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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.
| 101. |
If `L_(1)` and `L_(2)` are the lengths of two rods of coefficients of linear expansion `alpha_(1)` and `alpha_(2)` respectively the condition for the difference in lengths to be constant at all temperatures isA. `L_(1)alpha_(1) = L_(2)alpha_(2)`B. `L_(1)alpha_(2) = L_(2)alpha_(1)`C. `L_(1)alpha_(1) = L_(2)alpha_(2)`D. `L_(1)alpha_(2) = L_(2)alpha_(1) ` |
| Answer» Correct Answer - A | |
| 102. |
The length of two metallic rods at temperatures `theta` are `L_(A)` and `L_(B)` and their linear coefficient of expansion are `alpha_(A)` and `alpha_(B)` respectively. If the difference in their lengths is to remian constant at any temperature thenA. `L_(A)//L_(B) = alpha_(A)//alpha_(B)`B. `L_(A)//L_(B) = alpha_(B)//alpha_(A)`C. `alpha_(A) = alpha_(B)`D. `alpha_(A) alpha_(B) = 1` |
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Answer» Correct Answer - B Change in `L_(A)` = change in `L_(B)` i.e., `Delta L_(A) = L_(B)` `rArr alpha_(A)Delta T L_(A)=alpha_(B)Delta T L_(B)` or, `alpha_(A)L_(A)=alpha_(B)L_(B)`. |
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| 103. |
Two uniform metal rods one of aluminium of length `l_(1)` and another made of steel of length `l_(2)` and linear coefficients of expansion `alpha_(a)` and `alpha_(s)` respectively are connected to form a single rod of length `(l_(1) + l_(2))`, When the temperature of the combined rod is raised by `t^(@)C`, the length of each rod increases by the same amount then `(l_(l))/(l_(1) + l_(2))` isA. `(alpha_(s))/(alpha_(a) + alpha_(s))`B. `(alpha_(a))/(alpha_(a) + alpha_(s))`C. `(alpha_(a))/(alpha_(s))`D. `(alpha_(s))/(alpha_(a))` |
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Answer» Correct Answer - A `l_(1)alpha_(a)Delta_(1) = l_(2)alpha_(s)Deltat_(2)` |
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| 104. |
Coefficient of linear expansion of brass and steel rods are `alpha_(1)` and `alpha_(2)`. Length of brass and steel rods are `l_(1)` and `l_(2)` respectively. If `(l_(2) - l_(1))` is maintained same at all temperature, which one of the following relations holds good?A. `alpha_(1) l_(2) = alpha_(2) l_(1)`B. `alpha_(1) l_(2)^(2) = alpha_(2) l_(1)^(2)`C. `alpha_(1)^(2) l_(2) = alpha_(2) l_(1)`D. `alpha_(1) l_(1) = alpha_(2) l_(2)` |
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Answer» Correct Answer - D Change in length for both rods should be same `Delta l_(1) = Delta l_(2)` `l_(1)alpha_(1)Delta T = l_(2) alpha_(2) Delta T` `l_(1) alpha_(1) = l_(2) alpha_(2)`. |
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| 105. |
One end of conducting rod is maintained at temperature `50^(@)C` and at the other end ice is melting at `0^(@)C`. The rate of melting of ice is doubled if:A. the temperature is made `200^(@)C` and the area of cross-section of the rod is doubledB. the temperature is made `100^(@)C` and the length of the rod is made four timeC. area of cross-section of the rod is halved and length is doubled.D. the temperature is made `100^(@)C` and area of cross-section of rod and length both are doubled. |
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Answer» Correct Answer - D Rate of melting ice `prop` rate of heat transfer `(dQ//dt)` Further `(dQ)/(dt) = ("temperature difference")/((l//KA))` or `(dQ)/(dt) = ("temp. Diff.")/(l) xx A` If temp. Diff. `A` and `l` are all doubled, then `dQ//dt`, and hence rate of melting of ice, are doubled. |
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| 106. |
The two ends of a metal rod are maintained at temperature `100^(@)C` and `110^(@)C`. The rate of heat flow in the rod is found to be `4.0 J//s`. If the ends are maintained at temperature s `200^(@)C` and `210^(@)C`. The rate of heat flow will beA. `44.0 J//s`B. `16.8 J//s`C. `8.0 J//s`D. `4.0 J//s` |
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Answer» Correct Answer - D Rate of flow of heat `prop` temprature difference `4/Q = 10/10 rArr Q = 4 J//s`. |
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| 107. |
