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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.
| 501. |
A cylinderical vessel is filled with water up to height H. A hole is bored in the wall at a depth h from the free surface of water. For maximum range h is equal toA. `H`B. `3H//4`C. `H//2`D. `H//4` |
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Answer» Correct Answer - C `x = 2sqrt(h(H - 1))` for `x_(max) (dx)/(dh) = 0` or `h = (H)/(2)` |
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| 502. |
A small wooden ball of density `rho` is immersed in water of density `sigma` to depth h and then released. The height H above the surface of water up to which the ball will jump out of water isA. `(sigmah)/(rho)`B. `((sigma)/(rho)-1)h`C. hD. zero |
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Answer» Correct Answer - B Work done against Buyonancy = gain of gravitational potential energy `(B-W)h=WH` `H=((B)/(W)-1)h" "implies H=((sigma)/(rho)-1)h` |
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| 503. |
There is a metal cube inside a block of ice which is floating on the surface of water. The ice melts completely and metal falls in the water. Water level in the container A. RisesB. FallsC. Remains sameD. Nothing can be concluded |
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Answer» Correct Answer - B Now metal cube sink so water level falls. |
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| 504. |
A solid sphere of radius 5 cm floats in water. If a maximum load of 0. kg can be put on it withot wetting the load find the specific gravityof the material of the sphere. |
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Answer» Given, r=5cm, m=1 kg Here, `w_1+W_2=U` `rarr mg+Vxxrho_sxxg=vxxrho_wxxg` `[wehre rho_s=density of sphere in gm/cc]` `rarr(0.1)xx10^3+(4/3)xxpixx(5)^3xxrho_s` =(4/3)xxpixx(5)^3xx1` `rarr 100=(4/3)xxpixx125(1-rho_s)` `rarr 1-rho_s=(3xx1000/(4xxpixx125)=0.19` ltbr.gt `rarr rho_s=1-(0.19)` =0.81gm/cc=0.8gm/cc` So, specific gravity of the material is 0.8. |
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| 505. |
Find the ratio of the weights as measured by a spring balance, of a 1 kg block of iron nd a 1kg block of wood. Density of iron `=7800 kg m^-3` density of wood `=800 kg m^-3 and density of air =1.293 kgm^-3`. |
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Answer» Correct Answer - A Net weight of iron block `w_1=mg-V_(1rho_(air))xxg` `[m-(m/rho_1)rho_(air)]g` `=[1-(1/7800)xx1.293]xx(9.8)` Again net weight of wood `=W_w=mg-V_w.rho_(air)g` `=[m-(m/rho_w)rho_(air)]g` `rarr (1-1/800xx1.293)9.8` `rarr :. W_1/W_2=9.8((7800-1.293)/7800)/(9.8(800-1.293)/800)` `=(7800-1.293)/(800-1.293)x8/78=1.0015` |
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| 506. |
An ice cube of edge a is placed in an empty cylindrical vessel of radius 2a. Find the edge (in cm) of ice cube when it just leaves the contact with the bottom assuming that ice melts uniformly maintaining its cubical shape. Take a = 12 `pi` cm (Ice is lighter than water) |
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Answer» Correct Answer - B::C Let xrarr edge of ice block. when it just leaves contact with the bottom of the glass, hrarr Height of water method from ice. here W=U `rarr (x)^3xxrho_(ice)xxg=(x)^2xxhxxrho_wxxg` `rarr h=(0.9)x` Again volume of water formed from melting of ice is given by `(4)^3-(x)^3=pixx(r)^2xxh-x^I2h` [because amont of water `=rho(r_2-x_2)hrarr (4)^3-(x)^3=pixx(3)^2xxh-x^2h]` utting `h=0.9x` rarr (4)^3-(x)^3pixx(3)^2xx(0.9)x` solving the above equation we can get `x=2.26cm` |
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| 507. |
