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Physics 2 Marks Questions

PART - 2 CH - 10 Thermal Properties of Matter · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 23 questions.

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Question 12 Marks
When the filament of a bulb is given an energy of 32 watt then its temperature becomes 2000 K. Find that energy given to the filament when its temperature becomes 3000 K. Here the filament acts, as black-body.
Answer
From Stefan's law
$\begin{aligned}E & =\sigma T^4 \\\Rightarrow \quad \frac{E_2}{E_1} & =\left(\frac{T_2}{T_1}\right)^4 \\\Rightarrow \quad E_2 & =E_1\left(\frac{T_2}{T_1}\right)^4=32\left(\frac{3000}{2000}\right)^4 \\= & 32 \times \frac{81}{16} E_2=162 \text { Watt. }\end{aligned}$
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Question 22 Marks
If the absolute temperature of a black- body is double what would be its effect on emitted energy?
Answer
From Stefan's law the rate of emitted radiation energy
$\begin{aligned}& E=\sigma T^4 \\& E^1=\sigma(2 T)^4 \\\therefore \quad & \frac{E^1}{E}=\frac{\sigma(2 T)^4}{\sigma(T)^4}=\frac{16}{1} \\\Rightarrow \quad & E^1=16 E\end{aligned}$
i.e. the rate of emitted radiation energy would remain 16th times of the inittal rate.
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Question 32 Marks
The cross - sectional area of a non - conductor plate is $100 cm^2$ and its thickness is 2 cm . Its co-efficient of thermal conduction is $2 cal / s . cm$ ${ }^{\circ} C$. If the temperature difference between the two ends of the plate is $50^{\circ} C$ then calculate how much heat would flow through the plate in 10 hours?
Answer
Total quantity of heat flowing
$Q=KA\left(\frac{T_1-T_2}{d}\right) t$
$\begin{aligned}\text {Given :} \quad K & =2 cal / s cm{ }^{\circ} C \\& =2 \times 10^{-3} K.cal / sec\left(10^{-2} m\right)^{\circ} C \\& =2 \times 10^{-5} K cal / m \cdot sec .^{\circ} C \\A & =100 cm^2=100 \times 10^{-4} m^2 \\& =10^{-2} m^2 \\t & =10 \text { hours }=10 \times 3600=36000 sec\end{aligned}$
$\begin{aligned}T_1-T_2 & =50^{\circ} C, d=2 cm=2 \times 10^{-2} m \\Q & =2 \times 10^{-5} \times 10^{-2}\left(\frac{50}{2 \times 10^{-2}}\right) \times 36000 \\& =18 K . cal\end{aligned}$
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Question 42 Marks
Find out when a 50 m long steel rod of 10 $c m ^2$ cross-section is heated from $0^{\circ} C$ to $50^{\circ} C$ (i) increase in length of rod (ii) if both the ends of the rod are connected to rigid supports, then the force produced in it will know thermal stress. (Given $\alpha=1.2 \times 10^{-5}{ }^{\circ} C ^{-1}$ and $\left.Y =9.6 \times 10^7 N / m ^2\right)$
Answer
According to question
$\begin{aligned}L_0 & =50 m \\A_0 & =10 cm^2=10 \times 10^{-4} m^2\end{aligned}$
$\begin{aligned}\Delta T & =50^{\circ} C \\\alpha & =1.2 \times 10^{-5}{ }^{\circ} C^{-1} \\Y & =9.6 \times 10^7 Nm^{-2}\end{aligned}$
(i) We know that change in length
$\begin{aligned}\Delta L & =\alpha L_0 \Delta T \\& =1.2 \times 10^{-5} \times 50 \times 50 \\& =12\times 25 \times 10^{-4} \\& =300 \times 10^{-4}=0.03 m\end{aligned}$
(ii) Thermal stress produced in rod
$\begin{array}{l}=Y \alpha \Delta T \\=9.6 \times 10^7 \times 1.2 \times 10^{-5} \times 50 \\=9.6 \times 1.2 \times 50 \times 10^2 \\=5.76 \times 10^4 N / m^2\end{array}$
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Question 52 Marks
A layer of ice of 20 cm thickness accumulates on the surface of the water of a pond. The temperature of air is $-10^{\circ} C$. The thickness of the ice layer is 0.1 cm, how long will the increase take? The conductivity of ice is $2.1 \text {Joule}~ \text {sec} ^{-1} m^{-1} k ^{-1}$, latent heat $3.36 \times 10^5 \text {Joule}~\text {kg} ^{-1}$ and density $10^3$.