Two ends of a conducting rod of varying cross-section are maintained at `200^(@)C` and `0^(@)C` respectively. In steady state: A. temperature differences across `AB and CD` are equalB. temperature difference aross `AB` is greater than that across `CD`C. temperature difference aross `AB` is less than that across `CD`D. temperature difference may be equal or different depending upon thermal conductivity of the rod. |
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Answer» Correct Answer - C Rate of flow of heat `(dQ)/(dt)` of `H` is equal throughout the rod, temperature difference is given by `T.D = (H)` (thermal Resistance) or `T.D prop` theramal Resistance `R` Where `R = (l)/(KA) or R prop l/A` Area across `CD` is less. therefore `T.D` across `CD` will be more`. |
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| 108. |
What is the temperature of the steel-copper junction in the steady state system show in Fig. `7(e).17`? Length of steel rod = 15.0 cm, length of the copper rod = 10.0 cm, temperature of the furnace `=300^(@)C`, temperature of other end `0^(@)C`. The are of cross-section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel `=50.2 js^(-1) m^(-1) K^(-1)` and of copper `=3895 js^(-1) m^(-1) K^(-1)`). |
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Answer» The insulating material around the rods reduces heat loss from the sides of the rods. Therefore, heat flows only along the length of the rods. Consider any cross section of the rod. In the steady state, heat flowing into the element must equal the heat flowing out of it, otherwise there would be a net gain or loss of heat by the element and its temperature would not be steady. Thus in the steady state, rate of heat flowing across a cross section of the rod is the same at every point along the length of the combined steel-copper rod. Let T be the temperature of the steel-copper junction in the steady state. Then, `(K_(1)A_(1)(300-T))/(L_(1))=(K_(2)A_(2)(T-O))/(L_(2))` where 1 and 2 refer to the steel and copper rod respectively. For `A_(1) = 2 A_(2), L_(1) = 15.0 cm, L_(2) = 10.0 cm, K_(1) = 50.2 J s^(–1) m^(–1) K^( –1), K_(2) = 385 J s^(–1) m^(–1) K^( –1)`, we have `(50.2xx2(300-T))/(15)=(385T)/(10)` which gives `T=44.4^(@)C` |
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| 109. |
`10 gm` of ice at `-20^(@)C` is dropped into a calorimeter containing `10 gm` of water at `10^(@)C`, the specific heat of water is twice that of ice. When equilibrium is reached the calorimeter will contain:A. `20 gm` of waterB. `20 gm` of iceC. `10 gm` ice and `10 gm` of waterD. `5 gm` ice and `15 gm` of water |
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Answer» Correct Answer - C `Q_(1) = 10xx1xx10=100cal` `Q_(2) = 10xx0.5(0-(-20))+10xx80` `=(100+800)cal = 900cal` As `Q_(1) lt Q_(2)`, so ice will not completely melt and final temeprature = `0^(@)C` As heat given by water in cooling up to `0^(@)C` is only just sufficient to increase the temperature of ice from `-20^(@)C` to `0^(@)C` , hence mixture in equilibrium will consist of `10 gm` of ice and `10 gm` of water, both at `0^(@)C`. |
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| 110. |
Steam is passed into `54 gm` of water at `30^(@)C` till the temperature of mixture becomes `90^(@)C`. If the latent heat of steam is `536 cal//gm`, the mass of the mixture will beA. `80 gm`B. `60 gm`C. `50 gm`D. `24 gm` |
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Answer» Correct Answer - B Let mass of steam condensed = `M gm` `M xx 536 = mxx1xx(100-90)` `=54xx1xx(90-30)` `Mxx536=54xx60` `M = (54xx60)/(536) ~=6gm` Hence, mass of mixture = `54+6 = 60gm`. |
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| 111. |
Latent heat of ice 80 cal/gm . A man melts 60 g of ice by chewing in 1 minute . His power isA. 4800 WB. 336 WC. 1.33 WD. 0.75 W |
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Answer» Correct Answer - B |
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| 112. |
A brass disc fits snugly in a hole in a steel plate. Should you heat or cool this system to losen the disc from the hole ? given that `alpha _b gt alpha _Fe`.A. first heated then cooledB. first cooled then heatedC. is heatedD. is cooled |
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Answer» Correct Answer - D Since the cofficient of linear empansion of brass is greater than of steel. On cooling the brass contracts more, so it get loosened. |