Some question (Assertion-Reason type) are given below. Each question contains STATEMENT-1(Assertion) and STATEMENT-2(Reason) . Each question has 4 choices (A),(B),(C ) and (D) out of which ONLY ONE is correct. So select the correct choice : STATEMENT-1 1kg of cotton fibre will weight less lens in air when made more fluffy. STATEMENT-2 Weight of air in cotton will cancel out with the force of extra buoyancy acting on it.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - D Weight of fluffy cotton will include the weight of air trapped in it. It will however exactly cancel out by the extra buoyant force acting on the trapped air. |
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| 508. |
A horizontal jet of water coming out of a pipe of area of cross-section `20cm^(2)` hits a vertical wall with a velocity of `10ms^(-1)` and rebounds with the same speed. The force exerted by water on the wall is .A. the thrust exerted by the water on the wall will be doubledB. the thrust exerted by the water on the wall will be four timesC. the energy lost per second by water strikeup the wall will also be four timesD. the energy lost per second by water striking the wall be increased eight times. |
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Answer» Correct Answer - B::D Thrust = Rate of charge of linear momentum Thrust = mv where m is mass striking per sec. `=rho avxxvimplies " " "Thrust" = rhoav^(2)` Energy lost for second `=(1)/(2)mv^(2)" "=(1)/(2)rho av.v^(2)=(1)/(2)rhoav^(3)` so option (B,D) are correct. |
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| 509. |
Some question (Assertion-Reason type) are given below. Each question contains STATEMENT-1(Assertion) and STATEMENT-2(Reason) . Each question has 4 choices (A),(B),(C ) and (D) out of which ONLY ONE is correct. So select the correct choice : STATEMENT-1 If a body is floating in a liquid, the density of liquid is always greater than the density of solid. STATEMENT-2 Surface tension is the property of liquid surface.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - D A body can float into liquid even if the density of liquid is less then the density of solid. |
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| 510. |
Some question (Assertion-Reason type) are given below. Each question contains STATEMENT-1(Assertion) and STATEMENT-2(Reason) . Each question has 4 choices (A),(B),(C ) and (D) out of which ONLY ONE is correct. So select the correct choice : STATEMENT-1 Steel is more elastic than rubber. STATEMENT-2 When same deformation is produced in two identical bodies of these material greater restoring force develops in the steel body.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - A | |
| 511. |
Some question (Assertion-Reason type) are given below. Each question contains STATEMENT-1(Assertion) and STATEMENT-2(Reason) . Each question has 4 choices (A),(B),(C ) and (D) out of which ONLY ONE is correct. So select the correct choice : STATEMENT-1 In the steady flow of an ideal fluid, the velocity at any point is same for different fluid particles. STATEMENT-2 Steady fluid flow is the unaccelerated fluid flow.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - C | |
| 512. |
Some question (Assertion-Reason type) are given below. Each question contains STATEMENT-1(Assertion) and STATEMENT-2(Reason) . Each question has 4 choices (A),(B),(C ) and (D) out of which ONLY ONE is correct. So select the correct choice : STATEMENT-1 Two identical beakers contains water to the same level. A wooden block is floating in one of the beakers. The total weight of both beakers is same. STATEMENT-2 Volume of the displaced water is equal to the volume of the block.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - C | |
| 513. |
Some question (Assertion-Reason type) are given below. Each question contains STATEMENT-1(Assertion) and STATEMENT-2(Reason) . Each question has 4 choices (A),(B),(C ) and (D) out of which ONLY ONE is correct. So select the correct choice : STATEMENT-1 Falling raindrops acquire a terminal velocity. STATEMENT-2 A constant force in the direction of motion and a velocity dependent force opposite to the direction of motion, always result in the acquisition of terminal velocity.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - A | |
| 514. |
