^2=12 \pi r^2\)
Total increase in surface area \(=2 \times 12 \pi r^2=24 \pi r^2\)
Work done = change in surface energy
\(
=S \times 24 \pi r^2=24 \pi r^2 S
\)

Hint

Excess pressure inside a bubble is given by: \(P=\frac{4 T}{r}\).
When air is pushed into the bubble, it grows in size. Therefore, its radius increases. An increase in size causes the pressure inside the soap bubble to decrease as pressure is inversely proportional to the radius.

Hint

The smaller bubble has a greater inner pressure than the bigger bubble. Air moves from a region of high pressure to a region of low pressure. Therefore, air moves from the smaller to the bigger bubble.

Hint

Radius of the tube \(=r\)
Net upward force due to surface tension \(=\mathrm{S} \cos \theta \times 2 \pi r\)
Upward pressure \(=\frac{S \cos \theta \times 2 \pi r}{\pi r^2}=\frac{2 S \cos \theta}{r}\)
Net downward pressure due to atmosphere \(=P_o\)
\(\Rightarrow\) Net pressure at \(\mathrm{A}=P_o-\frac{2 S \cos \theta}{r}\)
Since \(\theta\) is small,
\(\cos \theta \approx 1\)
\(\Rightarrow\) Net pressure \(=P_o-\frac{2 S}{r}\)

Hint

Let the excess pressure inside the second bubble be \(P\).
\(\therefore\) Excess pressure inside the first bubble \(=2 \mathrm{P}\)
Let the radius of the second bubble be \(\mathrm{R}\).
Let the radius of the first bubble be \(\mathrm{x}\).
Excess pressure inside the 2 nd soap bubble:
\(
P=\frac{4 S}{R} \ldots \text { (1) }
\)
Excess pressure inside the 1st soap bubble:
\(
2 P=\frac{4 S}{x}
\)
From (1), we get :
\(
\begin{aligned}
& 2\left(\frac{4 S}{R}\right)=\frac{4 S}{x} \\
& \Rightarrow x=\frac{R}{2}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Volume of the first bubble }=\frac{4}{3} \pi x^3 \\
& \text { Volume of the second bubble }=\frac{4}{3} \pi R^3 \\
& \Rightarrow \frac{4}{3} \pi x^3=n \frac{4}{3} \pi R^3 \\
& \Rightarrow x^3=n R^3 \\
& \Rightarrow\left(\frac{R}{2}\right)^3=n R^3 \\
& \Rightarrow n=\frac{1}{8}=0.125
\end{aligned}
\)

Hint

The relationship between height \(h\) and radius \(r\) is given by:
\(
\mathrm{h}=\frac{2 S \cos \theta}{r \rho g}
\)
If \(\mathrm{S}, \theta, \rho\) and \(\mathrm{g}\) are considered constant, we have:
\(\mathrm{h} \propto \frac{1}{r}\)
This equation has the characteristic of a rectangular hyperbola. Therefore, curve (c) is a rectangular hyperbola.

Hint

Given :
\(
\begin{aligned}
& {l}=10 \mathrm{~cm} \\
& \alpha=45^0
\end{aligned}
\)
Rise in water level after the tube is tilted \(=\mathrm{h}\)
\(
\begin{aligned}
& \Rightarrow {l}=\mathrm{h} \cos 45^{\circ} \\
& \Rightarrow h=\frac{l}{\cos 45^0}=\frac{10}{\left(\frac{1}{\sqrt{2}}\right)}=10 \sqrt{2} \mathrm{~cm}
\end{aligned}
\)

Hint

Height of water column in capillary tube is given by:
\(
h=\frac{2 T \cos \theta}{r \rho g}
\)
A free falling elevator experiences zero gravity .
\(
\Rightarrow h=\frac{2 T \cos \theta}{r \rho 0}=\infty
\)
But, \(\mathrm{h}=20 \mathrm{~cm}\) (given)
Therefore, the height of the water column will remain at a maximum of \(20 \mathrm{~cm}\).

On freely falling elevator, the gravitational force on water will be zero therefore, surface tension will not be cancelled by any other force and, hence, water will be raised to full length of capillary tube.

Hint

\(
\text { Viscosity is one property of fluids. Fluids include both liquids and gases. }
\)

Hint

The force of viscosity arises from molecular interaction between different layers of fluids that are in motion. Molecular forces are electromagnetic in nature. Therefore, viscosity must also be electromagnetic.

Hint

Air has viscosity. During rainfall, the raindrops acquire acceleration due to gravity. However, the increase in velocity is hindered by the viscous force acting upwards. A gradual balance between the two opposing forces causes the raindrops to attain a terminal velocity, thus, falling with a uniform velocity.

