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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. |
Figure shows a large closed cylindrical tank containing water. Initially, the air trapped above the water surface has a height `h_(0)` and pressure `2p_(0)` where `rh_(0)` is the atmospheric pressure. There is a hole in the wall of the tank at a depth `h_(1)` below the top from which water comes out. A long vertical tube is connected as shown. Find the speed with which water comes out of the holeA. `1/(rho)[p_(0)-rhog(h_(1)-2h_(0))]^(1/2)`B. `[2/(rho)[p_(0)+rhog(h_(1)-h_(0))]]^(1/2)`C. `[3/(rho)[p_(0)+rhog(h_(1)+h_(0))]]^(1/2)`D. `[4/(rho)[p_(0)-rhog(h_(1)-h_(2))]]^(1/2)` |
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Answer» Correct Answer - B `2P_(0)=(h_(2)+h_(0))rhog+p_(0)` (since liquids at the same level have the same pressure) `P_(0)=h_(2)rhog+h_(0)rhog` `h_(2)rhog=P_(0)-h_(0)rhog` `h_(2)=(P_(0))/(rhog)-(h_(0)rhog)/(rhog)=(P_(0))/(rhog)-h_(0)` `KE` of the water `=` Pressure energy of the water at that layer `1/2mV^(2)=mxxP/(rho)` `V^(2)=(2P)/rho=2/(rho)[P_(0)+rhog(h_(1)-h_(2))]` `V=[2/(rho){P_(0)+rhog(h_(1)-h_(0))}]^(1/2)` We know `2P_(0)+rhog(h_(1)-h_(0))=P_(0)=rhogX` `impliesX=(P_(0))/(rhog)+(h_(1)-h_(0))=h_(2)+h_(1)` i.e., `X` is `h_(1)` metre below the top or `X` is `-h_(1)` above the top. |
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| 102. |
Two identical cylindrical vessel with their bases at the same level each contain a liquid of density `rho`. The height of the liquid in one vessel is `h_(1)` and in the other is `h_(2)` the area of either base is A. What is the work done by gravity is equalising the levels when the two vessels are connected?A. `2 rho Ag(h_(2)-h_(1))^(2)`B. `rho Ag(h_(2)-h_(1))^(2)`C. `(1)/(2) rho Ag(h_(2)-h_(1))^(2)`D. `(1)/(4) rhoAg(h_(2)-h_(1))^(2)` |
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Answer» Correct Answer - D Common height after they are connected can be determined by equating the volume. Hence `(A_(1)+A_(2))h=A_(1)h_(1)+A_(2)h_(2) " " (A_(1)=A_(2)=A)` `therefore " " h=(h_(1)+h_(2))/(2)` Work done by gravity `= - Delta U = U_(i)-U_(f)` `={Ah_(1) rho. g. (h_(1))/(2)+Ah_(2)rho.g.(h_(2))/(2)}-A(h_(1)+h_(2))rho. g. ((h_(1)+h_(2)))/(4)` `=(rho Ag)/(4)(h_(2)-h_(1))^(2)` |
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| 103. |
A given shaped glass tube having uniform cross-section is filled with water and is mounted on a rotatable shaft as shown in figure. If the tube is rotated with a constant angular velocity `omega` then : A. Water levels in both sections A and B go upB. Water level in Section A goes up and that in B comes downC. Water level in Section A comes down and that in B it goes upD. Water levels remains same in both sections |
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Answer» Correct Answer - A |
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| 104. |
Assertion : Railways tracks are laid on small sized wooden sleepers. Reason : Small sized wooden sleepers are used so that rails exert more pressure on the railway track. Due to which rail does not leave the trackA. 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 - D |
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| 105. |
A flat plate moves normally towards a discharging jet of water at the rate of `3 m//s`. The jet discharges the water at the rate of `0.1 m^(3)//s` and at the speed of `18 m8//s`. The force exerted on the plate due to the jet isA. `1800N`B. `2100N`C. `2450N`D. `1560N` |
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Answer» Correct Answer - C Relative velocity of the water jet with respect to the moving plate `=(18+3)=21m//s`. The mass of water hitting the plate per second `=arho(21)`, where `a` is the cross sectional area of the jet `F=` force normal to the plate `=` Rate of change of momentum along the direction of the water jet `=("Mass")xx("velocity")=[arho(21)]21=441arho` `:. 441xxrhoxx(0.1/18)` `=(441xx0.1xx1000)/1.8N=2450N` |
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| 106. |
Two bodies are in equilibrium when suspended in water from the arms of balance. The mass of one body is 36 g and its density is `9 g// cm^3` If the mass of the other is 46 g, its density in `g//cm^3` isA. `(4)/(3)`B. `(3)/(2)`C. `3`D. `5` |
