**Text:**Each of the gears \( A \) and \( B \) has a mass of 10 kg and a radius of gyration of 190 mm, while gear \( C \) has a mass of 2.5 kg and a radius of gyration of 80 mm. Consider that a couple \( M \) of constant magnitude 10 N·m is applied to gear \( C \).**Diagram Explanation:**The diagram shows three gears labeled \( A \), \( B \), and \( C \). - Gears \( A \) and \( B \) each have a diameter of 250 mm. - Gear \( C \) has a smaller diameter of 100 mm and is positioned below and to the right of gears \( A \) and \( B \).- A couple \( M \) is applied to gear \( C \), indicated by an arrow and labeled "10 N·m."**Instructions:**Determine the number of revolutions of gear \( C \) required for its angular velocity to increase from 450 rpm to 1800 rpm. (You must provide an answer before moving on to the next part.)The number of revolutions of gear \( C \) required for its angular velocity to increase from 450 rpm to 1800 rpm is [_______] rev.

Gear A IA = mAKa2 = IB IC = 2.5 x 0.082 = 0.016 kg-m2 = 10 x 0.1902 = 0.361 kg-m2WArA = WBrB = WCrCWA = rCrA x WC = WB∴ WA = WB = 100250 x WC∴ WA = WB = 0.4 WCInitially WC1 = 450 rpm = 47.12 rad/sec WA1 = 18.84 rad/sec = WB1 Final WC2 = 1800 rpm = 188.495 Rad/sec WA2 = WB2 = 75.398 Rad/secTotal energy of system initially T1 = ΣIW122 Final Position T2 = ΣIW222 V1-2 = MθcFrom work energy principle T1 + V1-2 = T212 IA x WA121 + 12 IC x WC12 + 12 IB x WB12 + 10 x θC = 12 IA x WA2 + 12 IC x WC2 + 12 IB x WB22 x 0.361 x 18.8422 + 0.016 x 47.1222 + 10 θC = 2 x 0.361 x 75.39822 + 0.016 x 188.49522146.016 + 10 θC = 2336.46θC = 219.04 rad/sec