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| 113. |
A steel rod of length `1 m` is heated from `25^@ "to" 75^@ C` keeping its length constant. The longitudinal strain developed in the rod is ( Given, coefficient of linear expansion of steel = `12 xx 10^-6//^@ C`).A. `-6 xx 10^(-6)`B. `-6 xx 10^(-5)`C. `-6 xx 10^(-4)`D. zero |
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Answer» Correct Answer - C Strain developed: `epsilon = alpha Delta T = (12 xx 10^(-6))(50)=6 xx 10^(-4)` Strain will be negative, as the rod is in a compressed state. |
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| 114. |
A steel rod of length `1 m` is heated from `25^@ "to" 75^@ C` keeping its length constant. The longitudinal strain developed in the rod is ( Given, coefficient of linear expansion of steel = `12 xx 10^-6//^@ C`).A. `6 xx 10^(-6)`B. `-6 xx 10^(-5)`C. `-6 xx 10^(-4)`D. zero |
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Answer» Correct Answer - C Strain developed: ` epsilon= alpha Delta T = (12 xx 10^(-6))(50)=6xx10^(-4)` Strain will be negative, as the rod is in a compressed state. |
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| 115. |
When the volumn of a gas is decreased at constant temperature the pressure increases because the moleculesA. strike the unit area of the walls of the container more often.B. strike the unit area of the walls of the container with higher speedC. strike the unit area of the wall of the container with lesser speed.D. move with more kinetic energy |
| Answer» Correct Answer - A | |
| 116. |
A gas is enclosed in a closed pot. On keeping this pot in a train moving with high speed , the temperature of the gasA. will increaseB. will decreaseC. will remain unchanged.D. increases or decrease depending on the chemical composition of gas |
| Answer» Correct Answer - C | |
| 117. |
Expansion during heating (a) occurs in solids only b) causes decrease of interatomic spacing (c) is due to increase of interatomic spacingA. only `(a)` is wrongB. (a),(b) & (c) are wrongC. (a) & (b) are wrongD. (a) ,(b) & (c) are correct |
| Answer» Correct Answer - C | |
| 118. |
Solids expand on heating becauseA. Kinetic energy of atom increasesB. Potential energy of atom increasesC. Total energy of atom increasesD. The potential energy curve is asymmetric about the equilibrium distance between neighbouring atoms |
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Answer» Correct Answer - D |
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| 119. |
The fraction of the volume of a glass flask must be filled with mercury so that the volume of the empty space may be the same at all temperature is `(alpha_("glass") = 9 xx 10^(-6)//^(0)C, gamma_(Hg) = 18.9 xx 10^(-5) // ^(0)C)`A. `(1)/(2)`B. `(1)/(7)`C. `(1)/(4)`D. `(1)/(5)` |
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Answer» Correct Answer - B `V_(1)gamma_(1) = V_(g)gamma_(g) rArr (V_(g))/(V_(l)) = (gamma_(l))/(gamma_(l)) rArr (V_(g))/(V_(l)) - 1 = (gamma_(l))/(gamma_(g)) - 1` |
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| 120. |
A glass flask of volume one litre at `0^(@)C` is filled, level full of mercury at this temperature. The flask and mercury are now heated to `100^(@)C`. How much mercury will spill out if coefficient of volume expansion of mercury is `1.82 xx 10^(-4)//^(@)C` and linear expansion of glass is `0.1 xx 10^(-4)//^(@)C` respectively?A. 21.2c cB. 15.2c cC. 2.12 c cD. 18.2c c |
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Answer» Correct Answer - B `DeltaV = V(gamma_(1) - gamma_(g))Deltat` |
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| 121. |
A glass flask with volume 200 `cm^(3)` is filled to the brim with mercury at `20^(@)C`. How much mercury overflows when the temperature of the system is raised to `100^(@)C`? The coefficient of linear expansion of the glass is `0.40xx10^(-5)K^(-1)`. Cubical expansion of mercury `=18xx10^(-5)K^(-1)`. |
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Answer» The coefficienct of volume expansion for the glass is `gamma_("glass")=3alpha_("glass")=1.2xx10^(-5)K^(-1)` The increase in volume of the glass flask is `DeltaV_("glass")=gamma_("glass")V_(0)DeltaT` `=(1.2xx10^(-5)K^(-1))(200cm^(3))(100^(@)C-20^(@)C)` `=0.19cm^(3)` The increase in volume of the mercury is `DeltaV_("mercury")=gamma_("mercury")V_(0)DeltaT` `=DeltaV_("mercury")=gamma_("mercury")V_(0)DeltaT` `=(18xx10^(-5)K^(-1))(200cm^(3))(100^(@)C-20^(@)C)=2.9cm^(3)` The volume of mercury that overflow is `DeltaV_("mercury")-DeltaV_("glass")=2.7cm^(3)` |