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Answer» Correct Answer - C Apparent weight `=V(rho-sigma)g = (m)/(rho) (rho-sigma)g` where m = mass of the body, rho = density of the body and `sigma` = density of water If two bodies are in equilibrium then their apparent weight must be equal. `therefore (m_(1))/(rho_(1))(rho_(1)-sigma)g = (m_(2))/(rho_(2))(rho_(2)-sigma)g implies (36)/(9)(9-1)=(48)/(rho_(2))(rho_(2)-1)g`. By solving we get `rho_(2) =3`. |
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| 107. |
Tow bodies are in equilibrium when suspended in water from the arms of balance. The mass of one body is 36 g and its density is `9 g// cm^3` If the mass of the other is 46 g, its density in `g//cm^3` isA. `(4)/(3)`B. `(3)/(2)`C. 3D. 5 |
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Answer» Correct Answer - C (c ) Apperent weight ` = V (rho - sigma) = (m_2)/(rho_2) (rho_2 - sigma)` `rArr (36)/(9) (9-1) = (48)/(rho_2) (rho_2 -1)` By solving we get `rho_2 = 3.` |
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| 108. |
Two bodies having volumes V and 2V are suspended from the two arms of a common balance and they are found to balance each other. If larger body is immersed in oil (density `d_(1)=0.9gm//cm^(3)`) and the smaller body is immersed in an unknown liquid, then the balance remain in equilibrium. The density of unknown liquid is given by :A. `2.4gm//cm^(3)`B. `1.8gm//cm^(3)`C. `0.45gm//cm^(3)`D. `2.7gm//cm^(3)` |
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Answer» Correct Answer - B `B_(1)=B_(2)" "impliesVrhog=2Vxx0.9gimpliesrho=1.8gm//c c` |
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| 109. |
Assertion : Deep inside a liquid density is more than the density on surface. Reason : Density of liquid increases with increase in depth.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
| Answer» Correct Answer - A | |
| 110. |
Assertion: In the figure shown v and R will increase if pressure above the liquid surface inside the chamber is increased. Reason: Value of v or R is independent of density of liquid.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
| Answer» Correct Answer - B | |
| 111. |
Assertion: Weight of solid in air is w and in water is `(2w)/(3)`.Then relative density of solid is `3.0`. Reason: Relative density of any solid is given by` RD=("Weight in air")/("Change in weight in water")`.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
| Answer» Correct Answer - A | |
| 112. |
Assertion: Surface tension `(T=(F)/(l))` is not a vector quantity. Reasson: Direction of force is specified.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
| Answer» Correct Answer - A | |
| 113. |
A non viscous liquid is flowing through a frictionless duct, cross-section varying as shown in figure. Which of the following graph represents the variation of pressure p along the axis of tube ?A. B. C. D. |
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Answer» Correct Answer - B As we know according to equation of continuity, when cross-section of duct decreases, the velocity of flow of liquid increases and in accordance with Bernoullis therem in a horizontal pipe. Gh=raph (b) correctly repersents the variation of pressure. |
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| 114. |
Siphon is a device to transfer liquid from a higher level to a lower level. The condition of working of a siphon is : A. `h_(2) gt h_(1)`B. `h_(2) = 2h_(1)`C. `h_(1)` should be less the height of corresponding liauid barometerD. `h_(1)` should be greater than the height of corresponding liquid barometer |
| Answer» Correct Answer - A::C | |
| 115. |
A ball of density d is dropped on to a horizontal solid surface. It bounces elastically from the surface and returns to its original position in a time `t_1`. Next, the ball is released and it falls through the same height before striking the surface of a liquid of density of `d_L` (a) If `dltd_L`, obtain an expression (in terms of d, `t_1` and `d_L`) for the time `t_2` the ball takes to come back to the position from which it was released. (b) Is the motion of the ball simple harmonic? (c) If `d=d_L`, how does the speed of the ball depend on its depth inside the liquid? Neglect all frictional and other dissipative forces. Assume the depth of the liquid to be large.A. `t_(2) gt t_(1)`B. `t_(2) gt t_(1)`C. the motion of the ball is not simple harmonicD. If `rho = rho_(L)`, then the speed of the ball inside the liquid will be independent of its depth |
