Answer: Let the volume of the iceberg be $V$. The weight of the iceberg is $\rho_{i} V g$. If $x$ is the fraction submerged, then the volume of water displaced is $x V$. The buoyant force is $\rho_{\mathrm{w}} x V g$ where $\rho_{\mathrm{w}}$ is the density of water.

$\rho_{i} V g=\rho_{\mathrm{w}} x V g$

$\therefore x=\frac{\rho_{i}}{\rho_{\mathrm{w}}}=0.917$

Answer: 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 $\rho_{\mathrm{w}}$ 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.,

$m_{\text {vessel }}+m_{\text {water }}+\rho_{\mathrm{w}} V g$.

Since the scale has been adjusted to zero without the block, the new reading is $\rho_{\mathrm{w}} V g$.

Answer: Let the density of water be $\rho_{w}$.

Then $\rho a L^{3}+\rho L^{3} g=\rho_{\mathrm{w}} x L^{3}(g+a)$

$\therefore x=\frac{\rho}{\rho_{\mathrm{w}}}$

Thus, the fraction of the block submerged is independent of any acceleration, whether gravity or elevator.

Answer: The height to which the sap will rise is

$h=\frac{2 T \cos 0^{\circ}}{\rho g r}=\frac{2\left(7.2 \times 10^{-2}\right)}{10^{3} \times 9.8 \times 2.5 \times 10^{-5}} \quad 0.6 \mathrm{~m}$

This is the maximum height to which the sap can rise due to surface tension. Since many trees have heights much more than this, capillary action alone cannot account for the rise of water in all trees.

Answer: If the tanker acclerates 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

$-\rho g \hat{\mathbf{y}}$ and $\quad-\rho a \hat{\mathbf{x}}$

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, $\tan \theta=a / g$

Answer: Let $v_{1}$ and $v_{2}$ be the volume of the droplets and $v$ of the resulting drop.

Then $v=v_{1}+v_{2}$

$\Rightarrow r^{3}=r_{1}^{3}+r_{2}^{3}=(0.001+0.008) \mathrm{cm}^{3}=0.009 \mathrm{~cm}^{3}$

$\therefore r \quad 0.21 \mathrm{~cm}$

$\therefore \Delta U=4 \pi T\left(r^{2}-\left(r_{1}^{2}+r_{2}^{2}\right)\right)$

$=4 \pi \times 435.5 \times 10^{-3}\left(0.21^{2}-0.05\right) \times 10^{-4} \mathrm{~J}$

$$ -32 \times 10^{-7} \mathrm{~J} $$

Answer: $R^{3}=N r^{3}$

$\Rightarrow r=\frac{R}{N^{1 / 3}}$

$\Delta U=4 \pi T\left(R^{2}-N r^{2}\right)$

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,

$$ \begin{aligned} & \Delta \theta=\frac{\Delta U}{m s}=\frac{4 \pi T\left(R^{2}-N r^{2}\right)}{\frac{4}{3} \pi R^{3} \rho s}, \text { where } \rho \text { is the density. } \ & \therefore \Delta \theta=\frac{3 T}{\rho s}\left(\frac{1}{R}-\frac{r^{2}}{R^{3}} N\right) \ & =\frac{3 T}{\rho s}\left(\frac{1}{R}-\frac{r^{2} R^{3}}{R^{3} r^{3}}\right)=\frac{3 T}{\rho s}\left(\frac{1}{R}-\frac{1}{r}\right) \end{aligned} $$

Answer: The drop will evaporate if the water pressure is more than the vapour pressure. The membrane pressure (water)

$$ \begin{aligned} & p=\frac{2 T}{r}=2.33 \times 10^{3} \mathrm{~Pa} \ & \therefore r=\frac{2 T}{p}=\frac{2\left(7.28 \times 10^{-2}\right)}{2.33 \times 10^{3}}=6.25 \times 10^{-5} \mathrm{~m} \end{aligned} $$

Answer: (a) 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. $(p+\mathrm{d} p) A-p A=-P g A d h$

$\Rightarrow \mathrm{d} p=-\rho g \mathrm{~d} h$

(b) Let the density of air on the earth’s surface be $\rho_{0}$, then

$$ \begin{aligned} & \frac{p}{p_{o}}=\frac{\rho}{\rho_{o}} \ & \Rightarrow \rho=\frac{\rho_{o}}{p_{o}} p \end{aligned} $$

$\therefore d p=-\frac{\rho_{o} g}{p_{o}} p d h$

$\Rightarrow \frac{d p}{p}=-\frac{\rho_{o} g}{p_{o}} d h$

$\Rightarrow \int_{p_{o}}^{p} \frac{d p}{p}=-\frac{\rho_{o} g}{p_{o}} \int_{o}^{h} d h$

$\Rightarrow \ln \frac{p}{p_{o}}=-\frac{\rho_{o} g}{p_{o}} h$

$\Rightarrow p=p_{o} \exp \left(-\frac{\rho_{o} g}{p_{o}} h\right)$

(c) $\ln \frac{1}{10}=-\frac{\rho_{o} g}{p_{o}} h_{o}$

$\therefore h_{o}=-\frac{p_{o}}{\rho_{o} g} \ln \frac{1}{10}$

$=\frac{p_{o}}{\rho_{o} g} \times 2.303$

$=\frac{1.013 \times 10^{5}}{1.29 \times 9.8} \times 2.303=0.16 \times 10^{5} \mathrm{~m}=16 \times 10^{3} \mathrm{~m}$

