InterviewSolution
Saved Bookmarks
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.
| 51. |
Define thermal capacity and water equivalent of a body. State their units and how are they related ? |
| Answer» Thermal capacity of a body is defind as the amount of heat required to raise the temperature of the whole body through `1^@C` or `1K.` Quantitatively, thermal capacity `=("heat given "(Delta Q))/("rise in temp "(Delta theta)) = (ms Delta theta)/(Delta theta) = mxxs` Where m is the mass and s is the specific heat. Unit of thermal capacity is cal `K^(-1) or JK^(-1).` Water equivalent of a body is defined as the mass of water in gram which would absord or evolve the same amount of heat as is done by the body in rising or falling through the same range of temperature. If `Delta Q` amount of heat is given to raise the temperature of w amount of water through temperature `Delta theta` then `Delta Q= w Delta theta or w = Delta q// Delta theta` The unit of water equivalent is gram . Quantitatively, thermal capacity of a body is equal to its water equivalent. | |
| 52. |
The specific heat capacity of a metal at low temperature (T) is given as `C_(p)(kJK^(-1) kg^(-1)) =32((T)/(400))^(3)` A 100 gram vessel of this metal is to be cooled from `20^(@)K` to `4^(@)K` by a special refrigerator operating at room temperaturte `(27^(@)C)` . The amount of work required to cool the vessel isA. equal to `0.002` kJB. greater than `0.148` kJC. between `0.148` kJ and `0.028` kJD. less than `0.028` kJ |
|
Answer» Correct Answer - A Here, `C_(p)(kJ K^(-1) kg^(-1)) =32 ((T)/(400))^(3) =32xx(T^(3))/((400)^(3))` , `m=100 gram =0.1 kg` , Workdone to cool the vessel = amount of heat taken out from metal while cooling it from 20 K to 4 K. `:. W = int_(20)^(4) mC_(p) dT = int_(20)^(4) 0.1xx(32(T^(3))/(400^(3)))dT` `=0.002 kJ` |
|
| 53. |
What is the relation between heat capacity and water equivalent of a substance? |
| Answer» The heat capacity and water equivalent of a substance are numerically equal but they have difference units of measurement. | |
| 54. |
In soldering, addition of flux makes soldering easily. Why ? |
| Answer» The addition of foux reduces the surface tension of molten tin, hence it spreads. | |
| 55. |
`600` kg. of water is to be pumped in a tank per minute under the perssure of `10` g. wt//sq. mm. Find the horse power needed. `1 H.P.=746` watts. |
|
Answer» Mass of water to be pumped, `m=600 kg`. Volume of water to be pumped, `V=(mass)/(density)=(600)/(10^(3))=0.06 m^(3)`, Pressure `=10` gwt//sq mm. `(10)/(1000)xx9.8 N//(10^(3))^(2) sq m` `=10xx9.8xx10^(3) N//m^(2)` `Power=("workdone")/("time")=("pressure"xx"volume")/("time")` `=((10xx9.8xx10^(3))xx0.6)/(60)=980 watt` `(980)/(746)hp=1.31 hp` |
|
| 56. |
The triple point of neon and carbon dioxide are `24.57K` and `216.55K` respectively. Express these temperature on the Celsius and Fahreaheit scales. |
|
Answer» Relation between Kelvin scale and Celsius scales is `T_(C)=T_(K) = 273.15` Where `T_(C), T_(K)` = temperature on Celsius and Kelvin scales respectively. For Neon `T_(C) = 24.57-273.15 = -248.58^(@)C` For `C_(O_2)` `T_(C) = 216.55-273.15 = -56.60^(@)C` Relation between Kelvin and Farenheight scalse is, `(T_(F)-32)/(180) = (T_(K)-273.15)/(100)` `T_(F) = (180)/(100) (T_(K)-273.15)+32` For Neon, `T_(F)=(180)/(100)(42.57-273.15)+32 = -415.44^(@)F` For `C_(O_2)` `T_(F) = (180)/(100)(216.55-273.15)+32 = -69.88^(@)F`. |
|
| 57. |
