5 Marks Questions

Question 15 Marks

We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say $10cm$. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their lengths remain constant. If $a_{iron} = 1.2 \times 10^{-5}/ K$ and $a_{brass} = 1.8 \times 10^{-5}/ K$, what should we take as length of each strip?

Answer

According to the problem, $L_1- L_b = 10cm$ where, $L_1$ = length of iron scale $L_b$ = Length of brass scale This condition is possible if change in length both the rods is remain same at all temperatures. Change in length of iron rod,$\Delta\text{L}=\alpha_{\text{I}}\text{L}_{\text{I}}\Delta\text{T}$
Change in length of brass rod,$\Delta\text{L}=\alpha_\text{B}\text{L}_\text{B}\Delta\text{T}$
As the cahange will equal in both the rods, so$\alpha_1\text{L}_1\Delta\text{T}=\alpha_\text{B}\text{L}_\text{B}\Delta\text{T}$
$\Rightarrow\alpha_1\text{L}_1=\alpha_\beta\text{L}_\beta$
$\Rightarrow\frac{\text{L}_1}{\text{L}_\text{B}}=\frac{\alpha_\text{B}}{\alpha_\text{I}}$
Here, $\alpha_\beta=1.8\times10^{-5}\text{K}^{-1},\alpha_\text{I}=1.2\times10^{-5}\text{K}^{-1}$$\therefore\ \frac{\text{L}_\text{I}}{\text{L}_\text{B}}=\frac{1.8\times10^{-5}}{1.2\times10^{-5}}=\frac{3}{2}$
$\text{L}_1=\frac{3}{2}\text{L}_\text{B}$
As, $\text{L}_\text{I}-\text{L}_\text{B}=10\text{cm}$$\therefore\ \frac{3}{2}\text{L}_\text{B}-\text{L}_\text{B}=10$
$\Rightarrow\frac{1}{2}\text{L}_\text{B}=10$
$\Rightarrow \text{L}_\text{B}=20{\text{cm}}$

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Question 25 Marks

Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of $57^{\circ} \mathrm{C}$ is drunk. You can take body (tooth) temperature to be $37^{\circ} \mathrm{C}$ and $\mathrm{a}=1.7 \times 10^{-5} /{ }^{\circ} \mathrm{C}$ bulk modulus for copper $=140 \times 10^9 \mathrm{~N} / \mathrm{m}^2$

Answer

According to the problem, decrease in temperature$(\Delta\text{t})=57-37=20^\circ\text{C}$
Coefficient of linear expansion$(\alpha)=1.7\times^{-5}/^{\circ}\text{C}$
Bulk modulus for copper $(\text{B})=140\times10^9\text{N/ m}^2$ Coefficient of cubical expansion,$(\gamma)=3\alpha=5.1\times10^{-5}/^\circ\text{C}$
Let initial volume of the cavity be V and its volume increases by $\Delta\text{V}$ due to increase in temperature.$\therefore\ \Delta\text{V}=\gamma\text{V}\Delta\text{t}$
$\Rightarrow\ \frac{\Delta\text{V}}{\text{V}}=\gamma\Delta\text{t}$
We know, $\text{B}=\frac{\text{stress}}{\text{volume strain}}$$\therefore$ Thermal stress $=\text{B}\times\Big(\frac{\Delta\text{V}}{\text{V}}\Big)=\text{B}(\gamma\Delta\text{T})$
$=\text{B}(3\alpha\Delta\text{T})\ \ (\because\gamma=3\alpha)$
$=140\times10^9\times3\times1.7\times10^{-5}\times20$
$=1.428\times10^8\text{Nm}^{-2}$
This is about $10^3$ times of atmospheric pressure.

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Question 35 Marks

During summers in India, one of the common practice to keep cool is to make ice balls of crushed ice, dip it in flavoured sugar syrup and sip it. For this a stick is inserted into crushed ice and is squeezed in the palm to make it into the ball. Equivalently in winter, in those areas where it snows, people make snow balls and throw around. Explain the formation of ball out of crushed ice or snow in the light of P–T diagram of water.

Answer

Given diagram shows the variation of pressure with temperature for water. When the pressure is increased in solid state (at 0°, 1 atm), ice changes into liquid state while decreasing pressure in liquid state (at 0°, 1 atm), water changes to ice.

When crushed ice is squeezed, some of it melts, filling up the gap between ice flakes upon releasing pressure. This water freezes, binding all ice flakes and making the ball more stable.

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Question 45 Marks

One day in the morning, Ramesh filled up 1/ 3 bucket of hot water from geyser, to take bath. Remaining 2/ 3 was to be filled by cold water (at room temperature) to bring mixture to a comfortable temperature. Suddenly Ramesh had to attend to something which would take some times, say 5-10 minutes before he could take bath. Now he had two options:

Fill the remaining bucket completely by cold water and then attend to the work.
First attend to the work and fill the remaining bucket just before taking bath. Which option do you think would have kept water warmer? Explain.

