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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.
| 351. |
An L shaped glass tube is kept inside a bus that is moving with constant acceleration. During the motion, the level of the liquid in the left arm is at 12 cm whereas of the tube is as shown. Assuming that the diameter of the tube is much smaller then levels of the liquid and neglecting effect of surface tension, acceleration of the bus will be `(g =10 m//s^2)` A. `1m//s^2`B. `2m//s^2`C. `4m//s^2`D. `5m//s^2` |
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Answer» Correct Answer - B (b) `tan theta = (a)/(g) = (h_2 - h_1)/(h_2 tan 45^@ + h_1 tan 45^@) = (4cm)/(20cm)` `rArr a = 2m//s^2` |
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| 352. |
A broad vessel, with a square base of edga s = 10 cm is saperated into two halves A and B, by a smooth vertical piston. A spring of spring constant k =1500 N//m is filled across the compertment A and the compertment B is filled with water to a height 20 cm. Find the compression in the spring. A. 1.3 cmB. 2.1 cmC. 3.9 cmD. 0.07 cm |
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Answer» Correct Answer - A (a) The force exerted on the spring, via the piston be water = (pressure at centroid of the pistion) xx (area of the pistion) `=(rho g (h)/(2)) ((ah))` `F = 1 xx 1000 xx (20^2)/(2) xx10 = 2 xx 10^5 dyne` Now, if x cm be the comperession of the spring Then for equilibrium F = Kx. ` 2 xx 10^6 = 1500 xx 10^3 (dyne)/(cm) xx x cm` x = 20//15 =1.3 cm |
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| 353. |
An open pan P filled with water of density `(rho_w)` is placed on a vertical rod, maintaining equilibrium. A block of density `rho` is placed on one side of the pan as shown. Water depth is more then height of the block. A. Equilibrium will be maintainer only if `rho lt rho_(W)`.B. Equilibrium will be maintainer only if `rho le rho_(W)`.C. Equilibrium will be maintainer for all realtlion between `rho` and `rho_(w)`.D. It is not possible to maintained the equilibrium |
| Answer» Correct Answer - B | |
| 354. |
If two soap bubble of different radii are in communication with each otherA. air flows from larger bubble into the smaller oneB. the size of the bubbles remains the sameC. air flows from the smaller bubble into the large one and the larger bubble grows at the expense of the smaller oneD. None of the above |
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Answer» Correct Answer - C Since, `Deltap prop 1//R` |
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| 355. |
The excess pressure due to surface tension in a spherical liquid drop of radius r is directly proportional toA. rB. `r^(2)`C. `r^(-1)`D. `r^(-2)` |
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Answer» Correct Answer - C `Deltap=(4T)/(r )` |
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| 356. |
A body having volume V and density `rho` is attached to the bottom of a container as shown. Density of the liquid is `d(gtrho)`. Container has a constant upward acceleration a. Tension in the string is A. `V[Dg-rho(g+a)]`B. `V(g+a)(d-rho)`C. `V(d-rho)g`D. none |
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Answer» Correct Answer - B `T=B-W" "impliesT=V(d-rho)g_("eff")" "impliesT=V(d-rho)(g+a)` |
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| 357. |
An air bubble in a water tank rises from the bottom to the top. Which of the following statements are true?A. Bubble rises upwards because pressure at the bottom is less than at the top.B. Bubble rises upwards because pressure at the bottom is greatger than that at the topC. As the bubble rises, its size remain same.D. As the bubble rises, its size decreases |
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Answer» Correct Answer - B PV= constant (Assumed isothermal process) |
