A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.
Answers
Answered by
125
Young's modulus of copper(Y1) = 110× 10^9 N/m²
Young's modulus of steel (Y2) = 190× 10^9 N/m²
Let D1 and D2 have diametres of copper and steel respectively.
A/C to question ,
Tension are same for both.
We know,
Young's modulus= stress/strain
Y = F/A(∆L/L)
Y = FL/(πd²/4)∆L
Hence, Y is inversely proportional to square of diametre .
So,
Y1/Y2 = d2²/d1²
d1/d2 = √{Y2/Y1}
= √{ 190×10^9/110×10^9}
=√{19/11}
= 1.31
Hence, d1 : d2 = 1.31 : 1
Young's modulus of steel (Y2) = 190× 10^9 N/m²
Let D1 and D2 have diametres of copper and steel respectively.
A/C to question ,
Tension are same for both.
We know,
Young's modulus= stress/strain
Y = F/A(∆L/L)
Y = FL/(πd²/4)∆L
Hence, Y is inversely proportional to square of diametre .
So,
Y1/Y2 = d2²/d1²
d1/d2 = √{Y2/Y1}
= √{ 190×10^9/110×10^9}
=√{19/11}
= 1.31
Hence, d1 : d2 = 1.31 : 1
Answered by
29
Answer:
Explanation:
Y=stress/strain
Y=force/area×strain
Y=f/πxd^2×strain
D^2=force/π×strain×Y
D^2propertional to 1/Y
D proportional to 1/√Y
D iron/D copper=√Y copper/√Y iron
=√11×10^10/√19×10^10
=√11/19
=1.31
Hope it's help you
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