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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.
| 201. |
In question number 7, the temperature of other interfaces (in `.^(@)C`) isA. `26.5, 0.5`B. `27.5, 1`C. `28.5, 2`D. `29.5, 3` |
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Answer» Correct Answer - A `T_(1) = 27-(Ht_(1))/(K_(1)A_(1)) = 27-0.52 = 26.48^(@)C = 26.5^(@)C` and `T_(2) = 0+ (Ht_(1))/(K_(1)A_(1)) = 0.52^(@)C` |
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| 202. |
A student records the initial length l, change in temperature `Delta T` and change in length `Delta T` of a rod as follows `{:("S.No.","l(m)",Delta T(.^(@)C),Delta l(m)),(1.,2,10,4xx 10^(-4)),(2.,1,10,4 xx 10^(-4)),(3.,2,20,2 xx 10^(-4)),(4.,3,10,6 xx 10^(-4)):}` If the first observation is correct, what can you say about observation 2, 3 and 4. |
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Answer» From the 1st observation `alpha = (Delta l)/(l Delta l) rArr = (4 xx 10^(-4))/(2 xx 10) = 2 xx 10^(-5) .^(@)C^(-1)` For 2nd obervation `Delta l = alpha l Delta T` `= 2 xx 10^(-5) xx 1 xx 10 =2 xx 10^(-4)m != 4 xx 10^(-4)m` (Wrong) For 3rd observation `Delta l =alpha l Delta T` `= 2 xx 10^(-5) xx 2 xx 20 = 8 xx 10^(-4)m != 2 xx 10^(-4)m` (Wrong) For 4th observation `Delta l = alpha l Delta T` `= 2 xx 10^(-5) xx 3 xx 10 = 6 xx 10^(-4)m = 6 xx 10^(-4) m` [i.e., observed value (Correct)] |
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| 203. |
A wall of dimensions 2.00 m by `3.50 m` has a single-pane window of dimensions `0.75 m` by `1.20 m` . If the inside temperature is `20^(@)C` and the outside temperature is `-10^(@)C` , effective thermal resistance of the opaque wall and window are `2.10 m^(2) K W^(-1)` and `0.21 m^(2) K W^(-1)` respectively. The heat flow through the entire wall will be .A. `215 W`B. `205 W`C. `175 W`D. `110 W` |
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Answer» Correct Answer - A The wall and the window are in parallel arrangement , so net heat flow is the sum of heat flow through the wall and the window. The temperature difference, `T_(H) - T_(C) = 30 K` Area of window, `A_(i) = (0.75) (1.20) = 0.90 m^(2)` Heat flow through window pane, `((dQ)/dt)_(1)= (T_(H) - T_(C))/(R_(1)) = (0.90)(30)/(0.21) = 128.6 W` The area of the wall, `A_(2) = (2.00)(3.50) - (0.75)(1.20) = 6.10 m^(2)` Heat flow through the wall, `((dQ)/dt)_(2) = (T_(H) - T_(C))/(R_(2)) = ((6.10)(30))/((2.10)) = 87 W` Net heat flo, `(dQ)/(dt) = 128.6 + 87 = 215.6 W` |
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| 204. |
Assertion : Water can be made to boil without heating. Reason : Boiling point is lowered by increasing pressure.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If the assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - C We know that liquid starts boling when its vapour pressure becomes equal to atmosperic pressure. Hence to boil the water with out heating we have to decrease the external pressure. |
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| 205. |
A calorimeter of mass `0.2 kg` and specific heat `900 J//kg-K`. Containing `0.5 kg` of a liquid of specific heat `2400 J//kg-K`. Its temperature falls from `60^(@)C` to `55^(@)C` in one minute. Find the rate of cooling.A. `5 J//s`B. `15 J//s`C. `100 J//s`D. `115 J//s` |
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Answer» Correct Answer - D Rate of cooling (here it is rate of loss of heat) `(dQ)/(dt) = (mc+W) (d theta)/(dt)=(m_(1)c_(1)+m_(c)c_(c))(d theta)/(dt)` `rArr (dQ)/(dt) = (0.5xx2400+0.2xx900)((60-55)/(60))` `115 (J)/(sec)`. |
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| 206. |
Three liquids A, B and C having same sepcific heats have masses m,2m and 3m. Their temperaures are 0,20 and `3theta` respectively. ltBrgt Q. What is the temperature of mixture, when A and C mixed?A. `(5)/(2)theta`B. `(5)/(3)theta`C. `(7)/(3)theta`D. `(13)/(5)theta` |
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Answer» Correct Answer - A |
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| 207. |
