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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.
| 1. |
Figure shows a cubical block of side 10 cm and relative density 1.5 suspended by a wire of cross sectional area `10^(-6) m^(2)`. The breaking stress of the wire is `7 xx 10^(6) N//m^(2)`. The block is placed in a beaker of base area `200 cm^(2)` and initially i.e. at t = 0, the top surface of water & the block coincide. There is a pump at the bottom corner which ejects `2 cm^(3)` of water per sec. Find the time at which the wire will break. |
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Answer» Correct Answer - 5 Breaking force `=` Breaking stress `xx` area of wire `= 7 xx 10^(6) xx 10^(-6) = 7N` weight of block `= vol xx density xx g` `10^(-3) xx (1.5 xx 10^(3)) xx 10` `= 15 N`. The upthrust when the liquid level has decended by `x cm`. `= 100 (10 - x). 10^(-6). 10^(3). 10` `= 10 - x` Newton. `:.` Net downward force on the block `= 15 - (10 - x) = 5 + x = "Tansion" T` `:.` The wire will break when `5 + x = 7`. i.e. when `x = 2cm` Let the level decends by `2 cm` in `t` time then `2t = (200 - 100).2` `t = 100 sec`. ........Ans. `20x = 10 :. x = 5` |
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| 2. |
If air is blown under one of the pans of a phusical balance in equilbrium, then the pan willA. not be disturbedB. go upC. go downD. becomes vertical |
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Answer» Correct Answer - C Pressure below the pan decreases on account increased velocity. |
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| 3. |
The pans of a physical balance are in equilibrium. Air is blown under the right hand pan, then the right hand pan willA. Move upB. Move downC. Move erraticallyD. Remain at the same level |
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Answer» Correct Answer - B |
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| 4. |
A ship made of iron can float in water but an iron needle sinks. Why ? |
| Answer» The weight of the displaced water by the immersed part of the ship balances the weight of the ship and as a result, the ship floats on water. On lthe other hand, the weight of the displaced water by the needle is less than the weight to the needle and consequently it sinks. | |
| 5. |
Small droplets of liquid are usually more spherical in shape than larger drops of the same liquid becauseA. Force of surface tension is equal and opposite to the force of gravityB. force of surface tension predominates the force of gravityC. force of gravity predominates the force of surface tensionD. force of gravity and force of surface tension act in the same direction and are equal |
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Answer» Correct Answer - B For small drops, force of surface tension predominates gravitational force and so, they are more spherical. |
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| 6. |
Two capillaries made of same material but of different radii are dipped in a liquid. The rise of liquid in one capillary is 2.2 cm and that in the other is 6.6 cm . The ratio of their radii isA. `9:1`B. `1:9`C. `3:1`D. `1:3` |
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Answer» Correct Answer - C As `h prop (1)/(r ) therefore (h_(1))/(h_(2)) = (r_(2))/(r_(1))` or `(r_(1))/(r_(2)) = (h_(2))/(h_(1)) = (6.6)/(2.2) = (3)/(1)` |
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| 7. |
A necklace weighs `50 g` in air, but it weighs `46 g` in water. Assume that copper is mixed with gold to prepare the necklace. Find how much copper is present in it. (Specific gravity of gold is `20` and that of copper is `10`.)A. 10gB. 20gC. 30gD. None |
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Answer» Correct Answer - C Let `m` be the mass of copper in neckless. Mass of gold = (50-m)` Volume of copper = `V_(1) =m//10` Volume of gold = `V_(2) = (50-m)/(20)` when immersed in water `rho_(w) = 1gm//cm^(3)` Decrease in weight = upthrust `(50-46)g = (V_(1)+V_(2))rho_(w)g` `implies 4 = m/10 + (50-m)/(20) implies m = 30 g`. |
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| 8. |
