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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.
| 151. |
A stirred liquid comes to rest after sometime. Why? |
| Answer» When stirred liquid is left, the diffenent layers of liquid will be moving with different layer of liqud will be moving with different velocities. Then an internal viscous force comes into play which will destroy the relative motion amongst the different layers. Dut to it, the stirred liquid coes to rest after some time. | |
| 152. |
Machine parts are jammed in winter. Why? |
| Answer» A lubricating oil is generally used between the various parts of a machine to reduce the friction. In winter, since the temperature is low, the viscosity of oil between the machine parts increases considerably, resulting in jamming of the machine parts. | |
| 153. |
A pendulum clock loses 12s a day if the temperature is `40^@C` and gains 4s a day if the temperature is `20^@C`, The temperature at which the clock will show correct time, and the co-efficient of linear expansion `(alpha)` of the metal of the pendulum shaft are respectively:A. `25^(@)C, alpha=1.85xx10^(-5) .^(@)C^(-1)`B. `60^(@)C, alpha=1.85xx10^(-4) .^(@)C^(-1)`C. `30^(@)C, alpha=1.85xx10^(-3) .^(@)C^(-1)`D. `55^(@)C, alpha=1.85xx10^(-2) .^(@)C^(-1)` |
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Answer» Correct Answer - A Time period of pendulum, `T=2pi sqrt((l)/(g))` `Delta T =(2pi)/(sqrt(g))xx(1)/(2)l^(-1//2) Delta l=(pi Delta l)/(sqrt(g l))` `:. (Delta T)/(T) =(pi Delta l//sqrt(gl))/(2pi sqrt(l//g)) =(1)/(2) (Delta l)/(l)` When clock gains 12 s a day, then `(12)/(T)=(1)/(2)alpha(40 - theta)` ....(i) When clock gain loses 4 s a day, then `(4)/(T) =(1)/(2)alpha(thata - 20)` ....(ii) Dividing (i) by (ii) we get `3=(40 - theta)/(theta - 20)` or `3theta - 60=40 - theta` or `4theta =100` or `theta =25^(@)C` Putting this value of `theta` in (i), we get `(12)/(T) =(1)/(2)alpha(40 - 25)=(1)/(2)alphaxx15` or `(12)/(24xx60xx60)=(1)/(2)alphaxx15` or `alpha=(12xx2)/(24xx60xx60xx15) =1.85xx10^(-5) .^(@)C^(-1)` |
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| 154. |
What is the power requird to raise 300 liters of water per minute through a height of 6 m using a pipe of diameter 2.5 cm ? `g = 10 m//s^2.` |
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Answer» Correct Answer - 600 watt. Workdone against gravity `W_1 = mgh = V rho gh` Workdonw againsty pressure difference is `W_2 = Delta PxxV= h rho g xx V = V rho gh.` Total workdonw, `W = W_1 + W_2 = 2 V rho gh` Power spend, `(W)/(t) = (2V rho gh)/(t) = (2xx(300 xx 10^(-3) xx 10^3 xx 10 xx6)/(60))` = 600 watt |
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| 155. |
The equation of continuity leads to |
| Answer» Correct Answer - mass | |
| 156. |
The difference between the two specific heat capacities (at constant pressure and volume) of a gas is `5000 J kg^(-1) K^(-1)` and the ratio of these specific heat capacities, i.e., `C_(V) and C_(p)`. |
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Answer» Given `C_(p) - C_(V) = 5000 J kg^(-1) K^(-1)` …(i) and `C_(p)//C_(V) = 1.6` or `C_(p) = 1.6 C_(V)` from (i) `1.6 C_(V) -C_(V) = 5000` or `0.6 C_(V) = 5000` or `C_(V) = 5000/0.6 = 8333.3 J kg^(-1)K^(-1)` `C_(p) = 1.6 C_(V) = 1.6 xx 5000/0.6 ` `=13333.3 J kg^(-1)K^(-1)`. |
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| 157. |
If for a liquid in a vessel force of cohesion is twice of adhesionA. the liquid will wet the solildB. the liquid will not wet the solidC. the miniscus will be convex upwardsD. the angle of contact will be obtuse |
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Answer» Correct Answer - B::C::D When force of cohesion is twice the force of adhesion the liquid dose not wet the sides of the vessel. The angle of contact will be obtuse and the shope of liquid of liquid meniscus is convex upwards. |
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| 158. |
The passengers flying in aeroplane are advised to remove the ink from their pens while going up in the aeroplane , why? |
| Answer» We know that as aeroplane goes up, the atmospheric pressure decreases with height. Since the ink present inside the pen is filled at the atmospheric pressure present on the surface of earth, so as the aeroplane gains height, the pressure inside the pen being more than outside, the ink tends to come out to equalise the pressure, which may spoil the cloths of the passengers. That is why the passengers are advinsed to remove the ink from their pens before taking flight. | |