A body weighs `W_(1)` in a liquid of density `rho_(1)` and `W_(2)` in a liquid of density `rho_(2)`. What is the weight of body in a liquid of density `rho_(3)`? |
|
Answer» let m be the mass of the body, V be its volume and `rho` be its density. So `V=m//rho`. Weight of body in air, `W=mg`. Effective weight of the body when completely immersed in a liquid of density `rho_(1)` is `W_(1) = mg - V rho_(1) g =mg - (m)/(rho) rho_(1)g` `= mg(1-(rho_1)/(rho))=W(1-(rho_1)/(rho))` ...(i) similarly, `W_(2) = W(1-(rho_2)/(rho))` ..(ii) and `W_(3) = W(1-(rho_3)/(rho))` ..(iii) Using (i),(ii) and (iii) , we get `W_(3) = W_(2) (rho_(3)-rho_(1))-W_(1)(rho_(3)-rho_(2))/(rho_(2)-rho_(1))`. |
|
| 58. |
An ornament weighing `50g` in air weighs only `46 g` is water. Assuming that some copper is mixed with gold the prepare the ornament. Find the amount of copper in it. Specific gravity of gold is `20` and that of copper is 10.A. 25 gB. 30 gC. 35 gD. 22 g |
|
Answer» Correct Answer - B Let m be the mass of copper in the ornament. The mass of gold in ornament is (50 -m) Volume of copper, `V_(1)=(mass)/(density)=(m)/(10)` Volume of gold, `V_(2)=(50-m)/(20)` When ornament is immersed in water of density `rho_(w)(=1 g//c.c.)`, then decrease in weight = upward thrust of water `(50-46)g =(V_(1)+V_(2))rho_(w)g` `4=((m)/(10)+(50-m)/(20))1` On solving, `m=30g` |
|
| 59. |
A metallic block weighs `15 N` in air. It weights `12 N` when immersed in water and `13 N` when immersed in another liquid. What is the specific gravity of the liquid? |
|
Answer» (i) Reletaive density of the block `=("weight of the block in air")/("loss in weight of block when immeresed in water") = (15)/(15-12) = 5` (ii) Relative density of the liquid, ` = ("loss of weight n liquid")/("Loss of weight of water") = (15-13)/(15-12)=2/3` |
|
| 60. |
If a drop of liquid breaks into smaller droplets, it result in lowering of temperature of the droplets. Let a drop of radius R, breaks into N small droplets each of radius r. Estimate the drop in temperature. |
|
Answer» Volume of a big drop of radius R = N xx volume of small drop let of radius r `:. (4)/(3) pi R^3 = N xx(4)/(3)pi r^3 or R^3 = N r^3 or N = R^3//r^3` Change in surface area `Delta U = ST xx` change in surface area `= Txx 4pip[R^2 - N r^2]` All this energy relased is at the cost of loweiring the temperature. Mass of the big drop of liquid, `m =(4)/(3) pi R^3 rho` Let s be the specific heat of liquid drop, `Delta theta` the decrease in temperature. Then `Delta theta = (Delta U)/(ms) = (Txx4pi(R^2 - Nr^2))/((4)/(3) pi R^3 rho)s = (3T)/(rho s) [(1)/(R) - (N r^2)/(R^3)] = (3T)/(rho s)[(1)/(R) - (R^3)/(r^3) xx(r^2)/(R^3)]` `= (3T)/(rho)[(1)/(R) - (1)/(r)]` |
|
| 61. |
A liquid drop of diameter 4 mm breaks into 1000 droplets of equal size. Calculate the resultant change in surface energy, the surface tension of the liquid is `0.07 Nm^(-1)` |
|
Answer» Correct Answer - `3.168 xx 10^(-5) J` Here, `R = 2xx10^(-3)m` , `n = 1000`. Let r be the radius of each small droplet formed. As, volume of big drop `= 1000 xx` volume of small droplet `(4)/(3) pi R^3 = 1000 xx(4)/(3) pi r^3` or `r= (R )/(10) = (2xx10^(-3))/(10) = 2xx10^(-4)m` Increase in surface area, `Delta A = 1000 4 pi r^2 - 4pi R^2 = 4pi (1000 r - R^2)` ` =4xx(22)/(7) [1000 (2xx10^(-4))^2 - (2xx 10^(-3))^2]` `= 4xx(22)/(7) [4 xx 10^(-5) - 0.4xx10^(-5)]` `=4xx(22)/(7) xx 3.6 xx10^(-5) m^2` Resultant increase in surface energy `= S.Txx Delta A = S xx Delta A` ` = 0.07 xx 4xx(22)/(7) xx3.6 xx 10^(-5)` `=3.168xx 10^(-5) J` |
|
| 62. |
