3 Marks Question

Question 13 Marks

Iceberg floats in water with part of it submerged. What is the fraction of the volume of iceberg submerged if the density of ice is $\rho_\text{i}=0.917\text{g cm}^{-3}?$

Answer

According to the problem, density of ice $(\rho_\text{ice})=0.917\text{g/ cm}^3,$ Density of water $(\rho_\text{w})=1\text{g/ cm}^3$ Let $V_i$ = Volume of iceberg, $V_w$ = Volume of water displaced by iceberg, Weight of iceberg, $\text{W}=\rho_\text{i}\text{V}_\text{i}\text{g}$ Upthrust, $\text{F}_\text{B}=\rho_\text{w}\text{V}_\text{w}\text{g}$ At equilibrium, Weight of the iceberg = Weight of the water displaced by the submerged part by ice$\Rightarrow\rho_\text{w}\text{V}_\text{w}\text{g}=\rho_\text{i}\text{V}_\text{i}\text{g}$
$\Rightarrow\frac{\text{V}_\text{w}}{\text{V}_\text{i}}=\frac{\rho_\text{i}}{\rho_\text{w}}=\frac{0.917}{1}=0.917$

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Question 23 Marks

Considering the pressure p to be proportional to the density, find the pressure p at a height h if the pressure on the surface of the earth is $p_0$.

Answer

$\therefore\ \text{p }\alpha\ \rho\text{ or }\frac{\text{p}}{\text{p}_0}=\frac{\rho}{\rho_0}\text{ or }\rho=\Big(\frac{\text{p}}{\text{p}_0}\Big)\rho_{0}\ ...(\text{ii})$$\because\ \text{from(iii)}\ \log\frac{\text{p}}{\text{p}_0}=-\frac{\rho_0\text{gh}}{\text{p}_0}$
Put $\text{p}=\frac{\text{p}_0}{10}(\text{given})$
$\Rightarrow\log\frac{\frac{\text{p}_0}{10}}{\text{p}_0}=\frac{-\rho_0\text{gh}}{\text{P}_0}$
$\Rightarrow\log\frac{1}{10}=\frac{-\rho_0\text{gh}}{\text{p}_0}$
$\Rightarrow10^{-1}=\frac{-\rho_0\text{gh}}{\text{p}_0}$
$\Rightarrow-\log10=\frac{-\rho_0\text{gh}}{\text{p}_0}$
Or $\text{h}=\frac{\text{p}_0}{\text{p}_0\text{g}}\log10$
$\text{h}=\frac{\text{p}_0\log_{10}10\times2.303}{\rho_0\text{g}}$
$\text{h}=\frac{1.031\times10^5\times2.303}{1.29\times9.8}=0.184\times10^{5}\text{m}$
$\text{h}=18.4\times10^{-3}\text{m}=18.4\text{lom}$

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Question 33 Marks

The sap in trees, which consists mainly of water in summer, rises in a system of capillaries of radius $r = 2.5 \times 10^{-5}m$. The surface tension of sap is $T = 7.28 \times 10^{-2}N m^{-1}$ and the angle of contact is $0°$. Does surface tension alone account for the supply of water to the top of all trees?

Answer

Radius of capillarity $\text{r}=2.5\times10^{-5}\text{m}$$\text{S}=\text{T}=7.8\times10^{-2}\text{Nm}^{-1}\ \text{g}=9.8\text{m/ s}^2$
$\theta=0^{\circ},\text{ p}=10^3\text{kg m}^3$
$\text{h}=\frac{2\text{s}\cos\theta}{\text{rpg}}=\frac{2\times7.28\times10^{-2}\cos0^\circ}{2.5\times10^{-5}\times10^3\times9.8}=\frac{2\times728\times10^{-2+5}}{25\times98\times10^3}$
$\text{h}=\frac{104}{175}\times\frac{10^{-3}}{10^3}=\frac{104}{175}=0.594\text{m}\cong6\text{m}$
Most of trees are of more than 0.6m height. So, capillary action alone connot account for the rise of water in all other tress.

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Question 43 Marks

The sufrace tension and vapour pressure of water at 20°C is $7.28 \times 10^{-2}N$ $m^{-1}$ and $2.33 \times 10^3$ Pa, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at 20°C?

Answer

The drop will evaporate if the water pressure on liquid, is greater than vapour pressure above the surface of liquid. Let a water droplet of radius R be formed without evaporation then. Vapour pressure = Excess pressure in a drop$\rho=\frac{2\sigma}{\text{R}}$ (onlt one surface in drop)
$\text{R}=\frac{2\times7.28\times10^{-2}}{\text{Vapour preessure}}$
$=\frac{2\times7.28\times10^{-2}}{2.33\times10^{3}}=\frac{1456\times10}{233\times10^5}$
$\text{R}=6.25\times10^{-5}\text{m}$

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Question 53 Marks

Pressure decreases as one ascends the atmosphere. If the density of air is $\rho,$ what is the change in pressure dp over a differential height dh?

Answer

consider a part (packet) of atmosphere of thickeness dh. As the pressure at a point in fluid is equal in all directions. so the pressure on upper layer is p acting downward and on lower layer is (p + dp) acting upward. Force due to pressure is balanced by Buoyant force by air,
(p + dp)A - pA = -Vpg
PA + dpA - pA = -Adhpg
dpA = -ppgdhA
dpp = -pgdh ....(i)
(image)
Negative sign shows that pressure decreases as height increases.

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