A metallic ball of radius 0.2 m is heated from 320 K to 522 K. What will be the increase in surface area of the ball?Given, coefficient of superficial expansion of the ball = 3.4 xx 10^(-5)K^(-1)

A metallic ball of radius 0.2 m is heated from 320 K to 522 K. What wi

1.	A metallic ball of radius 0.2 m is heated from 320 K to 522 K. What will be the increase in surface area of the ball?Given, coefficient of superficial expansion of the ball = 3.4 xx 10^(-5)K^(-1)
Answer» Solution :Given , RADIUS of ball at 320 K , `r_320 = 0.2 m` `DELTA T = 522 - 320 = 202 K` ` beta = 3.4 xx 10^(-5) K^(-1)` surface area of ball at 320 K, `S_320 = 4pi r_(320)^2` `= 4pi (0.2)^2 = (16pi)/(100) m^2` increase in surface area is `Delta S = S_(320) beta Detla T = (16pi)/(100) xx 3.4 xx 10^(-5) xx 202` ` = 3.452 xx 10^(-5) m^2`