91Ó°ÊÓ

1. Radius of solid cylinder,$r_1=2.5\text{ c³¾or}0.025\text{″¾}$

1. Length of solid cylinder,$h_1=5.0\text{ c³¾or}0.05\text{″¾}$
2. Radius of solid cylinder after stretching,$r_2=0.50\text{ c³¾or}0.005\text{″¾}$
3. Temperature of solid cylinder,$T={30}^{\text{0}}\text{C}or303\text{‿é}$
4. Temperature of surroundings,$T_{env}=50\text{°°ä´Ç°ù}323\text{K}$

Thermal radiation is the process of transferring heat through electromagnetic radiation that is produced by the thermal motion of matter particles. We use the concept of radiation. Using the equation of the rate at which an object emits energy via thermal radiation and the equation of the rate at which the object absorbs energy and substituting these two equations in the equation of net rate of energy exchange, we can calculate the cylinder’s net thermal radiation rate.

Formulae:

The total surface area of the cylinder, $A=2\mathrm{Ï€}{r}^2+2\mathrm{Ï€}rh$ …(¾±)

The rate at which the sphere emits thermal radiation, $P_{rad}=\mathrm{σ}\mathrm{Ï\mu }A{T}^4$ …(¾±¾±)

where,$\mathrm{σ}$is the Stefan–Boltzmann constant and is equal to$(5.67×{10}^{−8}{\text{ W/³¾}}^{\text{2}}{\text{K}}^{\text{4}})$

The net rate of the radiation, $P_{net}=P_{abs}−P_{rad}$ …(¾±¾±¾±)

The surface area of the cylinder is given by using equation (i) as:

$\begin{array}{c}{A}_1=2\mathrm{π}{(2.5×{10}^{−2}\text{m})}^2+2\mathrm{π}(2.5×{10}^{−2}\text{m})(5.0×{10}^{−2}\text{m})\\ =1.18×{10}^{−2}{\text{m}}^{\text{2}}\end{array}$

The rate at which an object emits energy via electromagnetic radiation is given by using equation (ii) as:

$P_{rad}=\mathrm{σ}∈A{T}^4$

And the rate at which an object absorbs energy via thermal radiation from its environment, which we take to be at uniform temperature$T_{env}$(in Kelvin’s), is given using equation (ii) as:

$P_{abs}=\mathrm{σ}\mathrm{Ï\mu }A{T}_{env}^4$

Because an object both emits and absorbs thermal radiation, its net rate of energy exchange due to thermal radiation, the net rate of energy radiation is given using equation (iii) as:

$\begin{array}{c}{P}_{net}=\mathrm{σ}\mathrm{Ï\mu }A(T_{env}^4−T^4)\\ =(5.67×{10}^{−8}{\text{ W/³¾}}^{\text{2}}{\text{K}}^{\text{4}})\left(0.85\right)(1.18×{10}^{−2}{\text{m}}^{\text{2}})({\left(323\text{‿é}\right)}^4−{\left(303\text{‿é}\right)}^4)\\ =1.4\text{ W}\end{array}$

Hence, net energy is absorbed by radiation.

$P_{net}$is positive if the net energy is being absorbed via radiation and negative if it is lost via radiation.

Let the new height of the cylinder be$h_2$.Since the volume V of the cylinder is fixed, we have the equation as:

$\mathrm{Ï€}{r}_1^2{h}_1=\mathrm{Ï€}{r}_2^2{h}_2$

We solve for the height$h_2:$

$\begin{array}{c}{h}_2={\left(\frac{r_1}{r_2}\right)}^2{h}_1\\ ={\left(\frac{2.5\text{cm}}{0.50\text{cm}}\right)}^2\left(5.0\text{cm}\right)\\ =125\text{ c³¾}\\ =1.25\text{″¾}\end{array}$

The corresponding new surface area$A_2$of the cylinder is given using equation (i) as:

$\begin{array}{c}{A}_2=2\mathrm{Ï€}{(0.50×{10}^{−2}\text{m})}^2+2\mathrm{Ï€}(0.50×{10}^{−2}\text{m})\left(1.25\text{″¾}\right)\\ =3.94×{10}^{−2}{\text{m}}^{\text{2}}\end{array}$

Consequently, the ratio of the rates of energy is given as:

$\begin{array}{c}\frac{P_2}{P_1}=\frac{A_2}{A_1}\\ =\frac{3.94×{10}^{−2}{\text{m}}^{\text{2}}}{1.18×{10}^{−2}{\text{m}}^{\text{2}}}\\ =3.3\end{array}$

Hence, the value of the required ratio is$3.3$