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| 358. |
A solid sphere of radius R made of a material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless pistion of area A floats on the surface of the liquid. When a mass M is placed on the piston to compress the liquid the fractional change in the radius of the sphere, `deltaR//R`, is .............A. `(Mg)/(AK)`B. `(Mg)/(2AK)`C. `(Mg)/(3AK)`D. `(Mg)/(4AK)` |
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Answer» Correct Answer - C `DeltaP=(Mg)/A` `|(DeltaV)/V|=(DeltaP)/K=(Mg)/(AK)` Now, as `V=4/3 pi R^(3)` or `V prop R^(3)` `:. (DeltaV)/V=3((DeltaR)/R)` or `(DeltaR)/R=1/3 ((DeltaV)/V)......(1)` `:.` from equation (1) `(DeltaR)/R=(Mg)/(3AK)` |
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| 359. |
State true/false: `S_1` : Hydrostatic pressure is a scalar quantity even though pressure is force divided by area and force is a vector quantity? `S_2`: A barometer made of a very narrow tube (see figure) is placed at normal temperature and pressure. The coefficient of volume expansion of mercury is `0.00018//.^@C` and that of the tube is negligible. The temperature of mercury in the barometer is now raised by `1^@C` but the temperature of the atmosphere does not change. Then, the mercury height in the tube remains unchanged. `S_3` A block of ice with a lead shot embedded in it is floating on water contained in a vessel. The temperature of the system is maintained at `0^@C` as the ice melts. When the ice melts completely the level of water in the vessel rises. A. TTFB. TFTC. TFFD. TTT |
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Answer» Correct Answer - C `S_(1)` : True Pressure is always acting normal to hence it is considered as scalar. `S_(2)` :False On increasing the temperature of mercury its density. Hence level of mercury in barometer tube will increases. `S_(3)`: False When ice melts level of water does not change. in case of lead, it was initially i.e., it would had displaced the water equal to weight of lead. so, volume of water displaced would be `V_(1)=m/(rho_(w))` (m=mass of lead) Now, when ice melts lead will sink and it would displace the water equal to volume of lead itself so, volume of water displaced in this case would be `V_(2)=m/(rho_(l))` Now, as `rho_(l) gt rho_(w), V_(2) gt V_(1)` or level will fall. |
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| 360. |
For a body immersed in a liquid, when the weight of the body is less than the upthrust then the body willA. float partially immersedB. sinkC. float full immersedD. both (a) and (c ) |
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Answer» Correct Answer - A For a body immersed in liquid when the weight of the body is less than the upthrust then the body will float partially immersed. |
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| 361. |
Water flows through a horizontal tube as shown in figure. If the difference of heights of water colun in the vertical tubes is 2 cm and the area of cross section at A and B are `4cm^2 and 2cm^2` respectively, find the rate of flow of water across any section. A. `130xx10^(-6)m^(3)//s`B. `146xx10^(-6)m^(3)//s`C. `160xx10^(-6)m^(3)//s`D. `170xx10^(-6)m^(3)//s` |
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Answer» Correct Answer - B `v_(A)a_(A)=v_(B)xxa_(B)=v_(A)xx4=v_(B)xx2v_(B)=2v_(A)`………i Again `1/2, rhov_(A)^(2)+rhogh_(A)+p_(A)=1/2rhov_(B)^(2)+rhogh_(B)+p_(B)` `implies1/2rhog_(A)^(2)+p_(A)=1/2rhov_(B)^(2)+p_(B)` (as `h_(A)=h_(B)`) `impliesp_(A)-p_(B)=1/2rho(v_(B)^(2)-v_(A)^(2))=1/2xx1xx(4v_(A)^(2)-v_(A)^(2))` `implies2x1xx1000=1/2xx1xx3v_(A)^(2)` `p_(A)=p_(B)=2cm` of water column `=2xx1xx1000dyn//cm^(2)` `:.v_(A)=sqrt(4000/3)=36.51cm//s` So, rate of flow `=V_(a)a_(A)=36.51xx4=146cm^(3)//s`. |
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| 362. |