| Answer» Correct Answer - A::C::D | |
| 116. |
A ball of density d is dropped on to a horizontal solid surface. It bounces elastically from the surface and returns to its original position in a time `t_1`. Next, the ball is released and it falls through the same height before striking the surface of a liquid of density of `d_L` (a) If `dltd_L`, obtain an expression (in terms of d, `t_1` and `d_L`) for the time `t_2` the ball takes to come back to the position from which it was released. (b) Is the motion of the ball simple harmonic? (c) If `d=d_L`, how does the speed of the ball depend on its depth inside the liquid? Neglect all frictional and other dissipative forces. Assume the depth of the liquid to be large. |
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Answer» Correct Answer - A::B::C::D In elastic collision with the surface, direction of velicity is reversed but its magnitude remains the same, Therefore, time of fall=time of rise, or, time of fall `=(t_(1))/(2)` Hence, velocity of the ball just before it collides with liquids is `v=g(t_(1))/(2)` Retardation inside the liquid `a=("upthrust-weight")/("mass")=(Vd_(L)g-Vdg)/(Vd)` `=((d_(L)-d)/(d))g` Time taken to come to rest under this retardation will be `t=v/a=(g t_(1))/(2a)=(g t_(1))/(2((d_(L)-d)/(d))g)` `=(dt_(1))/(2(d_(L)-d))` Same will be the time to come to back on the liquid surface. therefore, (a) `(t_(2))=` time the ball takes to came back to the position from where it was released `=t_(1)+2t=t_(1)+(dt_(1))/(d_(L)-d)` `=t_(1)[1+(d)/(d_(L)-d)]` or `t_(2) =(t_(1)d_(L))/(d_(L)-d)` (b) The motion of the ball is periodic but not simple harminic because the acceleration of the ball is g in air and `((d_(L)-d)/(d)) g` inside the liqudi which is not proportional to the displacenent, which is necessary and sufficient condition for SHM. (c ) When `(d_(L)=d)`, retardation or acceleration inside teh liquid becones zero (upthrust-weight). Therefore, the ball will continue to move with constant velocity `v=(gt_(1))/(2)` inside the liquid. |
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| 117. |
The ratio of radii of two bubbles is `2 : 1`. What is the ratio of excess pressures inside them ?A. `1 : 2`B. `1 : 4`C. `2 : 1`D. `4 : 1` |
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Answer» Correct Answer - A The excess pressure inside the bubble `p=(4T)/(r). "then" p_(1=(4T))/(r_(1))....(i)` `p_(2)=(4T)/(r_(2))....(ii)` From Eqs. (i) and (ii), we get `(p_(1))/(p_(2))=(4T//r_(1))/(4T//r_(2))=(r_(2))/(r_(1))=(1)/(2)` |
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| 118. |
Two solid spheres of same metal but of mass M and 8M fall simultaneously on a viscous liquid and their terminal velocitied are v and n v, then value of n isA. 16B. 8C. 4D. 2 |
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Answer» Correct Answer - C The terminal velocity is given by v, where `v prop r^(2)` (r=radius of the sphere). The mass of the sphere can be given by `m=(4)/(3)pi r^(3)rho`. Thus `m prop r^(3)` So, according to the question, `(m_(1))/(m_(2))=(1)/(8)=((r_(1))/(r_(2)))^(2) implies (r_(1))/(r_(2))=(1)/(2)` and since, `(v_(1))/(v_(2))=((r_(1))/(r_(2)))^(2)=((1)/(2))^(2)=(1)/(4)=(v)/(nv)=(1)/(4) implies n=4` |
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| 119. |
In a streamline flow, (i) the speed of a particle always remains same (ii) the velocity of a particle always remains same (iii) the kinetic energies of all the particles arriving at a given point are the same (iv) the momenta of all the particles arriving at a given point are the same.A. (i),(iii)B. (ii),(iii)C. (i),(iii)D. (iii),(iv) |
| Answer» Correct Answer - D | |
| 120. |
In a streamline flowA. the speed of a particle always remains sameB. the velocity of a particle always remains sameC. the kinetic energies of all particles arriving at a given point are the sameD. the potential eneergies of all the particles arriving at a given point are the same |