| 159. |
Why a clinical thermometer should not be sterilized by boiling water? |
| Answer» Generally, the range of clinical thermometer is from `95^(@)F` to `110^(@)F` and the boling point of water is `212^(@)F` . If a clinical thermoter is sterilised by boling water, the capiliary of thermometer will burst due to thermal expansion of mercury in the capillary tube of thermomter. | |
| 160. |
What is the physical signficance of the difference of two principal specific heat of a gas? |
| Answer» The difference in the two principal specific heats of a gas is equal to the amount of heat which in turns is equal to the work done by the gas during expansion at constant pressure. | |
| 161. |
How many specific heat does a gas passes? |
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Answer» A gas can passes inifinite value of specific heat which depends on the conditions of temperature and pressure. But for study we consider only two specific heats of a gas. (i) specific heat of a gas at constant pressure. (ii) specific heat of a gas at constant volume. |
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| 162. |
How many calories of heat are required for external work when one gram mole of a as is heated by `1^(@)C` at constat pressure? |
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Answer» By definition, energy spent in expansion. `=C_(p) - C_(upsilon) = R = 2` calories. |
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| 163. |
What are the factors on which the modulus of elesticity depends? |
| Answer» The modulus of elasticity depends upon (i) nature of material and (ii) type of stress used in producing the strain. | |
| 164. |
What basic principle is involved in the up and down motion of a fish in water and how is it achieved? |
| Answer» An object can sink or float on water if its density is more or less than that of water. Therefore, principle of adjustment of density is involved in the up and down motion of a fish in water. A fish can regulate its density by expanding or cntracting an inbuilt air sac taht changes its volume. The fish can move upward by increaseing its volume (which decreases its density) and downward by contraciting its volume (which increases its denstity). | |
| 165. |
two copper balls having masses 5 gm and 10 gm collide with a target with the same velocity. If the total energy is used in heating the balls. Which ball will attain higher temperature? |
| Answer» Rise in temperature of both the balls will be the same. | |
| 166. |
A metallic cube of side 8cm is under a tangential force. The top face of the cube is sheared through 0.15 mm with respect to the bottom face. Find (a) shearing stain (b) shearing stress and (c ) shearing force. Given , modulus of rigidity of the metal `=2.08 xx 10^(11) dyn e.//cm^(2)` . |
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Answer» Here, length of each side of cube L `8cm` Lateral displacement , `Delta L = 0.15 mm =0.015cm` (a) shearing strain, `theta = (Delta L)/(L) = (0.015)/(8)` `=0.001875`. (b) Shearing stress = `F/A = G theta` `(2.08 xx 10^(11)) xx 0.001875` `=3.9 xx 10^(8) dyn e//cm^(2)`. (c ) Area of face, `A = 8xx8 cm^(2)` As `F=F/(A theta) ir F=GA theta` `:.` Shearing force, `F=GA theta` `=(2.08 xx 10^(11)) xx (8 xx 8 ) xx 0.001875` `=0.25 xx 10^(11) dyn e = 2.5 xx 10^(10) dyn e`. |
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| 167. |
A wire gets heated when it is bent back and forth. Why? |
| Answer» When a wire is bent back and forth, the deformations in the wire are beyond the elastic limit. In this sitution the work done against the intertomic forces while bending back and forth a wire. Will no longer be stored totally in the form of potential energy. Due to it the crystalline structure of the wire will be affected and part of work done is corrected into heat energy. | |
| 168. |
What are the qualities of an ideal liquid? |
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Answer» An ideal liquid has the following qualities: (i) The liquid is perfectly incompressible. It means the density of the liquid remains constant irrespective of pressure. (ii) The liquid is non viscous, i.e., there are no tangential force between layers of liquid in relative motion. (iii) An ideal liquid can not withstand any shearing stress, howsoever small the stress may be . |
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| 169. |