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| 122. |
A glass flask of volume `200cm^(3)` is just filled with mercury at `20^(@)C`. The amount of mercury that will overflow when the temperature of the system is raised to `100^(@)C` is `(gamma_(glass)=1.2xx10^(-5)//C^(@),gamma_(mercury)=1.8xx10^(-4)//C^(@))`A. 2.15 `cm^(3)`B. 2.69 `cm^(3)`C. 2.52 `cm^(3)`D. 2.25 `cm^(3)` |
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Answer» Correct Answer - B |
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| 123. |
A glass flask of volume `200cm^(3)` is completely filled with mercury at `20^(0)C`. The amount of mercury that overflow when the flask is heated to `80^(0)C` (Coefficient of volume expansion of glass is `27 xx 10^(-6 //0)C, gamma` of mercury `0.18 xx 10^(-3) //^(0)C`A. `2.16cm^(3)`B. `0.032cm^(3)`C. `1.84cm^(3)`D. `2.40cm^(3)` |
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Answer» Correct Answer - C `DeltaV = V(gamma_(1) - gamma_(g))Deltat` |
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| 124. |
A metal ball immersed in alcohol weights `W_1` at `0^@C` and `W_2` at `50^@C`. The coefficient of expansion of cubical the metal is less than that of the alcohol. Assuming that the density of the metal is large compared to that of alcohol, it can be shown thatA. `W_(1) = W_(2)`B. `W_(1) gt W_(2)`C. `W_(1) lt W_(2)`D. `W_(1) ge W_(2)` |
| Answer» Correct Answer - C | |
| 125. |
When the temperature of a rod increases from t to `r+Delta t`, its moment of inertia increases from I to `I+Delta I`. If `alpha` is the value of `Delta I//I` isA. `Delta t//t`B. `Delta t//t`C. `alpha Delta t`D. `2alpha Delta t` |
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Answer» Correct Answer - D from `I = MR^(2), (DeltaI)/(I) = 2 xx (DeltaR)/(R), (DeltaI)/(I) = 2alphaDeltat` |
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| 126. |
`PQR` is a right angled triangle made of brass rod bent as shown. If it is heated to a high temperature the angle `PQR`. A. increasesB. decreases the density of the materialC. remain sameD. become `135^(@)` |
| Answer» Correct Answer - C | |
| 127. |
When a copper solid shpere is heated, its (a) moment of inertia increases (b) Elasticity decreases © The weight of a body in a liquid increasesA. Only `(b)` is trueB. `(a)` & `(b)` are correctC. `(a), (b)` & `(c)` are trueD. all the true |
| Answer» Correct Answer - C | |
| 128. |
A wire of length `60 cm` is bent to into a circle with a gap of `1 cm`. At its ends, on heating it by `100^(0)C`, the length of the gap increases to `1.02 cm`. `alpha of materials of wire isA. `2 xx 10^(-4//@)C`B. `4 xx 10^(-4//@)C`C. `6 xx 10^(-4//@)C`D. `1 xx 10^(-4//@)C` |
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Answer» Correct Answer - A `alpha = (l_(2) - l_(1))/(l_(1)Deltat)` (gap can be taken as `l_(1)`) |
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| 129. |
Statement-1: If an annular disk is heated uniformly then outer radius increases but inner radius decreases. Statement-2: Temperature gradient of a rod is constant in steady state. Staement-3: Density of metal decreases with increase in temperature.A. TTTB. TFTC. FTTD. FFF |
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Answer» Correct Answer - C |
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| 130. |
The substance which contracts on heating isA. silica glassB. ironC. inva steelD. aluminum |
| Answer» Correct Answer - A | |
| 131. |
If two balls of same metal weighing 5 gm and 10 gm strike with a target with the same velocity. The heat energy so developed is used for raising their temperature alone, then the temperature will be higherA. For bigger ballB. For smaller ballC. Equal for both the ballsD. None is correct from the above three |
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Answer» Correct Answer - C Energy = `1/2 mv^(2) = mc Delta theta , rArr Delta theta prop v^(2)` Temperature does not depend upon the mass of the balls. |
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| 132. |