In making an alloy, a substance of specific gravity `s_(1)` and mass `m_(1)` is mixed with another substance of specific gravity of the alloy isA. `[(s_(1)s_(2))/(m_(1)+m_(2))]`B. `[(s_(1)+s_(2))/(m_(1)+m_(2))]`C. `[(m_(1)l s_(1)+m_(2)l s_(2))/((m_(1)+m_(2)))]`D. `[(m_(1)+m_(2))/((m_(1)l s_(1)+m_(2)l s_(2)))]` |
|
Answer» Correct Answer - D Specific gravity of alloy `=("density of alloy")/("density of water") =("mass of alloy"//"volume of the alloy")/("density of water")` `s_(alloy)=((m_(1)+m_(2))//((m_(1))/(rho_(1))+(m_(2))/(rho_(2))))/(rho_(w))` `=((m_(1)+m_(2)))/((m_(1))/(rho_(1))+m_(2)/(rho_(2))rho_(w))=(m_(1)+m_(2))/(m_(1)/((rho_(1)//rho_(w)))+(m_(2))/((rho_(2)//rho_(w))))` `(m_(1)+m_(2))/((m_(1))/(s_(1))+m_(2)/(s_(2)))` |
|
| 63. |
When equal volumes of two metals are mixed together the specific gravity of alloy is 4. When equal masses of the same two metals are mixed together the specific gravity of the alloy becomes 3. find specific gravity of each metal? (specific gravity `=("density of substance")/("density of water"))` |
|
Answer» Correct Answer - 2 and 6 Let `m_(1),m_(2)` = masses of two metals, `V_(1), V_(2)` = volumes of two metals. If `rho` is the density of mixture of two metals, then `rho =(m_(1) + m_(2))//(V_(1) + V_(2))` When equal volume of two metals of density `rho_(1)` and `rho_(2)` are mixed, then `V_(1) = V_(2) = V` and `m_(1) =rho_(1) V` and `m_(2) = rho_(2) V` . If `rho` is the density of mixture of two metals, then `rho=((rho_(1)V+rho_(2)V))/(V+V) = (rho_(1) + rho_(2))/(2)` or `rho_(1) + rho_(2) =2rho =2xx4=8` ....(i) When equal masses are mixed, then `m_(1) =m_(2)=m` and `V_(1) =m//rho_(1)` and `V_(2) =m//rho_(2)` So, `rho=(m+m)/((m//rho_(1))+(m//rho_(2))) = (2rho_(1)rho_(2))/(rho_(1)+rho_(2)) =3` or `rho_(1)rho_(2) =(3)/(2) (rho_(1) + rho_(2)) =(3)/(2)xx8 =12` ...(ii) Solving (i) and (ii), we get, `rho_(1) =2` and `rho_(2) =6` |
|
| 64. |
A drop of liquid of radius `R=10^(2)`m having surface tension `S=(0.1)/(4pi)N//m^(-1)` divides itself into K identical drop. In this process the total change in the surface energy `Delta U =10^(-3)` J. If `K=10^(alpha)` , then the value of `alpha` is : |
|
Answer» Correct Answer - 6 Let r be the radius of each small drop. Then volume of big drop = K x volume of small drop. or `(4)/(3)pi R^(3) =Kxx(4)/(3)pi r^(3)` or `r=R K^(-1//3)` Change in surface energy `Delta U =S.T.` x change in surface area `=Sxx[K 4pi r^(2)-4 pi R^(2)]` `=Sxx[K 4 pi R^(2) K^(-2//3) - 4pi R^(2)]` `=4 pi R^(2) S[K^(1//3) - 1]` `:. 10^(-3) =4 pi (10^(-2))^(2)xx(0.1)/(4pi)[(10^(alpha))^(1//3) - 1]` or `100=10^(alpha//3) - 1` or `10^(alpha//3) =100+1 ~~10^(2)` or `(alpha)/(3)= 2` or `alpha =3xx2=6` |
|
| 65. |
How is the rise of liquid affected if top of the capillary is closed? |
| Answer» there will be a small rise in the capollary tube if top of capillary tube is closed , because the rise of liquid in capillarytube due to srface tension will be opposed by the downward force exerted by the comperssed air above the liquid in tube. This downward force increase in height of liquid column. Therefore a small rise of liquid column is possible in a capillary tube with closed top. | |
| 66. |
What is the importance of wetting agents used by dyers? |
| Answer» The wetting agents are used by dyers to decreases the angle of contact between the fabric and the dye so that the dye may well penetrate the fabric. | |
| 67. |
Why the tip of the nib of a pen is split? |