Velocity of water in a river isA. Same everywhereB. More in the middle and less near its banksC. Less in the middle and more near its banksD. Increase from one bank to other bank |
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Answer» Correct Answer - B |
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| 9. |
A large tank is filled with water to a height `H`. A small hole is made at the base of the tank. It takes `T_(1)` time to decrease the height of water to `(H)/(eta) (eta gt 1)`, and it takes `T_(2)` times to take out the rest of water. If `T_(1) = T_(2)`, then the value of `eta` isA. 2B. 3C. 4D. `2 sqrt(2)` |
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Answer» Correct Answer - C Time in which height of liquid falls from `H_1` to `H_2` `t prop (sqrt(H_1) - sqrt(H_2))` `(T_1)/(T_2) = (sqrt H - sqrt((H)/(eta)))/(sqrt((H)/(eta))- 0)` `1 = (1 - (1)/(eta))/((1)/(sqrt(eta)))` `(2)/(sqrt(eta)) = 1 rArr eta = 4`. |
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| 10. |
A large tank is filled with water to a height `H`. A small hole is made at the base of the tank. It takes `T_(1)` time to decrease the height of water to `(H)/(eta) (eta gt 1)`, and it takes `T_(2)` times to take out the rest of water. If `T_(1) = T_(2)`, then the value of `eta` isA. `2`B. `3`C. `4`D. `2sqrt(2)` |
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Answer» Correct Answer - C |
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| 11. |
A large tank is filled with water to a height `H`. A small hole is made at the base of the tank. It takes `T_(1)` time to decrease the height of water to `(H)/(eta) (eta gt 1)`, and it takes `T_(2)` times to take out the rest of water. If `T_(1) = T_(2)`, then the value of `eta` isA. 2B. 3C. 4D. `2sqrt(2)` |
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Answer» Correct Answer - C Apply result from Q. 16 of Ex-II `t=(A)/(A_(0))sqrt((2)/(g))lfloor(sqrt(H)-sqrt(x))rfloorimpliesT_(1)=(A)/(A_(0))sqrt((2)/(g))lfloor(sqrt(H)-sqrt(H//eta))rfloor" "…(1)` `T_(2)=(A)/(A_(0))sqrt((2)/(g))lfloorsqrt(H//eta)-sqrt(0)rfloor" "...(2)` From (1) and (2) `eta=4` |
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| 12. |
A vertical tank, open at the top, is filled with a liquid and rests on a smooth horizontal surface. A small hole is opened at the centre of one side of the tank. The area of cross-section of the tank is N times the area of the hole, where N is a large number. Neglect mass of the tank itself. The initial acceleration of the tank isA. `(g)/(2N)`B. `(g)/sqrt(2N)`C. `(g)/(N)`D. `(g)/(2sqrt(N))` |
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Answer» Correct Answer - C Force on tank is provided by change in linear movmentum to flow of water `F=rho((A)/(N))v^(2)" "impliesF=rho((A)/(N))(2g.(H)/(2))` `F=(rhoAHg)/(N)" "impliesF=m(g)/(N)" "impliesa=(g)/(N)` |
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| 13. |
Equal masses of water and a liquid of density 2g/cm3 are mixed together. The density of mixture is:A. `2//3`B. `4//3`C. `3//2`D. 3 |
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Answer» Correct Answer - B If two liquid of equal masses and different densities are mixed together then density of mixture `rho=(2rho_(1)rho_(2))/(rho_(1)+rho_(2)) = (2xx1xx2)/(1+2) = (4)/(3)` |
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| 14. |
Two substances of densities `rho_(1)` and `rho_(2)` are mixed in equal volume and the relative density of mixture is `4`. When they are mixed in equal masses, the relative density of the mixture is 3. the values of `rho_(1)` and `rho_(2)` are:A. `rho_(1) = 6 and rho_(2)=2`B. `rho_(1)=3 and rho_(2) = 5`C. `rho_(1) = 12` and `rho_(2) = 4`D. noneof these |
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Answer» Correct Answer - A When the substances are mixed in equal volumes then `V rho_(1)+Vrho_(2) = 2V xx 4` …(1) When the two substances are mixed in equal masses. Then, `(m)/(rho_1) + (m)/(rho_2) = (2m)/(2)` ..(2) From eq. (1) , `rho_(1)+rho_(2) = 8` from eq. (2) , `1/(rho_1)+1/(rhp_2) = 2/3` or `(rho_(1)+rho_(2))/(rho_(1)rho_(2)) = 2/3` ..(3) or `9/(rho_(1)rho_(2)) = 2/3 or rho_(1)rho_(2) = 12` ..(4) Now, `rho_(1) - rho_(2) = [(rho_(1)+rho_(2))^(2)-4 rho_(1)rho_(2)]^(1//2)` ..(5) solving eqs, (3) and (5) , we get `rho_(1) = 6 and rho_(2)=2`. |
