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Question

Ratio of the weights of a 1 kg block of iron and 1 kg block of wood as measured by a spring balance is:
Given: Density of iron =7800 kg/m3, density of wood =800 kg/m3 and density of air =1.293 kg/m3

A
1.5
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B
2.0015
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C
1.0015
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D
3.0015
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Solution

The correct option is C 1.0015
FBD of the block:


Applying equilibrium condition i.e F=0 along vertical direction. Considering air as a fluid, it will exert buoyant force on block.
T+Fbmg=0
T=mgFb
T=mgρagV ...(1)
[ρadensity of air]
From Eq.(i) for iron block:
T1=mgρagVi
where Vi volume of iron block
T1=(1×10)1.293×10×17800
[Vi=mρi]
T1=9.9983 N
Similarly, for wooden block ,
T2=mgρagVw
[Vw=mρw]
T2=(1×10)1.293×10×1800
T2=9.9838 N
On dividing T1 & T2,
T1T2=9.99839.9838
T1:T2=1.0015
Ratio of weights of iron and wooden block of mass 1 kg is 1.0015.

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