  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 N0.30 N0.35 N0.25 N

Solution

## The correct option is C 0.35 NLet 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.  Suggest corrections   