1.

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^-5per^@C`)A. `(omega_(0))/(1-0.002)`B. `(omega_(0))/(1+0.002)`C. `(omega_(0))/(1+0.004)`D. `(omega_(0))/(1-0.004)`

Answer» `I_(0) omega_(0) = I_(t) omegat`
`Mr_(0)^(2) omega_(0) = Mr_(0)^(2) (1+2 alpha Delta T) omega_(t)`
`omega_(t) = (omega_(0))/(1+0.004)`


Discussion

No Comment Found