SSC JE Mechanical Engineering 2013 Paper

Question : 114 of 200

Marks: +1, -0

A circular shaft can transmit a torque of 13 kN-m. If the torque is reduced to 12 kN-3, then the maximum value of bending moment that can be applied to the shaft is

1 kN-m
3 kN-m
5 kN-m
7 kN-m

Validate

Solution:

T_ e=√{m^2 +T^2}

13=√{12^2+m^2}

13^2=M^2+12^2

⇒

M^2=13^2-12^2=169-144

m = 5

KN- m

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