| Answer» The tip of the nib of a pen is split in order to provide a capillary which helps the ink to rise to the end of the nib and enables it to write continuously. | |
| 68. |
Statement I: A needle placed carefully on the surface of water may float, whereas the ball of the same material will always sink.Statement II: The buoyancy of an object depends both on the material and shape of the object.A. both Assertion and Reason are true and the Reason in the correct explanation of the Assertion.B. both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both Assertion and Reason are false. |
|
Answer» Correct Answer - C A needle placed carefully on the surface of water may float due to surface tension, as upward force due to surface tension balances the weight of the needle. But thes upward forces due to surface tension are very small as compared to weight of ball, also the weight of liquid displaced by the ball immersed in liquid is less than the weight of the ball, hence ball sinks into the liquid. Here, Asssertion is true but Reason is false because bouyancy of an object depends on the weight of the liquid displaced by the immersed part of the object in liquid. |
|
| 69. |
Water is depressed in a glass tube whose inner surface is cotated with paraffin wax. Why? |
|
Answer» Water is depressed in a glass tube whose inner surface is cotated with paraffin wax because of the fact that force of cohesion between molecules of water is larger than the force of adhesion between the moleculesa of water and wax. Therefore the water does not wet paraffin wax. The height upto which a liquid riese in a capillary tube is given by `h= ( 2S cos theta)/(r rho g)` Since water does not wet paraffin wax. `theta` is obutuse (i.e., `theta gt 90^(@))` , `cos theta` is negative. Hence h is -ve . therefore water is depressed in such a tube. |
|
| 70. |
Why does a gas not have a unique value of specific heat ? |
| Answer» This is because a gas can be heated under different conditions of pressure and volume. The amount of heat required to raise the temperature of unit mass through unit degree is different under different conditions of heating. | |
| 71. |
Find the greatest length of copper wire, that can hang without breaking. Breaking stress `=7.2 xx 10^(7) N//m^(2)`. Density of copper `7.2 g//c c. g=10 m//s^(2)`. |
|
Answer» Given, Breaking stress, `S=7.2 xx 10^(7) N//m^(2)`, Density of wire, `rho = 7.2 g//c c = 7.2 xx 10^(3) kg//m^(3)` Let l be the greatest length of wire that can hang without breaking and A be the area of cross section of the wire. Weight of wire = `(A l) rho g`. Breaking stress, `S=("weigth of wire")/("area of cross section")=(A l rho g)/(A)` or `l=(S)/(rho g) = (7.2 xx 10^(7))/((7.2 xx 10^(3)) xx 10) =1000m`. |
|
| 72. |
A gaeyser heats water flowing at the rate of 3 kg per minute from `27^(@)C` to `77^(@)C` . If the geyser operates on a gas burner, what is the rate of consumption of fuel if the heat of combusion is `4 xx 10^(4) J//g?` Given specific heat of water is `4.2 xx 10^(3) J//kg//K`. |
|
Answer» Here, `m= 3 kg//mi n`, `Delta T = 77 - 27 = 50^(@)C = 50 K` heat sphent to raise the temperature `Delta T` of water is `Q = mc Delta T` ` = 3 kg//mi n xx (4.2 xx 10^(3)) Jkg^(-1) K^(-1) xx 50 K` `= 63 xx 10^(4) J//mi n` Rate of consumption of fuel `=(Q)/("heat of compustionof fuel")` `=(63 xx 10^(4) J//mi n)/(4 xx 10^(4) Jg^(-1))` `= 15.75 g mi n^(-1)`. |
|
| 73. |