With the rise in temperature, which of the following forces can never increaseA. frictionB. elasticC. viscousD. surface tension force |
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Answer» Correct Answer - A::B::D With the rise in temperature, only viscosity of gases increases whereas force of surface tension decreases, elastic force decreases. Solid friction does not depend on temperature. |
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| 170. |
Water is coming out of a hole made in the wall of a fresh water tank . If the size of he hole is increasesd, (i) will velocity of efflux of water change? (ii) will the volume of the water coming out per second change? |
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Answer» (i) Velocity of efflux will remain same as it only depends upon the depth of orifice below the free surface of water. (ii) volume will increases , since volume of the liquid flowing out per second is equal to product of area of hole and velocity of liquid flowing out. |
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| 171. |
The maximum constant velocity acquired by the body while falling freely in a viscous medium is called. |
| Answer» Correct Answer - terminal velocity | |
| 172. |
Under what conditions, the velocity of a body falling in a medium (i) increases with time (ii) becomes constant with time (iii) becomes zero with time. |
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Answer» The velocity of a body falling in a medium. (i) Increases with time till the gravity pull on the body is more than the viscous force plus buoyant force of medium acting on body. (ii) becomes constant with time when gravity pull on body is equal to viscous force plus buoyant force of medium acting on body. (iii) becomes zero with time if body comes to rest in the medium . it will be so if the density of body is equal to the density of the medium. |
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| 173. |
A long a steamlineA. the velocity of a fluid particle remains constant.B. the velocity of all fluid particles crossing a given position is constant.C. the velocity of all fluid particles at a given instant is constant.D. the speed of a fluid particle remains constant. |
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Answer» Correct Answer - B A long a streamline, the velocity of every fluid particle fluid particle while crossing a given position is the same. |
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| 174. |
The velocity of efflux of an ideal liquid does not depend onA. the area of orificeB. the density of liquidC. the area of cross-section of the vesselD. the depth of point below the free surface of the liquid |
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Answer» Correct Answer - A::B::C Velocity of efflux `v =sqrt(2 gh)` , which is independent of area of orifice, area cross section of vessel and density of liquid. |
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| 175. |
The density of a substance at `0^@C 10 g//c c` and `100^@C,` its density is `9.7 g//c c.` The coefficient of linear expansion of the substance isA. `10^(-4), .^@C^(-1)`B. `10^(-2), .^@C^(-1)`C. `10^(-3), .^@C^(-1)`D. `10^(-5), .^@C^(-1)` |
| Answer» Correct Answer - A | |
| 176. |
Find the velocity of efflux of water from an oriffice near the bottom of a tank in which pressure is `500 gf//sq` cm above atmosphere. |
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Answer» Pressure at orifice, `P= 500 gf//sq cm` `= 500/1000 xx 9.8 xx (100)^(2) Nm^(-2)` `= 500 xx 9.8 Nm^(-2)` Let h be the depth of orifice below the surface. As, `P = h rho g` `:. H=P/(rho g) = (500 xx 98)/(10^(3) xx 9.8) = 5 cm` The velocity of effux, `upsilon = sqrt(2gh) = sqrt(2xx9.8xx5) = 9.893 ms^(-1)`. |
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| 177. |
The value of coefficient of volume expansion of glycerin is `5xx10^(4)K^(-1)` . The fractional change in temperature is :A. `0.010`B. `0.015`C. `0.020`D. `0.025` |
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Answer» Correct Answer - C Here, `gamma=5xx10^(-4)K^(-1)=5xx10^(-4) .^(@)C^(-1)` Let `rho_(2),rho_(1)` be the density of glycerin at temperature `t_(2) .^(@)C` and `t_(1) .^(@)C` respectively. Then `rho_(2) =rho_(1)[1-gamma(t_(2)-t_(1))]=rho_(1)-rho_(1)gamma(t_(2)-t_(1)` or `(rho_(1)-rho_(2))/(rho_(1))=gamma(t_(2)-t_(1))` Fractional change in the density of glycerine `(rho_(1)-rho_(2))/(rho_(1)) =gamma(t_(2)-t_(1)) =(5xx10^(-4))xx40` `=0.020` |
