An iron ball has a diameter of `6 cm` and is `0.010 mm` too large to pass through a hole in a brass plate when the ball and plate are at a temperature of `30^@ C`. At what temperature, the same for ball and plate, will the ball just pass through the hole?
Take the values of (prop) from Table `20 .2`.

An iron ball of diameter 6cm and is 0.01 mm too large to pass through a hole in a brass plate when the ball and plate are at a temperature of 20^(0)C . The temperature at which (both for ball and plate) the ball just passes through the hole is (alpha_(iron) = 12 xx 10^(-6 //0)C, alpha_(brass) = 18 xx 10^(-6 //0)C)

An iron ball of diameter 6cm and is 0.01 mm too large to pass through a hole in a brass plate when the ball and plate are at a temperature of 20^(@) C . The temperature at which (both for ball and plate) the ball just pass through the hole is (alpha_("iron") = 12 xx 10^(-6) //1^(@) C, alpha_("brass") = 18 xx 10^(-6) //1^(@) C)

An iron ball (coefficient of linear expansion =1.2xx(10^(-5)//^(@)C ) has a diameter of 6 cm and is 0.010 mm too large to pass through a hole in a brass plate (coefficient of linear expansion =1.9xx10^-5//^(@)C ) when the ball and the plate are both at a temperature of 30^@C At what common temperature of the ball and the plate will the ball just pass through the hole in the plate ?

A ball does not pass through a copper ring at room temperature. The same ball, after heating, passes through the same ring. It shows due to heating, the ball

A ball does not pass through a copper ring at room temperature. The same ball, after heating, passes through the same ring. It shows due to heating, the ball

A cylinder of diameter exactly 1 cm at 30^@C is to be slided into a hole in a steel plate. The hole has a diameter of 0.99970 cm at 30^@C . To what temperature must the plate be heated? For steel alpha = 1.1 xx 10^-5 ^@C^-1