Water flows through a horizontal tube as shown in figure. If the difference of heights of water colun in the vertical tubes is 2 cm and the area of cross section at A and B are `4cm^2 and 2cm^2` respectively, find the rate of flow of water across any section. |
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Answer» Correct Answer - A::C::D v_A=a_A=v_Bxxa_B` `rarr v_Axx4=v_Bxx2` `rarr v_B=2v_A`…i `-Again `1/2rhov_A^2+rhogh_A+p_A=1/2rhov+B^2+rhogh_B+p_B` `rarr 1/2rhov_A^2+p_A=1/2rho(v_B^2-v_A^2)` `rarr p_A-p_B=1/2rho(v_b^2-v_A^2)` `=1/2xx1xx(4v_A^2-v_A^2)` `rarr 2xx1xx1000=1/2xx1x3v_A^2` =[p_A-p_B=2cm` of water column `=2xx1xx1000dyne/cm^2]` `rarrv_A^2=sqrt(4000/3)=36.51cm/sec` `So, rate of flow =V_aa_A=36.50xx4` `=146cm^3/sec` |
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| 363. |
How much work will be done in increasing the diameter of a soap bubble from 2cm to 5cm? Surface tension solution is `3.0xx10^(-2)N//m.` |
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Answer» Correct Answer - A::C::D Soap bubble has two surface, Hence, `W=T Delta A (T=(Delta W)/(Delta A))` Here,` DeltaA= 2[4 pi {(2.5xx10^(-2))^(2)-(1.0xx10^(-2))^(2)}]` `=1.32xx10^(-2)m^(2)` `:. W=(3.0xx10^(-2))(1.32xx10^(-2))J` `=3.96xx10^(4) J`. |
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| 364. |
A liquid flows in a tube from left to right as shown in figure. `A_(1)` and `A_(2)` are the cross-section of the portions of the tube as shown. Then the ratio of speeds `v_(1)//v_(2)` will be A. `A_(1)//A_(2)`B. `A_(2)//A_(1)`C. `sqrt(A_(2))//sqrt(A_(1))`D. `sqrt(A_(1))//sqrt(A_(2))` |
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Answer» Correct Answer - B |
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| 365. |
Water flows through the tube shown in figure. The areas of cross section of the wide and the narrow portions of the tube are `5cm^2 and 2cm^2` respectively. The rate of flow of water through the tube is `500 cm^3s^-1` . Find the difference of mercury levels in the U-tube. |
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Answer» Correct Answer - A::C Given Rate of Glow `=500cm^3/sec` `rarrv_A=(500/5)=100cm/sec` `V_Aa_A=V_Ba_B` `rarr v_A/v_B=a_B/a_A=2/5` `rarr 5v_A=2v_B` `rarr v_B=(5/2)v_A`…………i `rarr 1/2rhov_A^2+rhogh_A+pA=1/2 rhov_B^2+rhogh_B+p_B` `rarr p_A-p_B=1/2[p(v_B^2-v_A^2)]` `rarr hxx13.6xx980=1/2xx1xx21/4(100)^2` [Using i] [because `p_A-p_B=?]` `rarr h=(21xx(100)^2)/(2xx13.6xx980xx4)` `=1.969cm` |
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| 366. |
We have three beakers A , B and C containing glycerine, water and kerosene respectively. They are stirred vigorously and placed on a table. The liquid which comes to rest at the earliest isA. GlycerineB. Water level in Section A goes up and that in B comes downC. KeroseneD. All of them at the same time |
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Answer» Correct Answer - A |
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| 367. |
A large wooden plate of area `10m^2` floating on the surface of river is made to move horizontally wilth a speed of `2ms^-1` by applying a tangential force. If the river is 1m deep and the water contact with the bed is stationary, find the tangential force needed to keep the plate moving. Coefficient of viscosity of water at the temperature of the river `=10^-2 poise.`A. velocity gradient is `2s^(-1)`B. velocity gradient is `1s^(-1)`C. force required to keep the plate moving with constant speed is 0.02ND. force required to keep the plate moving with constant speed is 0.01N |
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Answer» Correct Answer - A::C Velocity gradient `=(Delta v)/(Deltah)=(2m//s)/(1m)=2s^(-1)` `F=eta A(Delta v)/(Deltah)=(10^(-3))(10)(2)` `=0.02N`. |
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| 368. |
