A cube of wood of mass 0.5 kg and density ` 800 kg m^ (-3) ` is fastened to the free end of a vertical spring of spring constant ` k = 50 Nm^(-1) ` fixed at the bottom. Now, the entire system is completely submerged in water. The elongation or compression of the spring in equilibrium is ` (beta )/(2) cm `. Find the value of ` beta ` . (Given ` g = 10 ms ^(-2) ` )

Answer

5

Explanation

`rho_(s)ltrho_(w)` , So block tends to move up and thus the spring has elongation.
`F_(B)=mg+kx` (in equilibrium). `Vrho_(w)=g=Vrho_(s)g+kx`
`(0.5)/(800)xx1000xx10=(0.5)(10)+50x`
`x=0.025m=2.5cm=(5)/(2)cm" "Rightarrow" "beta=5`