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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. |
A black body at 20000 K, emits maximum energy at a wavelength of `1.56 mu m`. At what temperature will it emit maximum energy at a wavelength of `1.8 mu m` ? |
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Answer» Correct Answer - `[1733 K]` Here, `T_(1)=2000 K, lambda_(1)=1.56 mu m`, `T_(2)=?, lambda_(2)=1.8 mu m` `lambda_(1) T_(1)=lambda_(2)T_(2)` or `T_(2)=(lambda_(1)T_(1))/(lambda_(2))=(1.56xx2000)/(1.8)=1733 K` |
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| 102. |
Which state of matter has volume elasticity? |
| Answer» All the thre states of matter (i.e. Solids, liquids and gases ) have volume elasticity. | |
| 103. |
Two wires are made of same metal. The length of the first wire is half that of the second wire and its diameter is double that of the second wire. If equal loads are applied on both wires, find the ratio of increases in their lengths. |
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Answer» `Y=(F)/(pi (2 r^(2))^(2)) xx (l//2)/(Delta l_(1)) = (F)/(pi r^(2)) xx l/(Delta l_(2))` `:. (Delta l_(1))/(Delta l_(2)) = 1/8` |
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| 104. |
A wire is roplaced by another wire of same length and material but of twice diameter. (i) What will be the effect on the increases in its length under a given load? (ii) What will be the effect on the maximum load which it can bear? |
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Answer» (i) increases in length will be reduced to one fourth as `Delta l prop 1//r^(2)` (ii) Maximum bearable load becomes four times as breaking force `prop area (=pi r^(2))`. |
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| 105. |
Write two factors affecting viscosity. Which one is more viscous : Pure water or saline water? |
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Answer» Viscosity of a liquid depends on (i) its nature and (ii) its temperature. saline water is more viscous than pure water. |
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| 106. |
A boat floating in a tank is carrying passengers. If the passengers drink water, the water level of the tankA. risesB. fallsC. remains unchangedD. depends upon the atmospheric pressure |
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Answer» Correct Answer - C When passengers drink water, the load will increase. Thise will displace more water. But water displaced will be equal to the water drink by the passengers. Hence, the water level will not change. |
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| 107. |
Why does an iron needle float on clean water but sink when some detergent is added to this water? |
| Answer» Due to surface tension, the free surface of liquid at rest behaves likes a strectched membrane. When an iron needle floats on the surface of clean water, its weight is supported by the stretched membrane, when some detergent is added to this water, its surface tension decreases. As a result of it, the stretched membrane on the surface of water is weakend and is not able to support the weight of needle. Hence, needle sinks in such water. | |
| 108. |
An iron sphere has a radius `8 cm` at a temperature of `0^(@)C`. Calculate the change in volume of the sphere if is temperature is raised to `80^(@)C`. Coefficient of liner expansion of iron `=11xx10^(-6) .^(@)C^(-1)`. |
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Answer» Correct Answer - `[5.66cm^(3)]` Here, `R=8` cm, Volume of sphere at `0^(@)C` is `V_(0)=(4)/(3)pi R_(3)=(4)/(3)xx(22)/(7)xx8^(3)cm^(3)` Increase in temperature, `DeltaT=80-0-80 .^(@)C` Coefficient of cubical expansion of iron `gamma=3 alpha=3xx11xx10^(-6) .^(@)C^(-1)` Change in volume, `Delta V=V_(0) gamma Delta T` `=((4)/(3)xx(22)/(7)xx8^(3))xx(3xx11xx10^(-6)xx80=5.66cm^(3)` |
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| 109. |
Three metal rods of same length and same cross - sectional area are connected in parallel. If their conductivities area 70, 110, 180 `Wm^(-1) K^(-1)` respectively, the effective conductivity (in `Wm^(-1) K^(-1))` of the combination isA. 90B. 260C. 130D. 360 |