A soap bubble is blown with the help of mechanical pump at the mouth of a tube. The pump produces a cartain increase per minute in the volume of the bubble, irrespective of its internal pressure. The graph between the pressure inside the soap bubble and time `t` will beA. B. C. D. |
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Answer» Correct Answer - A `Deltap=(4T)/(r ) implies Deltap prop (1)/(r )` As radius of soap bubble increases with time. `therefore " " Delta p prop(1)/(t)` |
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| 369. |
A cylindrical tank has a hole of diameter 2r in its bottom. The hole is covered wooden cylindrical block of diameter 4r, height h and density `rho//3`. Situation I: Initially, the tank is filled with water of density `rho` to a height such that the height of water above the top of the block is `h_1` (measured from the top of the block). Situation II: The water is removed from the tank to a height `h_2` (measured from the bottom of the block), as shown in the figure. The height `h_2` is smaller than h (height of the block) and thus the block is exposed to the atmosphere. Find the minimum value of height `h_1` (in situation 1), for which the block just starts to move up?A. `h//3`B. `4h//9`C. `2h//3`D. `h` |
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Answer» Correct Answer - B Again considering equilibrium of wooden block Total downward force `=`Total upwards Wt. of block `+` force due to atmospheric pressure `=` Force due to pressure of liquid `+` Force due to atmospheric pressure `pi(16r^(2))rho/3+g+P_(0)pixx16r^(2)` `=[h_(2)rhog+p_(0)]pi[16-4r^(2)]+P_(0)xx4r^(2)` `pi(16r^(2))hrho/g g=h_(2)rhogxxpix12r^(2)` `16 h/3 =12h_(2)` `impliesh_(2)=4/9h` |
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| 370. |
A solid cone of height H and base radius `H/2` floats in a liquid of density `rho`. It is hanging from the ceiling with the help of a string. The force by the fluid on the curved surface of the cone is (`P_(0) =` atmospheric pressure) A. `piH^(2)((P_(0))/4+(rhogH)/3)`B. `piH^(2)((P_(0))/4+(rhogH)/6)`C. `(piH^(2))/4((P_(0))/4+rhogH)`D. `(piH^(2))/4(P_(0)+rhogH)` |
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Answer» Correct Answer - B `1/3pi(H/2)^(2)Hrhog=(P_(0)+rhogH)pi(H/2)^(2)-Fv` `F_(v)=P_(0)pi(H^(2))/4+rhog(H^(3))/4pi-(piH^(3)rhog)/12` `=P_(0)(piH^(2))/4+(rhogH^(3)pi)/6` |
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| 371. |
A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, R is equal toA. `(L)/(sqrt(2pi))`B. `2piL`C. `L`D. `(L)/(2pi)` |
| Answer» Correct Answer - A | |
| 372. |
When a hole is made in the side of a container holding water, water flows out and follows a parabolic trajectory. If a hole is made in the side of the container and the container is dropped in free fall (just before the water starts coming out), the water flow (Neglect effect of surface tension) A. diminishesB. stops altogetherC. goes out in a straight line.D. Curves upward. |
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Answer» Correct Answer - B When the container is at rest with respect to the earth, there is pressure on the walls due to the weight of the water. The pressure results from the contact force between the water and the container. In free fall, both the water and the container have acceleration of `g` and the contact force is zero, so removing part of a wall by making a hole produces no outward flow. (Note that some of the water is in contact with the air, which is not accelerating, so there is still atmospheric pressure on the water). |
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| 373. |
A light semi cylindrical gate of radius R is pivoted aat its mid point O, of radius R as shown in the figure holding liquid of density `rho`. The force F required to prevent the rotation of the gate is equal to A. `2piR^(3)rhog`B. `2rhogR^(3)l`C. `(2R^(2)lrhog)/(3)`D. None of these |