A vessel is completely filled with water at `32.2^(@)F`. Now it is cooled thenA. Water spills outB. Level of water remains samesC. level of water decreasesD. none |
| Answer» Correct Answer - A | |
| 208. |
Emissive power of an ideal black body at `127^(@)C` is E. the temperature at which it increases to 102% isA. 400 KB. 100 KC. 402 KD. 502 K |
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Answer» Correct Answer - C |
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| 209. |
A glass cylinder contains `m_0=100g` of mercury at a temperature of `t_0=0^@C`. When temperature becomes `t_1=20^@C` the cylinder contains `m_1=99.7g` of mercury The coefficient of volume expansion of mercury `gamma_(He)=18xx(10^(-5)//^(@)C` Assume that the temperature of the mercury is equal to that of the cylinder. The coefficient of linear expansion of glass `alpha` isA. `=10^(-4)//.^(@)C`B. `=10^(-3)//.^(@)C`C. `=10^(-5)//.^(@)C`D. `=10^(-2)//.^(@)C` |
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Answer» Correct Answer - C |
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| 210. |
The density of a substance at `0^(@)C` is `10 g//c.c.` and at `100^(0)C` its density os `9.7g//c.c.` The coefficient of linear expansion of the sunstance is.A. `10^(-4//@)C `B. `3 xx 10^(-4//@)c`C. `6 xx 10^(-4//@)c`D. `9 xx 10^(-4//@)C` |
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Answer» Correct Answer - A `gamma = (d_(0) - d_(1))/(d_(1)Deltat) alpha = (gamma)/(3)` |
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| 211. |
The density of a substance at `0^(@)C` is `10 g//c c` and at `100^(@)C`, its density is `9.7 g//c c`. The coefficient of linear expansion of the substance isA. `10^(2)`B. `10^(-2)`C. `10^(-3)`D. `10^(-4)` |
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Answer» Correct Answer - D The coefficient of volume expansion is given by `gamma = (Delta rho)/(rho_(1)xxDelta T)` where `rho_(1)` is density. `T` is temperature `gamma = (rho_(1)-rho_(2))/(rho_(1)(T_(2)-T_(1)))` `=(10-9.7)/(10(100-0)) = 3xx10^(-4)` Coefficient of linear expansion is `alpha = (gamma)/(3) = (3xx10^(4))/(3) = 10^(-4)`. |
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| 212. |
Show that the temperature of a planet varies inversely as the square root of its distance from the sum. |
| Answer» `Tprop(1)/(sqrt(r))` | |
| 213. |
An iron tank tas a capacity of V=50 litres of kerosene at `T_(0)=273K`. What amount of kerosene will flow out of the full tank, if it is brought in a room where the temperature is T=293K? [The density of kerosene `rho=0.8xx10^(3)kg//m^(3)`, coefficient of volume expansions of kerosene `gamma_(k)=10xx10^(-4)` and coefficient of linear expansio of iron `alpha_(i)=1.2xx10^(-5)`, all in SI units]. |
| Answer» Correct Answer - `Deltam=0.7712kg` | |
| 214. |
Find the rate of heat flow through a cross section of the rod shown in figure `(theta_(2)gttheta_(1))` . Thermal conductivity of the material of the rod is `K`. |
| Answer» `(pikr_(1)r_(2)(T_(1)-T_(2)))/(l)` | |
| 215. |
Which of the following graphs correctly shows variation of coefficient of volume expansion of copper as a function of temperature ?A. B. C. D. |
| Answer» Correct Answer - C | |
| 216. |
The variation of length of two metal rods A and B with change in temperature is shown in Fig. the coefficient of linear expansion `alpha_A` for the metal A and the temperature T will be?( alpha_(A)=3x10^(-6)/ C )A. `alpha_(A) = 3 xx 10^(-6 )//^(@)C, 500^(@)C`B. `alpha_(A) = 3 xx 10^(-6 )//^(@)C, 222.22^(@)C`C. `alpha_(A) = 27 xx 10^(-6) //^(@)C, 500^(@)C`D. `alpha_(A) = 27 xx 10^(-6) //^(@)C, 222.22^(@)C` |
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Answer» Correct Answer - D Slope of the line `A = (1006 - 1000)/(T) = (DeltaL)/(DeltaT)` `(6)/(T) = 1000mm alpha_(A)` ……..(1) similarly for `B` line `(2)/(T) = 1002mm alpha_(B)` …….(2) From `(1)` and `(2) alpha_(A) = 3alpha_(B)` |
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| 217. |