| Answer» Correct Answer - B | |
| 110. |
An iron sphere of radius `10 cm` is at temperature , `10^(@)C`. If the sphere is heated upto temperature `110^(@)C` , find the change in the volume of the sphere coefficient of linear expansion of iron = `11 xx 10^(-6) .^(@)C^(-1)` |
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Answer» Here, `r_(10) = 10 cm` , ` Delta T = 110 - 10 = 100^(@)C` `alpha = 11 xx 10^(-6) .^(@)C^(-1)` , `gamma = 3 alpha = 3 xx 11 xx 10^(-6) .^(@)C^(-1)` Initial volume, `V_(10) = 4/3 pi (r_(10))^(3) = 4/3 xx 3.14 xx (10)^(3) m^(3)` ltbr. Change in volume, `Delta V = V_(10) xx gamma xx Delta T` `=13.8 cm^(3)`. |
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| 111. |
A tooth cavity is filled with a copper having coefficient of linear expansion `1.7 xx 10^(-5) .^(@)C^(-1)` and bulk modules `1.4 xx 10^(11) Nm^(-2)`. The temperature of the tooth `3.^(@)C`. Calculate the thermal stress developed inside the tooth cavity when hot milk at temperature of `60 .^(@)C` is drunk. |
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Answer» Here, `alpha = 1.7 xx 10^(-5) .^(@)C_(1)` `B = 1.4 xx 10^(11) Nm^(-2)`, `Delta T = 60 - 35 = 25 .^(@) C` Thermal stress `= B gamma Delta T = (3 alpha) Delta T` `=(1.4 xx 10^(11)) xx (3 xx 1.7 xx 10^(-5)) xx 25` `=1.78 xx 10^(8) Nm^(-2)`. |
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| 112. |
How much metres can a 50 kg man climbs by using the energy from a slice of a bread which produces 420 kJ heat? Assuming that the human body effciency working is `30%`. Use `g=10 m//s^(2)`. |
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Answer» Let h be the height climbed by man. increase in PE of man = `mgh = 50 xx 10 xx h j` Heat produced , `H = 420 xx 1000 j` `=4.2 xx 10^(5) J` effciency of man = `30%`, so heat energy utilized by man in doing work is `=30/100 xx 4.2 xx 10^(5) = 12.6 xx 10^(4) J` Now, increase in PE = heat energy utilized `50 xx 10 xx h = 12.6 xx 10^(4)` or `h=(12.6 xx 10^(4))/(50 xx 10) = 252 m`. |
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| 113. |
The length of a steel wire is `l_(1)` when the stretching force is `T_(1)` and `l_(2)` when the stretching force is `T_(2)`. The natural length of the wire is |
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Answer» Let l and A be the original length and area of cross-section of the metal wire. Change in length in the first case = `(l_(1)-l)` Change in length in the second case` = (l_(2)-l)` `:. Y=(T_1)/(A) xx (l)/(l_(1)-l) = (T_2)/(A) xx (l)/(l_(2)-l)` or `T_(1)l_(2)-T_(1) l=T_(2) l_(1)-T_(2)l` or `l(T_(2)-T_(1)) = T_(2) l_(1)-T_(1)l_(2)` or `l=(T_(2)l_(1)-T_(1)l_(2))/(T_(2)-T_(1))`. |
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| 114. |
A wire of length L and radius a rigidlyl fixed at one end. On stretching the other end of the wire with a force F, the increase in its length is L, if another wire of same material but of length 2 L and radius 2 a is stretched with a force 2 F, the increase in its length will be |
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Answer» In first case , ` Y = (f//pi r^2)/(l//L) = (fL)/(pi r^2 l) ….. (i)` in second case, let x be the extension in the wire of radius 2r , length 2 L under a load 2 f, then `Y = (2f//pi(2r)^2)/(x//2L) = (fL)/(pi r^2 x) ….. (ii)` Since Y is same for two cases as the two wires are of same meterial, hence form (i) and (ii) `(fL)/(pi r^2 l) = (fL)/(pi r^2x) or x =l` |
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| 115. |
A `50` g lead bullet (specific heat `0.02`) is initially at `30^(@)C`. It is fired vertically upward with a speed of `840 ms^(-1)`. On returning to the starting level it strikes a cake of ice at `0^(@)C`. How much ice is melted ? Assume that all energy is spent in melting only. Latent heat of ice `=336 j g^(-1)`. |