Answer
Given :
Latent heat of ice $= L =3.36 \times 10^5$ Joule $kg ^{-1}$
Thickness of surface of ice $=20 cm=0.20 m$
Temperature under the ice layer $=-10^{\circ} C$
0.1 cm on 1 meter surface of ice ( 0.001 m )
Volume of thick ice layer $=1 \times 0.001$
$=.001 m^3$
Mass of ice $.001 m^3= m =0.001 \times 10^3=1 kg$
Energy lost to convert 1 kg of water into ice
$\begin{array}{l}=mL \\=1 \times 3.36 \times 10^5 \text { Joule }\end{array}$
Radiated energy from ice $1 m^3=\frac{ KA \left( T _2- T _1\right) \times t}{d}$
$\begin{aligned}3 .36 \times 10^5 & =\frac{2.1 \times 1(0-(-10)) \times t}{0.2} \\t & =\frac{3.36 \times 10^5 \times 0.2}{21} \\& =\frac{672}{21} \times 10^2=32 \times 10^2 \\& = 3200~ \text {sec} .\end{aligned}$
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Question 62 Marks
If the temperature of a substance is $-50^{\circ}\text C $ centigrade scale, then find its tempeature on (i) Fahrenheit and (ii) Kelvin scale.
Answer
We know that
$\begin{aligned} \frac{ T _{ C }}{100} & =\frac{ T _{ F }-32}{180} \\ \frac{-50}{5} & =\frac{ T _{ F }-32}{9} \\\therefore~ T_{ F } & =\frac{-50 \times 9}{5}+32 \\ T_{ F } & =-90+32=-58^{\circ} F \end{aligned}$
(ii) $\frac{ T _c}{100}=\frac{ T _{ K }-273}{373-273}$
$\begin{aligned} \frac{ T _{ C }}{100} & =\frac{ T _{ K }-273}{100} \\ T_{ K } & = T _{ C }+273 \\ & =-58^{\circ} C +273=215 K\end{aligned}$
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Question 72 Marks
The ratio of the coefficient of thermal conductivity of two different substances is 4: 3. If the thermal resistances of rods of equal thickness of these materials is to be kept the same, then what should be the ratio of their lengths?
Answer
If the length $l$, area of cross-sectional A of a rod and the thermal conductivity of its materials is K, then the thermal resistance R of the material of the rod will be :
$R =\frac{l}{ KA }$
Therefore, to keep the thermal resistance of two rods 1 and 2 are equal
$\frac{l_1}{K_1 A_1}=\frac{l_2}{K_2 A_2}$
If both the rods are of equal thickness i.e.
$\begin{array}{c}d_1=d_2 \\\frac{l_1}{l_2}=\frac{K_1}{K_2}\end{array}$
$\frac{ K _1}{K_2}=\frac{4}{3}$ then
$\frac{l_1}{l_2}=\frac{4}{3}$
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Question 82 Marks
The cross-sectional areas of two rods made of the same metal are equal. The length of one rod is 40 cm and the temperature difference at its ends is $120^{\circ}\text C $ and the length of the other rod is 60 cm and the temperature difference at its ends is $120^{\circ}\text C $. Give the necessary calculations and tell which rod will have the higher rate of heat conduction?
Answer
Let the thermal conductivity of the metal be K. The area of cross-section of each rod is A . The length of both the rods are $d_1$ and $d_2$ and the temperature difference at their ends are $\left( T _1- T _2\right)$ and $\left( T _3- T _4\right)$ respectively, then the temperature difference of the rate of heat conduction in them is $\left( T _1- T _2\right)$ and $\left( T _3- T _4\right)$ respectively then the rate of heat conduction will be :
$\begin{aligned}H_1 & =\frac{Q_1}{t_1}=\frac{KA\left(T_1-T_2\right)}{d_1} \ldots\ldots (1)\\H_2 & =\frac{Q_2}{t_2}=\frac{KA\left(T_3-T_4\right)}{d_2}\ldots\ldots (2) \\\therefore \quad \frac{H_1}{H_2} & =\left(\frac{T_1-T_2}{T_3-T_4}\right) \times \frac{d_2}{d_1} \\& =\frac{100}{120} \times \frac{60}{40}=\frac{5}{4}\end{aligned}$
The rate of heat conduction will be higher in the first rod.