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| 15. |
Two substances of densities `rho_(1)` and `rho_(2)` are mixed in equal volume and the relative density of mixture is `4`. When they are mixed in equal masses, the relative density of the mixture is 3. the values of `rho_(1)` and `rho_(2)` are:A. `rho_(1)=6` and `rho_(2)=2`B. `rho_(1)=3` and `rho_(2)=5`C. `rho_(1)=12` and `rho_(2)=4`D. None of these |
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Answer» Correct Answer - A When substances are mixed in equal volume then density `=(rho_(1)+rho_(2))/(2)=4 implies rho_(1)+rho_(2) = 8 " "…..(i)` When substance are mixed in equal masses then density `=(2rho_(1)rho_(2))/(rho_(1)+rho_(2)) = 3 implies 2rho_(1)rho_(2) = 3(rho_(1)+rho_(2))" "……(ii)` By solving (i) and (ii) we get `rho_(1) =6` and `rho_(2) =2`. |
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| 16. |
A body of density d is counterpoised by Mg of weights of density `d_1` in air of density d. Then the true mass of the body isA. `M`B. `M(1-(d)/(d_(2)))`C. `M(1-(d)/(d_(1)))`D. `(M(1-d//d_(2)))/((1-d//d_(1)))` |
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Answer» Correct Answer - D Let `M_(0)` = mass of body in vacuum. Apparent weight of the body in air = Apparent weight of standard weights in air implies Actual weight - upthrust due to displaced air = Actual weight - upthrust due to displaced air `implies M_(0)g-((M_(0))/(d_(1)))dg = Mg-((M)/(d_(2)))dg implies M_(0) = (M[1-(d)/(d_(2))])/([1-(d)/(d_(1))])` |
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| 17. |
A body of density d is conuterpoised by Mg of weights of density `d_1` in air of density d. Then the true mass of the body isA. MB. `M (1-(d)/(2))`C. `M (1 - (d)/(d_1))`D. `(M(1 - d//d_2))/((1- d//d_1))` |
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Answer» Correct Answer - D (d) Let `M_0 =` mass of body in vacuum. Apperent weight of the body in air =Apparent weight of standerd weight in air `rArr` Actual weight - upthrust due to displaced air = Actual weight - upthrust due to displaced air `rArr M_0g - ((M_0)/(d_1))dg = Mg -((M)/(d_2))dg` `rArr M_0 = (M[(1-(d)/(d_2)])/([1 -(d)/(d_1)])` |
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| 18. |
A boat floating in a water tank is carrying a number of large stones. If the stones are unloaded into water, what will happen to the water level?A. rises till half the number of stones are unladed and then begins to fallB. remains unchangedC. risesD. falls |
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Answer» Correct Answer - D e) When the stones are unloaded into water, the water level falls because the volume of the water displaced by stones in water will be less than the volumes of water displaced when stones are in the boat. |
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| 19. |
The pressure at the bottom of a tank containing a liquid does not depend onA. Acceleration due to gravityB. Height of the liquid columnC. Area of the bottom surfaceD. Nature of the liquid |
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Answer» Correct Answer - C |
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| 20. |
A body of density d is counterpoised by Mg of weights of density `d_1` in air of density d. Then the true mass of the body isA. `M`B. `M(1-d/(d^(2)))`C. `M(1-(d)/(d^(2)))`D. `(M (1-d//d_(2)))/((1-d//d_(1)))` |
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Answer» Correct Answer - D |
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| 21. |
The liquid in the capillary tube will rise if the angle of contact isA. `120^(@)`B. `90^(@)`C. obtuseD. acute |
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Answer» Correct Answer - D With acute angle, of contact, liquid will rise in capillary. |
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| 22. |
The water droplets in free fall are spherical due toA. gravityB. viscosityC. surface tensionD. inter-molecular attraction |
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Answer» Correct Answer - C Water droplets assumes minimum area due to surface tension and becomes spherical. |
