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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A sinker of weight `w_0` has an apparent weight `w_1` when weighed in a liquid at a temperature `t_1 and w_2` when weight in the same liquid at temperature `t_2`. The coefficient of cubical expansion of the material of sinker is `beta`. What is the coefficient of volume expansion of the liquid. |
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Answer» Let `theta=T_2-T_1` and `gamma=`coefficient of volume expansion of liquid. Let density of liquid at temperatures `T_1` and `T_2` be `rho_1` and `rho_2`, respectively. `impliesrho_1=rho_2(1+gammatheta)` .(i) Let `V_1` and `V_2` be the volumes of the sinker at temperatures `T_1` and `T_2`, respectively. `impliesV_2=V_1(1+gamma_stheta)` ..(ii) The loss in weight at `T_1=V_1rho_1gimpliesW_0-W_1=V_1rho_1g` ..(iii) The loss in weight at `T_2=V_2rho_2gimpliesW_0-W_2=V_2rho_2g`....(iv) Dividint. (iii) by Eq. (iv), `(W_0-W_1)/(W_0-W_2)=(V_1rho_1)/(V_2rho_2)` Using Eqs. (i) and (ii), `(W_0-W_1)/(W_0-W_2)=(1+gammatheta)/(1+gamma_stheta)` `implies1+gammatheta=(W_0-W_1)/(W_0-W_2)+gamma_s((W_0-W_1)/(W_0-W_2))theta` `impliesgamma=((W_2-W_1)/(W_0-W_2))(1)/(T_2-T_1)+((W_0-W_1)/(W_0o-W_2))gamma_s` |
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| 2. |
Consider two rods of same length and different specific heats (`S_1` and `S_2`), conductivities `K_1` and `K_2` and area of cross section (`A_1` and `A_2`) and both having temperature `T_1` and `T_2` at their ends. If the rate of heat loss due to conduction is equal thenA. `K_1A_1=K_2A_2`B. `K_2A_1=K_1A_2`C. `(K_1A_1)/(S_1)=(K_2A_2)/(S_2)`D. `(K_2A_1)/(S_2)=(K_1A_2)/(S_1)` |
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Answer» Correct Answer - A According to problem, rate of heat loss in both rods is equal, i.e.,`((dQ)/(dt))_(1)=((dQ)/(dt))_(2)` `implies(K_1A_1Deltatheta_1)/(l_1)=(K_2A_2Deltatheta_2)/(l_2)` `K_1A_1=K_2A_2[AsDeltatheta_1=Deltatheta_2=(T_1-T_2)` and `l_1=l_2` given] |
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| 3. |
If heat is supplied to a solid, its temperature (a) must increase (b) may increase (c) may remain constant (d) may decrease. |
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Answer» The correct answer is (b) (c) (b) may increase (c) may remain constant |
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| 4. |
If head a supplied to a solid, its temperatureA. must increaseB. may increaseC. may remain constantD. may decrease |
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Answer» Correct Answer - B::C may increase,may remain constant |
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| 5. |
A copper block of mass `60kg` is heated till its temperature is increased by `20^(@)C` Find the head supplied to the block specific head capacity of `= 0.09 cal g^(-1)^(@)C` |
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Answer» The head supplied is `Q = bs delta theta ` `= (60g)(0.09 cal g^(-1)^(@)C^(-1))(20^(@)C= 108 cal` The quantity ms is called the head cupacity of the body .Its unit is `JK^(-1)` The mass of water having the same head capacity as a given body is called water wquiualent of the body |
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| 6. |
In an industrial process 10 kg of water per hour is to be heated from `20^@C` to `80^@C`. To do this steam at `150^@C` is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at `90^@C`. How many kilograms of steam is required per hour (specific heat of steam`=1 cal//g^@C`, Latent heat of vapourization`=540 cal//g`)? |
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Answer» Correct Answer - `1 kg` |
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| 7. |
Water at `0^@C` was heated until it started to boil and then until it all changed to steam. The time required to heat water from `0^@C` to `100^@C` is 5 min and the time to change boiling water to steam is 28 minutes. If the flame supplied heat at a constant rate, the specific latent heat of vaporisation of water (neglecting heat losses, container etc. is (in `J//g`). (specific heat of water s `=4.2 J//g` )A. 540B. 2268C. 2352D. 2356 |