Some question (Assertion-Reason type) are given below. Each question contains STATEMENT-1(Assertion) and STATEMENT-2(Reason) . Each question has 4 choices (A),(B),(C ) and (D) out of which ONLY ONE is correct. So select the correct choice : STATEMENT-1 : Ratio of normal stress to volumetric strain is bulk modulus of given gas. STATEMENT-2 : Compressibility is the reciprocal of bulk modulus.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - B Force exerted by liquid on vessel is force exerted on bottom + force exerted on the walls. Force on bottom in three cases is same but force on walls is different. |
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| 515. |
Some question (Assertion-Reason type) are given below. Each question contains STATEMENT-1(Assertion) and STATEMENT-2(Reason) . Each question has 4 choices (A),(B),(C ) and (D) out of which ONLY ONE is correct. So select the correct choice : STATEMENT-1 A rain drop after falling through a certain distance attains a constant velocity. STATEMENT-2 The viscous force for spherical body is proportional to its speed. Hence after falling through a certain distance viscous drag and buoyant forces balance the gravitational force.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - A | |
| 516. |
Statement-1: As wind flows left to right and a ball is spinned as shown, there will be a lift of the ball. Statement-2: Decrease in velocity of air below the ball, increases the pressure more than that above the ball. A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Correct Answer - A | |
| 517. |
Two drops of the same radius are falling through air with a steady velcoity of `5 cm s^(-1).` If the two drops coalesce, the terminal velocity would beA. `10` cm per secB. `2.5` cm per secC. `5 xx (4)^(1//3)` cm per secD. `5 xx sqrt(2)`1 cm per sec |
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Answer» Correct Answer - C |
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| 518. |
A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference P . The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled isA. `P`B. `(3P)/(4)`C. `(P)/(2)`D. `(P)/(4)` |
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Answer» Correct Answer - D |
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| 519. |
We have two (narrow) capillary tubes T and T . Their lengths are l and l and radii of cross-section are r and r respectively. The rate of flow of water under a pressure difference P through tube T is `8 cm 3// sec`. If `l = 2l` and `r = r` what will be the rate of flow when the two tubes are connected in series and pressure difference across the combinatin is same as before `(= P)`A. `4 cm//sec`B. `(16//3) sec`C. `(8//17) cm//sec`D. None of these |
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Answer» Correct Answer - B |
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| 520. |
Two capillary tubes of lengths in the ratio `2 : 1` and radii in the ratio `1 : 2` are connected in series. Assume the flow of the liquid through the tube is steady. Then, the ratio of pressure difference across the tubes isA. `1 : 8`B. `1 : 16`C. `32 : 1`D. `1 : 1` |
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Answer» Correct Answer - C We know that the pressure `p alpha (r^(4))/(l)` `"Let" l_(1)=2l, l_(2)=l and r_(1)=r, r_(2)=2r` `"Now", (p_(1))/(p_(2))=(r_(1)^(4))/(l_(1))xx(l_(2))/(r_(2)^(4))Rightarrow (p_(1))/(p_(2))=(r^(4))/(2l)xx(l)/((2r)^4)` `(p_(1))/(p_(2))=(1)/(2xx(2)^(4)) Rightarrow (p_(1))/(p_(2))=(1)/(32)` |
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| 521. |
The height to which a cylindrical vessel be filled with a homogenous liquid, to make the average force with which the liquid presses the side of the vessel equal to the force exerted by the liquid on the bottom of the vessel, is equal to.A. Half of the radius of the vesselB. Radius of the vesselC. One-fourth of the radius of the vesselD. Three-fourth of the radius of the vessel |