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| 178. |
The force required by a man to move his limgs immersed in water is smaller than the force for the same movement in air. |
| Answer» The upthrust on the limbs of a man is more when immersed in water than in air. As a result of it, the effective weight of lims of a man is less in water than in air. Hence the force required to move his limbs is less in water than that in air. | |
| 179. |
A man while walking is carrying a fish in the bucket full of water in the other water and thinks, he is now carrying less-weight as weight of fish will reduce due to upthrust. What do you say about it. |
| Answer» When a man while walking places the fish in a bucket full of water, the weight of the fish is reduced due to upthrust of water but weight of water is increased by the same amount. As a result of it the total weight carried by the man remains unchanged. | |
| 180. |
A tall cylinder is filled with viscous oil. A round pebble is dropped from the top with zero initial velocity. From the plot shown in Fig. 7(EP).15, indicate the one that represent the velocity (v) of the people as a function of time (t).A. B. C. D. |
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Answer» Correct Answer - C When a round pebble is dropped from the top of a tall cylinder, filled with vissous oil the pebble acquires terminal velocity (i.e. constant velocity ) after some time. Hence option (c) is correct. |
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| 181. |
The shapes of cars and planes are streamlined. Why? |
| Answer» When a body moves through a fluid, its motion is opposed by the force of fluid friction, which increases with the speed of the body. When cars and planes, move through air their motion is opposed by the air friction, which in turn, depend upon the shape of the body . it is due to this reason that the cars or planes are given such shapes (known as stream-lined shaped) so that air friction is minimum. Rather the movement of air layer on the upper and lower side of streamlined shaped body provides a lift which helps in reducing the solid friction, resulting in rncreasing the speed of the car. | |
| 182. |
Under a constant pressure head, the rate of streamlined volume flow of a liquid through a capillary tube is V. if the length of the capillary tube is double and diameter of the bore is halved, find the rate of flow of the liquid through the capillary tube. |
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Answer» `V = (pi p r^(4))/(8eta l) or V prop (r^4)/(l)` `:. (V_1)/(V) = ((r//2)^(4))/(r^4) xx l/(2l) = 1/32 or V_(1) =V/32`. |
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| 183. |
In case of a rod of length l and radius r fixed at one end, angle of share `phi` is related to angle of twiest `theta` by the relation, `theta` = ………. |
| Answer» Correct Answer - `theta = l phi//r` | |
| 184. |
Compute the bulk modulus of water from the following data : initial volume = 100.0 litre, pressure increase = 100.0 atmosphere. Final volume - 100.5 litre . (1 atmosphere = `1.013 xx 10^(5) Pa)`. Compare the ulk modulus of water that of air (at constant temperature). explain in simple terms why the ratio is so large. |
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Answer» Here, `V = 100 litre = 100 xx 10^(-3) m^(3) , p = 100 atm. = 100 xx 1.013 xx 10^(5) Pa`. `V =Delta V = 100.5litre or Delta V = (V + Delta V) -V = 100.5-100 = 0.5 litre = 0.5 xx 10^(-3)m^(3)` We know that bulk moduls, `B = (pV)/(Delta V) = (100 xx 1.013 xx 10^(5) xx 100 xx 10^(-3))/(0.5 xx 10^(-3)) = 2.026 xx 10^(9) Pa`. Bulk modulus of air at constant temperature = atmospheric pressure = `1.013 xx 10^(9) Pa` `("Bulk modulus of water")/("Bulk modulus of air") = (2.026 xx 10^(9))/(1.013 xx 10^(5)) = 2.0 xx 10^(4)` It is so because gases are much more compressibel than those of liquids. the molecules in gases are very poorly coupled to their neighbours as compared to those of gases. |
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| 185. |
A concrete sphere of radius `R` has cavity of radius `r` which is packed with sawdust. The specific gravities of concrete and sawdust are respectively `2.4 and 0.3` for this sphere to float with its entire volume submerged under water. Ratio of mass of concrete to mass of swadust will be |
