A sphere and a cube of same material and same total surface area are placed in the same evaculated space turn by turn after they are heated to the same temperature. Find the ratio of their initial rates of cooling in the enclosure.

Options

A. `sqrt((pi)/(6)):1`
B. `sqrt((pi)/(3)):1`
C. `(pi)/(sqrt(6)):1`
D. `(pi)/(sqrt(3)):1`

Answer

A

Explanation

`Q_(1)=Q_(2)` as surface area and material is same
So, `m_(1)s((dQ)/(dt))_(1)m_(2)s((dQ)/(dt))_(2)`
So, `So,((dQ)/(dt))_(1)//((dQ)/(dt))_(2)=(m_(2))/(m_(1))=(V_(2))/(V_(1)) "but " 4pir^(2)=6a^(2)`
so , `r=(sqrt((6)/(4pi)))a" so",((4)/(3)pir^(3))/(a^(3))=(4)/(3)pi((6)/(4pi))^(3//2)`
or `,((dQ)/(dt))_(1)//((dQ)/(dt))_(2)=(a^(3))/((4)/(3)pir^(3))=((4pi)^(1//2).3)/((6)^(3//2))=(sqrt(pi))/(sqrt(2)*sqrt(3))=sqrt((pi)/(6))`