Hint

The viscous force acts tangentially between two parallel layers of a liquid. In terms of force on a material, it is analogous to a shearing force.

Hint

The density of wood is less than that of water.When a piece of wood is immersed deep inside a long column of water and released, it experiences a buoyant force that gives it an upward acceleration. The velocity of wood increases as its motion is accelerated by the buoyant force. However, the viscous drag force acts simultaneously to oppose its upward motion. As a result, the initial acceleration decreases and the wood rises with a decreasing upward acceleration.

Hint

In vacuum, no viscous force exists. The sphere therefore, will have constant acceleration because of gravity. An accelerated motion implies that it won’t have uniform velocity throughout its motion. In other words, there will be no terminal velocity.

Hint

Initially, when the ball starts moving, its velocity is small. Gradually, the velocity of the ball increases due to acceleration caused by gravity. However, as the velocity increases, the viscous force acting on the ball also increases. This force tends to decelerate the ball. Therefore, after reaching a certain maximum velocity, the ball slows down.

Hint

(c) The surface molecules acquire air and liquid molecules in their sphere of influence.
(d) The surface molecules have different magnitudes of forces pulling them from the top and the bulk. So, they are affected by a net finite force in one direction.

Hint

Height of the liquid in the capillary tube is given by :
\(
\begin{aligned}
& \mathrm{h}=\frac{2 S \cos \theta}{r \rho g} \\
& \mathrm{~h}=\text { Height } \\
& \mathrm{S}=\text { Surface tension } \\
& \mathrm{r}=\text { Inner radius of the tube } \\
& \rho=\text { Density of the liquid } \\
& \mathrm{g}=\text { Acceleration due to gravity }
\end{aligned}
\)
a) \(\theta\) and \(\rho\) depend upon the material of the capillary tube and the liquid .
\(b\) ) \(\mathrm{h}\) is dependent on the length of the tube. If the length is insufficient, then \(\mathrm{h}\) will be low .
d) \(r\) is the inner radius of the tube.

Hint

The angle of contact between a solid and a liquid depends upon the molecular forces of both the substances. Therefore, it depends upon the material of the solid and the liquid.

Hint

Let the height of the liquid-filled column be \(L\).
Let the radius be denoted by \(R\).
Total perimeter of the curved part \(=\) semi – circumference of upper area \(=\pi r\)
Total surface tension force \(=\pi R S\)
Total perimeter of the flat part \(=2 \mathrm{R}\)
Total surface tension force \(=2 \mathrm{RS}\)
Ratio of curved surface force to flat surface force \(=\frac{\pi R S}{2 R S}=\frac{\pi}{2}\)

Hint

The rise in a capillary tube is given as, \(\mathrm{h}=2 \mathrm{~S} \cdot \cos \theta / r \rho g\) The capillary tube has a certain inner radius r. So \(r, \rho\) and \(g\) are not zero. For \(h\) to be zero either surface tension \(\mathrm{S}\) may be zero or the contact angle \(\theta=90^{\circ}\) so that \(\cos \theta=0\). But we can not conclude that \(\mathrm{S}\) must be zero or \(\theta\) must be \(90^{\circ}\). Hence the option (c) and (d) are true.

Hint

since there is no gravity, tnere will be no inıtıal acceleratıon in downward direction. Hence the option (a) is incorrect.

\(20 \mathrm{~m} / \mathrm{s}\) is the terminal velocity. At this velocity, the viscous force is equal to the weight of the sphere (assuming the buoyant force to be negligible because air has a very low density). i.e \(F_v=m g\) at \(v=20 \mathrm{~m} / \mathrm{s}\)
When sphere is thrown downwards at a speed of \(20 \mathrm{~m} / \mathrm{s}\), once again, viscous force \(F_v=m g\) will act vertically upwards.
\(\mathrm{W}=\mathrm{mg}=0\) [since gravity free space]
\(
\Rightarrow \mathrm{a}=\frac{F_v}{m}=\frac{m g}{m}=g
\)
The sphere gets an initial acceleration \(a=9.8 \mathrm{~m} / \mathrm{s}^2\) in upward direction.
\(\therefore\) Hence option (b) is correct.

\(F_v\) acting in opposite direction of velocity will cause retardation in the sphere. Hence, its velocity will decrease continously.
\(F_v \propto v\)
\(\Rightarrow F_v \downarrow\) as \(v \downarrow\)
Hence, magnitude of acceleration (retardation) also reduces as time passes.
\(\therefore\) Option (c) is correct

Since the viscous force \(\left(F_v\right)\) continously opposes the motion of the sphere, it will eventually stop i.e at some instant \(v=0\) will be achieved.
Hence, the option (d) is also correct.