Mechanical Properties of Solids - Result Question 6

7. If the ratio of diameters, lengths and Young’s modulus of steel and copper wires shown in the figure are p,q and s respectively, then the corresponding ratio of increase in their lengths would be

[NEET Kar: 2013]

(a) 7q(5sp)

(b) 5q(7sp2)

(c) 7q(5sp2)

(d) 2q(5sp)

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Answer:

Correct Answer: 7. (c)

Solution:

  1. (c) From formula,

Increase in length ΔL=FLAY=4FLπD2Y

ΔLSΔLC=FSFC(DCDS)2YCYSLSLC=75×(1p)2(1s)q=7q(5sp2)

If a wire of length l is suspended from a rigid support and A is the area of cross-section of the wire, then mass of the wire, m=Alp. Then extension in the wire due to its own weight can be find as follows.

Young Modulus, Y=mgA×(l/2)Δl

Here, L=l/2 because the weight of wire acts at the mid point of wire.

Extension in the wire due to its own weight Δl=l2pg2Y



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