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Answer» Correct Answer - C In streamlined flow, velocity of all the particles arriving at a common point remains same with time (both in magnitude and direction). So, the `KE=(1)/(2) mv^(2)`= a constant for all the particles as all fluid particles have identical mass also. |
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| 121. |
A vessel with water is placed on a weighing pan and reads `600g`. Now a ball of 40g and density `0.80g//c c` is sunk into the water with a pin as shown in fig. keeping it sunk. The weighing pan will show a reading A. `600 g`B. `550 g`C. `650 g`D. `632 g` |
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Answer» Correct Answer - C Upthrust on ball = weight of displaced water `V sigma g = ((m)/(rho))sigmag= (40)/(0.8)xx1xxg = 50g " Dyne or " 50gm` As the ball is sunk into the water with a pin by applying downward force eual to upthrust on it. So reading of weighing pan = weight if water + downward force against upthrust |
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| 122. |
The spring balance A reads 2 kg with a block m suspended from it. A balance B reads 5kg when a beaker with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass in inside the liquid in the beaker as shown in the figure. In this situation: A. the balance A will read more than `2kg`B. the balance B will read more than `5kg`C. the balance A will read less than `2kg` and B will read more than `5kg`D. the balance A and B will read `2kg` and `5kg` repectively |
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Answer» Correct Answer - B::C Due to Bouyancy effective weight of block of `2kg` reduces so reading of spring balance. A reduces but weight on balance B increases so its reading increases. |
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| 123. |
Water stands at a depth of `15 m` behind a reservoir dam. A horizontal pipe `4 cm` in diameter passes through the dam `6 m` below the surface of water as shown. There is a plug which secures the pipe opening. Then find the friction between the plug and pipe wall. |
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Answer» As plug secures the pipe opening the force of friction between plug and pipe wall `F=A(p_(2)-p_(1))` But `p_(1)=p_(2)` and `p_(2)=p_(0)+hrhog` `F=Ahrhog` `F=pi(2xx10^(-2))^2xx6xx10^(3)xx9.8implies F=74n` |
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| 124. |
A vessel with water is placed on a weighing pan and reads `600g`. Now a ball of 40g and density `0.80g//c c` is sunk into the water with a pin as shown in fig. keeping it sunk. The weighing pan will show a reading A. 600gB. 550gC. 650gD. 632g |
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Answer» Correct Answer - C Upthrust on ball = weight of displaced water `= V sigma g = ((m)/(rho)) sigma g = (40)/(0.8) xx 1xx g = 50g` As the ball is sunk into the water with a pin by applying downward force equal to upthrust on it. So reading of weighing pan `=weight of water + downward force against upthrust =600+50 = 650 gm`. |
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| 125. |
Two solids `A` and `B` floats in water. It is observed that `A` floats with half of its volume immersed and `B` Floats with `2//3` of its volume immersed. The ration of densities of `A` and `B` isA. `4:3`B. `2:3`C. `3:4`D. `1:3` |
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Answer» Correct Answer - C `V_(A)rho_(A) = 1/2 V_(A)xx1 or rho_(A) = 1/2` Again, `V_(B)rho_(B) = 2/3 V_(B) xx 1 xx rho_(B) = 2/3` `(rho_A)/(rho_B) = 1/2 xx 3/2 = 3/4`. |
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| 126. |
Assertion: A dam for water reservoir is built thicker at bottom than at the loop. Reason : Pressure of water is very large at the bottom.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertionC. if assertion is true but reason is false.D. if both assertion and reason are false. |
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Answer» Correct Answer - A With increase in depth the pressure increases from the formula `P prop h` therefore, the force perpendicular to the wall of dam increases. Hence, the dam must have greater strength at base than at top. Due to this dam are made ticker at the base than at the top. |