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Question 92 Marks
The shapes and thickness of the walls of two containers made of different metals are similar in all respects. If equal quantities of ice filled in both the containers melt completely in 10 minute and 25 minutes respectively, then find the ratio of the coefficient thermal conductivity of the materials of the containers.
Answer
Let us assume that the thermal conductivities of the metals of the cantainers are $k _1$ and $k _2$. The area of both the containers is A , the thickness of both is d , let the heat in the containers with thermal conductivity be $k _1$ and for time $t_2$, heat is passed in a container with thermal conductivity be $k _2$. The amount of heat flowing in each container will be equal.
$i.e.$
$Q =\frac{ K _1 A\left( T _1- T _2\right) t_1}{d}=\frac{ K _2 A\left( T _1- T _2\right) t_2}{d}$
Where $T _1$ and $T _2$ is the temperature of the outermost and innermost surface of the container wall.
From this equation,
$K _1 t_1= K _2 t_2$
$\therefore \quad \frac{ K _1}{K_2}=\frac{t_2}{t_1}=\frac{25 min}{10 min}=\frac{5}{2}$
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Question 102 Marks
A pan filled with hot food cools from $94^{\circ}\text C $ to $86^{\circ}\text C $ is 2 minutes when the room temperature is at $20^{\circ}\text C $. How long will it take to cool from $71^{\circ}\text C $ to $69^{\circ}\text C $?
Answer
The average temperature of $94^{\circ} C$ and $86^{\circ} C$ is $90^{\circ} C$ which is $70^{\circ} C$ more than room temperature. In these conditions the temperature of the vessel decreases in 2 minutes ( $94^{\circ} C -86^{\circ} C =8^{\circ} C$ ).
From Newton's law of cooling we know that
$\begin{array}{c}\frac{\text { Temperature difference }}{\text { Time }}=K \Delta T \\\frac{8^{\circ} C}{2 \text { minute }}=K\left(70^{\circ} C\right)\ldots\ldots (1)\end{array}$
The mean temperature of $69^{\circ} C$ and $71^{\circ} C$ is $70^{\circ} C$ which is $50^{\circ} C$ more than room temperature. In this state K is same as the ground state.
Hence,
$\frac{2^{\circ} C}{\text { Time }}=K\left(50^{\circ} C\right)\ldots\ldots (2)$
Dividing both equation
$\begin{aligned} \frac{8^{\circ} C / 2 \text { minute }}{2^{\circ} C / \text { time }} & =\frac{70^{\circ} C }{50^{\circ} C }=\frac{7}{5} \\ & =\frac{8 \times \text { time }}{2 \times 60 \times 2}=\frac{7}{5}\end{aligned}$
$\begin{array}{ll}\Rightarrow & \text { Time }=\frac{7 \times 2 \times 60 \times 2}{5 \times 8}=42 \text { second } \\ \Rightarrow & \text { Time }=42 \text { second }\end{array}$
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Question 112 Marks
In an electric store heat keeps on coming out continously, yet its temperature becomes stable after some time. Why?
Answer
When electric current is passed through an electric stove, the stove starts heating slowly and gradually due to which its temperature increases. When heated, heat starts radiating from it. When the rate of heat radiated by the store becomes equal, then the temperature of the stove becomes stable.
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Question 122 Marks
The temperature of star emitting blue light would be less or more than the temperature of sun? Also give arguments in favour of your answer.
Answer
According to Wein's law of radiation
$\begin{array}{l}\lambda_{ m } T =\text { Constant } \\ \text { or } T \propto \frac{1}{\lambda_{ m }}\end{array}$
In the light emitted by the sun the intensity of yellow light is maximum. So the sun can be considered to be a yellow coloured star. The wavelength of blue coloured star is less than that of yellow colour. So the temperature of blue coloured star will be more than the temperature of the sun.
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Question 132 Marks
The radius of an object A is twice the radius of object B. If both have the same surface and temperature then how many times will the rate of cooling of A be the rate of cooling of B?
Answer
The rate of cooling of A will be 4 times that of B. Since the rate of cool i.e. the rate of energy emission is directly proportional to the area of the object while the area is directly proportional to the square of the radius. Therefore, if the area of A is double that of B, the rate of occurrence of A will be 4 times that of B.
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Question 142 Marks
Star A emits green coloured light and star B blue coloured. Among the two temperature of which are is more?