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Answer» Correct Answer - C Let heat is supplied at constant rate P, For `0^@C` to `100^@C` `Pxx(5xx60)=mxx4.2xx(100-0)` .(i) For boiling: `Pxx(28xx60)=mL` ..(ii) From (i) and (ii) `L=(4.2xx100)xx(28)/(5)=2352 J//g` |
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| 8. |
A ball is droped as a floor from a height of `2.0m` After the collsion it rises up to a height of `1.5m` Assume thet `40%` of the machanical energy lost goes as thermal energy into the ball.Calculate the rise in the temperature of the ball in the colliation headcapacity of the ball is `800J K^(-1)` |
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Answer» Correct Answer - A::B::C Let the mass of hall be m kg and `V_(1) = sqrt(2gh) = sqrt(40)` `V_(2)= sqrt(2gh) = sqrt(30)` So change in K.E. `= (1)/(2)xx mxx 40- ((1)/(2)m)xx 30 = ((10)/(2))m` `= 5m` That is utilisedto increases in temperature of the ball `((40)/(100)) xx (10)/(2)m = m xx 800 xx Delta t` `rArr Delta t = (1)/(400) = 0.0025` `= 2.5 xx 10^(-3)^(@)C` |
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| 9. |
Name three fossil fuels that emit carbon dioxide into the atmosphere |
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Answer» Coal, petroleum, natural gas. |
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| 10. |
With the increase in carbon dioxide in the atmosphere the acidity of oceans will : (a) decrease (b) remain unaffected (c) increase (d) none of these |
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Answer» (d) none of these |
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| 11. |
What is carbon tax? Who shall pay this tax? |
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Answer» The tax calculated on the basis of: carbon emission from industry, number of employee hour and turnover of the factory is called carbon tax. This tax shall be paid by industries. This will encourage the industries to use the energy efficient techniques. |
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| 12. |
The global warming has resulted:(a) the increase in yield of crops(b) the decrease in sea levels(c) the decrease in human deaths(d) the increase in sea levels |
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Answer» The increase in sea levels. Explanation: Due to global warming, the average temperature of the Earth has increased and has lead to the melting of ice around both the poles. This melting of ice has lead to an increase in the level of water in sea. |
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| 13. |
State the effect of enhancement of green house effect. |
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Answer» The effect of enhancement of greenhouse effect are: 1. The variable change in the climate in different parts of the world has created difficulty and forced the people and animals to migrate from one place to another place. 2. It has affected the blooming season of the different plants. 3. The climate changes have shown the immediate effect on simple organism and plants. 4. It has affected the world's ecology. 5. It has increased the heat stroke deaths. |
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| 14. |
Global warming will result in : (a) increase in agricultural production (b) decrease in the level of sea water (c) decrease in disease caused by bacteria (d) increase in the level of sea water |
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Answer» (d) increase in the level of sea water |
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| 15. |
State three ways to minimize the global warming. |
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Answer» Avoid Deforstation By Conservation of water. Usage of fossil fuel should be reduced. |
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| 16. |
What is meant by global warming? |
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Answer» Global warming means the increase in the average effective temperature of earth's surface due to an increase in the amount of greenhouse gases in its atmosphere. |
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| 17. |
State the impact of global warming on life on the earth. |