Which of the following is the largest rise in temperature?A. `1^(@)F`B. `1^(@)R`C. `1K`D. `1^(@)C` |
| Answer» Correct Answer - B | |
| 218. |
The variation of density of a solid with temperature is give by the formulaA. `d_(2) = (d_(1))/(1 + gamma (t_(2) - t_(1))`B. `d_(2) = (d_(1))/(1 - gamma (t_(2) - t_(1))`C. `d_(2) = (d_(1))/(1 - 2gamma (t_(2) - t_(1))`D. `d_(2) = (d_(1))/(1 + 2gamma (t_(2) - t_(1))` |
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Answer» Correct Answer - A `d_(1) = (d_(0))/((1 + gammaDeltat))` |
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| 219. |
A hole is drilled in a copper sheet. The diameter of the hole is `4.24 cm` at `27.0^(@)C`. What is the change in the diameter of the hole when the sheet is heated to `227^(0)C`? `alpha` for copper `= 1.70 xx 10^(-5)K^(-1)`A. `1.44 xx 10^(-2)cm`B. `14.4 xx 10^(-2)cm`C. `144 xx 10^(-2)cm`D. `0.144 xx 10^(-2)cm` |
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Answer» Correct Answer - A `L_(2) - L_(1) = L_(1)alphaDeltatT` |
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| 220. |
The range of clinical thermometer isA. `37^(@)C` to `42^(@)C`B. `95^(@)F` to `110^(@)F`C. `90^(@)F` to `112^(0)F`D. `95^(0)C` to `104^(@)C` |
| Answer» Correct Answer - B | |
| 221. |
A bimetal made of copper and iron strips welded together is straight at room temprature. It is held vertically so that the iorn strip is towards the left hand and coper strip is then heated. The bimetal strip willA. remain straightB. bend towards rightC. bend towards leftD. have no change |
| Answer» Correct Answer - C | |
| 222. |
Write the dimensional formul of Thermal stress. |
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Answer» Thermal stress=`(F)/(A)` [Thermal stress]=[`ML^(-1)T^(-2)`] |
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| 223. |
Find the thermal resistance of an aluminium rod of length 0.20 m and area of cross section `1xx10^(-4)m^(2)`. The heat current is along the length of the rod. [Thermal conductivity of aluminium`=200Wm^(-1)K^(-1)`] |
| Answer» `R=(x)/(KA),R=(0.20)/((200)(10^(-4)))=10KW^(-1)`. | |
| 224. |
A copper block of mass 0.06 kg is heated till its temperature is inreased by `20^(@)C`. Find the heat supplied to the block. [specific heat of copper`=9xx10^(-2)calg^(-1).^(@)C^(-1)`] |
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Answer» `Q=msDeltaT` `=(0.06)xx10^(3)(9xx10^(-2))(20^(@)C)=108` cal |
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| 225. |
A pendulum clock loses 12 s a day if be the temperature is `40^(@)C` and gains 4 s a day if the temperature is `20^(@)C`. The temperature at which the clock will show correct time and the coefficient of linear expansion `alpha` of the pendulum shaft are, respectivelyA. `25^(0)C,alpha = 1.85 xx 10^(-5) // ^(0)C`B. `60^(0)C,alpha = 1.85 xx 10^(-4) // ^(0)C`C. `30^(0)C,alpha = 1.85 xx 10^(-3) // ^(0)C`D. `55^(0)C,alpha = 1.85 xx 10^(-2) // ^(0)C` |
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Answer» Correct Answer - A Let at temperature `theta`, clock gives correc time `DeltaT = ((1)/(2)alphaDeltatheta)T, T = 1day = 86400s` `12 = (1)/(2)alpha(40-theta)T` ……(i) `4 = (1)/(2)alpha(theta - 20)T` …….(ii) `i//ii rArr theta = 25^(0)C` substitung `theta` in equation `(ii)`, we get `4 = (1)/(2)alpha(25 - 20) xx 86400` `alpha = (1)/(5 xx 86400) = 1.85 xx 10^(-5 //0)C` |
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| 226. |
A child running a temperature of `101^F` is given and antipyrin (i.e. a madicine that lowers fever) which cause an increase in the rate of evaporation of sweat from his body. If the fever is brought down to `98^@F` in 20 min., what is the averatge rate of extra evaporation caused, by the drug ? Assume the evaporation mechanism to the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about `580 cal. g^(-1).` |
