(i) A = 4\(\times\)10^-6m² ; I = 0.5 m
ΔT = 100 - 0 = 100°C = 100 k
Y =10¹¹Nm^-2, a = 10^-5 k^-1
Change in length,
Δl = lαΔT
= 0.5\(\times\)10⁵ \(\times\)100
= 5\(\times\)10^-4m
as \(y=\frac{stress}{\Delta l/l}\)
stress = \(y=\frac{y\Delta l}{l}=y\times \alpha \Delta T\)
= 10¹¹\(\times\)10^-5\(\times\)100
= 10⁸ Nm^-2
Stretching force,
F = stress\(\times\)area = 10⁸ \(\times\)4\(\times\)10^-2
= 4510²N
but F = Mg  ∴ 4\(\times\)10^-2
\(M=\frac{4\times 10^{2}}{g}=\frac{4\times 10^{2}}{10}=40kg\)
(ii) Energy stored in the rod
\(=\frac{1}{2} F \times \Delta l\)
\(= \frac{1}{2} \times 4\times 10^{2} \times 5 \times 10^{-4}\) 
=10 J

The situation in diagram
Let A & B be a rod of length 2m suppose a weight w is hung at C at a distance x from A. Let T₁ & T₂ be the tension in the steel & brass rods.
(i) stress in steel wire = \(\frac{T_{1}}{A_{1}}\)
stress in brass wire = \(\frac{T_{2}}{A_{2}}\) as both the stresses are equal, so
\(\frac{T_{1}}{A_{1}}\)=\(\frac{T_{2}}{A_{2}}\) or \(\frac{T_{1}}{T_{2}}=\frac{A_{1}}{A_{2}}\)

\(= \frac{0.1}{0.2}=\frac{1}{2}\)
Now moments about C are equal system is in equilibrium
\(T_{1}=T_{2}(2-x) \ or \ \frac{T_{1}}{T_{2}}=\frac{2-x}{x}\)
\(\frac{1}{2}=\frac{2-x}{x}\)
x = 4-2x
3x = 4 or \(x=\frac{4}{3}=1.33 m\)
(ii) Now \(y= \frac{stress}{stram}\)
∴ strain = stress/ y
strain in steel wire = \(\frac{T_{1}/A_{1}}{y}\)
strain in brass wire = \(\frac{T_{2}/A_{2}}{y_{2}}\)
Now, \(\frac{T_{1}}{A_{1}y_{1}}=\frac{T_{2}}{A_{2}y_{2}}\)
\(\frac{T_{1}}{T_{2}}=\frac{A_{1}y_{1}}{A_{2}y_{2}}\)
=\(\frac{0.1 cm^{2}\times 20\times 10^{10}Nm^{-2}}{0.2 m^{2}\times 10\times 10^{10}Nm^{-2}}=1\)
again, T₁x = T₂(2-x)
\(1=\frac{2-x}{2}\) 
x = 2 - x
2x = 2 or x = 1m

Let V be the total volume of body
When body is floating in water, then
\(V {\rho_{body}} g =0.6\)
\(V \rho _{water }g \ \ (or) \rho _{water} = \frac{\rho _{body}}{0.6}\)
When body is floating in oil, then
\(V \rho_{body} g = 0.4 V \rho \ oil \ g\)
or \(\rho_{oil}=\frac{\rho_{body}}{0.4}\)
relative density of oil
\(= \frac{\rho_{oil}}{\rho _{water}}= \frac{\rho _{body }/0.4}{\rho_{body}/0.6}=\frac{6}{4}\)=1.5

(i) In the two wires, the wire A is more ductile. Because, the material of wire A have large plastic range of extension than the wire B.
(ii) In the two wires, the slope of graph in wire A is greater than the slope of graph in wire B, so wire A has greater young's modulus.
(iii) Wire A is stronger than wire B because it can withstand more load without breaking. For Wire A, the breakeven point is higher.
(iv) WireB is more brittle. Because the material of wire B have small range of extension than the wire A.