| Answer» Correct Answer - D | |
| 374. |
A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, R is equal toA. `2piL`B. `(L)/(sqrt(2pi))`C. `L`D. `(L)/(2pi)` |
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Answer» Correct Answer - B |
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| 375. |
A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, R is equal toA. `2piL`B. `(L)/sqrt(2pi)`C. LD. `(L)/(2pi)` |
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Answer» Correct Answer - B Velocity of efflux when the hole is at depth h, `v=sqrt(2gh)` Rate of flow of water from square hole `Q_(1) = a_(1)v_(1) = L^(2) sqrt(2gy)` Rate of flow of water from circular hole `Q_(2) = a_(2)v_(2) = piR^(2) sqrt(2g(4y))` and according to problem `Q_(1) = Q_(2) implies L^(2) sqrt(2gy) = piR^(2) sqrt(2g(4y)) implies R = (L)/sqrt(2pi)` |
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| 376. |
A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, R is equal toA. `(L)/sqrt(2pi)`B. `2 pi L`C. `L`D. `(L)/(2pi)` |
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Answer» Correct Answer - A Volume flowing per sec `V=av` `L^(2)sqrt(2gy)=pir^(2)sqrt(2g(4y))" "impliesL^(2)=2piR^(2)` `R=(L)/sqrt(2pi)` |
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| 377. |
A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, R is equal toA. `L/sqrt(2r)`B. `2piL`C. `sqrt(2/r).L`D. `L/(2pi)` |
| Answer» Correct Answer - D | |
| 378. |
Figure shows a semi-cylindrical massless gate pivoted at the point `O` holding a stationary liquid of density `rho`. The length of the cylinder is `l`. Calculate, the horizontal force exerted by the liquid on the gate. |
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Answer» The force exerted by liquid on the gate is `f_(H)=P_(av)xx`vertical projection area of the gate `=((P_(A)-P_(B))/2)xx(2Rl)` `=((Rrhog+3Rrhog)/2)xx2Rl=4rhogR^(2)l` |
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| 379. |
When a hole is made in the side of a container holding water, water flows out and follows a parabolic trajectory. If a hole is made in the side of the container and the container is dropped in free fall (just before the water starts coming out), the water flow (Neglect effect of surface tension) A. diminishesB. stops altogetherC. goes out in a straight lineD. curves upwards |
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Answer» Correct Answer - A (Easy) When the container is at rest with respect to the Earth, there is pressure on the walls due to the weight of the water. The pressure results from the contact force between the water and the container. In free fall, both the eater and the container have acceleration of g, and the contact force is zero, so removing part of a wall by making a hole produces no outward flow. (Note that some of the water is in contact with the air, which is not accelerating, so there is still atmospheric pressure on the water). |
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| 380. |
The tension in a string holding a solid block below the surface of a liquid (of density greater than that of solid) as shown in figure is `T_(0)` when the system is at rest. What will be the tension in the string if the system has an upward acceleration a? .A. `T(1-a/g)`B. `T(1+a/g)`C. `T(a/g-1)`D. `a/gT` |
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Answer» Correct Answer - B Let `v` be the volume of the solid block of density `rho.` Let `rho_(1)` be the density of water.Weight of body `=vrhog`. When the body is immersed in water. Tension in the string`=` Upward thrust `-` Weight of the body `implies T=vrho_(1)g-vrhog=vg(rho_(1)-rho_(2))` When the lift is moving upwards with acceleration a the tension in the string is `T=v(rho_(1)-rho)(g+a)` From eqn i and ii `T=T_(0)(1+a/g)` |
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| 381. |
