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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-


  1. 0.20 N
  2. 0.30 N
  3. 0.35 N
  4. 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,
FbMg=0
Fb=Mg
ρgV=ρVg
where VVolume of water displaced;VVolume 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=0.030.024 m
y=0.006 m
On applying equilibrium condition along vertical direction,
Fs+Fb(M+m)g=0
ky+ρgV(ρVg+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
W0.35 N
Maximum weight that can be put on block without wetting it is 0.35 N.

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