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Answer» The effect of enhancement of greenhouse effect are: 1. The variable change in the climate in different parts of the world has created difficulty and forced the people and animals to migrate from one place to another place. 2. It has affected the blooming season of the different plants. 3. The climate changes have shown the immediate effect on simple organism and plants. 4. It has affected the world's ecology. 5. It has increased the heat stroke deaths. |
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| 18. |
The gap between any two rails each of length l laid on a railway trach equals x at `27^@C` When the temperature rises to `40^@C` the gap closes up. The coefficient of linear expansion of the material of the rail is `alpha`. The length of a rail at `27^@C` will beA. `(x)/(26alpha)`B. `(x)/(13alpha)`C. `(2x)/(13alpha)`D. none of these |
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Answer» Correct Answer - B `l=x=lmuT` `l=(x)/(13alpha)` |
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| 19. |
A metal block of density `6000 kg m^(-3)` and mass `1.2 kg` is respended throught a spring of spring constant `200Nm^(-1)` .The spring - block system is dipped in water kept in a vessel .The water has a mass of `250g` and the block is at a height `40cm` above the bottom of the vassel .If the support to the spring is broken , what will be the rise in the temperature of the water specific beat capacity of the block is `250J kg^(-1)K^(-1)` and that of water is `4200J kg^(-1)K^(-1)` Head capacities of the vessel and the spring are nogligible |
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Answer» Correct Answer - C volume of the block `= (12)/(600) = 2 xx 10^(-4)m^(3)` When the mass in diped in water the block experience a huoyant force and spring experience PE whih is counteracted by its own weight `Kx + Vpg = mg` `rArr 200tau + 2 xx 10^(-1) xx 1000 xx 10= 12` `rArr x = ((12 - 2))/(200)` `= (10)/(200)= 0.05` Now the heat is equally transfered to both block and water `So, (1)/(2)Kx^(2) + mgh - Vpgh = m_(1)s_(1) Delta theta + m_(2)s_(2) Delta theta` rArr (1)/(2) xx 200 xx 0.0025 + 1.2 xx 10 xx ((40)/(100))` `= ((260)/(1000)) xx 6200 xx Delta t + 1.2 xx 250 xx Delta t` `rArr 0.25 + 4.8 - 0.8= 1092 = 200 Delta t` rArr 1392 Delta t = 4.25` `rArr Delta t = (4.25)/(1392) = 0.0030531` `= 3 xx 10^(-3)^(@)C` |
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| 20. |
Liquid nitrogen has a boiling point of `-195.81^@C` at atmospheric pressure. Calculate this temperature (a) in degrees Fahrenheit and (b) in kelvin. |
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Answer» We can use Eq. (i) to convert degree celsius into Fahrenheit and kelvin. (a) Temperature in Fahrenheit is given by `F_F=(9)/(5)T_C+32^@F=(9)/(5)(-195.81)+32=-320.46^@F` (b) Temperature in Kelvin `T_k=273.15K-195.81K=77.3K` |
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| 21. |
1 g ice of 0℃ melts to form 1 g water at 0℃. State whether the latent heat is absorbed or given out by ice. |
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Answer» Latent heat is absorbed by ice. |
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| 22. |
A solid of mass 80g and at 80°C melts completely to form liquid at 80°C by absorbing 640J of heat energy. What is the sp. latent heat of fusion of solid? |
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Answer» Mass of solid m = 80g at 80°C on melting 80g of liquid at 80°C mL = 640 80L = 640 Latent heat of solid = L = 640/80 = 8Jg-1 |
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| 23. |
A certain mass of a solid exists at its melting temperature of `20^(@)C`. When a heat Q is added `(4)/(5)` of the material melts. When an additional Q amount of heat is added the material transforms to its liquid state at `50^(@)C`. Find the ratio of specific latent heat of fusion (in `J//g`) to the specific heat capacity of the liquid (in `J g^(-1) .^(@)C^(-1)`) for the material. |
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Answer» Correct Answer - 60 |
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| 24. |
Due to thermal expansion with rise in temperature:A. metallic scale reading becames lesser that true valueB. pendulum clock becomes fastC. a floating body sinks a little moreD. the weight of a body in a liquid increases |
