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Mechanical Properties of Solids

Modulus of Elasticity and Hooke's Law

A 6 {~kg} weight is fastened to the end of a steel wire of uns

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Medium

Solving time: 3 mins

A $6 kg$ weight is fastened to the end of a steel wire of unstretched length $60 cm$ . It is whirled in a vertical circle and has an angular velocity of $2 rev / s$ at the bottom of the circle. The area of cross-section of the wire is $0.05 cm^{2}$ . Calculate the elongation of the wire when the weight is at the lowest point of the path. Young's modulus of steel $= 2 \times 1 0^{11} N / m^{2}$ .

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Book:

Mechanics Volume 2 (DC Pandey)

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Henry Discussed

6 kg

weight is fastened to the end of a steel wire of unstretched length

60 cm

. It is whirled in a vertical circle and has an angular velocity of

2 rev / s

at the bottom of the circle. The area of cross-section of the wire is

0.05 cm^{2}

. Calculate the elongation of the wire when the weight is at the lowest point of the path. Young's modulus of steel

= 2 \times 1 0^{11} N / m^{2}

8 mins ago

Discuss this question LIVE

8 mins ago

Text solutionVerified

T - m g T Now, Δ l = m l ω^{2} = m l (2 π f)^{2} = m g + 4 π^{2} m l f^{2} = 6 \times 9.8 + 4 π^{2} (6) (0.6) (2)^{2} = 628 N = \frac{Tl}{A Y} = \frac{( 628 ) ( 0.6 )}{( 0.05 \times 1 0 ^{- 4} ) ( 2 \times 1 0 ^{11} )} = 3.8 \times 1 0^{- 4} m

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Question Text	A $6 kg$ weight is fastened to the end of a steel wire of unstretched length $60 cm$ . It is whirled in a vertical circle and has an angular velocity of $2 rev / s$ at the bottom of the circle. The area of cross-section of the wire is $0.05 cm^{2}$ . Calculate the elongation of the wire when the weight is at the lowest point of the path. Young's modulus of steel $= 2 \times 1 0^{11} N / m^{2}$ .
Topic	Mechanical Properties of Solids
Subject	Physics
Class	Class 11
Answer Type	Text solution:1
Upvotes	76

Text solutionVerified

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