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  • A unifform solid cart Iron sphire is rotating. With angular speed  about a diameter. If its temperature is now increased by <img alt="53^{\circ

A unifform solid cart Iron sphire is rotating. With angular speed w_{o} about a diameter. If its temperature is now increased by 53^{\circ} \mathrm{C} What will be its new angular speed \text { (Given } \alpha_{\text {cost }}=10^{-5} \text { per }{ }^{\circ} \mathrm{C} \text { ) }
 

Option: 1

1.1 \omega_{0}


Option: 2

1.01 \omega_{0}


Option: 3

0.986 \omega_{0}


Option: 4

0.999 \omega


Answers (1)

best_answer

As there is no ext torque i.e \tau _{ext}= 0

I_{1} \omega_{1}=I_{2} \omega_{2}=\text { constant }

I_{1}=\frac{2 MR_{1}^{2}}{5}                R_{2}=R_{1}(1+\alpha \Delta T)

I_{2}=\frac{2}{5} M R_{2}^{2}                 \begin{aligned} &=R_{1}\left(1+10^{-5} \times 50\right) \\ R_{2} &=R_{1}\left(1+0.5 \times 10^{-3}\right) \\ R_{2}^{2} &=R_{1}^{2}\left(1+0.5 \times 10^{-3}\right)^{2}=R_{1}^{2}(1.001) \end{aligned}

\begin{aligned} &I_{1} \omega_{1}=I_{2} \omega_{2} \\ &\frac{R_{1}^{2}}{R_{2}^{2}}=\frac{\omega_{2}}{\omega_{1}} \Rightarrow \omega_{2}=\left(\frac{1}{1.001}\right) \omega_{1}=\frac{\omega_{0}}{1.001}=0.991 \omega_{0} \end{aligned}

 

Posted by

Deependra Verma

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