Question

A wooden block of density 860 kg/m${}^3$ at $0^{0}C$ is floating on benzene liquid of density 900 kg/m${}^3$ at $0^{0}C$ . The temperature at which the block just submerge in benzene is $[{\gamma }_{wood}=8\times {10}^{-5}{/}^{0}c,{\gamma }_{benzene}=12\times {10}^{-4}{/}^{0}c]$

• A. 24${}^0$ c
• B. 42${}^0$ c
• C. 16${}^0$ c
• D. 32${}^0$ c

Answer 1

Answer: (B)

42${}^0$ c

It would submerge when the density is same.
$\therefore \frac{860}{1+{\gamma }_{wood}T}=\frac{900}{1+{\gamma }_{benzene}T}$
$\Rightarrow 43+0.0516T=45+0.0036T$
$\Rightarrow 0.048T=2$
$\therefore T={42}^{0}C$