For meaurements of very high temperature say around `5000^(@)C`(of sun), one can use:A. Gas thermometerB. Platinum resistance thermometerC. Vapour pressure thermometerD. Pyrometer (Radiation thermometer) |
| Answer» Correct Answer - D | |
| 133. |
Melting and Boiling point of water on Fahrenheit scale of temperature respectivelyA. (a) `212^(0)F, 32^(0)F`B. (b) `32^(0)F, 212^(0)F`C. (c) `0^(0)F, 100^(0)F`D. (d) `32^(0)F, 132^(0)F` |
| Answer» Correct Answer - B | |
| 134. |
An aluminium sphere of `20cm` diameter is headed from `0^(@)C` to `100^(@)c`. Its volume changes by (given that the coefficient of linear expanison for aluminium `(alpha_(Al) = 23 xx 10^(-6//0)C)` |
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Answer» Given `d = 20cm` `V = V_(0) (1 + gammat) = V_(0)(1 + 3alphat)(since gamma = 3alpha)` Change in volume `= V - V_(0) = 3V_(0)alphat` `= 3 xx (4)/(3)pi((d)/(2))^(3) xx 23 xx 10^(-6) xx 100` `= 3 xx (4)/(3)pi ((0.2)/(2))^(3) xx 23 xx 10^(-6) xx 100` `= 28.9c c (1c c = 10^(-6)m^(3))` |
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| 135. |
A bar of iron is 10 cm at `20^(@)C`. At `19^(@)C` it will be `(alpha_(Fe)=11xx10^(-6)//.^(@)C)`A. `11xx10^(-6)`cm longerB. `11xx10^(-6)cm` cm shorterC. `11xx10^(-5)cm` shorterD. `11xx10^(-5)cm` longer |
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Answer» Correct Answer - C |
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| 136. |
When a strip made or iron `(alpha_(1))` and copper `(alpha_(2)),(alpha_(2) gt alpha_(1))` is heatedA. its length does not changeB. it gets twistedC. it bends with iron on concave sideD. it bends with iro on convex side |
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Answer» Correct Answer - C |
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| 137. |
If the boiling point of water is `95^(@)F`, what will be reduction at celsius scale?A. `7^(@)C`B. `65^(@)C`C. `63^(@)C`D. `35^(@)C` |
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Answer» Correct Answer - D |
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| 138. |
The portion of the curve representing the state of matter denotes A. Change from gaseous state to liquid stateB. The liquid state of matterC. Gaseous state of matterD. all of these |
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Answer» Correct Answer - B |
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| 139. |
The portion `AB` of the indicator diagram representing the state of matter denots A. The liquid state of matterB. Gaseous state of matterC. Change from liquid to gaseous stateD. Change from gaseous state to liquid state |
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Answer» Correct Answer - A The volume of matter in portion `AB` of the curve is almost constant and pressure is decreasing. These are the characteristics of liquid state. |
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| 140. |
Which of the following methods will enable the volume of an ideal gas to be made four timesA. double the absolute temperature and pressureB. halve the absolute temperature and double the pressure.C. quadruple the absolute temperature at constant pressureD. quarter the absolute temperature at constant pressure |
| Answer» Correct Answer - C | |
| 141. |
The dimensions of thermal resistance areA. `M^(-1)L^(-2)T^(3)K`B. `ML^(2)T^(-2)K^(-1)`C. `ML^(2)T^(-3)K`D. `ML^(2)T^(-2)K^(-2)` |
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Answer» Correct Answer - A |
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| 142. |
The temperature of water at the surface of a deep lake is `2^@C`. The temperature expected at the bottom isA. `2^(@)C`B. `3^(@)C`C. `4^(@)C`D. `1^(@)C` |
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Answer» Correct Answer - C |
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| 143. |
The mass, specific heat capacity and the temperature of a solid are `1000g,(1)/(2) (cal)/g ^(@)C` and `80^(@)C` respectively. The mass of the liquid and the calorimeter are `900g` and `200g`. Initially,both are at room temperature `20^(@)C` Both calorimeter and the solid are made of same material. In the steady state, temperature of mixture is `40^(@)C`, then specific heat capacity of the unknown liquid.A. `0.25 cal//gm^(@)C`B. `0.5 cal//gm^(@)C`C. `1 cal//gm^(@)C`D. `1.5 cal//gm^(@)C` |
