A) 4.0 mm
B) zero
C) 5.0 mm
D) 3.0 mm.
Correct Answer: D
Solution :
Let \[\rho \]and \[\sigma \]be the density of the liquid and material of the load respectively. In first case, the extension in the wire is \[x=Mg/k=V\rho g/k\] ?.(i) When the load is immersed in the liquid, up thrust + internal force due to extension in wire = weight of the load \[\Rightarrow V\sigma g+k{{x}_{1}}=V\rho g\Rightarrow {{x}_{1}}=Vg(\rho -\sigma )/k\] Using (i) and (ii), \[{{x}_{1}}=\frac{Vgx}{Vg\rho }(\rho -\sigma )=x\left( 1-\frac{\sigma }{\rho } \right)=4\times \left( 1-\frac{2}{8} \right)=3mm\]
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