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Answer» Initial temperature of the body of the child, `T_(1) = 101^(@)F` Final temperature of the body of the child, `T_(2) = 98^(@)F` Change in temperature, `DeltaT=[(101-98)xx(5)/(9)]^(@)C` Time taken to reduce the temperature, t = 20 min Mass of the child, m = 30 `kg = 30 × 10^(3) g` Specific heat of the human body = Specific heat of water = c `=1000 cal//kg// ^(@)C` Latent heat of evaporation of water, `L = 580 cal g^(–1)` The heat lost by the child is given as: `Delta theta=mcDeltaT` `=30xx1000xx(101-98)xx(5)/(9)` =50000 cal Let`m_(1)` be the mass of the water evaporated from the child’s body in 20 min. Loss of heat through water is given by: `Delta theta=m_(1)L` `:. m_(1)=(Delta theta)/(L)` `=(50000)/(580)=86.2 g` `:. " Average rate of extra evaporation caused by the drug " =(m_(1))/(t)` `=(86.2)/(200)=4.3g//min` |
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| 227. |
If a cylinder of diameter `1.0cm` at `30^(@)C` is to be slid into a hole of diameter `0.9997 cm` in a steel plate at the same temperature, the minimum required rise in the temperature of the plate is: (Coefficient of linear expansion of steel`= 12 xx 10^(-6//@)C`)A. `25^(@)C`B. `35^(@)C`C. `45^(@)C`D. `55^(@)C` |
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Answer» Correct Answer - A `r_(2)=r_(1)(1+alphaDeltat) impliesr_(2)-r_(1)=r_(1)alphaDeltat` |
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| 228. |
If temeratre of two shpheres of same size but made of different materials changes by `DeltaT` thenA. both expands equallyB. Sphere with greater `alpha` expands or contracts more than other.C. sphere with greater `alpha` expands or conteats less than other.D. both contracts equally. |
| Answer» Correct Answer - B | |
| 229. |
Fraunhofer line of the solar system is an example ofA. line emission spectrumB. emission of band spectrumC. line absorption spectrumD. none of the above |
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Answer» Correct Answer - C Fraunhofer lines are a set of spectral lines. These lines were originally observed as dark features in the optical spectrum of the sun. It was discovered by kirchhoff that each chemical element was associated with a set of spectral lines and deduced that the dark lines in the solar spectrum were caused by absorption by those elements in the upper layers of the sun. Hence, Fraunhofer line of the solar system is an example of line absorption specturm. |
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| 230. |
Refer to the plot of temperature versus time (figure) showing the changes in the state if ice on heating (not to scale). Which of the following is correct ? .A. The region AB represent ice and water in thermal equilibriumB. At B water starts boilingC. At C all the water gets converted into steamD. C to D represents water and steam in equilibrium at boiling point |
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Answer» During the process AB temperature of the system is `0^(@)C` Hence, it represents phase change that is transformation of ice into water while temperature remains `0^(@)C`. BC represents rise in temperature of water from `0^(@)C` to `100^(@)C` (at C). Now, water starts converting into steam which is represents by CD. |
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| 231. |
If a given mass of a gas occupies a volume `100cc` at one atmospheric pressure and a temperature of `100^(0)C`. What will be its volume at `4` atmospheric pressure, the temperature being the same?A. `100cm^(3)`B. `400cm^(3)`C. `25cm^(3)`D. `200cm^(3)` |
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Answer» Correct Answer - C `P prop (1)/(V)` |
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| 232. |
On centigrade scale the temperature of a body increases by 30 degrees. The increase in temperature on fahrenheit scale isA. `50^(@)C`B. `40^(@)C`C. `30^(@)C`D. `54^(@)C` |
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Answer» Correct Answer - D |
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| 233. |
The coeefficient of volume expansion of liquid is `gamma`. The fractional change in its density for `DeltaT` rise in tempeature isA. `gammaDeltaT`B. `(DeltaT)/(gamma)`C. `1+ gamma Deltat`D. `1- gamma DeltaT` |