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Answer» Correct Answer - `[52.875 g]` When bullet falls back to its original level its downward velocity is same as given upward velocity at that location, i.e. ,`840 ms^(-1)` . KE of bullet on returening to the initial level `W=(1)/(2)mv^(2) =(1)/(2)xx((50)/(1000))xx(840)^(2)=17640 j` This KE will be converted into heat energy on striking the cake of ice. So heat produced `Q_(1)=W=17640j` Heat lost by bullet in cooling from `30^(@)C` to `0^(@)C` is `Q_(2)=mc Delta T=50xx0.02xx(30-0)=30 cal =30xx4.2 j=126 j` Total heat given by bullet to icec is `Q=Q_(1)+Q_(2)=17640+126=17766 j` Let m be the mass of ice melted. Then `17766=mL` or `m=(17766j)/(Ljg^(-1))=(17766)/(336)g =52.875 g` |
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| 116. |
The upward force acting on the body immersed in a fluid is called…….. Force. |
| Answer» Correct Answer - buoyant | |
| 117. |
Do you think a body at higher temperature contains more heat? Explain. |
| Answer» It is not true always because the heat possessed by a body `Q (=ms Delta theta)` depends upon the mass of body (m) its specific heat (s) and rise in temperature. `(Delta theta)`. | |
| 118. |
A pendulum clock shows correct time at certain time at certain temperature. At a higher temperature the clockA. loses timeB. gains timeC. neither gains nor loses timeD. firstly gains and then loses |
| Answer» Correct Answer - A | |
| 119. |
Certain amount of heat is given to 100 g of copper to increase its temperature by `21^@C`. If same amount of heat is given to 50 g of water, then the rise in its temperature is (specific heat capacity of copper ` = 400 J kg^(-1) K^(-1)` and that for water ` = 4200 J kg^(-1) K^(-1))`A. `4^@C`B. `5.25^@C`C. `8^@C`D. `10.5^@C` |
| Answer» Correct Answer - A | |
| 120. |
The ratio of densites or iron at `10^@C` is (alpha of iron `= 10xx10^(-6), .^@C^(-1))`A. 1.003B. 1.0003C. 1.006D. 1.0006 |
| Answer» Correct Answer - D | |
| 121. |
A copper wire of length l increases in length by 0.3 % on heating from `20^@C` to `40^@C.` Then percentage change in area of copper plate of dimensions `3 lxx 2l` on heating from `20^@C` to `40^@C` isA. `0.05%`B. `0.3%`C. `0.4%`D. `0.6%` |
| Answer» Correct Answer - D | |
| 122. |
Pressure is a scalar or vector quantity. |
| Answer» Pressure is a scalar quantity because at one level inside the liquid , the pressure is exerted equally in all directions , which shows that a definite direction is not associated with hydrostatic pressure. | |
| 123. |
Why are the bridge declared unsafe after long use? |
| Answer» A bridge during its use undergoes alternating stresses and strains for a large number of times each day, depending upon the movement fo vehicles on it. When a bridge is used for long times. It lises its elastic strength. Therefore, the amount of strain in the bridge for a given stress will become large and ultimately, the bridge may collapse. That is why the bridge are declared unsafe after long use. | |
| 124. |
The bags and suitcases are proveded with broad handles. Why? |
| Answer» So that the small pressure is exerted while carrying them. | |
| 125. |
What is 1 torr? |
| Answer» One torr is the unit of atmospheric pressure where 1 torr= one mm of mercury column. | |
| 126. |
Under which condition (i) the centre of buoyancy coincides with the center of gravity (ii) the centre of bouyancy does not coincide with the centre of gravity. |
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Answer» (i) For a solid body of uniform density, the centre of gravity coincides with the centre of buoyancy. (ii) For a solid body having different densities over different parts, its centre of gravity does not coincide with the centre of buoyancy. |
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| 127. |
What is buoyancy and centre of buoyancy? |
| Answer» The upward thrust acting on the body immersed in a liquid is called buoyant force or simply bouyancy. The centre of bouyancy is the centre of gravity of the displaced liquid by the body when immersed in a liquid. | |
| 128. |
The ratio stress/strain remains constant for small deformation. What will be the effect on this ratio when the deformation made is very large? |
| Answer» When the deforming force is applied beyond elastic limit, the strain produced is more than that has been observed within elastic limit. Due to it, the ratio stress/strain will decrease. | |