A man is carrying a block of a certain substance (of density `1000 kgm^-3`) weighing `1 kg` in his left hand and a bucket filled with water and weighing `10 kg` in his right hand. He drops the block into the bucket. How much load does he carry in his right hand now.A. 9 kgB. 10 kgC. 11 kgD. 12 kg |
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Answer» Correct Answer - C Total weight in right hand `= 10 + 1 = 11 kg`. |
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| 382. |
A block of ice is floating in a liquid of specific gravity 1.2 contained in the beaker. What will be the effect on the level of liquid in the beaker when the whole ice melts?A. remains sameB. risesC. lowersD. (1),(2) or (3) |
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Answer» Correct Answer - B Volume of liquid displaced by floating ice `V_1 = (M)/(sigma_l) = (M)/(1.2)` Volume of water formed by melting ice `V_2 = (M)/(sigma_omega) = (M)/(1.0)` `V_2 gt V_1`, level will rise. |
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| 383. |
A block of ice is floating in a liquid of specific gravity 1.2 contained in the beaker. What will be the effect on the level of liquid in the beaker when the whole ice melts?A. Remain sameB. RisesC. LowersD. (a), (b) or (c) |
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Answer» Correct Answer - B |
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| 384. |
A block of ice in which a piece of stone is embedded is floating on water contained in a beaker. When all the ice melts the level of water in the beakerA. risesB. fallsC. remains unchangedD. None of the above |
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Answer» Correct Answer - B Volume oif liquid displaced is more than the volume of water formed. So, the level fo water in the beaker falls. |
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| 385. |
In figure, block `A` hangs by a cord form spring balance `D` and it submerged in a liquid `C` contained in a beaker `B`. The mass of the beaker is `1 kg`. The mass of the liquid is `1.5 kg`. Balance `D` reads `7.5 kg`. The volume of block `A` is `0.003 m^(3)`. The mass per unit volume of the liquid is A. `2500 kg m^(-3)`B. `5000 kg m^(-3)`C. `1 kg m^(-3)`D. `(5000)/(3) kg m^(-3)` |
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Answer» Correct Answer - D `7.5g = 1 g +1.5g+Downward reaction of block or `5g = | Downward reaction force|` `=|upthrust|=0.003 rho_(1)g` `:. Rho_(1) = 5/0.003 kgm^(-3) = 5000/3 kg m^(-3)`. |
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| 386. |
The hydraulic press shown in the figure is used to raise the mass M through a height of `0.5 cm` by performing `500J` of work at the small piston. The diameter of the large piston is `10cm`, while that of the smaller one is `2 cm`. The mass M is A. `100kg`B. `10^(6)kg`C. `10^(3)kg`D. None of these |
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Answer» Correct Answer - D Diameter of left hand side is `(1)/(5)` times. So area will be `(1)/(25)` times. `Deltap_(LHS) =Delta p_(RHS)implies (F)/((A//25))=((Mg))/(A)` `:. F=(Mg)/(A)` Equating the volumes Displacement of `LHS=5` times Displacement of RHS `:. S=(5)(0.5)=2.5cm` `:. (500)=((Mg)/(25))(2.5xx10^(-2))` `:. M=5xx10^(4)kg` (with `g=10m//s^(2)`. |
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| 387. |
The velocity of the liquid coming out of a small hole of a vessel containing two different liquids of densities `2rho` and `rho` as shown in the figure is A. `sqrt(6gh)`B. `2sqrt(gh)`C. `2sqrt(2gh)`D. `sqrt(gh)` |
| Answer» Correct Answer - B | |
| 388. |
The velocity of the liquid coming out of a small hole of a vessel containing two different liquids of densities `2rho` and `rho` as shown in the figure is A. `sqrt(6gh)`B. `2sqrt(gh)`C. `2sqrt(2gh)`D. `sqrt(gh)` |
| Answer» Correct Answer - B | |