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Answer» Correct Answer - A::C::D a. If `theta^(`)gttheta`, Then Scale reading`lt`true value b. `Deltat=(1)/(2)alphaDeltat` If `theta^(`)gttheta` or `theta` is positive i.e., clock will lose time i.e., will become slow. Hence not correct. c. With rise in temperature upthrust decreses. As weight remains same, hence, a floating body sinks a littele more. d. As upthrust on the body decreases with rise in temperature hence, weight of a body in a liquid`(W_0-th)` increases. |
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| 25. |
During heat exchange, temperature of a solid mass does not change. In this process, heatA. in not being supplied to the massB. is not being taken out from the massC. may have been supplied to the massD. may have been taken out from the mass |
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Answer» Correct Answer - C::D This situation arises during melting (alternative (c )) or freezing (alternative (d)) of the mass. |
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| 26. |
The volume of a solid decreases by 0.6% when it is cooled through `50^(@)C`. Its coefficient of linear expansion isA. `4xx10^(-6) K`B. `5xx10^(-5) K`C. `6xx10^(4)K`D. `4xx10^(-5) K` |
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Answer» Correct Answer - D `(DeltaV)/V=0.6/100=gammaxxDelta theta` `gamma=0.6/100xx1/50xx2/2` `3 alpha=0.6/100xx2/100=1.2xx10^(-4)` `alpha=4xx10^(-5) .^(@)//C` |
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| 27. |
A heavy machine rejects a liquid at `60^(@)C` which is to be cooled to `30^(@)C` before it is fed back to the machine. The liquid rejected by the machine is kept flowing through a long tube while it is cooled by 60 liter water surrounding the tube. The initial temperature of the cooling water is `10^(@)C` and it is `20^(@)C` when it is changed after 1 hour. Calculate the amount of liquid that passes through the tube in one hour. Specific heat capacity of the liquid and water are `0.5 calg^(-1) .^(@)C^(-1)` and `1.0 calg^(-1).^(@)C^(-1)` respectively. |
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Answer» Correct Answer - `40 kg` |
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| 28. |
Some heat is provided to a body to raise its temperature by 25°C. What will be the corresponding rise in temperature of the body as shown on the kelvin scale? |
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Answer» The rise in temperature on kelvin scale will be 25k. |
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| 29. |
S.I. unit of thermal capacity is : (a) Jkg-1 (b) kJKg-1 (c) Jkg-1K-1 (d) caloC-1 |
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Answer» S.I. unit of thermal capacity is Jkg-1 K-1. |
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| 30. |
A lead piece of mass `25 g` gives out `1200` calories of heat when it is cooled from `90^@ C` to `10^@ C`. What is its (i) specific heat (ii) thermal capacity (iii) water equivalent. |
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Answer» Mass of lead piece `(m) = 25 g = 0.025 kg` Heat energy given out `(dQ) = 1200 xx 4.2 J` (i) specific heat `S = (1)/(m)(dQ)/(d theta)` =`(1)/(0.025) xx (1200 xx 4.2)/(80) = 2520 JKg^(-1) K^(-1)` (ii) Thermal capacity `= mS = 0.025 xx 2520 = 63 J//K` (iii) Water equivalent `(63)/(4200) Kg = 0.015 Kg`. |
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| 31. |
How should 1 kg of water at `5^@C` be divided into two parts so that if one part turned into ice at `0^@C`, it would release enough heat to vapourize the other part? Latent heat of steam`=540 cal//g` and latent heat of ice `=80 cal//g`. |
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Answer» Let the mass be divided into x grams (for ice) and (1000`-x`) grams (for vapour). Heat released by x grams of water `=x xx1xx5+x xx80` Heat absorbed by `(1000-x)` grams of water `=(1000-x)xx1xx95+(1000-x)xx540` Assuming that the conversion of the other part takes place at `100^@C`. `85x=95(1000-x)+540(1000-x)` or `x=882g` Thus the mass is to be divided into 882 g for conversion into ice and 118 g for conversion into vapour. |