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Answer» Correct Answer - B |
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| 522. |
The velocity of kerosene oil in a horizontal pipe is `5 m//s`. If `g = 10m//s^(2)` then the velocity head of oil wlill beA. `1.25 m`B. `12.5 m`C. `0.125 m`D. `125 m` |
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Answer» Correct Answer - A |
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| 523. |
In the following fig., the flow of liquid through a horizontal pipe is shown. Three tubes A, B and C are connected to the pipe. The radii of the tubes `A,B and C` at the junction are respectively `2 cm, 1 cm and 2 cm`. It can be said that the A. Height of the liquid in the tube A is maximumB. Height of the liquid in the tubes A and B is the sameC. Height of the liquid in all the three tubes is the sameD. Height of the liquid in the tubes A and C is the same |
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Answer» Correct Answer - D |
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| 524. |
A solid shell loses half, its weight in water. Relative density of shell is 0.5 what fraction of its volume is hollow ?A. `(3)/(5)`B. `(2)/(5)`C. `(1)/(5)`D. `(4)/(5)` |
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Answer» Correct Answer - A Loss of weight = Upthrust `W/2" = total volume"xxrho_(w)xxg` `or" "((V-V_(C))5xxrho_(w)xxg)/2=Vxxrho_(w)xxg" "(V_(C)"=Volume of cavity")` `or" "V-V_(C)=2/5VrArrV_(C)=3/5V` |
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| 525. |
A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 20mm and 1mm respectively. The upper end of the container is open to the atmosphere. If the piston is pushed at a speed of `5mms^-1`, the air comes out of the nozzle with a speed ofA. `0.1ms^(-1)`B. `1ms^(-1)`C. `2ms^(-1)`D. `8ms^(-1)` |
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Answer» Correct Answer - C `A_(1)V_(1) = A_(2)V_(2), A_(1) = 400 A_(2)` `400 (5 xx 10^(-3)) = V_(2), rArr V_(2) = 2m//s` |
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| 526. |
The excess pressure inside an air bubble of radius `r` just below the surface of water is `P_(1)`. The excess pressure inside a drop of the same radius just outside the surface is `P_(2)`. If `T` is surface tension thenA. `P_(1)=2P_(2)`B. `P_(1)=P_(2)`C. `P_(2)=2P_(1)`D. `P_(2)=0,P_(1)ne0` |
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Answer» Correct Answer - B Excess pressure inside a bubble just below the surface of water `P_(1)=(2T)/(r )` and excess pressure inside a drop `P_(2) = (2T)/(r ) " "therefore P_(1)=P_(2)` |
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| 527. |
Water is following in a river. If the velocity of a layer at a distance 10 cm from the bottom is 20n cm/s. Find the velocity of layer at a height of 40 cm from the bottom.A. 10 m/sB. 20 m/sC. 30 m/sD. 80 m/s |
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Answer» `:. F=-etaA(dv)/(dx)` `:. " " xpropv` `(v_(2))/(v_(1))=(x_(2))/(x_(1))or (v_(2))/(20) =(40)/(10)` `v_(2)=80 cm//s` |
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| 528. |
The realativen humidity on day when partial pressure of water vapour is `0.12xx10^(6)` Pa at `12^(@)` C is (Take vapour pressure of water at this temperatue as `0.06xx10^(5) Pa)`A. 0.7B. 0.4C. 0.75D. 0.25 |
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Answer» Relative humidity at a given temperature (R) `=("partial pressue of water pressue")/("vapour pressure of water") ` `(0.012xx10^(5))/(0.016xx10^(5))` `=0.75=75%` |
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| 529. |
What are the factors, which effect the atmospheric pressure at a place ? |
| Answer» Height of the atmospherer, density of the atmosphere and the acceleration due to gravity | |
| 530. |
A ball of relative density 0.8 falls into water from a height of 2 m. The depth to which the ball will sink is (neglect viscous forces)A. 8 mB. 2 mC. 6 mD. 16 m |