Answer

According to the Newton’s law o‘f cooling, the rate of loss of heat is directly proportional to the difference of temperature. Or we can say which gives a consequence about rate of fall of temperature of a body with respect to the difference of temperature of body and surroundings. The first option would have kept water warmer because by adding hot water to cold water, the temperature of the mixture decreases. Due to this temperature difference between the mixed water in the bucket and the surrounding decreases, thereby the decrease in the rate of loss of the heat by the water.
In second option, the hot water in the bucket will lose heat quickly. So if he first attend to the work and fill the remaining bucket with cold water which already lose much heat in 5-10 minutes then the water become more colder as comparison with first case.

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Question 55 Marks

A thin rod having length $L_0$ at $0°C$ and coefficient of linear expansion $\alpha$ has its two ends maintained at temperatures $\theta_1$ and $\theta_2,$ respectively. Find its new length.

Answer

When temperature of a rod varies linearly, then average temperature of the middle point of the rod can be taken as mean of temperatures at the two ends. According to the diagram, According to the diagram,

$\theta=\frac{\theta_1+\theta_2}{2}$
Let temperature varies linearly in the rod from its one end tp other end from $\theta_1$ to $\theta_2$ Let $\theta$ be the temperature of the mid-point of the rod. Therefore, average temperature of the mid-point of the rod is,$\Rightarrow\theta=\frac{\theta_1+\theta_2}{2}$
Using relation, $\text{L}=\text{L}_0(1+\alpha\theta)$ Or $\text{L}=\text{L}_0\Big[1+\alpha\Big(\frac{\theta_1+\theta_2}{2}\Big)\Big]$

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Question 65 Marks

A rail track made of steel having length 10m is clamped on a raillway line at its two ends. On a summer day due to rise in temperature by 20°C , it is deformed as shown in figure. Find x(displacement of the centre) if $\alpha_\text{steel}=1.2\times10^{-5}/{^\circ\text{C}}.$

Answer

Diagram shows the deformation of a railway track due to rise in temperature.

Applying Pythagoras theorem in right angled triangle,$\text{x}^2=\Big(\frac{\text{L}+\Delta\text{L}}{2}\Big)^2-\Big(\frac{\text{L}}{2}\Big)^2$
$\text{x}=\sqrt{\Big(\frac{\text{L}+\Delta\text{L}}{2}\Big)^2-\Big(\frac{\text{L}}{2}\Big)^2}$
$=\sqrt{\Big(\frac{\text{L}}{2}\Big)^2+\frac{2\text{L}\Delta\text{L}}{4}+\Big(\frac{\Delta\text{L}}{2}\Big)^2-\Big(\frac{\text{L}}{2}\Big)^2}$
$=\frac{1}{2}\sqrt{\big(\text{L}^2+\Delta\text{L}^2+2\text{L}\Delta\text{L}\big)-\text{L}^2}$
$=\frac{1}{2}\sqrt{\big(\Delta\text{L}^2+2\text{L}\Delta\text{L}\big)}$
As increase in length $\Delta\text{L}$ is very small, therefore, neglecting $(\Delta\text{L})^2,$ we get$\text{x}=\frac{\sqrt{2\text{L}\Delta\text{L}}}{2}$
But $\Delta\text{L}=\text{L}\alpha\Delta\text{t}$ According to the problem, L = 10m$\alpha=1.2\times10^{-5}{^\circ\text{C}^{-1}},\Delta\text{T}=20{^\circ\text{C}}$
Substituting value of $\Delta\text{L}$ in Eq. (i) from Eq. (ii)$\text{x}=\frac{1}{2}\sqrt{2\text{L}\times\text{L}\alpha\Delta\text{t}}=\frac{1}{2}\text{L}\sqrt{2\alpha\Delta\text{t}}$
$=\frac{10}{2}\times\sqrt{2\times1.2\times10^{-5}\times20}$
$=5\times\sqrt{4\times1.2\times10^{-4}}$
$=5\times2\times1.1\times10^{-2}$
$=0.11\text{m}=11\text{cm}$

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Question 75 Marks

We would like to make a vessel whose volume does not change with temperature (take a hint from the problem above). We can use brass and iron $\big(\beta_\text{ubrass}=6\times10^{-5}/\text{K and }\beta_\text{uiron}=3.55\times10^{-5}\text{K}\big)$ to create a volume of $100cc$. How do you think you can achieve this.