| 129. |
What does the slops of stress versus strain graph indicate? |
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Answer» The slpoe of stress (on y-axis) and strain (x-axis ) gives modulus of elasticity. The slope of stress (on x-axis) and strain (y - axis) gives the reciprocal of modulus of elasticity. |
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| 130. |
Pressure inside two soap bubbles are `1.01` and `1.02` atmospheres. Ratio between their volumes isA. 16B. 8C. 4D. 2 |
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Answer» Correct Answer - B Excess pressure in one bubble, `p_(1)=1.01 - 1=0.01 =(4S)/(r_(1))` …(i) Excess pressure in other bubble, `p_(2)=1.02 - 1=0.02=(4S)/(r_(2))` ….(ii) Dividing (i) by (ii) `(0.01)/(0.02)=(r_(2))/(r_(1))` or `(r_(1))/(r_(2))=2` `(V_(1))/(V_(2))=(r_(1)^(3))/(r_(2)^(3))=(2)^(3) =8` |
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| 131. |
Two solild spheres manufactured of the same material freely fall down in the air. One sphere has a dimeter twice as large as the other. The force due to air resistance is proportional to the cross-section area of a moving object and is quadration function of the speed of an object and is quadration function of the speed of an object. In sometime after the beginning of motion in the presence of air resistance, the velocity of each sphere become constant. It is called the terminal velocity `v_(big)//v_(small)` isA. `(1)/(sqrt(2))`B. `(1)/(2)`C. `sqrt(2)`D. `2` |
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Answer» Correct Answer - C An object will attain terminal velocity when force of gravity on object is equal to the force of the air resistance, i.e., Mg `=k v^(2) A` or `(4)/(3)pi r^(3)rho g = kv^(2)pi r^(2)` …..(i) where M is the mass of the sphere, r is its radius and `rho` is the density of the material of sphere and k is a constant of proportionality. From (i), `v prop sqrt(r)` `(v_(big))/(v_(small)) =sqrt((r_(big))/(r_(small))) =sqrt((2r)/(r)) =sqrt(2)` |
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| 132. |
Fig, shown a thin film supporting a small weight = `4.5xx10^(-2)N`. What is the weight supported by a film of the same liquid at the same temperature in fig. explain your answer physically. . |
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Answer» (a) Here, length of the film supporting the weight = 40cm =0.4 m Total weight supported (or force) =`4.5xx10^(-2)N`, Film has two surfaces, `:.`Surface tension, `S = (4.5xx10^(-2))/(2xx0.4) = 5/625xx10^(-2)Nm^(-1)` Since the liquid is same for all the cases(a), (b) and (c),and temperature is also same, therefore, surface tension for cases (b) and (c) will also be the dame = `5.625xx10^(-2)`. in fig., length of the film supporting the weight is also the same as that of (a), hence the total weight supported in each case is `4.5xx0^(-2)N`. |
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| 133. |
If I is the moment of inertial of a disc about an axis passing through its centre then find the change in moment of inertial due to small change in its moment of inertia due to small change in its temperature `Delta t`. `alpha` is the coefficient of linear expansion of disc. |
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Answer» Moment of inertial of a disc of mass M, radius R about an axis passing through its centre and perpendicular to the plane of disc is `I = 1/2 MR^(2)` Then `Delta I = M/2 2 RDelta R = MRDeltaR` `:. (Delta I)/(I) = (MRDelta R)/(MR^(2)//2) = (2DeltaR)/(R)` But `(Delta R)/(R) = alpha Delta t` therefore, `(Delta I)/(I) = 2 alpha Delta t or Delta I = 2 I alpha Deltat`. |
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| 134. |
What happens if water vapour at a pressure of 0.003 atmosphere is cooled to `0^(@)C`? |
| Answer» The pressure corresponding to triple point of water is 0.46 cm of Hg (=0.006 atm). From the study of P-T phase diagram, we note that at a pressure lower than 0.006 atm. Water vapour condenses directly into ice. As the given water vapour is at a pressure 0.003 atm. which is less than the pressure of triple point of water, hence, these water vapours condense directly into ice. | |