| 389. |
A U-tube of base length l filled with same volume of two liquids of densities `rho " and " 2rho` is moving with an acceleration a on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then height h is given by A. `(al)/(g)`B. `(3al)/(2g)`C. `(2gl)/(3g)`D. `(al)/(2g)` |
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Answer» Correct Answer - B Writing presure equation from one end to other end of tube. `p_(0)+2rho gh-2rho. (l)/(2). A-rho(l)/(2). A-rho gh=p_(0)` `therefore " " h=(3al)/(2g)` |
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| 390. |
A ball of density `rho` is dropped from a height on the suraca of a non-visous liquid of dinsity `2rho`. Choose the correct options.A. Motion of ball is periodic but not simple harmonicB. Acceleration of ball in air and in liquid are equalC. Magnitude of upthrust in the liquid is two times the weight of ballD. Net force on ball in air and in liquid are equal and opposite, |
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Answer» Correct Answer - A::C::D In air, `a_(1)=g` (downwards) In liquid, `a_(2) =("upthrust-weight")/("mass")` `=((V)(2 rho)(g)-(V)(rho)(g))/((V_(rho))` `=g`(upwards)` `:. a_(1)!=a_(2)`. |
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| 391. |
An iceberg of density `900kg//m^(3)` is floating in water of density `1000 kg//m^(3)`. The percentage of volume of ice cube outside the water isA. 0.2B. 0.8C. 0.1D. 0.9 |
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Answer» Correct Answer - C Percentage volume outside the water `=(rho_("water")-rho_("ice"))/rho_("water")xx100=(1000-900)/1000xx100=10%` |
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| 392. |
An iceberg of density `900kg//m^(3)` is floating in water of density `1000 kg//m^(3)`. The percentage of volume of ice cube outside the water isA. `20%`B. `35%`C. `10%`D. `25%` |
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Answer» Correct Answer - C |
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| 393. |
The area of cross sectioin of the two arms of a hydraulic press is `1cm^(2)` and `20 cm^(2)` respectively. A force of `10 N` is applied on the water in the thinner arm. What force should be applid on the water in the thicker arms so that water may remain in equilibrium? |
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Answer» In equilibrium, the pressures at the two surfaces should be equal as they lie in the same horizontal level. If the atmospheric pressure is `P_(0)` and a force `F` is applied to maintain the equilibrium the pressur just below the pistons are `P_(M)=P_(0)+(10N)/(1cm^(2))` and `P_(N)=P_(0)+F/(10cm^(2))` These pressures should be the same i.e. `P_(M)=P_(n)`which gives `F=200N`. |
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| 394. |
An iceberg is floating partially immersed in sea water. The density of sea water is `1.03 g cm^(-3)` and that of ice is`0.92 g cm^(-3)`. The approximate percentage of total volume of iceberg above the level of sea water isA. `8`B. `11`C. `34`D. `89` |
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Answer» Correct Answer - B `(V-/_V)xx1.03xxg=Vxx0.92xxg` or `(V-/_V)/V=0.92/1.03` or `(/_V)/V=1-0.92/1.03=0.11/1.03` or `(/_V)/Vxx100=11/1.03~~11` |
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| 395. |
In a simple hydraulic press, the cross sectional area of the two cylinders is `5xx10^(-1)m^(2)` and `10^(-2)m^(2)` respectively. A force of `20 N` is applied at the small plunger. a. What is the pressure produced in the cylinders? b. What is the thrust exerted on the large plunger? c. How much work is done by the operator, if the smaller plunger moves down `0.1m`? |