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| 23. |
Due to capillary action, a liquid will rise in a tube, if the angle of contact isA. acute obtuseB. obtuseC. `90^(@)`D. zero |
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Answer» Correct Answer - A `h=(2T cos theta)/(r d g)`. It `theta` is less than `90^(@)`, then h will be positive. |
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| 24. |
A cylinder of mass M and density `d_(1)` hanging from a string, is lowered into a vessel of cross-sectional area A, containing a liquid of density `d_(2) (d_(2) lt d_(1))` until it is fully immersed. The increase in pressure at the bottom of the vessel isA. `(Md_(2)g)/(d_(1)A)`B. `(Mg)/(A)`C. `(Md_(1)g)/(d_(2)A)`D. zero |
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Answer» Correct Answer - A Increase in pressure `=("Force exetted by the cylinder on the liquid ")/A=("Upthrust")/A` |
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| 25. |
A tank 5 m high is half filled with water and then is filled to top with oil of density `0.85 g//cm^3` The pressure at the bottom of the tank, due to these liquids isA. `1.85 g//cm`B. `89.25 g//cm`C. `462.5 g//cm`D. `500 g//cm` |
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Answer» Correct Answer - C |
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| 26. |
Some liquid is filled in a cylindrical vessel of radius R. Let `F_(1)` be the force applied by the liquid on the bottom of the cylinder. Now the same liquid is poured into a vessel of uniform square cross-section of side R. Let `F_(2)` be the force applied by the liquid on the bottom of this new vessel. (Neglect atmosphere pressure). ThenA. `F_(1)=piF_(2)`B. `F_(1)=F_(2)/pi`C. `F_(1)=sqrt(pi)F_(2)`D. `F_(1)=F_(2)` |
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Answer» Correct Answer - D Force on base = mg of liquid implies `F_(1)=F_(2)` |
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| 27. |
What will be the length of mercury column in a barometer tube, when the atmospheric pressure is 76 cm of mercury and the tube is inclined at an angle of `30 ^(@)` with the horizontal direction ? |
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Answer» Here, h=76 cm `theta=30^(@)` If l is the length of mercury column is the barometer tube then `(h)/(l)=sin 30^(@) implies (76cm)/(l)=(1)/(2)` `implies " " l=2xx76=152 cm` |
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| 28. |
A capillary tube is attached horizontally to a constant pressure head arrangement. If the radius of the capillary tube is increased by `10%`, then the rate of flow of the liquid shall change nearly byA. `+10%`B. `+46%`C. `-10%`D. `-40%` |
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Answer» Correct Answer - B `V=(Ppir^(4))/(8etal) implies (V_(2))/(V_(1)) = ((r_(2))/(r_(1)))^(4) implies V_(2) = V_(1) ((110)/(100))^(4) = V_(1)(1.1)^(4) = 1.4641 V` `(DeltaV)/(V) = (V_(2)-V_(1))/(V) = (1.4641V-V)/(V) = 0.46 " or " 46%`. |
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| 29. |
A capillary tube of area of cross-section A is dipped in water veritcally. Calculate the amount of heat evolved as the water rises in the capillary tunbe upto height h. The density of water is `rho`A. `(Arhogh^(2))/(2)`B. `Agh^(2)rho`C. `2Agh^(2)rho`D. None of these |
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Answer» (a) Heat evolved =Potential energy of water `=(mgh)/(2)=(rhoAhgh)/(2)=(rhoAh^(2)g)/(2)` |
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| 30. |
A fixed thermally conducting cylinder has a radius R and height `L_0`. The cylinder is open at the bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is `P_0`. The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then beA. `P_(0)`B. `(P_(0))/(2)`C. `(P_(0))/(2) + (Mg)/(piR^(2))`D. `(P_(0))/(2) - (Mg)/(piR^(2))` |
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Answer» Correct Answer - A Since it is open from the top, the pressure will be `P_(0)` |
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| 31. |
A fixed thermally conducting cylinder has a radius R and height `L_0`. The cylinder is open at the bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is `P_0`. The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then beA. `P_(0)`B. `(P_(0))/2`C. `(P_(0))/2-(Mg)/(piR^(2))`D. `(P_(0))/2-(Mg)/(piR^(2))` |