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Answer» Correct Answer - C `S_(2) = ((1000)(1//2)(80^(@)-40^(@)))/(900(40^(@)-20^(@)))-((200)(1//2))/(900)` `=1 cal//gm^(@)C`. |
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| 144. |
Which of the following is the example of ideal black body?A. KajalB. Black boardC. A pin hole in a boxD. None of these |
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Answer» Correct Answer - C When light incident on pin hole enters into the box and suffers successive reflection at the inner wall. At each reflection some energy is absorbed. Hence the ray once it enters the box can never come out and pin hole acts like a perfect black body. |
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| 145. |
A rod of length 2 m is at a temperature of `20^@` C. find the free expansion of the rod, if the temperature is increased to `50^@`C, then find stress produced when the rod is (i) fully prevented to expand, (ii) permitted to expand by 0.4mm. `Y=2xx10^(11)N//m^(2), alpha=15xx10^(-6//^(@))C`.A. `9 xx 10^(7) N // m^(2)`B. `4.5 xx 10^(7) N // m^(2)`C. `5 xx 10^(7) N // m^(2)`D. `3 xx 10^(7) N // m^(2)` |
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Answer» Correct Answer - A `:. "Thermal stress = strain" xx Y = (Deltal)/(l) xx Y = (alphaDeltat)Y` |
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| 146. |
An equilateral triangle `ABC` is formed by joining three rods of equal length and `D` is the mid-point of `AB`. The coefficient of linear expansion for `AB` is `alpha_(1)` and for `AC` and `BC` is `alpha_(2)`. The relation between `alpha_(1)` and `alpha_(2)`, if distance `DC` remains constant for small changes in temperture is A. `alpha_(1) = alpha_(2)`B. `alpha_(1) = 4alpha_(2)`C. `alpha_(2) = 4alpha_(1)`D. `alpha_(1) = (1)/(2)alpha_(2)` |
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Answer» Correct Answer - B Before and after changing the tempreature, `sqrt(l_(2)^(2) - (l_(1)^(2))/(4)) = sqrt([l_(2)(1 + alphat)]^(2) - (1)/(4)[l_(1)(1 + alphat)]^(2))` and `l_(1) = l_(2)` |
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| 147. |
A cube of edge `(L)` and coefficient of linear expansion `(alpha)` is heated by `1^(0)C`. Its surface area increases byA. `6alphaL^(2)`B. `8alphaL^(2)`C. `12alphaL^(2)`D. `2alphaL^(2)` |
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Answer» Correct Answer - C `DeltaA = AbetaDeltat,` |
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| 148. |
Two identical containers connected by a fine caillary tube contain air at `N.T.P` if one of those containers is immersed in pure water, boiling under noramal pressure then new pressure isA. `76 cm` of `Hg`B. `152 cm` of `Hg`C. `57 cm` of `Hg`D. `87.76 cm` of `Hg` |
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Answer» Correct Answer - D `(P_(1)V_(1))/(T_(1)) = `(P_(1)V_(1))/(T_(1)) = P_(1)((V_(1))/(T_(1)) + (V_(1))/(T_(1)))` |
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| 149. |
Good absorbers of heat areA. Poor emittersB. Non-emittersC. Good emittersD. Highly polished |
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Answer» Correct Answer - C |
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| 150. |
At `50^(@)C` , a brass rod has a length 50 cm and a diameter 2 mm . It is joined to a steel rod of the same length and diameter at the same temperature . The change in the length of the composite rod when it is heated to `250^(@)C` is (Coefficient of linear expansion of brass = `2.0 xx 10^(-5)"^(@) C^(-1)` , coefficient of linear expansion of steel = `1.2 xx 10^(-5) "^(@) C^(-1)`)A. `0.28` cmB. `0.30` cmC. `0.32` cmD. `0.34` cm |
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Answer» Correct Answer - C Change in length of the brass rod is `DeltaL_(b) = alpha_(b) L_(b) DeltaT` `= 2.0 xx 10^(-5).^(@)C^(-1) xx 50 cm xx (250 .^(@)C - 50 .^(@)C)` `= 0.2 cm` Change in length of the steel rod is `DeltaL_(S) = alpha_(S)L_(S)DeltaT` `= 1.2 xx 10^(-5).^(@)C^(-1) xx 50 cm xx (250 .^(@)C - 50^(@)C)` `= 0.12 cm` `:.` Change in length of the combined rod `= DeltaL_(b) xx DeltaL_(S) = 0.2 cm + 0.12 cm = 0.32 cm` |
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