Q1: The maximum shear stress in a solid shaft of circular cross-section having diameter d subjected to a torque T is τ. If the torque is increased by four times and the diameter of the shaft is increased by two times, the maximum shear stress in the shaft will be

ANS:C - τ/2

T/IP = t/r = Gπ/L.

Ip= πd^4/32.
t= shear stress.


Now, new Torque = 4T and new diameter = 2d.
So new Ip= π(2d)^4/32.

t= (4T/new Ip)*r.
Also,
Here r is new i.e 2d/2 = d,
So r=d.

Old r is d/2,

Upon solving you get 8T/πd^3.

Also upon solving the initial case you get
16T/πd^3.

Clearly, the new conditions have shear stress equal to half of the one in the first case.