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JEE Foundation Course
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NCERT Solutions
The angle of twist of a closely helical spring under an axial torque is given by
$$\frac{64 Tdn}{ED^4}$$
$$\frac{32 Tdn}{ED^4}$$
$$\frac{32 TDn}{ED^4}$$
$$\frac{64 TDn}{Ed^4}$$
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