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Answer» Correct Answer - A Velocity of ball just before striking water surface, `v=sqrt(2gh)` or `v=sqrt(4g) m s^(-1) " " (h=2m)` or `a=(V rho_(w)g-Vrho g)/(V rho)=((1-0.8)/(0.8))g=g//4` Now, depth to which ball will sink is `d=(v^(2))/(2a)=(4g)/(2g//4)=8m` |
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| 531. |
A ball whose density is `0.4 xx 10^3 kg//m^3` falls into water from a height of 9 cm. To what depth does the ball sink ?A. 9 cmB. 6 cmC. 4.5 cmD. 2.25 cm |
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Answer» Correct Answer - B (b) The velocity of ball before entering the water surface `v = sqrt(2gh) = sqrt(2gxx9)` When ball enters into water, due to upthrust of water the velocity of ball decreases (or retarded) The retardation, a = `(apparent weight)/(mass of hall)` `(V(rho = simga)g)/(V rho) = ((rho = sigma)/(rho))g` `= ((0.4 - 1)/(0.4))xxg =- (3)/(2)g` If h be the depth upto which ball sink, then, `0 - v^2 = 2xx((-3)/(2)g)xxh` `rArr 2g xx 9 = 3gh :. h = 6 cm. |
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| 532. |
A ball whose density is `0.4zxx1^(3) kg//m^(3)` falls into water from a height of 9 cm.To what depth does the balll sink?A. 9 cmB. 6 cmC. `4.5 cm`D. `2.25 cm` |
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Answer» The velocity of ball before entering the water surface `v=sqrt(2gh)=sqrt(2gxx9)` When ball enters into water, due to upthrust of water, the velocity of ball decreases (of retarded) The retardaion, `a=("apparent weight")/("mass of ball")` `(V(rho-sigma)g)/(Vrho)=((rho-sigma)g)/(rho)` `=((0.4)-1)/(0.4)g=-(3)/(2)g` If h be the depth upto which ball sin k, then `o-v^(2)=2xx((-3)/(2)g)xxh` `rArr " " 2g xx9=3gh` `:. " " h=6 cm ` |
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| 533. |
A body weighs `50 g` in air and `40 g` in water. How much would it weigh in a liquid of specific gravity `1.5`A. 65 gB. 45 gC. 30 gD. 35 g |
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Answer» Correct Answer - D Loss of weight in water = upthrust in water `50 g - 40 g = rho_omega Vg rArr V = 10 cm^3` Loss of weight in liquid `= rho_l Vg = 1.5 rho_omega xx 10 g = 15 g` Weight in liquid `= 50 g - 15 g = 35 g`. |
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| 534. |
The work done in increasing the size of a soap film from `10cmxx6cm` to `10cmxx11cm` is `3xx10^-4` Joule. The surface tension of the film isA. `1.5xx10^(-2)Nm^(-1)`B. `3.0xx10^(-2) Nm^(-1)`C. `6.0xx10^(-2)Nm^(-1)`D. `11.0xx10^(-2) Nm^(-1)` |
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Answer» Correct Answer - B `A_(1) = 10xx6 = 60cm^(2) = 60xx10^(-4)m^(2), A_(2) = 10xx11 = 110 cm^(2) = 110xx10^(-4) m^(2)` As the soap film has two free surfaces `therefore W = T xx2Delta A` `implies W = Txx2xx(A_(2)-A_(1)) implies T (W)/(2xx50xx10^(-4)) =(3xx10^(-4))/(2xx50xx10^(-4))= (3xx10^(-2)) N//m` |
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| 535. |
A glass beaker of mass 400 kg floats in water with the open end just touching the surface of water and half of the beaker filled with water. The inner volume of the beaker is `500 cm^3` What is the density of the meterial of the beaker ?A. `1.52 g cm^(-3)`B. `2.67 g cm^(-3)`C. `3.01 g cm^(-3)`D. `3.87 g cm^(-3)` |
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Answer» Correct Answer - B (b) Mass of water in the beaker = 250 xx1 = 250 g Mass of beaker + Mass of water in it = mass of the volume fo water displaced = 400 + 250 = 650g So volume of water displaced ` =650 cm^3` As the inner volume ` =500 cm^3` So volume of the meterial of beaker ` =650 - 500 = 150 cm^3` `:. Density = (400)/(150) = 2.67 g//cm^3` |
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| 536. |
A piece of solid weighs 120 g in air, 80 g in water and 60 g in a liquid, then the relative density of the solid, and that fo liquid areA. 3,2B. `2,(4)/(3)`C. `3,(3)/(2)`D. 4,3 |
