JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Elasticity & SHM

The three elastic constants are tied together by the single relation $\frac{1}{Y}=\frac{1}{9K}+\frac{1}{3\eta }$. To isolate the bulk modulus, clear the fractions in $Y(3K+\eta )=9K\eta$ to get $3KY+Y\eta =9K\eta$, then collect the terms containing $K$: $Y\eta =9K\eta -3KY=K(9\eta -3Y)$. Dividing by $(9\eta -3Y)$ gives $K=\frac{Y\eta }{9\eta -3Y}$, which is the correct rearrangement of the standard relation between Young's modulus, the bulk modulus and the modulus of rigidity.