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Answer» Correct Answer - A When the piston is pulled out slowly, the pressure drop produced inside the cylinder is almost instantaneously neutralized by the air entering from outside into the cylinder (through the small hole at the top). Therefore, the pressure inside the cylinder is `P_(0)` throughout the slow pulling process. |
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| 32. |
A small spherical monoatomic ideal gas bubble `(gamma=5//3)` is trapped inside a liquid of density `rho` (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is `T_0`, the height of the liquid is H and the atmospheric pressure `P_0` (Neglect surface tension). As the bubble moves upwards, besides the buoyancy force the following forces are acting on itA. only the force of gravityB. The force due to gravity and the force due to the pressure of the liquidC. The force due to gravity, the force due to the pressure of the liquid and the force due to viscosity of the liquidD. The force due to gravity and the force due to viscosity of the liquid |
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Answer» Correct Answer - D As the bubble moves upwards besides the buoyancy force (cause of which is pressure difference) only force of gravity and force of viscosity will act. |
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| 33. |
A ball of radius r and density r falls freely under gravity through a distance h before entering water. Velocity of ball does not change even on entering water. If viscosity of water is `eta` the value of h is given by` A. `2/9r^(2)((1-rho)/(eta))g`B. `(2)/(81)r^(2)(rho - 1)/(eta) g`C. `2/(81) r^(4)((rho - 1)/(eta))^(2)g`D. `2/9r^(4)((rho - 1)/(eta))^(2) g` |
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Answer» Correct Answer - C |
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| 34. |
A ball of radius r and density r falls freely under gravity through a distance h before entering water. Velocity of ball does not change even on entering water. If viscosity of water is `eta` the value of h is given byA. `(2)/(9) r^(2)((l-rho)/(eta))g`B. `(2)/(81)r^(2)((rho-1)/(eta))g`C. `(2)/(81)r^(4)((rho -1)/(eta))^(2)g`D. `(2)/(9)r^(4)((rho -1)/(eta))^(2)g` |
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Answer» Correct Answer - C Velocity of ball when it strikes the water surface `v=sqrt(2gh)" "…..(i)` Terminal velocity of ball inside the water `v=(2)/(9)r^(2)g((rho-1)/(eta))" "……(ii)` Equating (i) and (ii) we get `sqrt(2gh)=(2)/(9) (r^(2)g)/(eta) (rho-1) implies h=(2)/(81) r^(4) ((rho-1)/(eta))^(2)g` |
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| 35. |
A spherical ball of radius `3.0xx10^(-4) m` and density `10^(4) kg//m^(3)` falls freely under gravity through a distance h before entering a tank of water. If after entering the water the velocity of the ball does not change, find h. Viscosity of water is `9.8xx10^(6) N-s//m^(2)`. |
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Answer» Correct Answer - A::C Before entering the water the velocity of ball is `sqrt(2g)`. If after entering the water this velocity does not change then this value should be equal to the terminal velocity. Therefore, `sqrt(2gh)=2/9 (r^(2)(rho-sigma)g)/(eta)` `:. h={(2)/(9) (r^(2)(rho-sigma)g)/(eta)}^(2)` `=(2)/(81) xx(r^(4)(rho-sigma)^(2)g)/(eta^(2))` `=(2)/(81)xx((3xx10^(-4))^(4)(10^(4)-10^(3))^(2)xx9.8)/(9.8xx10^(-6))^(2)` `=1.65xx10^(3)m`. |
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| 36. |
A small sphere falls from rest in a viscous liquid. Due to friction, heat is produced. Find the relation between the rate of produced heat and the radius of the sphere at terminal velocity. |
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Answer» Correct Answer - D Terminal velocity `(v_T)=(2(r^2)g)/(9 eta)(rho_S-rho_L)` and viscous force `F=6 pi eta rv_(T)` Rate of production of heat (Power): as viscous force is the only dissipative force. Hence, `(dQ)/(dt)=Fv_(T)=(6 pi eta rv_(T))(v_(T))` `(6 pi etarv_(T)^2)` `=6 pi etar{(2)/(9)(r^(2)g)/(eta) (rho_(S)-rho_(L))}^(2)` `=(8 pi g^(2))/(27 eta) (rho_(S)-rho_(L))^(2)r^(5)` or `(dQ)/(dt)prop r^(5)`. |