A blacksmith fixes iron ring on the rim of the wooden wheel of a bulkock cart. The diameter of the rim and the ring are `5.243 m` and `5.231 m` respectively at `27^(@)C`. To what temperature should the ring be heated so as to fit the rim of the wheel ? Coefficient of linear expansion of iron =` 1.20 xx 10^(-5) K^(-1)` |
|
Answer» Here, `L_(T_1) = 5.231 m`, `L_(T_2) = 5.243 m, T_(1) = 27^(@)C , T_(2) = ?` As `alpha = (L_(T_2)-L(T_1))/(L_(T_1) xx alpha) + T_(1)` `=(5.243 - 5.231)/(5.231 xx 1.2 xx 10^(5)) + 27` `= 191.1 + 27 = 218. 1 ~~ 218^(-1) C` |
|
| 74. |
The length of a metal wire is l when the tension is F and x l when the tension si y F. Then the natural length of the wire isA. `((x -y)l)/(x - 1)`B. `((y - x)l)/((y -1))`C. `((x - y)l)/((x +1))`D. `((y - x)l)/(y +1)` |
| Answer» Correct Answer - B | |
| 75. |
Two springs have their force constant `K_1 and K_2.` Both are strectched tilll their elastic energies are equal. If stretching forces are `F_1 and F_2,` then `F_1 : F_2` siA. `K_1 : K_2`B. `K_2 : K_1`C. `sqrtK_1// sqrtK_2`D. `K_1^2 // K_2^2` |
| Answer» Correct Answer - C | |
| 76. |
Two identical rods are connected between two containers. One of them is at `100^(@)C` containing water and another is at `0^(@)C` containing ice. If rods are connected in parallel then the rate of melting of ice is `q_(1)g//s` . If they are connected in series then teh rate is `q_(2)g//s` . The ratio `q_(2)//q_(1)` isA. `1//8`B. `1//4`C. `1//2`D. `4` |
|
Answer» Correct Answer - B The resultant conductivity of combination of two rods , in parallel is, `K_(p) =K_(1) + K_(2) =K + K=2 K` in secries is, `K_(s) =(K_(1)K_(2))/(K_(1) + K_(2)) = (KxxK)/(K+K) = (K)/(2)` `q_(1) =((2K) A(100-0))/(x) = (2KAxx100)/(x)` `q_(2) =((K)/(2)) (A(100-0))/(x) = (KAxx100)/(2x)` `:. (q_(2))/(q_(1)) = (KAxx100)/(2x)xx(x)/(2KAxx100) =(1)/(4)` |
|
| 77. |
Put a piece of chalk into water. The chalk will emit bubbles in all directions . Explain this phenomenon. |
| Answer» A chalk has pores in all directions which act as narrow capollaries. When a piece of chalk is immersed in water, the water enters into these capillaries and force the air out in the form of bubbles in water. | |
| 78. |
The phenomenon of refreezing of ice on reducing the pressure from ice is called………… |
| Answer» Correct Answer - relagation | |
| 79. |
What is 1 bar? |
|
Answer» 1 bar is the unit of atmospheric pressure used for metreological purposes where ` 1" bar" = 10^(5) Pa`. |
|
| 80. |
The plots of intensity (I) of radiation versus wavelength `(gamma)` of thee black bodies at temperatures `T_(1), T_(2)` and `T_(3)` are shown in Fig.7(CF).25. Then, A. `T_(3)gtT_(2)gtT_(1)`B. `T_(1)gtT_(2)gtT_(3)`C. `T_(2)gtT_(3)gtT_(1)`D. `T_(1)gtT_(3)gtT_(2)` |
|
Answer» Correct Answer - D For black body radiation, `lambda_(m)T=a` constant =b or `lambda_(m) prop (1)/(T)`. The wavelength `lambda_(m)` is less for curve `T_(1)`, more for curve `T_(3)` and maximum for curve `T_(2)`. Hence, `T_(1)gtT_(3)gtT_(2)` |
|
| 81. |
Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of `57^(@)C` is drunk. You can take body (tooth) temperature to be `37^(@)C` and `alpha_(Cu) = 1.7 xx 10^(-5)//^(@)C` bulk modulus for copper `B_(Cu) = 140 xx 10^(9) N//m^(2)`. |
|
Answer» Volumetric strian in tooth cavity ` = Delta V//V` Let `gamma` be the coefficient of volume expansion with the change in temperature `Delta theta.^@C,` the cange in volume is `Delta V = gamma V Delta theta` or `(Delta V)/(V) = gamma Delta theta` Thermal stress in tooth cavity ` = B xx` volumetric strain `= B xx gamma Delta theta = B xx 3 alpha Delta theta (:. gamma = 3 alpha)` `= (140 xx 10^9) xx (3xx1.7 xx 10^(-5)) xx 20 = 1.428 xx 10^8 N//m^2` |