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Answer» In a hydraulic press a force `F_(1)` applied to the smaller plunger creats a pressure `(F_(1))/(A_(1))` in the liquid and this pessure is transmitted equally throughout the liquid and acts on the larger plunger. The thrust acting on te larger plunger upwards due to tis pressure is `F_(2)=A_(2)((F_(1))/(A_(1)))` Hence, the thrust on `A_(2)` is magnified by `(A_(2))/(A_(1))` times `a. (F_(1))/(A_(1))=20/(5xx10^(-4))=40000N//m^(2)` b. `F_(2)=F_(1)xx((A_(2))/(A_(1)))=20xx(10^(-2))/(5xx10^(-4))=400N` work done by force `F_(1)=F_(1)d_(1)=20xx0.1=2J` |
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| 396. |
The relative density of ice is 0.9 and that of sea water is 1.125. What fraction of the whole volume of an iceberg appears above the surface of the sea ?A. `1//5`B. `2//5`C. `3//5`D. `4//5` |
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Answer» Correct Answer - A Fraction immersed`=rho_("solid")/rho_("liquid")=(0.9)/(1.125)=0.8` Therefore, fraction outside the water = 0.2=1/5 |
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| 397. |
A piston of cross-sectional area `100 cm^(2)` is used in a hydraulic pressure to exert a force of `10^(7)` dyne on the water. The cross-sectional area of the other piston which support a truck of mass 2000 kg isA. `100 cm^(2)`B. `10cm^(2)`C. `2 xx 10^(4) cm^(2)`D. `2 xx 10 cm` |
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Answer» Correct Answer - C |
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| 398. |
The flow of blood in a large artery of a anesthetized dog is diverted through a venturimeter. The wider part of the meter has a cross-sectional area equal to that of the artery. Ie., `10mm^(2)` . The narrower part has an area `5 mm^(2)`. The pressure drop in the artery is `22 Pa`. Density of the blood pressure drop in the artery is `22Pa`. Density of the blood pressure drop in the artery is `22Pa`. Density of the blood is `1.06xx10^(3)kgm^(-3)`. The speed of the blood in the artery isA. `0.12 ms^(-1)`B. `0.62 ms^(-1)`C. `0.24 ms^(-1)`D. `0.42 ms^(-1)` |
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Answer» Correct Answer - A Here, `A = 10 mm^(2), a= 5mm^(2), rho=1.06 x 10^(2) kg m^(-3)` `=1060 kg m^(-3), P_(1) -P_(2) = 22Pa` Speed of the fluid through the wide neck is `v = sqrt((2(P_(1)-P_(2)))/(rho)) {(A/a)^(2)-1}^(-1//2)` or `A/a = 10/5 = 2` `:. V = sqrt((2xx22)/(1060(2^(2)-1))) = 0.12 ms^(-1)`. |
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| 399. |
Iceberg floats in sea water with a part of it submerged. The percentage fraction of the ice berg submerged is (density of ice = `0.9 g cm^(-3)`, density of sea water = `1.1 g cm^(-3)`)A. `18%`B. `12%`C. `10%`D. `8%` |
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Answer» Correct Answer - A `V` be the volume of the iceberg, `x be the volume out of sea water. The iceberg is floating in sea water then `C rho_(ice) g = (V-x)rho_(sea water) g` or `V xx 0.9 xx g = (V-x)1.1g` or `eta = 2.0 xx 10^(-5)kg m^(-1)s^(-1)` or `0.2 V = 1.1 x` `:. (x)/(V) = (0.2)/(1.1) ` Percentage of fraction of the volume of iceberg above the level of sea water `x/V xx 100 = (0.2)/(1.1) xx 100 = 18%` |
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| 400. |
For the system shown in figure, the cylinder on the left, at L, has a mass of 600 kg and a cross-sectional area of `800 cm^(2)`. The piston on the right, at S, has cross-sectional area `25 cm^(2)` and negligible weight. If the apparatus is filled with oil `(rho=0.78 g//cm^(3))`, what is the force F required to hold the system is equilibrium? . |
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Answer» Correct Answer - A::C Let `A_(1)`=area of cross-section on LHS and `A_(2)`=area of cross-section on RHS Equating the pressure on two sides of the dotted line. Then, `(F)/(A_(2))+ rho gh =(Mg)/(A_(1)) (M=600kg)` `:. F=A_(2) [(Mg)/(A_(1))-rhogh]` `=(25xx10^(-4)) [(600xx9.8)/(800xx10^(-4))-780xx9.8xx8]` `=31N`. |
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