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| 37. |
Assertion : Water flows faster than honey. Reason : The coefficient of viscosity of water is less than honey.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true and the reason is not the correct explanation of the assertion.C. If assertion ture but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - A |
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| 38. |
An air bubble rises from the bottom of a lake of large depth. The rising speed of air bubble willA. go on increasing till it reaches surfaceB. go on decreasing till it reaches surfaceC. increases in two beginning, then will become constantD. be constant all throughout. |
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Answer» Correct Answer - C Velocity becomes constant terminal velocity is attanined. |
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| 39. |
A man is sitting in a boat which is floating in a pond. If the man drinks some water from the pond, the level of water in the pond decreases.A. increasesB. decreasesC. remains unchangedD. increases are decreases depends upon the weight of man |
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Answer» Correct Answer - C The volume of water a man drinks, will be equal to additional volume of water displaced by the boat , during to which the level of water will remain same. |
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| 40. |
The vertical sections of the wing of a fan are shown. Maximum upthrust is inA. B. C. D. |
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Answer» Correct Answer - A Air streams have to cover a larger distance about the surface as compared to the air streams below the surface. So, velocity above the surface is large and consequently the pressure is less. On the other hand velocity below the surface is small and consequent the pressure is large. This explains maximum upthrust in figure. |
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| 41. |
Two vessels A and B of cross sections as shown contain a liquid up to the same height. As the temperature rises, the liquid pressure at the bottom (neglecting expansion of the vessels ) will : A. Increase in A , decrease in BB. increase in B , decrease in AC. Increase in both A and B but more in AD. Increase in both A and B equally |
| Answer» Correct Answer - A | |
| 42. |
Fig. represents vertical sections of four wings moving horizontally in air. In which case the force is upwardsA. B. C. D. |
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Answer» Correct Answer - A |
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| 43. |
Two unequal soap bubbles are formed one on each side of a tube closed in the middle by a tap. What happens when the tap is opened to put the two bubbles in communication?A. No air passes in any direction as the pressures are the same on two sides of the tapB. Larger bubble shrinks and smaller bubble increases in size till they bexome equal in sizeC. Smaller bubble gradually collapses and the bigger one increases in sizeD. None of the above |
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Answer» Correct Answer - C Pressure inside a soap bubble is given by, `P=P_(0)+(4T)/(r)` ltbRgt So, pressure inside a smaller soap bubble will be more and air will flow from smaller drop to bigger drop. |
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| 44. |
A metal sphere of radius 1 mm and mass 50 mg falls vertically in glycerine. Find (a) the viscous force exerted by the glycerine on the sphere when the speed of the sphere is 1 cm s-1, (b) the hydrostatic force exerted by the glycerine on the sphere and (c) the terminal velocity with which the sphere will move down without acceleration. Density of glycerine `=1260kgm^-3` and its coefficient of viscosity at room temperature = 8.0 poise. |
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Answer» Correct Answer - A::B::C (a) `F_(r)=6pietarv` `=6 pi (0.8)(10^(-3))(10^(-2))` `=1.5xx10^(-4)N` (b) Hydrostatic force= upthrust `=((4)/(3) pir^(3))rho g=(4)/(3) pi (10^(-3))^(3)xx1260xx9.8` `=5.2xx10^(-5) N` (c ) At terminal velocity `w=`Upthrust+viscous force or, `(50xx10^(-3)xx9.8)=(5.2xx10^(5)` `+6pi (0.8)(10^(3))v_(T)` Solving we get `v_(T)=32.5 m//s`. |