Answer
According to Wien's displacement law
$\begin{array}{l}\lambda_{ m } T =\text { constant } \\ \text { and } T \propto \frac{1}{\lambda_{ m }}\end{array}$
Since the wavelength of blue colour is less than of green colour, so the temperature of the star emitting blue light will be more i.e. the temperature of star B would be more than that of star A.
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Question 152 Marks
When the temperature of an object is increased slowly, why does it appear red at first?
Answer
According to Wien's law related to radiations
$\lambda_{ m } T =$ Constant
According to this if temperature T is less then the wavelength of emitted radiation $\lambda_m$ would be more. In the visible light the wavelength of red light is maximum. So when the object is heated slowly then the wavelength of emitted radiation at low temperature will be maximum so on heating an object the radiation waves with big wavelengths (of red colour in visible light) are emitted first, and the object appears red at first.
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Question 162 Marks
Two spheres of the same external radius and made of the same material are heated to the same temperature one of them is hollow and the other is a solid sphere. Whose temperature will fall faster when cooled?
Answer
Both have the same external area. For this reason the rate of cooling will be the same but the rate of fall of temperature is inversely proportional to the mass. Therefore the temperature of hollow sphere will fall faster than that of a solid sphere.
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Question 172 Marks
Of the three stars in the sky, one is visible white the other red and the third blue. Which of these will have minimum temperature and which will have maximum temperature?
Answer
The temperature of red star will be minimum and the temperature of blue star will be maximum. The intensity of white radiation is maximum in the yellow part. This part is between in red and blue.
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Question 182 Marks
If a drop of water falls on a very hot iron, it does not evaporate for a long time, why?
Answer
The main reason for this is that due to the formation of a thin layer of water vapour, the water drop becomes insulated from the not iron which is a insulator is nature. The drop does not evaporate for very long. If the drop does not fall on a very hot iron, it comes in direct contact with the iron and vaporizes immediately.
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Question 192 Marks
A wool blanket keeps our body warm. By wrapping the same blanket over the ice, it keeps the ice cold. How would you explain this contradiction?
Answer
Wool is a bad conductor of heat. Apart from this, wool traps air inside it self and air is a bad conductor, hence no heat can be lost through convection. Woollen blanket keep the body warm by preventing heat loss from our body and hence our body remains warm.
In the case of ice, heat from outside is prevented from reaching the ice by the blanket and thus the ice remains cold.
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Question 202 Marks
Define heat conduction and coefficient of thermal conductivity.
Answer
Heat conduction : The process of energy transfer from one part of a solid to another at lower temperature without actual movement of molecules in called heat conduction.
Coefficient of Thermal Conductivity : The flow of heat energy per second from opposites faces of unit positive when the opposite faces are at a temperature difference of one $\text K$ or $1^{\circ}\text C $ at a unit distance is called coefficient of thermal conductivity.
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Question 212 Marks
If you have boiling water and cold milks for tea and you want to drink tea after some time, then to drink hot tea, should you add the milk at the same time or should you mix it just before drinking the tea, explain.
Answer
To drink hot tea, cold milk's should be mixed before hand because in this case the tea will be hotter to drink. The reason for this is that the rate of cooling will reduce.
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Question 222 Marks
Write Newton's law of cooling and explain the necessary condition for it and also explain the cooling curve.
Answer
Newton's law of cooling : According to Newton's law of cooling, if the heat excess is less than the surrounding environment, then the net energy emitted per second by the object or rate of cooling of the object is proporitonal to the heat excess. This law was propounded by Newton through experiment.
Necessary conditions : (1) The value temperature difference of the object and the surroundings should be less than the absolute temperature of the surroundings i.e. $\text T_0 = \text {Absolute temperature of surrounding}$
(2) Heat loss from the object should be only through radiation.
(3) This law is obtained from Stefan, Boltzmann law which applies for black-body so the radiating surface should be black.
Cooling curve : The graph between the temperature of the body and time is called cooling curve. The gradient of the tangent at any point on the curve gives the rate of temperature.
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Question 232 Marks
What are the necessary conditions of Newton's law of cooling?
Answer
Necessary conditions of Newton's law of colling : Newton's law of cooling should have the following conditions :
(1) The value of temperature difference should be less.
(2) The cooling process should happen slowly.
(3) The surface from which radiation is emitted should be a black-body surface and the heat loss of the object should occur only through radiation.
(4) The air should be calm.
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