1.50 cm length of pianos wire with a diameter of 0.25 cm is stretched by attaching a 10 kg mass to one end. How far is the wire stretched? Answer (1.5 x 10⁻⁴m) 



Given Data:

Length of the pianos wire = L =  1.50 cm = 0.015 m 

Diameter = D =  0.25 cm = 0.0025 m

Radius = r = D/2 = 0.0025/2 m = 0.00125 m

Mass attached = m = 10 kg 

Young's modulus of steel = Y = 20 ｘ10¹⁰ N m⁻²

value of gravitation acceleration = g = 9.81 m s⁻²


To Find:

Change in length = ΔL = ?



Solution:


The formula for Young's modulus

Y = StressStrainStressStrain -----------(1)

Where, Stress = FAFA  = mgㄫr²

[F = mg  and A = ㄫ r²] 


and 

Strain = ΔLL

So equation (1)


Y = mgㄫr² ｘLΔL

or

ΔL = mgㄫr² ｘLY

By putting the corresponding values


ΔL = 10kgｘ9.81ms⁻²3.1416ｘ(0.00125m)2 ｘ0.015m20ｘ10¹⁰Nm⁻²

ΔL = 98.1kgms⁻²4.908ｘ10⁻⁶m² ｘ0.015m20ｘ10¹⁰Nm⁻²

ΔL = 1.4715kgm²s⁻²98.16ｘ10⁴N

ΔL = 0.014991 ｘ10⁻⁴ m

or

ΔL = 1.4991 ｘ10⁻⁶ m  -------------------Ans.




************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.