|
| 82. |
When bits of comphor are dropped on water, they move helter-skelter. Why? |
| Answer» When bits of camphor are dropped on water, they lower the surface tension where they are dipping. Since the shape of camphor bits is irregular, unequal forces of surface tension act on them. That is why they move helter-skelter or erratically on the surface of water. | |
| 83. |
Animals curl into a ball, when they feel very cold. |
| Answer» When the animals feel cold, they curl their bodies into a ball so as to decrease the surface are of their bodies. As total energy radiation by a body varies directly as the surface are of the body, the loss of heat due to radiation would be reduced. | |
| 84. |
What is the temperature of steel-copper jumction in the steady state of the system shown in fig. Length of the steel rod = 30.0 cm, length of the copper rod = 20.0 cm, temperature of the furnace = `300^(@)C`, temperature of cold end `=0^(@)C` . The area of cross-section of the steel rod is twice that of the copper rod. thermal conductivity of steel = `50 .2 Js^(-1) m^(-1) .^(@)C^(-1)` and of copper = `358 Js^(-1) m^(-1) .^(@)C^(-1)`. |
|
Answer» In the steady state, let T be the temperature of steel-copper junction. Then heat flowing per second of steel = heat flowing per second of copper. `(K_(1)A_(1)(300-T))/(x_1) = (K_(2)A_(2)(T-0))/(x_2) or (50.2 xx 2A_(2) xx (300-T))/(30) = (385 xxA_(2)(T-0))/(20)` or, `100.4 (300-T) xx 20 = 385 T xx 30` On solving , `T = 44.4^(@)C` |
|
| 85. |
What should be the length of steel and copper rods at `0^(@)C` that the length of steel rod is 5 cm longer than copper at all termperature? Given `alpha_(Cu) = 1.7 xx 10^(5) .^(@)C^(-1)` and `alpha_(steel) = 1.1 xx 10^(5) .^(@)C^(-1)`. |
|
Answer» Let l be the length of copper rod at `0^(@)C`. The length steel rod at `0^(@)C = l + 5` If `Delta T ` is the rise in temperature , then as per question. increaes in length of copper rod = increase in length of steel rod `:. L xx 1.7 xx 10^(-5) xx Delta T = (l+5) xx 1.1 xx 10^(-5) xx Delta T` or `1.7 l = 1.1 l + 5.3` OR `0.6 L = 5.5` or ` l = 5.5//0.6 = 9.17 cm` Thus length of copper rod, `l = 9.17 cm` Length of steel rod = `l+5 = 9.17 + 5` ` = 14.17 cm`. |
|
| 86. |
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)`). |
|
Answer» Correct Answer - `[44.4^(@)C]` Let `T .^(@)C` be the temperature of the steel-copper junction at steady state. So rate of heat conducted through steel = rate of heat conducted through copper `:. (K_(s)A_(s)(T_(1)-T_(2)))/(x_(1))=(K_(c)A_(c)(T_(1)-T_(2)))/(x_(2))` or `((50.2)xx(2A)xx(300-T))/((15.0xx10^(-2)))` `=((385)xxAxx(T-0))/((10.0xx10^(-2)))` On solving , we get `T=44.4^(@)C` |
|
| 87. |
The outer face of a rectangular slab of equal thickness of iron and brass are maintained at `100^(@)C` and `0^(@)C`, respectively. Find the temperature of the interface. The conductivities of iron and brass are 14 and `126 Wm^(-1) K^(-1)` respectively. |
|
Answer» Correct Answer - `[10^(@)C]` If T is temperature of the interface of iron and brass slabs, then `(dQ)/(dt)=K_(1)(A(100-T))/(Delta x) =(K_(B)A(T-0))/(Delta x)` or `K_(1)(100-T)=K_(B)T` or `(100-T)/(T)=(K_(B))/(K_(1))=(126)/(14)=9` or `T=10^(@)C` |
|
| 88. |
A piece of iron of mass `100g` is kept inside a furnace for a long time and then put in a calorimeter of water equivalent `10g` containing `240g` of water at `20^(@)C`. The mixture attains an equilibrium temperature of `60^(@)C`. Find the temperature of the furnace. Specific heat capacity of iron `= 470J kg^(-1).^(@)C^(-1)` |
