3 Marks Question

Question 13 Marks

100 g of water is supercooled to -$10^\circC$. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze?
[Sw = 1cal/ $g/^\circC$ and $L^W$Fusion = 80cal/ g]

Answer

Water mass = 100g At -$10^\circC$ ice and water mixture exists. Heat required (given out) by $10^\circ\text{Cice to }0^\circ\text{C ice}=\text{ms}\Delta\text{t}$$=100\times1\times[0-(-10)]$
$\text{Q}=1000\text{cal}$
Let gm ice melted Q = ml$\text{m}=\frac{\text{Q}}{\text{L}}=\frac{1000}{80}=12.5\text{g}$
So, there is m $= 12.5g$ water and ice in mixture. Hence temperature of mixture remains $0^\circC.$

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Question 23 Marks

Find out the increase in moment of inertia I of a uniform rod $($coefficient of linear expansion $\alpha)$ about its perpendicular bisector when its temperature is slightly increased by $\Delta\text{T}.$

Answer

I of rod its axis along perpendicular bisector $=\frac{1}{12}\text{ML}^2$$\Delta\text{L}=\alpha\text{L}\Delta\text{T}$
$\text{I}'=\frac{1}{12}\text{M}(\text{L}+\Delta\text{L})^2$
$=\frac{1}{12}\text{M}(\text{L}^2+\Delta\text{L}^2+2\text{L}\Delta\text{L})$
Neglecting $\Delta\text{L}^2$ due to very small term.$\text{I}'=\frac{\text{M}}{12}(\text{L}^2+2\text{L}\Delta\text{L})$
$=\frac{\text{ML}^2}{12}+\frac{\text{ML}\Delta\text{L}}{6}\times\frac{2\text{L}}{2\text{L}}$
$\text{I}'=\frac{\text{ML}^2}{12}+\frac{\text{ML}^2}{12}\cdot\frac{2\Delta\text{L}}{\text{L}}$
$=\text{I}+\text{I}\cdot\frac{2\alpha\text{L}\Delta\text{T}}{\text{L}}$
$\text{I}'=\text{I}(1+2\alpha\Delta\text{T})$
So, new moment of inertia increased by $(2\text{I}\alpha\Delta\text{T}).$

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Question 33 Marks

A student records the initial length l, change in temperature $\Delta\text{T}$ and change in length $\Delta\text{l}$ of a rod as follows:

S.NO.	l(m)	$\Delta\text{T(C)}$	$\Delta\text{l}(\text{m})$
$1.$	$2	$10	$4 \times 10^{-4}$
$2.	$1	$10	$4 \times 10^{-4}$
$3.	$2	$20	$2 \times 10^{-4}$
$4.	$3	$10	$6 \times 10^{-4}$

If the first observation is correct, what can you say about observations 2, 3 and 4.

Answer

If the first observation is correct, hence from the $1^{\text {st }}$ observation we get the coefficient of linear expansion.
$\alpha=\frac{\Delta l}{1 \times \Delta T}=\frac{4 \times 10^{-4}}{2 \times 10}=2 \times 10^{-50} C^{-1}$
\$text { For } 2^{\text {nd }} \text { observation, } \Delta l=\alpha l \Delta T=2 \times 10^{-5} \times 1 \times 10$
$=2 \times 10^{-4} m \neq 4 \times 10^{-4} m$
Which is incorrect, For $3^{\text {rd }}$ observation, $\Delta l =\alpha l \Delta T =2 \times 10^{-5} \times 2 \times 20$
$=8 \times 10^{-4} m \neq 2 \times 10^{-4} m$
which is incorrect. For $4^{\text {th }}$ observation, $\Delta l =\alpha l \Delta T =2 \times 10^{-5} \times 3 \times 10=6 \times 10^{-4} m$ [i.e., observed value is correct]

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Question 43 Marks

Calculate the temperature which has same numeral value on celsius and Fahrenheit scale.

Answer

Let the required temperature is $x^0C = x^0F$$\frac{\text{C}}{100}=\frac{\text{F}-32}{180}$
$\Rightarrow\frac{\text{x}}{5}=\frac{\text{x}-32}{9}$
$\Rightarrow5\text{x}-160=9\text{x}$
$\Rightarrow-9\text{x}+5\text{x}=160$
$\Rightarrow-4\text{x}=160$
$\Rightarrow\text{x}=\frac{160}{-4}=-40^\circ$
$\therefore-40^\circ\text{F}=-40^\circ\text{C}$

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Question 53 Marks

Why does a metal bar appear hotter than a wooden bar at the same temperature? Equivalently it also appears cooler than wooden bar if they are both colder than room temperature.

Answer

It is due to facts that conductivity of metal bar is very high as of wood. So the rate of transferring the heat in metal is very large than in wood.
The specific heat of metal is very low as compared to wood, so metal requires very smaller quantities of heat than wood to change each degree of temperature.

So due to larger conductivity and smaller specific heat, metals become more colder when placed in colder region as compared to wood and become more hot when placed in hot region.

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