SSC JE Mechanical Engineering 2012 Paper

Question : 183 of 200

Marks: +1, -0

A circular shaft of length L and diameter D is fixed in one end. The other end of the shaft is subjected to a twisting moment T. The angle of rotation of the free end with respect to the fixed end is θ. What will be the angle of rotation of the free end if length is doubled and diameter is reduced to

D /2

16 θ
32 θ
$θ/2$
2 θ

Validate

Solution:

T/J={Gθ}/l⇒θ_1={Tl}/{GJ}

={T×l}/{G×({πd^2}/32)}={32Tl}/{G×πd^4}

when

l=2l,d=d/2

θ_2={T×2l}/{G×({πd^2}/{32×16})}= 32 ×(θ_1)

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