|
Answer» Correct Answer - `[953.6^(@)C]` Here,`m_(1)=100 g, w=10 g, m_(2)=240 g` `T_(1)=20^(@)C`, Equilibrium temp. `T=60^(@)C` Let, `T_(2)` be the team. Of furnce. Heat gained by water and cal. = Heat lost by iron `(m_(2)+w)(T-T_(1))=m_(1) s(T_(2)-T)` `(240+10)(60-20)=100xx(470(T_(2)-60))/(4.2xx10^(3))` `T_(2)-60=(250xx40xx4.2xx10^(3))/(100xx470)=893.6` `T_(2)=893.6+60=953.6^(@)C` |
|
| 89. |
If two liquids of same mass but densities `rho_1` and `rho_2` respectively are mixed, then the density of the mixture is: |
| Answer» `(2 rho_1 rho_2)/((rho_1 + rho_2))` | |
| 90. |
A solid uniform ball having volume V and density `rho` floats at the interface of two unmixible liquids as shown in Fig. 7(CF).7. The densities of the upper and lower liquids are `rho_(1)` and `rho_(2)` respectively, such that `rho_(1) lt rho lt rho_(2)`. What fractio9n of the volume of the ball will be in the lower liquid. A. `(rho-rho_(2))/(rho_(1)-rho_(2))`B. `(rho_(1))/(rho_(1)-rho_(2))`C. `(rho_(1)-rho)/(rho_(1)-rho_(2))`D. `(rho_(1)-rho_(2))/(rho_(2))` |
|
Answer» Correct Answer - C Let `V_(1)` and `V_(2)` be the volumes of the ball in the upper and lower liquids respectively. So `V_(1)+V_(2)=V` A ball = upthrust on the ball due to two liquids `V rho g=V_(1)rho_(1)g+V_(2)rho_(2)g` or `V rho= V_(1)rho_(1)+V_(2)rho_(2)` or `V rho= V_(1)rho_(1)+(V-V_(1))rho_(2)` or `V_(1)=((rho-rho_(2))/(rho_(1)-rho_(2)))V` `:.` Fraction in the upper liquid `(V_(1))/(V)=(rho-rho_(2))/(rho_(1)-rho_(2))` Fraction in the lower liquid `=1-(V_(1))/(V)` `=1-(rho-rho_(2))/(rho_(1)-rho_(2))=(rho_(1)-rho)/(rho_(1)-rho_(2))` |
|
| 91. |
A square lead slab of side 50 cm and thickness 10 cm is subjected to a shearing force (on its narrow face) of `9xx10^4N`. The lower edge is riveted to the floor. How much will the upper edge be displaced? (Shear modulus of lead`=5.6xx10^9Nm^-2`) |
|
Answer» Here, `L-50 cm = 50 xx 10^(-2)m` , `G=5.6 xx 10^(9)Pa, F=9.0 xx 10^(4)N`. area of the face on which force is applied , `A=50 xx 10 =500 sq cm =0.05m^(2)` If `Delta L` is the displacement of the upper edge of the slab due to tangential force F applied, then `G=(F//A)/(Delta L//L)` or `Delta L=(FL)/(GA) = (9 xx 10^(4)xx50 xx 10^(-2))/(5.6 xx 10^(9) xx 0.05)` `=1.6 xx 10^(4)m`. |
|
| 92. |
Black body radiation is white and a hole in the cavity of a radiator is a black body. Explain. |
|
Answer» A black body absorbs radiations of all wavelenghts falling on it. When it is heated to a suitable temperature it emits radiations of all wavelengths . The combination of radiations of all visible wavelength makes white light. Hence the black body radiation is white. A hole in the cavity of a radiator does not reflect any radiation incident on it but absorbs them all. so it works as a black body. |
|
| 93. |
Two wires of equal cross section but one made of steel and the other of copper, are joined end to end. When the combination is kept under tension, the elongations in the two wires are found to be equal. Find the ratio of the lengths of the two wires. Young modulus of steel `=2.0xx10^11Nm^-2` and that of copper `=1.1xx10^1Nm^-2` |
|
Answer» Correct Answer - 20//11 `Y_s = (F)/(A) xx (l_s)/(Delta l_s) , Y_c = (F)/(A) xx (l_c)/(Delta l_c),` `So, (Y_s)/(Y_c) = (l_s)/(l_c) xx (Delta l_c)/(Delta l_s) = (l_s)/(l_c) ( :. Delta l_c = Delta l_s)` `:. (l_s)/(l_c) = (Y_s)/(Y_c) = (2.0xx 10^(11))/(1.1xx 10^(11)) = (20)/(11)` |
|
| 94. |
Mercury does not cling to glass. Why? |
