Question

In the given figure a block of mass M and density $\rho$ is immersed in liquid of density $\sigma$. The block is attached to the spring from top and string from bottom. Elongation in the spring when string is tight is x and spring constant is $\frac{Mg}{3x}$.

$\begin{array}{cccccc}& Column1& & Column2& & Column3\\ & & & \left(Tensiondevelopedinstring\right)& & \left(Forcegeneratedinspring\right)\\ \left(I\right)& If\rho =2g/cc& \left(i\right)& T=\frac{4Mg}{3}& \left(P\right)& F=\frac{Mg}{3}\\ & \sigma =1g/cc& & & & \\ \left(II\right)& If\rho =1g/cc& \left(ii\right)& T=\frac{Mg}{3}& \left(Q\right)& F=0\\ & \sigma =1g/cc& & & & \\ \left(III\right)& If\rho =1g/cc& \left(iii\right)& T=\frac{Mg}{2}& \left(R\right)& F=\frac{Mg}{2}\\ & \sigma =2g/cc& & & & \\ \left(IV\right)& If\rho =1.2g/cc& \left(iv\right)& T=0& \left(S\right)& F=\frac{4Mg}{3}\\ & \sigma =0.8g/cc& & & & \end{array}$
Mark the correct combination.

• A. (II) (iv) (Q)
• B. (II) (iv) (P)
• C. (III) (i) (P)
• D. (I) (iii) (P)

Answer 1

The correct option is C (III) (i) (P)

$\rho =2g/cc\sigma =1g/cc{W}_{app}=Mg(1-\frac{\sigma }{\rho })=\frac{Mg}{2}$

Due to this elongation in spring is
$k{x}^{\prime }=\frac{Mg}{2}\frac{Mg}{3x}{x}^{\prime }=\frac{Mg}{2}{x}^{\prime }=\frac{3x}{2}\therefore T=0andF=\frac{Mg}{2}$
II. $W_{app}=0T=F=\frac{Mg}{3x}.x=\frac{Mg}{3}$
III. x' = x
$\therefore F=\frac{Mg}{3x}.x=\frac{Mg}{3}T=\frac{Mg}{3}+Mg=\frac{4}{3}Mg$

IV. $W_{app}=Mg(1-\frac{\sigma }{\rho })$
$=\frac{Mg}{3}F=\frac{Mg}{3}$
T = 0