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Answer» Correct Answer - C (c ) Relative density of solid ` =(120)/(120 - 80) = 3` Relative density of liquid `= (120 -60)/(120 -80) = (60)/(40) = (3)/(2).` |
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| 537. |
A piece of solid weighs 120 g in air ,80 g in water and 60 kg in a liquid . The relative density of the solid and that of the liquid are respectivelyA. `2/3`B. `4/5`C. `13/15`D. `15/13` |
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Answer» Correct Answer - A Specific gravity of body = `150/30 = 5` Volume of body = `150./5 = 30 cm^(3)` Now, `(150-130)g = 30 xx rho_(1) xx g` `implies rho_(l) = 20/30 = 2/3`. |
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| 538. |
An irregular piece of metal weighs 10.00g in air and 8.00g whaen submerged in water (a) Find the volume of the metal and its density. (b) If the same piece of metal weighs 8.50g when immersed in a particular oil. What is the density of the oil? |
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Answer» Correct Answer - A::B::C (a) Relative density of metal `=("weight in air" )/("change in weight in water" )=(10/2)=5` `:.` Density of metal`=5rho_(w)=5000kg//m^(3)` Now, volume `=("mass")/("density")` `=(10xx10^(-3))/(5000)` `2xx10^(-6)m^(3)` (b) change in weight upthrust on `100%` volume of solid or, `Delta w =V_(s)rho_(l)g` `:. Delta w prop rho_(l)` `Delta w_(l)/(Delta w_(w))=(rho_(l))/(rho_(w))` or,`rho_(l)= ((Deltaw_(i))/(Deltaw_(w))) rho_(w) =((1.5)/(2)) (1000)` `= 750 kg//m^(3)` |
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| 539. |
A massless conical flask filled with a liquid is kepth on t a table in a vacuum the force exerted by the liquid on the bse of the flask is `W_(1)`. Th force exerted by the flask on the table is `W_(2)`A. `W_1 =W_2`B. `W_1 gtW_2`C. `W_1 ltW_2`D. The force exerted by the liquid on the walls of the flask is `(W_1 -W_2)` |
| Answer» Correct Answer - B::C | |
| 540. |
A massless conical flask filled with a liquid is kept on a table in a vacuum. The force exerted by the liquid on the base of the flask is `W_1`. The force exerted by the flask on the table is `W_2`. (i) `W_1 = W_2` (ii) `W_1 gt W_2` (iii) `W_1 lt W_2` (iv)The force exerted by the liquid on the walls of the flask is `(W_1 - W_2)`. .A. (i),(iv)B. (ii),(iv)C. (iii),(iv)D. (iii) |
| Answer» Correct Answer - B | |
| 541. |
A hollow sphere of volume V is floating on water surface with half immersed in it. What should be the minimum volume of water poured inside the sphere so that the sphere now sinks into the water ?A. `V//2`B. `V//3`C. `V//4`D. `V` |
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Answer» Correct Answer - A |
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| 542. |
Solve the prievious problem if the lead piece is fastened on the top surface of the block and the block is to float with its upper surface just dipping into water. |
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Answer» `m_w=200g, rho_w=0.8gm/cc, `rho_(pb)=11.3gm/cc` `Again Mg=w` `rarr (m_w+m_(pb))g=V_wxxrhoxxg` [rho=density of water] `rarr 200+m_(pb)=200/0.8x1` `rarrm_(pb)=250-200=50gm` |
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| 543. |
A pump is designed as a horizontal cylinder with a piston of areal `A` and and outlet orifice of areal a arranged mear the cylinder axis. Find the velocity of out flow of the liquid from the pump if the piston moves with a constant velocity under the action of a constant force F. The density of the liquid is `rho`.A. `sqrt((F)/(Arho))`B. `sqrt((2F)/(A rho))`C. `sqrt((A rho)/(F))`D. `sqrt((A rho)/(2F))` |
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Answer» Correct Answer - B `Deltap=(F)/(A)`. Difference in pressure energy is equal to difference in kinetic energy. `therefore (F)/(A)=(1)/(2)rho v^(2) " or" v=sqrt((2F)/(rho A))` |
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| 544. |