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| 32. |
How should `2kg` of water at `60^(@)C` be divided so that when one part of it is turned into ice at `0^(@)C` it would give out sufficient heat to vaporize the other part? (Specific latent heat of fusionof ice `=336xx10^(3)J kg^(-1)` and specific latent heat of vaporization of steam `=2250xx10^(3)Jkg^(-1)`) |
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Answer» Correct Answer - `0.4144kg` |
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| 33. |
The melting point of ice `0^@C` at `1 atm`. At what pressure will it be `-1^@ C` ? (Given, `V_(2)-V_(1) =(1-(1)/(0.9)) xx 10^(-3) m^(3)`). |
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Answer» Here `Delta T = (-1-0) =-1,T = 273 +0 = 273 K` and `V_(2)-V_(1)=(1-(1)/(0.9)) xx 10^(-3) m^(3)` (given) `L = 80 cal//g` we have, `(Delta P)/(Delta T) = (L)/(T(V_(2)-V_(1)))` `(Delta P)/((-1)) = (80 xx 4.2 xx 10^(3))/(273(1-(1)/(0.9))xx 10^(-3))` `:. Delta P = 110.8 xx 10^(5) N//m^(2) = 110.8 atm` `P_(2) -P_(1) =110.8 atm rArr P_(2) = 110.8 +P_(1) = 111.8 atm`. |
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| 34. |
A thermal insulated vessel contains some water at `0^(@)C`. The vessel is connected to a vaccum pump to pum out water vapour. This results in some water getting frozen. It is given latent heat of vaporization of water at `0^(@)C = 21 xx 10^(5) J//kg` and latent heat of freezing of water `=3.36 xx 10^(5) J//kg`. the maximum percentage amount of water vapour that will be solidified in this manner will be:A. `86.2 %`B. `33.6 %`C. `21 %`D. `24.36 %` |
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Answer» Correct Answer - A Let `m_(1)` mass is vaporised and `m_(2)` mass gets solidified Then heat taken in vaporisation = heat given during or `m_(1)(21 xx 10^(5)) = (m_(2))(3.36 xx 10^(5))`. |
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| 35. |
It is observed that the temperature of the surrounding starts falling when the ice in a frozen lake starts melting. Give a reason for the observation. |
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Answer» Every kilogram of ice at 0°C on melting to form water 0°C needs 336 x 103J of heat energy as its specific latent heat is 336 x 103J. This heat energy is supplied by the surrounding of the lake, which in turn results in the fall in temperature. |
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| 36. |
The temperature of the surrounding starts falling when ice in a frozen lake starts melting. Give reason. |
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Answer» The reason is that the heat energy required for melting the frozen lake is absorbed from the surrounding atmosphere. As a result, the temperature of the surrounding falls and it became very cold. |
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| 37. |
A circular ring (centre O) of radius a, and of uniform cross section is made up of three different metallic rods AB, BC and CA (joined together at the points A, B and C in pairs) of thermal conductivityies `alpha_1`,`alpha_2` and `alpha_3` respectively (see diagram). The junction A, B and C are maintained at the temperatures `100^@C`,`50^@C` and `0^@C`, respectively. All the rods are of equal lengths and cross sections. Under steady state conditions, assume that no heat is lost from the sides of the rods. Let `Q_1`,`Q_2` and `Q_3` be the rates of transmission of heat along the three rods AB, BC and CA. ThenA. `Q_1=Q_2=Q_3` and all are transmitted in the clockwise senseB. `Q_1` and `Q_2` flow in clockwise sense and `Q_3` in the anticlockwise sense.C. `Q_1:Q_2:Q_3::alpha_1:alpha_2:2alpha_3`D. `(Q_1)/(alpha_1)+(Q_2)/(alpha_2)=(Q_3)/(alpha_3)` |
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Answer» Correct Answer - B::C::D The flow of heat will always be in the direction of the temperature gradient from higher to lower temperature Hence `Q_1` in rod AB, `Q_2` in rod BC will both be in clockwise sense while `Q_3` in CA will be in anti-clockwise sense Also we have if L is the length of each rod and A its area of cross-section `Q_1=(alpha_1A(100-50))/(L)=(50alpha_1)(A)/(L)` `Q_2=(alpha_2A(50-0))/(L)=(50alpha_2)(A)/(L)` `Q_3=(alpha_3A(100-0))/(L)=(100alpha_3)(A)/(L)` Hence `Q_1:Q_2:Q_3::alpha_1:alpha_2:2alpha_3` Also `(Q_1)/(alpha_1)+(Q_2)/(alpha_2)=(50(A)/(L))+(50(A)/(L))=(100A)/(L)=(Q_3)/(alpha_3)` |
