A solid cylinder of radius r₁=2.5 ~cm , length l₁=5.0 ~cm and temperature 40^ C is suspended in an environment of temperature 60^ C . The thermal radiation transfer rate for cylinder is 1.0 ~W . If the cylinder is stretched until its radius becomes r₂=0.50 ~cm , the thermal radiation transfer rate is changed to

Step-by-step solution

Keeping nature of surface and temperature of body and surroundings same, rate of heat transfer by radiation depends on area of body, Rate of heat transfer Surface area of body Q₂ Q₁ = S₂ S₁ Q₂=Q₁ S₂ S₁ (i) Now, volume remains same in stretching, so r₁^2 l₁= r₂^2 l₂ l₂= r₁^2 r₂^2 l₁= (2.5)^2 (0.5)^2 5=125 ~cm So, area before stretching, S₁=2 r₁ (l₁+r₁ ) and area after stretching, S₂=2 r₂ (l₂+r₂ ) Hence by Eq. (i), we get Q₂=Q₁ S₂ S₁ =Q₁ 2 r₂ (l₂+r₂ ) 2 r₁ (l₁+r₁ ) =Q₁ r₂ (l₂+r₂ ) r₁ (l₁+r₁ ) = 1 0.5 10⁻²(125+0.5) 10