Question

A wooden block of density at 860 kg/m³ at 0℃ is floating on benzene liquid of density 900 kg/m³ at 0℃ The temperature at which the block just submerge in benzene is\[\left[ {{\gamma _{wood}} = 8 \times {{10}^{ - 5}}{/^ \circ }_C,{\gamma _{benzene}} = 12 \times {{10}^{ - 4}}{/^ \circ }_C} \right]\]
A) 24℃
B) 42℃
C) 16℃
D) 32℃

Answer 1

Hint: It would submerge when the density is the same.

Complete step by step solution:
Given,
Density of wood at 0℃(ρw) = 860 kg/m³
Density of benzene at 0℃(ρb) = 900 kg/m³
\[
  {\gamma _{wood}} = 8 \times {10^{ - 5}}{/^ \circ }_C \\
  {\gamma _{benzene}} = 12 \times {10^{ - 4}}{/^ \circ }_C \\
\]
Initial temperature(T1) =0℃
Let T2 be the temperature at which the boat sinks in benzene.
∴∆T= T2- T1
The boat sinks in benzene when its weight = benzene displaced
Mass = volume x density
∴Vw ρw g = Vb ρb g (where Vw = volume of boat and Vb = volume of benzene)
or, Vw ρw = Vb ρb
(V=$\dfrac{1}{{1 + \gamma \Delta T}}$)
$
   \Rightarrow {\rho _w}\dfrac{1}{{1 + {\gamma _w}\Delta T}} = {\rho _b}\dfrac{1}{{1 + {\gamma _b}\Delta T}} \\
   \Rightarrow 860\dfrac{1}{{1 + \left( {8\times{{10}^{ - 5}}} \right)\Delta T}} = 900\dfrac{1}{{1 + \left( {12\times{{10}^{ - 4}}} \right)\Delta T}} \\
   \Rightarrow 43 + 0.0516\Delta T = 45 + 0.0036\Delta T \\
   \Rightarrow 0.048\Delta T = 2 \\
   \Rightarrow \Delta T = 41.6 \\
$
So, ∆T= T2- T1
T2=∆T- T1= (41.6-0) ℃=41.6℃
∴ The boat sinks in benzene when its temperature is 42℃ (approx.)

Correct Answer: B) 42℃

Note: The coefficient of volume expansion of an object at constant pressure is defined as the fraction of its volume at 0°C by which the volume of a fixed mass of object expands per degree Celsius rise in temperature.