(i) This is because a body with large reflectivity is poor absorber of heat, and poor aborbers of heat are emitters.
(ii) When we touch a brass tumber on a chilly day, heat flows from our body to the tumbler quickly (as thermal conductivity of brass is very high) and as a result, it appears colder.
(iii) On the other hand, as the wood a bad conductor, heat does not flow to the wooden tray from our body, on touching it.

\(F=-\eta A\frac { dv }{ dx } \)
n magnitude, \(\eta =\frac { F }{ A } \frac { dv }{ dx } \)
\(=\frac { \left( 2.5N \right) \quad \left( 2.5\times { 10 }^{ -3 }m \right) }{ \left( 2.5\times { 10 }^{ -4 }{ m }^{ 2 } \right) \left( 3\times { 10 }^{ -2 }{ ms }^{ -1 } \right) } \)
= 0.083\(\times\)10³ Nm^-2s

Let each side of the cube be l. The volume occupied by 3 cm depth of cube,
V  = (3cm) \(\times\) l² = 3l²cm
According to the principle of floatation, we have
V\(\rho\)g = mg \(\Rightarrow\) V\(\rho\) = m
\(\rho\) is density of water = 1000 kg m^-3
\(\Rightarrow\) (3l²\(\times\)10^-2m) \(\times\)(1000 kgm^-3) = 300\(\times\)10^-3 kg
\({ l }^{ 2 }=\frac { 300\times { 10 }^{ -3 } }{ 3\times { 10 }^{ -2 }\times 1000 } { m }^{ 2 }\Rightarrow { l }^{ 2 }=100\times { 10 }^{ -4 }{ m }^{ 2 }\)
l = 10\(\times\)10^-2 m = 10 cm
Therefore, volume of cube V = l³ = 1000 cm³

Since, the diameter of the pistons are given, we can calculate the radius of the piston
r = \(\frac{D}{2}\)
Area of smaller piston, A₁ = \(\pi{ \left( \frac { 5 }{ 2 } \right) }^{ 2 }=\pi { \left( 2.5 \right) }^{ 2 }\)
Area of larger piston,  A₂ = \(\pi{ \left( \frac { 60 }{ 2 } \right) }^{ 2 }=\pi { \left( 30 \right) }^{ 2 }\)
\({ F }_{ 2 }=\frac { { A }_{ 2 } }{ { A }_{ 1 } } \times { F }_{ 1 }=\left( 50N \right) \times { \left( \frac { 30 }{ 2.5 } \right) }^{ 2 }=7200N\)
This means, with the force of 50 N, the force of 7200 N can be lifted.

Radius of the sphere R = 1.5 cm
mass m = 0.038 kg
Volume of the sphere V = \(\frac{4}{3}\pi{R^2}\)
= \(\frac{4}{3}\)\(\times\)(3.14)\(\times\)(1.5\(\times\)10^-2)³ = 1.413\(\times\)10^-5m³
Therefore, density
\(\rho =\frac { m }{ V } =\frac { 0.038kg }{ 1.413\times { 10 }^{ -5 }{ m }^{ 3 } } =2690{ kg \ m }^{ -3 }\)
Hence, the specific gravity of the sphere
\(=\frac { 2690 }{ 1000 } =2.69\)

\({ a }_{ 1 }v_{ 1 }{ =30a }_{ 2 }{ v }_{ 2 }\Rightarrow { \pi { r }_{ 1 }^{ 2 }v }_{ 1 }=30{ \pi { r }_{ 2 }^{ 2 }v }_{ 2 }\)
\({ v }_{ 2 }=\frac { 1 }{ 30 } { \left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) }^{ 2 }{ v }_{ 1 }\Rightarrow { v }_{ 2 }=\frac { 1 }{ 30 } \times { \left( \frac { 0.8\times { 10 }^{ -2 }m }{ 0.4\times { 10 }^{ -2 }m } \right) }^{ 2 }\times \left( 0.33{ ms }^{ -1 } \right) \)
v₂ = 0.044 m s^-1

Capillary descent, cos140 = cos(90+50)–sin50 = –0.7660
\(h=\frac { 2Tcos\theta }{ r\rho g } =\frac { 2\times ({ 0.465N \ m }^{ -1 })({ cos \ 140 }^{ 0 }) }{ \left( 2\times { 10 }^{ -3 }m \right) \left( 13.6\times { 10 }^{ 3 } \right) \left( { 9.8 }ms^{ -2 } \right) } \)
\(=\frac{2 \times 0.456 \times(-0.7660)}{2 \times 13.6 \times 9.8} \)
\(=\frac{-0.6986}{266.56}=-2.62 \times 10^{-3} \mathrm{~m}\)
where, negative sign indicates that there is fall of mercury (mercury is depressed) in glass tube.

From equation (7.34), we have 
h∝\(\frac { 1 }{ r } \Rightarrow hr=\)constant
Consider two capillary tubes with radius r₁ and r₂ which on placing in a liquid, capillary rises to height h₁ and h₂, respectively. Then,
h₁r₁ = h₂r₂ = constant
\(\Rightarrow { h }_{ 2 }=\frac { { h }_{ 1 }{ r }_{ 1 } }{ { r }_{ 2 } } =\frac { \left( 2\times { 10 }^{ -2 }m \right) }{ \frac { r }{ 3 } } \Rightarrow { h }_{ 2 }=6{ \times 10 }^{ 2 }m\)

The excess of pressure inside the soap bubble is
\(\Delta P={ P }_{ 2 }-{ P }_{ 1 }=\frac { 4T }{ R } \)
But \(\Delta P={ P }_{ 2 }-{ P }_{ 1 }=\rho gh\Rightarrow \rho gh=\frac { 4T }{ R } \)
\(\Rightarrow\) Surface tension,
\(T=\frac { \rho ghR }{ 4 } =\frac { (800)(9.8)(4\times { 10 }^{ -3 })\left( 2\times { 10 }^{ -2 } \right) }{ 4 } \)=T= 15.68\(\times\)10^-2 N m^-1

A soap bubble has two free surfaces,
therefore increase in surface area \(\Delta\)A = A₂ − A₁ = 2(100 - 50)\(\times\)10^-4m² = 100\(\times\)10^-4m².
Since, work done W = T\(\times\) \(\Delta\)A \(\Rightarrow\) T \(=\frac { W }{ \Delta A }\)
\( =\frac { 2.4\times 10^{ -4 }J }{ 100\times { 10 }^{ -4 }{ m }^{ 2 } } =2.4\times { 10 }^{ -2 }Nm^{ -1 }\)