| Answer» It is because the force of adhesion between mercury and glass molecules is less than the force of cohesion between mercury molecules. | |
| 95. |
what is the effect of impurities on the surface tension of liquid. |
| Answer» The presence of impurities either on the liquid surface or dissolved in it, considerably affect the force of surface tension, depeneding upon the degree of contamination. A highly soluble substance like sodium chloride when dissolved in water, increases the surface tension of water. But the sparingly soluble substance like phenol when dissolved in water, reduces the surface tension of water. | |
| 96. |
what is the effect of solute on the surface tension of liquid? |
| Answer» In case the solute is very easily soluble, then the surface tension of liquid increases. For example, when salt is dissolved in water, the surface tension increases. If the solute is less soluble, then the surface tension of liquid decreases. For example, by adding soap or phenol in water, its surface tension decreases. | |
| 97. |
When wax is rubbed on cloth becomes water proof, why? |
| Answer» The capillaries formed in threads disappear when wax is rubbed on cloth. | |
| 98. |
A room heater is made of 10 polished thin walled tubes of copper, each one meter long are 5cm in diameter. If hot water at `70^(@)C` circulates constantly through the tubes, calculate the amount of heat radiated in an hour in a room where the average temperature is `15^(@)C` . Emissitivity of copper = `4 xx 10^(-2)cal//degree//sec//sq. metre`. |
|
Answer» Here, no. of tubes , n=10 Length of each tube, `l=1 metre` Radius of each tube, `r = 5/2 cm = 5/200 m = 1/40 m` Surface area of each tube = `2 pi r l` and total area of all the tubes `A = 2 pi r l n` `= 2 xx 22/7 xx 1/40 xx 1 xx 10m^(2) = 11/7 m^(2)` Temperature of hot water `= 70^(@)C` Average room temperature = `15^(@)C` `:.` Difference of temperature, `theta = 70-15 = 55^(@)C` Emissivity, `epsilon = 4 xx 10^(-2) cals//sec//degree//sq.metre` Time, `t =1 hour = 3600 secs` Assuming that heat energy `radiated//sec//area` is directly proportional to temp. diff. `theta` the quantity of heat (Q) radiated = `A.theta . t epsilon` `:. Q = 11/7 xx 55 xx 3600 xx 4 xx 10^(-2)` `=1.245 xx 10^(4)` calories |
|
| 99. |
Why felt is used for thermal insulation in preference to air? |
|
Answer» Though air is a bad conductor of heat, it transfers more heat by convection. In felt, actual movement of air is stopped by trapping air inbetween its fine fibers .Therefore, convection is cheacked. Due to it, the felt becomes a better thermal insulator than air. |
|
| 100. |
A block of wood is floating on water at `0^(@)C` with a certain volume `V` above water level. The temperature of water is slowly raised to `20^(@)C` . How does the volume `V` change with the rise of temperature ? |
| Answer» If `rho` is the density of water in which the block is floating with volume `upsilon` in water, then weight of block, `W= upsilon rho g`. We know that with the rise in temperature water contracts from `0^(@)C` to `4^(@)C` and then expands form `4^(@)C` to onwards (due to anomolous expansion of water). Thus the density of water will increase from `0^(@)C` to `4^(@)C` and then will decrease. Therefore, to provide proper upthrust (=weight of wood) for floating the volume `upsilon` will decrease from `0^(@)C` to `4^(@)C` and then will increase from `0^(@)C` to `4^(@)C` and then will decrease with further increase in temperature. | |