A uniform solid brass sphere is rotating with angular speed `omega_(0)` about a diameter. If its temperature is now increased by `100^(@)C`. What will be its new angular speed? (Given `alpha_(B)=2.0xx10^(-5) per^(@)C`)

Options

A. `1.1omega_0`
B. `1.01omega_0`
C. `0.996omega_0`
D. `0.824omega_0`

Answer

C

Explanation

Due to increase in temperature, radius of the sphere changes.
Let `R_0` and `R_(100)` be radius of sphere at `0^@C` and `100^@C`, respectively. `R_(100)=R_0(1+alphaxx100)`
Squaring both the sides and neglecting higher terms
`R_(100)^2=R_0^2(1+2alphaxx100)`
By the law of consevation of angular momentum
`I_1omega_1=I_2omega_2`
`implies(2)/(5)MR_0^2omega_1=(2)/(5)MR_(100)^2omega_2`
`impliesR_0^2omega_1=R_0^2(1+2xx2xx10^-5xx100)omega_2`
`impliesomega_2=(omega_1)/((1+4xx10^-3))=(omega_0)/(1.004)=0.996omega_0`