Question

A solid sphere of radius 1 m and density 1000 kg/m3 is attached to one end of a massless spring of force constant 5000 N/m. The other end of the spring is connected to another solid sphere of radius 1 m and density 3000 kg/m3. The complete arrangement is placed in a liquid of density 2000 kg/m3 and is allowed to reach equilibrium. The correct statement(s) is(are):

- The net elongation of the spring is 8π3 m
- The net elongation of the spring is 4π3 m
- The light sphere is partially submerged
- The light sphere is completely submerged

Solution

The correct options are

**A** The net elongation of the spring is 8π3 m

**D** The light sphere is completely submerged

Let us suppose, the net elongation of the spring is x.

According to question,

For heavier sphere, on applying ∑F=ma

kx+Fb−Mg=0 [Fb→ buoyancy force]

⇒kx+2ρgV−3ρgV=0 [V→ volume of sphere]

⇒kx=ρgV ... (1)

⇒x=ρgVk

⇒x=ρgk×43πR3

⇒x=1000×105000×43×π×13

⇒x=8π3 m

For lighter sphere, on applying ∑F=ma

F′b−kx−mg=0 [F′b→ buoyancy force]

⇒F′b=kx+mg

=ρgV+ρgV [from(1)]

⇒F′b=2ρgV ... (2)

Also, force of buoyancy on lighter sphere is maximum when it is fully submerged and is equal to 2ρgV.

From (2), as force of buoyancy is maximum,

therefore, it is completely submerged.

Let us suppose, the net elongation of the spring is x.

According to question,

For heavier sphere, on applying ∑F=ma

kx+Fb−Mg=0 [Fb→ buoyancy force]

⇒kx+2ρgV−3ρgV=0 [V→ volume of sphere]

⇒kx=ρgV ... (1)

⇒x=ρgVk

⇒x=ρgk×43πR3

⇒x=1000×105000×43×π×13

⇒x=8π3 m

For lighter sphere, on applying ∑F=ma

F′b−kx−mg=0 [F′b→ buoyancy force]

⇒F′b=kx+mg

=ρgV+ρgV [from(1)]

⇒F′b=2ρgV ... (2)

Also, force of buoyancy on lighter sphere is maximum when it is fully submerged and is equal to 2ρgV.

From (2), as force of buoyancy is maximum,

therefore, it is completely submerged.

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