Q: A homogeneous solid cylinder of length L(L<H/2), cross-sectional area 5A is immersed such that it floats with its axis vertical at the liquid-liquid interface with length 4L in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure Po∗. Then density D of solid is given by :

Weight of cylinder = Upthrust to both liquidsV×D×g=(5A×43L)×d×g+(5A×4L)×2d×g or, (5A×L)×D×g=4A×L×d×g or, 5D=4d or D=45d

Q: A body floats in water with 40% of its volume outside water. When the same body floats in oil, 60% of its volume remains outside oil. The relative density of oil is

Using the law of buoyancy, we have Vσg=0.6Vσ1g, for the part of body outside oil and Vσg=0.4Vσ2g, for the part of body outside water, hence we get1=0.4σ20.6σ1So we have, σ1σ2=46=23=1.5

Q: A small spherical droplet of density d is floating exactly half immersed in a liquid of density ρ and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet):

FBD of droplet,B+F=mgB=(32πR3)ρgF=T(2πR)m=d(34πR3)(32πR3)ρg+T(2πR)=d(34πR3)gT(2πR)=(32πR3)g[2d−ρ]R=(2d−ρ)g3T

Q: A cork of mass 10 g is floating on water. The net force on the cork is:

When the cork is floating, its weight is balanced by the upthrust. Therefore, net force on the cork is zero.

Q: A bob of mass ' m ' suspended by a thread of length ℓ undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density 41 times that of the bob and the length of the thread is increased by 1/3rd of the original length, then the time period of the simple harmonic oscillations will be :

T=2πgℓgell =g−4g=43g∴T′=2π3g/44ℓ/3=2π9g16ℓT′=34T

Q: A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [Take, density of water =103 kg/m3 ] (a) 30.1 kg (b) 46.3 kg (c) 87.5 kg (d) 65.4 kg

Exp. (c)When only block is floating on water, 30% of its volume is in water as shown belowBy Archimedes' principle,weight of block = weight of displaced water⇒Vρbg=(30%V)ρwgwhere, ρb= density of block, and ρw= density of water So, V= volume of block. ρb=0.3ρwNow, let a mass m is placed over block to just keep the cube fully immersed in water.Now, by Archimedes' principle,weight of water displaced = weight of block + weight of mass m⇒Vρwg=Vρbg+mg⇒V(ρw−ρb)=m⇒V×0.7ρw=mHere, V=(side)3=(0.5)3 m3[∵ρb=0.3ρw]ρw=103 kg m−3Substituting these values in the above relation, we get m=(0.5)3×0.7×103=87.5 kg

Q: A solid sphere of radius R and density ρ is attached to one end of a massless spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density 3ρ. The complete arrangement is placed in a liquid of density 2ρ and is allowed to reach equilibrium. The correct statement(s) is (are)

On small sphere34πR3(ρ)g+kx=34πR3(2ρ)g ...(i)On second sphere (large)34πR3(3ρ)g=34πR3(2ρ)g+kx ...(ii)By Eqs. (i) and (ii), we getx=3k4πR3ρg

Q: A body of weight w1 when floats in water displaces an amount of water w2. Then w1<w2. Is this statement true or false?

In floating condition, weight of liquid displaced = weight of solid

Q: The work done in blowing a bubble of volume V is W, then what is the work done in blowing a soap bubble of volume 2V ?

If volume is doubled, then radius becomes (2)31 times. or W=T(8πR2)W∝R2 Therefore W′=(2)32 W

Question 1

A 20 N metal block is suspended by a spring balance. A beaker containing some water is placed on a weighing machine which reads 40 N. The spring balance is now lowered so that the block gets immersed in the water. The spring balance now reads 16 N. The reading of the weighing machine will be

Accepted Answer

44 N Upthrust exerted by the water on the block = Change in the reading of the spring balance=(20−16)N=4 N Downthrust =4 N Actual weight of the beaker containing water =40 N ∴ Effective weight =(40+4)N=44 N

Question 2

A homogeneous solid cylinder of length L(L<H/2), cross-sectional area 5A​ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length 4L​ in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure Po∗​. Then density D of solid is given by :

Accepted Answer

Weight of cylinder = Upthrust to both liquidsV×D×g=(5A​×43​L)×d×g+(5A​×4L​)×2d×g or, (5A​×L)×D×g=4A×L×d×g​ or, 5D​=4d​ or D=45​d

Question 3

A body floats in water with 40% of its volume outside water. When the same body floats in oil, 60% of its volume remains outside oil. The relative density of oil is

Accepted Answer

Using the law of buoyancy, we have Vσg=0.6Vσ1​g, for the part of body outside oil and Vσg=0.4Vσ2​g, for the part of body outside water, hence we get1=0.4σ2​0.6σ1​​So we have, σ1​σ2​​=46​=23​=1.5

Question 4

A small spherical droplet of density d is floating exactly half immersed in a liquid of density ρ and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet):

Accepted Answer

FBD of droplet,B+F=mgB=(32​πR3)ρgF=T(2πR)m=d(34​πR3)(32​πR3)ρg+T(2πR)=d(34​πR3)gT(2πR)=(32​πR3)g[2d−ρ]R=(2d−ρ)g3T​​

Question 5

A cork of mass 10 g is floating on water. The net force on the cork is:

Accepted Answer

When the cork is floating, its weight is balanced by the upthrust. Therefore, net force on the cork is zero.