| 135. |
At the triple point of water wheater the relative amounts of ice, water and vapour fixed or not. Explain. |
| Answer» Ar the triple point of water, the temperature and pressure are fixed. For water, triple point temperature is 273.16K (or `0.01^(@)C` ) and triple point pressure is 0.46 cm of Hg column. However, the relative amounts of three phases at triple point of water are not unique. The relative amount of three phases can be varied by adding or taking out heat from the system. | |
| 136. |
The triple point of water is a standard fixed point in mordern thermometry. Why? |
| Answer» This is because the triple point of a substance is unique i.e., it oc cures only at one paticular set of value of pressure and temperature. | |
| 137. |
What is meant by triple point ? Give the values fo triple point pressure and triple point temperature of water. |
| Answer» It is point in the phase diagram, representing a particular pressure and temperature at which the solid, liquid and vapout phases of the substance can co-exist. Triple point pressure of water is 0.46cm fo mercury colum or 0.006 atm. And triple point temperature of water is 273.16 K or `0.01^@C.` | |
| 138. |
How has the Physics of heat been utillsed by fire figthers? |
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Answer» The fire fighters have to enter burning buildings to save lives and property. For this the knowledge of Physics of heat for them plays an important roles because. (i) Knowing the types of fumes and radiations emitting from the present fire will help them to decide the type of spray (only water spray or fine mist spray) to be used to control the fire. (ii) The type of clothes to be used during the fire fighting so that they are not harmed by fire. (iii) The type of instruments to be used during spray and their placing to control the fire. (iv) the best technique to be used to bring out the person trapped in fire. |
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| 139. |
Can temperature be assigned to a vacuum? |
| Answer» Ac cording to classical physics, temperature is defined in terms of average kinetic energy of particles of the object. Since there are no particle in vacuum, hence temperature in a pure vacuum is underfined. | |
| 140. |
A body suspended from a spring balanc ce is immersed in water. If the coefficient of cubicacl expansion of water is twice that of the suspended body, then on heating the water, the reading on the spring balance decreases, increases or reamains the same, Explain. |
| Answer» The reading of the spring balance which measures the apparent weight will increase. Infact, on heating, the ture weight (=mg) of the body remains the same, but volume increases and density decreases proportionality. But for weight of water displaced, volume of the body increases and densiy of the water decreases at double the rate. Due to it, the upward thrust will become less and hence apparent weight will become more. | |
| 141. |
Calculate the rate of loss of heat through a glass window of area `1000 cm^(2)` and thickness 0.4 cm when temperature inside is `37^(@)C` and outside is `-5^@)C` . Coefficient of thermal conductivity of glass is `2.2 xx 10^(-3) cal s^(-1) cm^(-1) K^(-1)`. |
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Answer» Here, `A =1000 cm^(2), Delta x = 0.4 cm`, `T_(1) = 37^(@)C = 273 + 37 = 310 K` `T_(2) = 273 - 5 = 268 K`, `K = 2.2 xx 10^(-3) cal s^(-1) cm^(-1) K^(-1)` Rate of loss of heat `=(Delta Q)/(Delta t) = (KA Delta T)/(Delta x) = (KA(T_(1)-T_(2)))/(Delta x)` `=((2.2 xx 10^(-3)) xx 1000 xx (310 - 268))/(0.4)` `=231 cal s^(-1)`. |
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| 142. |
A metal rod of length 20 cm and diamter 2 cm is convered with a non conducting substance. One of its ends is maintained at `100^(@)C` , while the other end is put at `0^(@)C`. It is found that 25 g ice melts in 5 min. calculate the coefficient of thermal conductivity of the metal. Latent that of ice = `80 cal g^(-1)`. |