A cylinder is fitted with a piston, beneath which is a spring, as in the figure. The cylinder is open at the top. Friction is absent. The spring is `3600 N//m`. The piston has a negligible mass and a radius of 0.025m (a) when air beneath the piston is conpletely pumped out, how much does the atmospheric pressure cause the spring to compress? (b) How much work does the atmospheric pressure do in compressing the spring? |
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Answer» Correct Answer - A::B::C::D (a) `p_(0)A=kx` `:. x=(p_(0)A)/(k) ((p_(0))(pir^(2)))/(k)` `=((1.0xx10^(5))(pi)(0.025)^(2))/(3600)` `=0.055m=5.5 cm`. (b) Work done by atmospheric pressure `W=(1)/(2)kx^(2)=(1)/(2) (3600)(0.055)^(2)` `=5.445 J` |
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| 545. |
Assume that a drop of liquid evaporates by decreases in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is `rho` and L is its latent heat of vaporization. |
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Answer» Correct Answer - A Decreses in surface energy=heat required in vaporization. `:. T(dS)=L(dm)` `:. T(2)(4 pi r)dr=L(4 pi r^(2)dr)rho` `:. r=(2T)/(rho L)` The correct option is (d). |
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| 546. |
A capillary tube whose inside radius is `0.5mm` is dipped in water having surface tension `7.0xx10^(-2) N//m`. To what height is the water raised above the normal water level? Angle of contact of water with glass is `0^(@)`. Density of water is `10^(3)kg//m^(3) and g=9.8 m//s^(2)`. |
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Answer» Correct Answer - B::C `h=(2T costheta)/(r rho g)` Substituting the proper values, we have `h =((2)(7.0xx10^(-2))cos 0^(@))/((0.5 xx 10^(-3)) (10^(-3)) (9.8))` `=2.86xx10^(-2)m=2.86cm`. |
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| 547. |
A film of water is formed between two straight parallel wires each `10 cm` long and at a seperation of `0.5 cm`. Calculate the work required to increase `1mm` distance between the wires. Surface tension of water `=72xx10^(-3)N//m`. |
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Answer» Correct Answer - A::D `W=T(Delta A )` `=T(2ld)` `=(72xx10^(3))(2)(0.1)(10^(-3))` `=14.4xx10^(-6)J`. |
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| 548. |
Calculate the energy released when 1000 small water drops each of same radius `10^(-7)m` coalesce to form one large drop. The surface tension of water is `7.0xx10^(-2)N//m`. |
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Answer» Correct Answer - A::B Let r be the radius drops and `R` of bigger one. Equanting the initial and final volume, we have `(4)/(3) piR^(3)=(1000)((4)/(3) pi r^(3))` or, `R=10r=(10)(10^(-7))m` or, `R=10^(-6)m` Further, the water drops have only one free surface. Therefore, `Delta A=4pi R^(2)-(1000)(4 pi r^(2))` `=4pi [(10^(-6))^(2)-(10^(3))(10^(-7))^(2)]` `=-36pi (10^(-12))m^(2)` Here, negative sign implies that surface area is decreasing. Hence, energy released in the process. `U=T|Delta A|=(7xx10^(-2))(36 pi xx10^(-12))J` `=7.9xx10^(-12) J`. |
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| 549. |
What is ratio of surface energy of 1 small drop and 1 large drop, if 1000 small drops combined to form 1 large dropA. `1 : 100`B. `1 : 1000`C. `1 : 10`D. `0.1 : 1` |
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Answer» Correct Answer - A Let, R =radius of bigger drop and r=radius of smaller drop Then, `4/3 piR^(3)=1000 xx 4/3 rArr r^(3) rArr E_(1)/E_(2)=1/100` |
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| 550. |
Two small drops of mercury each of radius r form a single large drop. The ratio of surface energy before and after this change isA. `2 : 2^(2//3)`B. `2^(2//3) : 1`C. `2 : 1`D. `1 :2` |
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Answer» Correct Answer - A Suppose R be the radius of bigger drop, Then by equating volumes, we have `2(4//3pir^(3))=(4//3)piR^(3)" or "R=(2)^(1//3)r` Now, surface energy `prop` surface area `therefore" "U_(1)/U_(2)=A_(1)/A_(2)=(2r^(2))/R^(2)=2/2^(2//3)` |
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