Question 6

A bob of mass ' m ' suspended by a thread of length ℓ undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density 41​ times that of the bob and the length of the thread is increased by 1/3rd  of the original length, then the time period of the simple harmonic oscillations will be :

Accepted Answer

​T=2πgℓ​​gell ​=g−4g​=43g​∴T′=2π3g/44ℓ/3​​=2π9g16ℓ​​T′=34​T​

Question 7

A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [Take, density of water =103 kg/m3 ]
(a) 30.1 kg
(b) 46.3 kg
(c) 87.5 kg
(d) 65.4 kg

Accepted Answer

Exp. (c)When only block is floating on water, 30% of its volume is in water as shown belowBy Archimedes' principle,weight of block = weight of displaced water⇒Vρb​g=(30%V)ρw​gwhere, ρb​= density of block, and ρw​= density of water So, V= volume of block. ρb​=0.3ρw​Now, let a mass m is placed over block to just keep the cube fully immersed in water.Now, by Archimedes' principle,weight of water displaced = weight of block + weight of mass m⇒Vρw​​g=Vρb​​g+mg⇒V(ρw​−ρb​)=m⇒V×0.7ρw​=mHere, V=(side)3=(0.5)3 m3[∵ρb​=0.3ρw​]ρw​=103 kg m−3Substituting these values in the above relation, we get m=(0.5)3×0.7×103=87.5 kg

Question 8

A solid sphere of radius R and density ρ is attached to one end of a massless spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density 3ρ. The complete arrangement is placed in a liquid of density 2ρ and is allowed to reach equilibrium. The correct statement(s) is (are)

Accepted Answer

On small sphere34​πR3(ρ)g+kx=34​πR3(2ρ)g ...(i)On second sphere (large)34​πR3(3ρ)g=34​πR3(2ρ)g+kx ...(ii)By Eqs. (i) and (ii), we getx=3k4πR3ρg​

Question 9

A body of weight w1​ when floats in water displaces an amount of water w2​. Then w1​<w2​. Is this statement true or false?

Accepted Answer

In floating condition,
weight of liquid displaced = weight of solid

Question 10

The work done in blowing a bubble of volume V is W, then what is the work done in blowing a soap bubble of volume 2V ?

Accepted Answer

If volume is doubled, then radius becomes (2)31​ times.
 or ​W=T(8πR2)W∝R2​
 Therefore W′=(2)32​ W

Question 11

A block of wood has a mass of 25 g. When a 5 g metal piece with a volume of 2 cm3 is attached to the bottom of the block, the wood barely floats in water. What is the volume V of the wood?

Accepted Answer

Weight = upthrust
∴(25+5)g=(V+2)ρw​g
Putting ρw​=1 g/cm3
We get, V=28 cm3

Question 12

What is the minimum volume of a block of wood (density =850 kg/m3 ) if it is to hold a 50 kg woman entirely above the water when she stands on it?

Accepted Answer

Weight of both (block + woman)
= upthrust on 100% volume of block ∴50g+V(850)g=V(1000)g∴V=0.33 m3​

Question 13

A cubical block of ice floating in water has to support a metal piece weighing 0.5 kg. What can be the minimum edge of the block so that it does not sink in water? Specific gravity of ice =0.9.

Accepted Answer

Weight of ice block + weight of metal piece = upthrust on 100% volume of ice cube.
Let a= side of ice cube. Then,
∴ or ​a3=(5×10−3)m3a≈0.17 m or 17 cm​[(a3)×900×g+0.5 g]=(a3)(1000)(g)

Question 14

The time period of a simple pendulum is T. Now the bob is immersed in a liquid of density σ. If density of material of bob is ρ, what will be the new time period of the pendulum.

Accepted Answer

Buoyant force due to liquidThe net force on the bob is⇒F=mg−VρgFor small oscillations​⇒FR​=(mg−Vρg)sinθ As [sinθ=lx​]∴FR​=(mg−Vρg)lx​⇒a=−(g−mVρg​)lx​​As we know, mass = density × volume⇒m=σ×VSubstitute the value of m in the above equation, we get​⇒a=−(g−VσVρg​)lx​⇒a=−g(1−σρ​)lx​​The general equation of SHM is,⇒a=−w2x(2)By equating the above equation, we get⇒ω=g(1−σρ​)​⇒T=ω2π​⇒T=2πg(1−σρ​)l​​

Question 15

A vessel contains oil ( density =0.8gm/cm3 ) over mercury (density =13.6gm/cm3 ). A uniform sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of sphere in gm/cm3 is:

Accepted Answer

For floating we useWeight = Buoyant forceVρm​g=2V​ρHg​g+2V​ρoil ​ρm​​=2ρHg​+ρoil ​​=213.6+0.8​=214.4​=7.2​

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