Answer

Volume of vessel, $V_0 = 100cm^3 = 10^{-4}$ = constant Volume of iron vessel $(V_I)$ - Volume of brass rod $(V_B) = 10^{-4}m^3 \Rightarrow V_I - V_B = 10^{-4}m^3$ This condition is possible if $\beta_\text{I}\text{V}_\text{I}\Delta\text{T}=\beta_\text{B}\text{V}_\text{B}\Delta\text{T}$
$\therefore\ \text{V}_{\text{I}}=\Big(\frac{\beta_\text{B}}{\beta_\text{I}}\Big)\text{V}_\text{B}$
$=\Big(\frac{6\times100^{-5}}{3.55\times10^{-5}}\Big)\text{V}_\text{B}$
$=1.69\text{V}_\text{B}$
From equation (i),$1.69\text{V}_\text{B}-\text{V}_\text{B}=10^{-4}$
$\Rightarrow\text{V}_\text{B}=\frac{10^{-4}}{0.69}=1.449\times10^{-4}\text{m}^3$
$\Rightarrow\text{V}_\text{B}=\frac{10^{-4}}{0.69}=1449\times10^{-4}\text{m}^3$
$\Rightarrow\text{V}_\text{I}=1.69\text{V}_\text{B}=1.69\times144.9=244.9\text{cm}^3$
Therefore, an iron vessel with a volume of $249.9cm^3$ fitted with a brass rod of volume $144.9cm^3$ will serve as a vessel of volume $100cm^3$, which will not change with temperature.

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Question 85 Marks

According to Stefan’s law of radiation, a black body radiates energy $\sigma=\text{T}^4$ from its unit surface area every second where T is the surface temperature of the black body and $\sigma=5.67\times10^{-8}\text{w}/\text{m}^2\text{K}^4$ is known as Stefan’s constant. A nuclear weapon may be thought of as a ball of radius $0.5m$. When detonated, it reaches temperature of 106K and can be treated as a black body.

Estimate the power it radiates.
If surrounding has water at $30C°$, how much water can $10\%$ of the energy produced evaporate in $1s$?

$\big[\text{s}_w=4186.0\text{J/ kgK and L}_v=22.6\times10^5\text{J/ kg}\big]$

If all this energy U is in the form of radiation, corresponding momentum is $\rho=\frac{\text{U}}{\text{c}}$ How much momentum per unit time does it impart on unit area at a distance of 1km?

Answer

$\text{E}=\sigma\text{T}^4$ per second per sq. m

Total E = radiated from all surface area A per will be power radiated by nuclear weapon, $\text{P}=\sigma\text{AT}^4$
$\sigma=5.67\times10^{-8}\text{W/m}^2/{\text{ K}}^4,\text{ R}=0.5\text{m},\text{ T}=10^6\text{K}$
$\text{P}=5.67\times10^{-8}\times(4\times\pi\text{R}^2)(10^6)^4$
$=5.67\times4\times3.14\times0.5\times0.5\times10^{-8}\times10^{24}$
$=5.67\times4\times3.14\times10^{24-8}\times1.00$
$\text{P}\cong18\times10^{16}\text{ Watt}=1.8\times10^{17}\text{J/s}\ ...(\text{i})$

$\therefore\ \text{P} =18\times10^{16}\text{watt}$

10% of this power is required to evaporate water
$\text{E}=\frac{10}{100}\times18\times10^{16}\text{ Watt }=1.8\times10^{16}\text{J/ s}$
Energy required by mkg water at 30°C to evaporate 100°C.
E required to heat up water from 30°C to 100°C
+E required to evaporate water into vapour
$=\text{mS}_w(\text{T}_2-\text{T}_1)+\text{mL}=\text{m}(\text{S}_w(\text{T}_2-\text{T}_1)+\text{L})$
$1.8\times10^{16}=\text{m}\big[4180(100-30)+22.6\times10^5\big]$
$=\text{m}\big[4186\times70+22.6\times10^5\big]$
$\text{m}(2.93020+22.6)\times10^5=1.8\times10^{16}$
$\text{m}25.5\times10^5=1.8\times10^{16}$
$\text{m}=\frac{1.8\times10^{16}}{25.5\times10^{5}}\cong7\times10^9\text{kg}$

Momentum per unit time $\text{p}'=\frac{\text{U}}{\text{c}}=\frac{1.8\times10^{17}}{3\times10^{8}}=0.6\times10^{9}$

From (i) $\text{P}'=6\times10^{8}\text{kg ms}^{-2}$
P per unit time per unit at a distance 1km $=\frac{6\times10^{8}}{4\pi\text{R}^2}$
$\therefore\ \text{p}=\frac{6\times10^8}{4\times3.14\times(10^3)^2}$
$\frac{6\times10^8}{4\times3.14\times10^6}=\frac{6\times100}{12.56}$
$=47.77\text{kg ms}^{-2}\text{/m}^2$
P per sec at 1km away on $m^2 = 47.8N/ m^2$

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