Question

A wooden stick of length $100\text{cm}$ is floating in water while remaining vertical. The relative density of the wood is $0.7$. Then the apparent length of the stick when viewed from the top [close to the vertical line along the stick] is $\text{cm}$. Take refractive index of water to be $\frac{4}{3}$

Answer 1

Applying Archimedes principle,
Force of buoyancy is equal to weight of water displaced. So,
$\left[V_{\text{immersed}}\right]{\rho }_{w}g=\left[V_{\text{net}}\right]{\rho }_{\text{wood}}g$
$\Rightarrow V_{\text{immersed}}=\frac{{\rho }_{\text{wood}}}{{\rho }_{\text{water}}}.\left[V_{\text{net}}\right]=0.7V_{\text{net}}$
$30\%$ of the length of stick will lie outside the water. Then the lower end will be at a distance $70\text{cm}$ from the surface of the water.
Apparent depth of the object in denser medium as seen from the rarer medium is $D_{apparent}=\frac{D_{actual}}{\mu }$

Apparent depth of the lower end from water surface is,
$h=\frac{70}{\left(\frac{4}{3}\right)}\text{cm}=52.5\text{cm}$
$\therefore$ Apparent length of stick is = length of stick above water surface+Apparent depth of the lower end from water surface $\Rightarrow 30\text{cm}+52.5\text{cm}=82.5\text{cm}$