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प्रश्न
A cylinder of length 1.5 m and diameter 4 cm is fixed at one end. A tangential force of 4 × 105 N is applied at the other end. If the rigidity modulus of the cylinder is 6 × 1010 Nm-2 then, calculate the twist produced in the cylinder.
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उत्तर
Torsion of a cylinder τ = `(πη"r"^4Φ)/(2"l")`
Twist produced in the cylinder Φ = `(τ xx (2"l"))/(πη"r"^4) = (("F" xx "l") xx (2"l"))/(πη"r"^4)`
= `(4 xx 10^5 xx 2 xx (1.5)^2)/(3.14 xx 6 xx 10^10 xx (2 xx 10^-2)4)`
= `(18 xx 10^5)/(3.01 xx 10^-6 xx 10^10)`
Φ = 59.80
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