Question

A solid steel shaft has to transmit 100 KW at 160 r.p.m. Taking allowable shear stress as 70 MPa,
find the suitable diameter of the shaft. The maximum torque transmitted in each revolution exceeds the
mean by 20%.

Answer 1

Step-by-step explanation:

Revolution per minute is 160 RPM. Allowable shear stress is 70 Mpa. Maximum torque is 20% more than the mean torque. Thus, the diameter of the shaft is 80.5 mm.

Answer 2

Step-by-step explanation:

Divide Rs 760 among A,B,C such that A gets 5/6 of what B gets and the ratio between the shares of B & C is 3:4 . [Hint:Let B's share be Rs 3x and C's share be Rs 4x. Then,A's share=Rs(5/6×3x) =Rs 5x/2.

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