The temperature of a thin uniform ciruclar disc ,of 1 m diameter is increased by `10^(@)C`. The percenrtage increase in moment of the disc about an axis passing through its centre and perpendicular to the circular face ( linear coefficient of expansion = `11xx10^(-6//@C)

Options

A. 0.0055
B. 0.011
C. 0.022
D. 0.044

Answer

C

Explanation

Coefficient of Areal expansion, `beta = (A_(2)-A_(1))/(A_(1)(Delta t))`
`DeltaA = A_(1)(2alpha)(Deltat)`
`implies DeltaA = pi (0.5)^(2) (2 xx 11 xx 10^(-6)) (10)`
`Delta A = 172.78 xx 10^(-6) m^(2)`
Total area of the disc `A' = A + Delta A `
` implies A' = pi (0.5)^(2) + (172.78 xx 10^(-6))`
`implies A' = 0.786`
`therefore r^(3) = 0.500055 `
Percentage increase in moment of inertia ,
`(T-I)/(I) = ((0.500055)^(2) - (0.5) ^(2) ) / (0.5)^(2)xx100 = 0.022 %`