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| 38. |
Two bodies at different temperature are mixed in a calorimater.Which of the following quantities remain conserved?A. Sum of the temperature of the two bodiesB. Total heat of the two bodiesC. Total internal energy of the two bodiesD. Internal energy of each body |
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Answer» Correct Answer - C Total internal energy of the two bodies |
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| 39. |
A copper collar is to fit tightly about a steel shaft that has a diameter of 6 cm at `20^@C`. The inside diameter of the copper collar at the temperature is `5.98cm` Q. The tensile stress in the copper collar when its temperature returns to `20^@C` is `(T=11xx10^10N//m^(2))`A. `1.34xx10^5 N//m^(2)`B. `3.68xx10^-12 N//m^(2)`C. `3.68xx10^8 N//m^(2)`D. `1.34xx10^-12 N//m^2` |
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Answer» Correct Answer - C Tensile stress `=TalphaDeltatheta` `=11xx10^(10)xx17xx10^-6xx(217-20)` ltbr. `=3.68xx10^(+8)(N)/(m^2)` |
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| 40. |
A copper collar is to fit tightly about a steel shaft that has a diameter of 6 cm at `20^@C`. The inside diameter of the copper collar at the temperature is `5.98cm` Q. To what temperature must the copper collar be raised to that it will just slip on the steel shaft, assuming the steel shaft remains at `20^@C` ? `(alpha_("copper")=17xx10^(-6)//K`)A. `324^@C`B. `21.7^@C`C. `217^@C`D. `32.4^@C` |
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Answer» Correct Answer - C `l_1=l_0(1+alphat)` `(2pixx6)/(2)=l_0(1+alphat)` `(2pixx5.98)/(2)=l_0(1+20alpha)` `(6)/(5.98)=(1+alphat)/(1+20alpha)=((1+17xx10^-6t))/(1+20xx17xx10^-6)` `t=216.8^@Capprox217^@C` |
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| 41. |
In the continuous flow method of Callender and Barnes the the potential difference across the wire was 3 volts and the current 2 amperes. The temperature of in flowing water was `20^(@)C` and that of out flowing water `22.7^(@)C` and `300g` of water were collected in 10 minutes. When the p.d. was increased to 3.75 volts and the current to 2.5 amperes, the flow was adjusted to maintain the same temperature difference and `240g` of water were collected in 5 minutes. Calculate the mean specific heat capacity of water at the mean temperature `21.35^(@)C`. |
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Answer» Correct Answer - `4167J kg^(-1)K^(-1)` |
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| 42. |
How does the (a) average kinetic energy (b) average potential energy of molecules of a substance change during its change in phase at a constant temperature? |
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Answer» (a) Average kinetic energy does not change. (b) Average potential energy increases. Explanation: When a substance is heated at constant temperature (i.e. during its phase change state), the heat supplied makes the vibrating molecules gain potential energy to overcome the intermolecular force of attraction and move about freely. This means that the substance changes its form. However, this heat does not increase the kinetic energy of the molecules, and hence, no rise in temperature takes place during the change in phase of a substance. This heat supplied to the substance is known as latent heat and is utilized in changing the state of matter without any rise in temperature. |
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| 43. |
If the length of a cylinder on heating increases by `2%`, the area of its base will increase byA. `0.5%`B. `2%`C. `1%`D. `4%` |
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Answer» Correct Answer - D Length is increased by 2% `(DeltaA)/A=Bxx4 theta=2(alphaxx Deltatheta)` Area will increases by 4% |
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| 44. |
A copper collar is to fit tightly about a steel shaft that has a diameter of 6 cm at `20^@C`. The inside diameter of the copper collar at the temperature is `5.98cm` If the breaking stress of copper is `230 N//m^2`, at what temperature will the copper collar break as it cools?A. `20^@C`B. `47^@C`C. `94^@C`D. `217^@C` |