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| 45. |
A cubical metal block of edge 12 cm floats in mercry with one fifth of the height inside the mercury. Water poured till the surface of the block is just immersed in it. Find the height of the block is just immersed in it. Find the height of the water column to be poured. Specific gravity of mercury =13.6. |
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Answer» Correct Answer - A::C::D Given `x=12cm ` =length of the edge of the block `P_(Hg)=13.6gm/cc `Given that initially `1/5` of block is inside mercury. `Let Pbrarr density of block in gm/cc. `:(x)^3xxrho_bxxg=(x)^2x(x/5)xxrho_(Hg)xxg` `rarr (12)^3xxrho_bxxg=(12^2xx12/5xx13.6` `rho_b=13.6/g gm/cc` After water pured let x=height of water column. `V_b=V_(Hg)+V_w(12)^3` Where `V_(Hg)` and `V_w` are volumes of block inside mercury and water respectively. `:. (V_bxxrho_bxxg)=(V_(Hg)xxrho(Hg)xg)+(V_wxxrho_wxxg)` `rarr (V_(Hg)xxV_m)xx13.6/5` `=V_(Hg)xx13.6xxV_wxx1` `rarr (12)^3xx13.6/5=(12-x)xx13.6+(x)` `rarr 12.6x=13.6(12-12/5)` `=(13.6)xx(9.6)` `rarr x=((9.6)xx(13.6))/(12.6)=10.4cm |
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| 46. |
A cubical block of copper of side `10cm` is floating in a vessel containing mercury. Water is poured into the vessel so that the copper block just gets submerged. The height of water column is (`rho_(Hg) = 13.6 g//c c , rho_(Cu) = 7.3 g//c c , rho_(water) = 1gm//c c)`A. 1.25 cmB. 2.5 cmC. 5 cmD. 7.5 cm |
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Answer» Correct Answer - C Let h =height to of water column then `rho_(w) gh+rho_(Hg)g(10-h)=rho_(Cu)g^(10)` `rArr h+13.6(10-h)=73` `rArr 63=12.6 h rArr h=5 cm ` |
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| 47. |
body of density `rho` is dropped from height h into a liquid having density `sigma(sigmagt rho)` .If the body just touches the of the container, then the distance of fallen would be proprotional to (Neglect viscous forces)A. `(h)/(sigma-rho)`B. `(h)/(sigma+rho)`C. `hxx(sigma-rho)`D. `(hrho)/(sigma-rho)` |
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Answer» Bouyabt force `Fprop(sigma-rho)` `:.` Decleration `prop(sigma-rho)rArra=-k(sigma-rho)` Now, as initial velocity , `u=sqrt(2gh)` and final velocity .v=o Distance fellen by the body , `s=(2gh)/(2xxk(sigma-rho))" " [:. s=(u^(2))/(2a)]` `rArr" " sprop(h)/(sigma-rho)` |
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| 48. |
Water rises in a vertical capillary tube up to a length of 10 cm. If the tube is inclined at `45^@`, the length of water risen in the tube will beA. 10 cmB. `10sqrt(2)cm`C. `5sqrt(2) cm`D. 20 cm |
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Answer» Correct Answer - B When capillary tube is tilled through an angle `theta`, then length of water risen in capallary. `l=h/(cos theta)` Here, `h=10 cm, theta=45^(@)` `l=10/(cos 45^(@)) =10 sqrt(2)` |
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| 49. |
For a stream line flow of water following statements are given below `(a)` Two streamlines do not intersect each other. `(b)` streamlines must be straight. `(c )` streamlines flow is more likely for liquids with low density and height viscosity. `(d)` streamlines flow is more likely for liquids with high density and low viscosity.A. `(a)` and `(b)` are true.B. `(a)` `(b)` and `(c )` are true.C. `(a)` and `(c )` are true.D. All are true. |
| Answer» Reason : Streamlines may be straight or curved. | |
| 50. |
Calculate for the rise of water in a capillary tube when kept veritcal in water whose radii is 1/4th of that capillry tube which when kept veritcal water rise in it upto a height of 3 mmA. 12 mmB. 10 mmC. 4 mmD. 3 mm |
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Answer» `h=(2Scostheta)/(rgp)` `rArr " " hr=(2Scostheta)/(rhog)` For the experiment `S,theta,rho,g` are same `hr=(2Scostheta)/(rhog) ` =constant `h_(1)r_(1)=h_(2)r_(2)` `h_(2)=((r)/(r//4))3=13 mm` |
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