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A 14.5 Kg Mass, Fastened to the End of a Steel Wire of Unstretched Length 1.0 M, is Whirled in a Vertical Circle with an Angular Velocity of 2 Rev/S at the Bottom of the Circle. the Cross-sectional Area of the Wire is 0.065 Cm2. Calculate the Elongation of the Wire When the Mass is at the Lowest Point of Its Path - CBSE (Science) Class 11 - Physics

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Question

A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.

Solution 1

Mass, m = 14.5 kg

Length of the steel wire, l = 1.0 m

Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s

Cross-sectional area of the wire, a = 0.065 cm2 = 0.065 × 10-4 m2

Let Δl be the elongation of the wire when the mass is at the lowest point of its path.

When the mass is placed at the position of the vertical circle, the total force on the mass is:

F = mg + mlω2

= 14.5 × 9.8 + 14.5 × 1 × (12.56)2

= 2429.53 N

Young's modulus = Stress/Strain

`Y = (F/A)/trianglel = F/A  l/trianglel`

`:.trianglel = (Fl)/(AY)`

Young’s modulus for steel = 2 × 1011 Pa

`trianglel = (2429.53xx1)/(0.065xx10^(-4)xx2xx10^11)`

`=>trianglel = 1.87 xx 10^(-3) m`

Hence, the elongation of the wire is 1.87 × 10–3 m.

Solution 2

Here, m = 14.5 kg; l = r = 1 m; v = 2 rps; A = 0.065 x 10-4 m2 Total pulling force on mass, when it is at the lowest position of the vertical circle is F = mg + mr w2 = mg + mr 4,π2 v2

`=14.5 xx 9.8 + 14.5 xx 1 xx 4 xx (22/7)^2 xx 2^2`

=142.1 + 2291.6 = 2433.9 N

`Y = F/A xx l/(trianglel)`

or `trianglel = (Fl)/(AY)= (2433.7xx1)/(0.065 xx 10^(-4)xx(2xx10^11)) = 1.87 xx 10^(-3) m = 1.87 m`

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 NCERT Solution for Physics Textbook for Class 11 (2018 to Current)
Chapter 9: Mechanical Properties of Solids
Q: 11 | Page no. 244
Solution A 14.5 Kg Mass, Fastened to the End of a Steel Wire of Unstretched Length 1.0 M, is Whirled in a Vertical Circle with an Angular Velocity of 2 Rev/S at the Bottom of the Circle. the Cross-sectional Area of the Wire is 0.065 Cm2. Calculate the Elongation of the Wire When the Mass is at the Lowest Point of Its Path Concept: Elastic Moduli - Young’s Modulus.
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