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Answer» Correct Answer - 4 Let R be the radius of the whole sphere. Let `rho_1, rho_2` be the specific gravity of concrete and saw dust respectively. Ac cording to principle of floatation, weight of whole sphere = upthrust `(4)/(3) pi (R^3 - r^3) rho_1 +(4)/(3) pi r^3 rho_2 = (4)/(3)pi R^3 xx1` or `R^3 rho_1 - r^3 rho_1 + r^3 rho_2 = R^3` or `R^3 (rho_1 - 1) =r^3(rho_1- rho_2)` or `(R^3)/(r^3) = ((rho_1 -rho_2))/((rho_1 - rho_1))` `or (R^3 - r^3)/(r^3) = ((rho_1 - rho_2) - (rho_1 - 1))/((rho_1 - 1))` or `((R^3 - r^3) rho_1)/(r^3 rho_2) = ((1 - rho_2)/(rho_1 - 1))xx (rho_1)/(rho_2)` `:.("Mass of concrete")/("Mass of saw dust") = ((4)/(3)pi (R^3 - r^3)rho_1)/((4)/(3)pi r^3 rho_2)` `=((R^3 - r^3)rho_1)/(r^3 rho_2) = ((1 - rho_2) rho_1)/((rho_1 - 1)rho_2)` `=((1 -0.3)/(2.4 - 1))xx(2.4)/(0.3) = 4` |
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| 186. |
Explain the effect of (i) density (ii) temperature and (iii) pressure on the viscosity of liquids and gases. |
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Answer» (i) In case of liquid , viscosity increase with increase in density and for gases, it decreases with increase in density. (ii) with the rise in temperature, the viscosity of liquid decrease while that of gases increases. (iii) With ithe increases in pressure, the viscosity of liquids (expect water) increases while that of gases in practically independent of pressure. The viscosity of water decreases with the increase in pressure. |
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| 187. |
Explain the difference between solid friction and viscosity. |
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Answer» The solid friction and viscosity both oppose the relative motion but they differ from each other as discussed below: SOLID FRICTION: (i) solid friction between two solids is independent of the area of solid surfaces in contact. (ii) It is independent of the relative velocity of one body on the surface of another body. (iii) It is directly proportional to the normal reaction between the surfaces in contact. VISCOSITY (VISCOUS DRAG): (i) Viscous drag between the layers is directly proportional to the area of of liquid layers. (ii) It is directly proportional to the relative velocity between the two layers of the liquid. (iii) It is independent of the normal reaction between two layers of the liquid. |
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| 188. |
Explain why small drops of mercury are spherical and large drops become flat? |
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Answer» In case of a small and the force of surface tension plays a vital role. Therefore, the free surface of drop tends to have minimum surface area. For a given volume, the shere has minimum area. Hence the small mercury drops are of spherical shape. In case of large mercury drop, the gravitational pull becomes more effective than the surface tension and exerts downwards pull on the drop so that its centre of gravity may lie at the lowest possible position. Hence the lagre drop of mercuty becomes elliptical or flat in shape from middle. |
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| 189. |
The heat of combustion of ethane gas at 373 k cal per mole. Assume that 50% of heat is useful, how many be burnt litres to convert 60 kg of water at `20^(@)C` to steam at `100^(@)C` ? One mole of gas oc cupies 22.4 litre at S.T.P. Latent heat of steam = `2.25 xx 10^(6) Jkg^(-1)`. |
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Answer» Heat required to convert 60 kg of water at `20^(@)C` into steam at `100^(@)C` `=ms_(w) DeltaT + mL` `= 60 xx 10^(3) xx (100 -20) + 60 xx (2.25 xx 10^(6))/(4.2) cal` `= 4.8 xx 10^(6) + 32.14 xx 10^(6) = 36.94 xx 10^(6) cal` Since `50%` of heat is useful, so that heat produced is `Q = 100/50 xx 36.94 xx 10^(6) = 73.88 xx 10^(6) cal` Heat of combustion of ethane gas `=373 kcal mol^(-1) = 373 xx 10^(3) cal mol^(-1)` No. of moles of ethane required to be burnt for heat `Q is , n = (73.88 xx 10^(6))/(373 xx 10^(3)) = 198 mol e` Volume of ethane needed to be burnt ` 22.4 xx n` ` = 22.4 xx 198` ` =4435.2 litres`. |
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| 190. |
Calculate heat of combustion of coal, when 0.5kg of coal on burning raise the temperature of 50 liters of water from `20^(@)C` to `90^(@)C` |
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Answer» Here, mass of coal, `M = 0.5 kg`, mass of water `m = (50 xx 1000) c c xx 1 g//c c` `=50 xx 10^(3) g` `Delta T = 90^(@) - 20^(@) = 70^(@)C` Specific heat of water `s=10^(3) cal. Kg^(-1) .^(@)C^(-1)` Total heat produced = `ms Delta T` `(50 xx 10^(3)) xx 1 xx 70 cal. = 35 xx 10^(5) cal`, Heat of combustion of coal `(35 xx 10^(5))/(M) = (35 xx 10^(5))/(0.5) = 7 xx 10^(6) cal, kg^(-1)`. |