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Answer» Here, length of the rod, `Delta x = 20 cpm = 20 xx 10^(-2)m`, Diameter = 2cm radius `= r = 1 cm =10^(-2) m`, Area of cross section `A = pi r^(2) = pi (10^(-2))^(2) = 10^(-4) pi sq. m,` `Delta T = 100 - 0 = 100^(@)C`, Mass of ice melted , `m= 25 g , L = 80 cal.g^(-1)` `:.` Heat conducted, `Delta Q = m L = 25 xx 80 = 2000 cal.` `= 2000 xx 4.2 J` `Delta t = 5 min = 300 s` As, `(Delta Q)/(Delta t) =( KA (Delta T))/(Delta x)` `:. K = (Delta Q//Delta t)/(A Delta T//Delta x) = (Delta Q Delta x)/(Delta t.A Delta T)` `=(2000 xx 4.2 xx 20 xx 10^(-2))/(300 xx 10^(-4)pi xx 100)` `=1.78 Js^(-1)m^(-1) .^(@)C^(-1)`. |
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| 143. |
A metal tube and a rod of same length same material and same outer diameter are given same amount of heat. Which will show less expansion and why ? |
| Answer» For a metal tube and a rod of same length same meterial and f same outer diameter, the mass or rod is more then that of tube. When same amount of heat is given to metal tube and a rod, then the temperature rise for rod is less, hence expansion for roed is less then that of tube. | |
| 144. |
A wire 2m in length suspended vertically stretches by 1mm when a mass of 20 kg is attached to the lower end. The elastic potential energy gained by wire is ………… (use `g= 10 m//s^2)` |
| Answer» 0.1 J [ Elastic pot. Energy,`U = (1)/(2) F Delta l = (1)/(2) xx (20 xx 10) xx (10^(-3) = 0.1 J)].` | |
| 145. |
Water can rise up to a height of 10 cm in a capillary tube. If a capillary of same diameter but of length 6 cm is held vertically in water, will the water come out in the form of a fountain? Explain. |
| Answer» No, when the size of the capillary is less than the height to which the water can rise, the radius of curvature of water meniscus adjusts itself so that there is no over flowing of water but the product of height of liquid and radius of curvature of the meniscus remains a finite constant. | |
| 146. |
What is the reason that a constant driving force is always required for the maintenance of the flow of oil through the pipe lines in the oil refineries? |
| Answer» When oil is flowing through a pipe, it flows in the from of layers. The motion of oil layers get retarded after travellig certain distance due to viscous, force acting between the oil layers. Therefore, to maintain a constant flow of oil in the pipe, a constant driving force is required . | |
| 147. |
Why oils of different viscosity are used in different seasons? |
| Answer» The temperature of surrounding air is different in different seasons. In summer the temperature of surrounding is more than that in winter. The viscosity of oil decreases with rise in temperature. Hence viscosity of oil is less in summer than in water. therefore, more viscous oils are used in summer and less viscous oils in winter. | |
| 148. |
Discuss the effect of temperature on the viscosity of liquids and gases. |
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Answer» The viscosity of liquid decreases rapidly with rise of temperature. For example , in case of water, the viscosity at `50^(@)C` is found to be about half of its value at `10^(@)C` . In case of castor oil, the viscostiy at `20^(@)C` is about `2//5th` of its value at `10^(@)C`. The viscosity of all the gases increases with the rise in temperature. |
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| 149. |
Surface energy per unit area of liquid surface is called…………… |
| Answer» surface tension of liquied. | |
| 150. |
The sides of a horizontal pipe carrying dirty water get dirty. Explain. |
| Answer» when dirty water is flowing through the horizontal pipe, the velocity of the water is maximum along the axis of pipe, i.e. a relative motion exists between the various layers of the liquid flowing through the pipe. As a result of it, a viscous drag comes into play which opposes the relative motion between the adjacent layers. Due to it, the dirt particles presuent in dirty water move outwards and get struck on the walls of pipe, making the pipe dirty. | |