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| 127. |
Assertion: The flow of fluid is said to be steady if at any given point, the velocity of each passing fluid particle remains constant. Reason: The path taken by a fluid particle under a steady flow is a streamline.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertionC. if assertion is true but reason is false.D. if both assertion and reason are false. |
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Answer» Correct Answer - B A streamlined is the actual path followed by the precession of paricles in a steady flow, which may be straight or curved such that tangent to it at any point indicates the direction of flow of liquid at that point. |
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| 128. |
A 50 kg girl wearing heel shoes balances on a single heel. The heel is circular with a diameter 1 cm. The pressure exerted by the heel on the horizontal floor is `(Take g = 10 m s^(-2))`A. `6.4 xx 10^4 Pa`B. `6.4 xx 10^5 Pa`C. `6.4xx 10^6 Pa`D. `6.4 xx 10^7Pa` |
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Answer» Correct Answer - C (c ) Here, m = 50 kg, D = 1 cm ` 1-^(-2)m, g = 10 ms^(-1)` :. Pressure exerted by the heel on the horizontal floor is `p = (F)/(A) = (mg)/(pi(D//2)^2) = (4mg)/(pi D^2)` `=(4xx 50 kgxx 10 ms^(-2))/(3.14 xx (10^(-2)m)^2)` `6.4 xx 10^6 Pa` |
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| 129. |
A block of silver of mass `4 kg` hanging from a string is immersed in a liquid of relative density `0.72`. If relative density of silver is `10`, then tension in the string will beA. 37.12 NB. 42.34 NC. 73 ND. 21.15 N |
| Answer» Correct Answer - A | |
| 130. |
Figure shows water filled in a symmetrical container. Four pistons of equal area `A` are used at the four openings to keep the water in equilibrium. Now an additional force `F` is applied at each piston. The increase in the pressure at the centre of the container due to this addition is A. `F/A`B. `(2F)/A`C. `(4F)/A`D. 0 |
| Answer» Correct Answer - B | |
| 131. |
A closed tube is filled with `AB=2m` `BC=4cm` water `(rho=10^(3)kg//m^(3))`. It rotating about an axis shown in figure with an angular velocity `omega=2 rad//s`. Find, `p_(A)-p_(C)`. |
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Answer» Correct Answer - A::B::C::D Pressure decreases in moving towards the axis of rotation and increases in moving away from the axis `(Delta p= +-(rho omega^(2)x^2)/(2))` ` :. p_(A)gtp_(B)` and `p_(B)gtp_(C)` `p_(A)-p_(C)=(p_(A)-p_(B))+(p_(B)-p_(C))` `=(+(rho omega^(2)x_(1)^(2))/(2))+((-rho omega^(2)x_(2)^(2))/(2))` Here, `x_(1)=AB=2m` and `x_(2)=BC=4m` Substituting teh values we have, `p_(A)-p_(C)=((10^(3))(2)^(2)(2)^(2))/(2)-((10^(3))(2)^(2)(4)^(2))/(2)` `=-2.4xx10^(3) N//m^(2)`. |
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| 132. |
A uniform long tube is bent into a circle of radius R and it lies in vertical plane. Two liquids of same volume but densities `rho " and " delta` fill half the tube. The angle `theta` is A. `tan^(-1) ((rho-delta)/(rho+delta))`B. `tan^(-1)((rho)/(delta))`C. `tan^(-1)((delta)/(rho))`D. `tan^(-1)((rho+delta)/(rho-delta))` |
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Answer» Correct Answer - A From the figure, we have `delta gR(cos theta+sin theta)=rho gR(cos theta - sin theta)` `implies delta cos theta+ delta sin theta= rho cos theta- rho sin theta` `implies sin theta(delta+rho)=cos theta(rho-delta) implies tan theta=(rho-delta)/(rho+delta)` |
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| 133. |
A liquid of density `rho` comes out with a velocity v from a horizontal tube of area of cross-section A. The reaction force exerted by the liquid on the tube is F :A. `F prop v`B. `F prop v^(2)`C. `F prop A`D. `F prop rho` |
| Answer» Correct Answer - A::B::C | |
| 134. |
A tube in vertical plane is shown in figure. It is filled with a liquid of density `rho` and its end `B` is closed. Then the force exerted by the fluid on the tube at end `B` will be : [Neglect atmospheric pressure and assume the radius of the tube to be negligible in comparison to `l`] A. `P_(atm) A_0`B. `(P_(atm)+rhogl)A_0`C. `(P_(atm)+2rhogl)A_0`D. `(P_(atm)+4rhogl)A_0` |
| Answer» Correct Answer - C | |
| 135. |