| Answer» Correct Answer - A | |
| 234. |
The coefficient of volume expansion of glycerine is `49 xx 10^(-5)//^(@)C`. What is the fractional change in its density (approx.) for `30^(@)C` rise in temperature? |
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Answer» Coefficient of volume expansion of glycerin, `α_(V) = 49 × 10^(–5) K^(–1)` Rise in temperature, `ΔT = 30^(@)C` Fractional change in its volume `=(DeltaV)/(V)` This change is related with the change in temperature as: `(DeltaV)/(V)=a_(V)DeltaT` `V_(T2)-V_(T1)=V_(T1)a_(V)DeltaT` `(m)/(P_(T2))-(m)/(P_(T1))=(m)/(P_(T1))a_(v)DeltaT` Where, m = Mass of glycerine `P_(r1)`=Initial density at `T_(1)` ∴Fractional change in the density of glycerin = 49 ×10–5 × 30 = 1.47 × 10–2`P_(r_(2))"= Final density at" T_(2)` `(P_(r_(1))-P_(r_(2)))/(P_(r_(2)))a_(v)DeltaT` Where, `(P_(r_(1))-P_(r_(2)))/(P_(r_(2)))`=Fraction change in density ∴Fractional change in the density of glycerin `= 49 ×10^(–5) × 30 = 1.47 × 10^(–2)` |
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| 235. |
A crystal has linear coefficients `0.00004//^(@)C, 0.00005//^(@)C, 0.00006//6^(@)C`. Coefficient of cubical expansion of the crystal isA. `0.000015//^(@)C`B. `0.00015//^(@)C`C. `0.00012//^(@)C`D. `0.00018//^(@)C` |
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Answer» Correct Answer - B `gamma = alpha_(x) + alpha_(y) + alpha_(z)` |
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| 236. |
The Mechanical Equivalent Of HeatA. Has the same dimensional formula as heatB. has the same dimensional formula as workC. Has the same dimensional formula as energyD. Is dimensionless |
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Answer» Correct Answer - 4 |
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| 237. |
Distance between two places is `200km`. `alpha` of metal is `2.5 xx 10^(-5//@)C`. Total spece that must be left between steel rails to allow a change of temperature from `36^(0)F` to `117^(0)F` isA. `2.25km`B. `0.225km`C. `22.5km`D. `0.0225km` |
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Answer» Correct Answer - B `l_(2) - l_(1) = l_(1)alphaDeltat` |
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| 238. |
A metal sheet with a cricular hole in heated. The holeA. Gets largerB. Gets smallerC. Remains of the same sizeD. Gets deformed |
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Answer» Correct Answer - A |
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| 239. |
The holes through which the fist plates are fitted to join the rails are ovalin shape becauseA. bolts are in oval shapeB. to allow the movement of rails in the direaction of length due to change in temperature.C. to make the fitting easy and tightD. only oval shape holes are possible |
| Answer» Correct Answer - B | |
| 240. |
A ring shaped piece of a metal is heated, if the material expands, the hole willA. contractB. expandC. remain sameD. expand or contract depending on the width |
| Answer» Correct Answer - B | |
| 241. |
A semicircular metal ring subtends an angles of `180^(0)` at the centre of the circle. When it is heated, this angleA. remains constantB. increase slightlyC. decrease slightlyD. become `360^(@)` |
| Answer» Correct Answer - A | |
| 242. |
Two absolute scale `X` and `Y` have triple points of water defined to be `300 X` and `450 Y`. How are `T_(X)` and `T_(Y)` related to each other ?A. `T_(A) = T_(B)`B. `T_(B) = (3)/(2)T_(A)`C. `T_(B) = (2)/(3)T_(A)`D. `T_(B) = (3)/(4)T_(A)` |
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Answer» Correct Answer - B Size fo the degree on absolute scale `A =` Size of the degree on absolute Scale `B` `((276.16)T_(A))/(200) = ((276.16)T_(B))/(300)` |
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| 243. |
A Centigreade scales `A` and `B` have triple points of water defined to be `200 A` and `300 B` (given triple point of water is `= 276.16 K`). The relation between `T_(A)` and `T_(B)` isA. `30^(0)`B. `40^(0)`C. `60^(0)`D. `80^(0)` |