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Answer» Correct Answer - C `alphaLDeltaT+L((F)/(A))/(alphaY)=0` `DeltaT=((F)/(A))/(alphaY)=(230xx10^6(N)/(m^2))/((17xx10^-6K^-1)(110xx10^9(N)/(m^2))` `=-123K=-123^@C` Final temperature at which the copper collar breaks `T_f=T_i+DeltaT=217^@C-123^@C=94^@C` |
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| 45. |
A container has a square cross-section of `10 cm xx 10 cm`. A cubical ice block of side length 6 cm is floating in water in the container. Water level in the container is 6 cm high. The ice block is at a temperature of `0^(@)C` and the water is at `16.15^(@)C`. Assume that heat exchange take place between the ice block and water only. What length of ice block will remain submerged in water when the system reaches thermal equilibrium? Assume that the ice block maintains its cubical shape as it melts. Take - density of ice `= 0.9 g//c c`, density of water = `1.0 g//c c` Specific heat capacity of water `= 1 cal g^(-1) .^(@)C^(-1)`, Specific latent heat of fusion of ice `= 80 cal g^(-1)` |
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Answer» Correct Answer - `4.5 cm` |
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| 46. |
What energy change would you expect to take place in the molecules of a substance when it undergoes.1. a change in its temperature ?2. a change in its state without a ny change in its temperature? |
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Answer» 1. The energy of the molecules of a body in creases with the rise in temperature and decreases with the fall of temperature. 2. Since, the temperature remains constant there is no change in the kinetic energy of the molecules. The energy given to substance to change the state of the substances increases potential energy of the molecules. |
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| 47. |
Water in lakes and ponds do not freeze at once in cold countries. Give a reason is support of your answer. |
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Answer» The specific latent heat of fusion of ice is sufficiently high (=336Jg-1), and so to freeze water, a large quantity of heat has to be withdrawn. Hence, it freezes slowly and thus keeps the surroundings moderate |
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| 48. |
Fig 11. 11 shows the variation in temperature with time when some wax cools from the liquid phase to the solid phase. (i) In which part of the curve, the wax is in liquid phase? (ii) What does the part QS of the curve represent? (iii) In which part of the curve, the wax will be the in the liquid as well as solid phase? (iv) In which part of the curve, the wax is in solid phase? |
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Answer» (i) In part PQ, the wax is in liquid phase. (ii) In the part QS, temperature remains constant with time, and hence, this part of the curve represents freezing. (iii) In part QS, the wax will be in the liquid as well as solid phase. (iv) In part ST, the wax is in solid phase. |
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| 49. |
Ice of mass 600 kg and at a temperature of `-10^@C` is placed in a copper vessel heated to `350^@C`. The resultant mixture is 550 g of ice and water. Find the mass of the vessel. The specific heat capacity of copper `(c)=100 cal//kg-K` |
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Answer» The final temperature is obviously `0^@C`. `600xx10^-3xx0.5xx10+(500-550)xx10^-3xx80` kcal `=mxx0.1xx350` kcal or `m=200xx10^-3kg` or `200g`. |
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Ice with mass `m_(1)=600g` and at temperature `t_(1)=-10^(@)C` is placed into a copper vessel heated to `t_(2)=350^(@)C`. As a result, the vessel now contains `m_(2)=550g` of ice mixed with water. Find the mass of the vessel. The specific heat of copper `=0.1"cal"//C^(@)g` and sp. heat of ice `=0.5"cal"//gC^(@)`. Latent heat of ice `=80"cal"//g`. |
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Answer» Correct Answer - `200g` |
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