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| 191. |
A tank of volume `0.2 m^(3)` contains helilum gas at a temp. of `300` K and pressure `10^(5) N//m^(2)` . Find the amount of heat required to raise the temp. to `500` K. The molar heat capacity of helium at constant volume is `3.0 cal//mol e-K` . Neglect any expansion in the volume of the tank. Take `R=8.31 j//mol e -K`. |
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Answer» Correct Answer - `[4813.2 cal.]` Amount of gas in moles `n=(pV)/(RT)=(10^(5)xx0.2)/(8.31xx300)=8.022` `:.` The amt. of heat required, `Delta Q=nC_(v) Delta T` `=8.022xx3.0xx(500-300)=4813.2 cal` |
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| 192. |
The state of the rod, in which temperature of each part becomes constant and there is no further absorption of heat anywhere in the rod during propagation of heat is called………. |
| Answer» Correct Answer - steady state | |
| 193. |
What is temperatrue gradient? |
| Answer» The fall in temperature in a body per unit distance is called the temperature gradiant. | |
| 194. |
The rate of flow of heat per unit area per unit temperature gradient across the solid is called……….. |
| Answer» coefficient of thermal conductivity | |
| 195. |
A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass M. It is suspended by a string in a liquid of density `rho` where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is |
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Answer» Buoyant force on the cylinder = weight of liquid displaced = `V rho g` downward force on the top of cylinder ` =(h rho g) pi R^(2)` Force on the bottom of cylinder is `F =` Bouyant force + downward force on the top of cylinder `=V rho g +(h rho g) pi R^(2) = rho g (V + pi R^(2)h)`. |
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| 196. |
A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass is M. It suspended by a string in a liquid of density `rho` where it stays vertical. The upper surface of cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder bby the liquid is A. MgB. Mg - h `rho` gC. `Mg + pi R^(2)h rho g`D. `rho g(V+pi R^(2)h)` |
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Answer» Correct Answer - D The net upward force on the bottom of cylinder = wt. of liquid displaced by cylinder + thrust on the upper surface of cylinder due to h column of lliquid. `V rho g+h rho gxxpi R^(2)` `=rho g(V+pi R^(2)h)` |
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| 197. |
A tank of volume `0.2 m^3` contains helium gas at a temperature of `300 K` and pressure `1.0 xx 10^5 N m^(-2)`. Find the amount of heat required to raise the temperature to `400 K`. The molar heat capacity of helium at constant volume is `3.0 cal K^(-1 ) mol^(-1)`. Neglect any expansion in the volume of tank. |
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Answer» Correct Answer - `[2400 cal.]` For monoatomic gas `C_(v)=(3)/(2)R, n_(1)=1 mol e` For diatomic gas, `C_(v)=(5)/(2)R, n_(2)=3 mol e` Average molar specific heat of mmixture at constant volume is, `C_(v)=(n_(1)(C_(v))_(1)+n_(2)(C_(v))_(2))/(n_(1)+n_(2))=(1xx(3)/(2)R+3xx(5)/(2)R)/((1+3))` `C_(v)=(9)/(4)xx8.31xx300=8 mol e` `:.` Amount of heat requuired `Delta Q=nC_(v) Delta T=8xx3.0xx(400-300)` `Delta Q=2400 cal` |
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| 198. |
Is the specific heat of water greater than that of sand? |
| Answer» Yes , infact, specific heat of water is maximum. | |
| 199. |
The earth constantly receives heat radiation from the sun and gets warmed up. Why does the earth not get as hot as the sun is ? |
| Answer» Due to the large distance of the earth from the sun, it receives only a small part of heat radiation radiated by the sun. Further, there oc curs loss of heat from the surface of the earth due to radiation and convection. Hence the earth cannot becomes as hot as the sun. | |
| 200. |
Is it necessary that all black coloured objects should be considered black bodies ? |
| Answer» No, it is not necessary that all black coloured object should be considered as black bodies. For example , if we take a black surface which is highly polished, it will not behave as a perfect black body . On the other hand, the sun, which is a shining hot sphere, behaves as a perfect black body. | |