A tube in vertical plane is shown in figure. It is filled with a liquid of density `rho` and its end `B` is closed. Then the force exerted by the fluid on the tube at end `B` will be : [Neglect atmospheric pressure and assume the radius of the tube to be negligible in comparison to `l`] |
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Answer» Correct Answer - B Pressure exerted by fluid at closed end `B` is `P = rhogl` `:.` force exerted by fluid at closed end `B` is `F PA = lrhogA_(0)` |
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| 136. |
A horizontal tube of uniform cross-sectional area `A` is bent in the form of `U` as shown in figure. If the liquid of density `rho` enters and leaves the tube with velcity v, then the extermal force `F` requried to hold the bend stationary is A. `F=0`B. `rhoAv^(2)`C. `2rhoAv^(2)`D. `(1)/(2) rhoAv^(2)` |
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Answer» Correct Answer - C `F=(Delta p)/(Delta t)=((Delta m)/(Delta t))(Delta v)` `=rho((Delta V)/(Delta t))(2v)` `=(rho)(Av)(2v)=2rhoAv^(2)`. |
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| 137. |
A liquid of density `rho` comes out with a velocity `v` from a horizontal tube of area of cross-section `A`. The reaction force exerted by the liquid on the tube is `F`. Choose the incorrect option.A. `F prop v`B. `F prop v^2`C. `F prop A`D. `F prop rho` |
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Answer» Correct Answer - A Reaction force `F = rho A v^2`. |
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| 138. |
A metallic sphere with an internal cavity weighs `40 gwt` in air and `20 gwt` in water. If the density of the material with cavity be `8g per cm^(3)` then the volume of cavity is :A. ZeroB. `15 cm^(3)`C. `5 cm^(3)`D. `20 cm^(3)` |
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Answer» Correct Answer - B Weight of sphere in air = 40 gwt weight of sphere in water = `20 gwt` Loss in weight = (40-20)gf = 20 gwt` Weight of water displaced = loss of weight =20 gwt` Mass of water displaced = `20 g` Volume of water displaced = `20cm^(3)` Actual volume of sphere = Volume of water displaced =`20cm^(3)` Volume of material in the sphere = 40/8 = 5 cm^(3)` Volume of cavity = (20-5) = 15cm^(3)`. |
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| 139. |
A metal block having an internal cavity weight 110 g in air and 80 g in water. If the density of metal is 5.5 g//cc, then the volume of cavity is :A. 30ccB. 20 ccC. 10 ccD. 5 cc |
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Answer» Correct Answer - C (c ) Weight of block in air =110 g Weight of block in water = 80 g Loss of weight in water = 110 - 80 = 30 g this is equal to weight of water displaced , so volume of water displaced = 30 cc (which is the apperent volume of block) Real volume of the meterial of block `= (110)/(5.5) = 20 cc` :. Volume of cavity =30 cc - 20cc =10cc |
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| 140. |
A piece of gold weighs 10g in air and 9g in water. What is the volume of cavity? (Density of gold =`19.3 g cm^(-3)`)A. 0.282 ccB. 0.482 ccC. 0.682ccD. None of these |
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Answer» Correct Answer - B Actual volume of gold = `10//19.3 = 0.518 cm^(3)` Now, if `V` be the weight of gold in water = weight of `V` volume of water `implies 10-9 = V xx 1 = 1 c c` Volume of cavity = `1-0.518 = 0.482 C C`. |
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| 141. |
A liquid is kept in a cylindrical vessel which is being rotated about a vertical axis through the centre of the circular base. If the radius of the vessel is `r` and angular velocity of rotation is `omega`, then the difference in the heights of the liquid at the centre of the vessel and the edge is.A. `(romega)/(2g)`B. `(r^(2)omega^(2))/(2g)`C. `sqrt(2gromega)`D. `(omega^(2))/(2gr^(2))` |
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Answer» Correct Answer - D |
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| 142. |
A steel block having an internal cavity weighs 234g in air and 197g in water. If the density of steel is `7.8 g cm^(-3)` then the volume of the cavity isA. `5cm^(3)`B. `7 cm^(3)`C. `9 cm^(3)`D. `11 cm^(3)` |