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

Answer: (a) $1 \mathrm{~kg}$ of water requires $\mathrm{L}_{v} \mathrm{k}$ cal

$\therefore M_{\mathrm{A}} \mathrm{kg}$ of water requires $M_{\mathrm{A}} \mathrm{L}_{v} \mathrm{k}$ cal

Since there are $N_{\mathrm{A}}$ molecules in $M_{\mathrm{A}} \mathrm{kg}$ of water the energy required for 1 molecule to evaporate is

$$ \begin{aligned} u= & \frac{M_{A} L_{v}}{N_{A}} \mathrm{~J} \ = & \frac{18 \times \not 940 \times 4.2 \times 10^{3}}{\not 6 \times 10^{26}} \mathrm{~J} \ = & 90 \times 18 \times 4.2 \times 10^{-23} \mathrm{~J} \ & 6.8 \times 10^{-20} \mathrm{~J} \end{aligned} $$

(b) Consider the water molecules to be points at a distance $d$ from each other.

$N_A$ molecules occupy $\frac{M_A}{\rho_w} l$

Thus, the volume around one molecule is $\frac{M_{A}}{N_{A} \rho_{w}} l$

The volume around one molecule is $d^{3}=\left(M_A / N_A \rho_w\right)$

$$ \begin{aligned} \therefore d & =\left(\frac{M_{A}}{N_{A} \rho_{w}}\right)^{1 / 3}=\left(\frac{18}{6 \times 10^{26} \times 10^{3}}\right)^{1 / 3} \ & =\left(30 \times 10^{-30}\right)^{1 / 3} \mathrm{~m} \quad 3.1 \times 10^{-10} \mathrm{~m} \end{aligned} $$

(c) $1 \mathrm{~kg}$ of vapour occupies $1601 \times 10^{-3} \mathrm{~m}^{3}$.

$\therefore 18 \mathrm{~kg}$ of vapour occupies $18 \times 1601 \times 10^{-3} \mathrm{~m}^{3}$

$\Rightarrow 6 \times 10^{26}$ molecules occupies $18 \times 1601 \times 10^{-3} \mathrm{~m}^{3}$

$\therefore 1$ molecule occupies $\frac{18 \times 1601 \times 10^{-3}}{6 \times 10^{26}} \mathrm{~m}^{3}$

If $d^{\prime}$ is the inter molecular distance, then

$$ \begin{aligned} & d^{3}=\left(3 \times 1601 \times 10^{-29}\right) \mathrm{m}^{3} \ & \therefore \quad d^{\prime}=(30 \times 1601)^{1 / 3} \times 10^{-10} \mathrm{~m} \ & =36.3 \times 10^{-10} \mathrm{~m} \end{aligned} $$

(d) $F\left(d^{\prime}-d\right)=\mathrm{u} \Rightarrow F=\frac{u}{d^{\prime}-d}=\frac{6.8 \times 10^{-20}}{(36.3-3.1) \times 10^{-10}}=0.2048 \times 10^{-10} \mathrm{~N}$

(e) $\quad F / d=\frac{0.2048 \times 10^{-10}}{3.1 \times 10^{-10}}=0.066 \mathrm{~N} \mathrm{~m}^{-1}=6.6 \times 10^{-2} \mathrm{~N} \mathrm{~m}^{-1}$

Answer: Let the pressure inside the balloon be $P_{i}$ and the outside pressure be $P_{\mathrm{o}}$ $P_{i}-P_{o}=\frac{2 \gamma}{r}$

Considering the air to be an ideal gas

$P_{i} V=n_{i} R T_{i}$ where $V$ is the volume of the air inside the balloon, $n_{\mathrm{i}}$ is the number of moles inside and $T_{i}$ is the temperature inside, and $P_{o} V=n_{o} R T_{o}$ where $\mathrm{V}$ is the volume of the air displaced and $n_{\mathrm{o}}$ is the number of moles displaced and $T_{\mathrm{o}}$ is the temperature outside.

$n_{\mathrm{i}}=\frac{P_i V}{R T_{i}}=\frac{M_i}{M_A}$ where $M_i$ is the mass of air inside and $M_A$ is the molar mass of air and $n_{\mathrm{o}}=\frac{P_{O} V}{R T_{O}}=\frac{M_{O}}{M_{A}}$ where $M_{\mathrm{o}}$ is the mass of air outside that has been displaced. If $W$ is the load it can raise, then $W+M_{i} g=M_{o} g$ $\Rightarrow W=M_{o} g-M_{i} g$

Air is $21 \% O_2$ and $79 \% N_2$

$\therefore$ Molar mass of air $M_{\mathrm{A}}=0.21 \times 32+0.79 \times 28=28.84 \mathrm{~g}$.

$\Rightarrow W=\frac{M_{A} V}{R}\left(\frac{P_{O}}{T_{o}}-\frac{P_{i}}{T_{i}}\right) g$

$=\frac{0.02884 \times \frac{4}{3} \pi \times 8^{3} \times 9.8}{8.314}\left(\frac{1.013 \times 10^{5}}{293}-\frac{1.013 \times 10^{5}}{333}-\frac{2 \times 5}{8 \times 313}\right) \mathrm{N}$ $\frac{0.02884 \times \frac{4}{3} \pi \times 8^{3}}{8.314} \times 1.013 \times 10^{5}\left(\frac{1}{293}-\frac{1}{333}\right) \times 9.8 \mathrm{~N}$

$=3044.2 \mathrm{~N}$.