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Answer» Correct Answer - B `(F - 32)/(180) = (C - 0)/(100)` |
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| 244. |
If l is the length, A is the area of cross-section and K the thermal conductivity, then the thermal resistance of the block is given byA. `(Kl)/(A)`B. `(l)/(KA)`C. `(AK)/(l)`D. `(A)/(Kl)` |
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Answer» Correct Answer - B `Q=(KA(theta_(1)-theta_(2)))/(l) or Q/l = (theta_(1)-theta_(2))/(l//KA)` Compare with `I = V/R` So, thermal resistance = `l/(KR)`. |
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| 245. |
If the coefficient of conductivity of aluminium is `0.5 cal cm^(-1) s^(-1).^(@)C^(-1)`, then the other to conductor `10 cal s^(-1) cm^(-2)` in the steady state, the temperature gradient in aluminium must beA. `5^(@)C//cm`B. `10^(@)C//cm`C. `20^(@)C//cm`D. `10.5^(@)C//cm` |
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Answer» Correct Answer - C `(d theta)/(dt) = 1/K (dQ)/(Adt)`. |
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| 246. |
300 grams of water at `25^@ C` is added to 100 grams of ice at `0^@ C.` The final temperature of the mixture is _____ `^@C`A. `-(5)/(3).^(@)C`B. `-(5)/(2).^(@)C`C. `-5^(@)C`D. `0^(@)C` |
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Answer» Correct Answer - D |
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| 247. |
The surface temperature of the sun is `T_(0)` and it is at average distance `d` from a planet. The radius of the sun is `R`. The temperature at which planet radiates the energy isA. `T_(0) sqrt((R)/(2d))`B. `T_(0) sqrt((2R)/(d))`C. `T_(0) sqrt((R)/(d))`D. `T_(0) ((R)/(d))^(1//4)` |
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Answer» Correct Answer - A `r =` Radius of planet `T_(1)`= Temperature of planet at which energy is radiated Power radiated by Sun, `P=4 pi R^(2)sigma T_(0)^(4)` Energy recived by planet = `(P)/(4 pi d^(2)) xx pi r^(2) = (R^(2)T_(0)^(4)pi r^(2)sigma)/(d^(2))` Energy radiated by planet = `4 pi r^(2)sigma T_(1)^(4)` For thermal equilibrium, `(R^(2)T_(0)^(4)pi r^(2)sigma)/(d^(2)) = 4 pi r^(2)sigma T_(1)^(4)` `R^(2)T_(0)^(4) = 4d^(2)T_(1)^(4)` `T_(1) = T_(0)sqrt((R)/(2d))`. |
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| 248. |
The solar constant for a planet is `sum`. The surface temperature of the sun is T K. if the sun subtends an angle `theta` at the planet, thenA. `sum propT^(4)`B. `sumpropT^2`C. `sumproptheta^(2)`D. `sumproptheta` |
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Answer» Correct Answer - A::C |
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| 249. |
A solid sphere of iron at `2^(@)C` is lying at the bottom of a bucket full of water at `2^(@)C`. If the temperature of the water is increased to `3^(@)C`, the buoyant force on the sphere due to water willA. increaseB. Be unchangedC. DecreaseD. Increase or decrease depends upon the numerical values of coefficient of expansion of water and iron. |
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Answer» Correct Answer - A As the temperature of water is increased from `2^(@)C` to `3^(@)C` the density of water increase (remember anomalous behaviour of water), also the volume of sphere increase. Therefore buoyant force on sphere due to water shall increase. |
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| 250. |
Two identical metal balls at temperature `200^(@)C` and `400^(@)C` kept in air at `27^(@)C`. The ratio of net heat loss by these bodies isA. `1//4`B. `1//2`C. `1//16`D. `(473^(4) - 300^(4))/(673^(4) - 300^(4))` |
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Answer» Correct Answer - D If temperature of surrounding is considered, then net loss of energy of a body by radiation `Q = A epsilon(T^(4)-T_(0)^(4)) rArr Q prop (T^(4)-T_(0)^(4))` `=(Q_1)/(Q_2) = (T_(1)^(4)-T_(0)^(4))/(T_(2)^(4)-T_(0)^(4))` `=((273+200)^(4)-(273+27)^(4))/((273+400)^(4)-(273+27)^(4))` `=((473)^(4)-(300)^(4))/((673)^(4)-(300)^(4))` . |
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