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Answer» Correct Answer - B Weight of water displaced by steel block = Loss of weight of block in water `=(234-197)gf = 37gf` Volume of water displaced = `37 cm^(3)` This is the volume of the block with cavity. Volume of cavity = (37-30)cm^(3) = 7 cm^(3)`. |
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| 143. |
A siphon in use is demostrated in the following in siphon is `1.5 gm//c c`. The pressure differece between the point `P` and `S` will be A. `10^(5)N//m`B. `2 xx 10^(5)N//m`C. zeroD. Infinity |
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Answer» Correct Answer - C As the both point are at the surface of liquid and these points are in the open atmosphere. So both point possess similar pressure and equal to 1 atm. Hence the pressure difference will be zero. |
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| 144. |
A uniformly tapering vessel is filled with a liquid of density `900 kg // m`. The force that acts on the base of the vessel due to the liquid is `(g = 10 ms^(-2))` A. `3.6 N`B. `7.2 N`C. `9.0 N`D. `14.4 N` |
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Answer» Correct Answer - B |
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| 145. |
A siphon in use is demonstrated in the following in siphon is `1.5 gm//c c`. The pressure difference between the point `P` and `S` will be A. `10 Nm`B. `2 xx 10 Nm`C. ZeroD. Infinity |
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Answer» Correct Answer - C |
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| 146. |
A candle of diameter `d` is floating on a liquid in a cylindrical container of diameter `D(D lt lt d)` as shown in figure. If is burning at the rate of `2 cm//h` then the top of the candle will : A. Remain at the same heightB. Fall at the rate of `1 cm//"hour"`C. Fall at the rate of `2 cm//"hour"`D. Go up the rate of `1 cm//"hour"` |
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Answer» Correct Answer - B |
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| 147. |
A wooden sylinder floats vertically in water with half of its length immersed. The density of wood is:A. Equal of that of waterB. Half the density of waterC. Double the density of waterD. The question is incomplete |
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Answer» Correct Answer - B |
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| 148. |
A wooden block of volume `1000cm^(3)` is suspended from a spring balance its weight is 12 N in air. It is suspended in water such that half of the block is below the surface of water. The reading of spring balance isA. `10 N`B. `9 N`C. `8 N`D. `7 N` |
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Answer» Correct Answer - D Reading of the spring balance = Apparent weight of the block = Actual weight - upthrust `= 12 - V_("in")sigmag` `=12-500xx10^(-6)xx10^(3)xx10=12-5 =7 N`. |
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| 149. |
The work done to split a liquid drop id radius `R` into `N` identical drops is (taken `sigma` as the surface tension of the liquid)A. `4piR^(2)(N^(-1//3)-1)sigma`B. `4piR^(2)N sigma`C. `4piR^(2)(N^(-1//2)-1)`D. None of these |
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Answer» Correct Answer - A `N((4)/(3) pir^(3))=(4)/(3) piR^(3)` `:. r=R((1)/(N))^(1//3)` `W=sigma (Delta A)` `= sigma (A_(f)-A_(i))` `=sigma [N(4 pi r^(2))- 4 pi R^(2)]` `=4 pi sigma [NR^(3)((1)/(N))^(2//3)-R^(2)]` `=4sigma pi R^(2) (N^(-1//3)-1)`. |
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| 150. |
If the container filled with liquid gets accelerated horizontally or vertically, pressure in liquid gets changed. In liquid (`a_(y)`) for calculation of pressure, effective `g` is used. A closed box horizontal base `6 m` by `6 m` and a height `2m` is half filled with liquid. It is given a constant horizontal acceleration `g//2` and vertical downward acceleration `g//2`. Water pressure at the bottomof centre of the box is equal to (atmospheric pressure `=10^(5)N//m^(2)` density of water `=1000 kg//m^(3),g=10m//s^(2)`)A. `1.1MPa`B. `0.11MPa`C. `0.101MPa`D. `0.011MPa` |
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Answer» Correct Answer - B `P=P_(a)+rhogh=10^(5)-1000x10xx1` `=(10^(5)+10^(4))N//m^(2)=0.11MPa` |
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