Question

The lower surface of the cube floating in water just touches the free end of a vertical spring fixed at the bottom of the vessel. The maximum weight that can be put on the block without wetting it is-

- 0.20 N
- 0.30 N
- 0.35 N
- 0.25 N

Solution

The correct option is **C** 0.35 N

Let the mass has kept on the block be m .

FBDs of block in initial and final cases:

From initial case :

On applying equilibrium condition along vertical direction,

⇒Fb−Mg=0

⇒Fb=Mg

⇒ρgV=ρ′V′g

where V→Volume of water displaced;V′→Volume of cube

⇒1000×10×(0.03)2×x=800×(0.03)3×10

⇒x=0.024 m ..(1)

From final case:

Block will be completely immersed and mass kept over it will not get wet. i.e depth of immersion will be 3 cm

∴ compression in spring = Change in depth of immersion

⇒y=0.03−0.024 m

∴y=0.006 m

On applying equilibrium condition along vertical direction,

Fs+F′b−(M+m)g=0

⇒ky+ρgV′−(ρ′V′g+mg)=0

⇒(50×0.006)+(1000×10×(0.03)3)=(800×(0.03)3+m)×10

⇒0.57=(0.0216+m)×10

⇒m=0.0354 kg

Now, weight, W=mg=0.0354×10

∴W≃0.35 N

Maximum weight that can be put on block without wetting it is 0.35 N.

Let the mass has kept on the block be m .

FBDs of block in initial and final cases:

From initial case :

On applying equilibrium condition along vertical direction,

⇒Fb−Mg=0

⇒Fb=Mg

⇒ρgV=ρ′V′g

where V→Volume of water displaced;V′→Volume of cube

⇒1000×10×(0.03)2×x=800×(0.03)3×10

⇒x=0.024 m ..(1)

From final case:

Block will be completely immersed and mass kept over it will not get wet. i.e depth of immersion will be 3 cm

∴ compression in spring = Change in depth of immersion

⇒y=0.03−0.024 m

∴y=0.006 m

On applying equilibrium condition along vertical direction,

Fs+F′b−(M+m)g=0

⇒ky+ρgV′−(ρ′V′g+mg)=0

⇒(50×0.006)+(1000×10×(0.03)3)=(800×(0.03)3+m)×10

⇒0.57=(0.0216+m)×10

⇒m=0.0354 kg

Now, weight, W=mg=0.0354×10

∴W≃0.35 N

Maximum weight that can be put on block without wetting it is 0.35 N.

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