A wire of cross-sectional area A is stretched horizontally between two clamps loaded at a distance 2l metres from each other. A weight w kg suspended from the midpoint of the wire. The strain produced in the wire, (if the vertical distance through which the midpoint of the wire moves down x

Question

A wire of cross-sectional area A is stretched horizontally between two clamps loaded at a distance 2l metres from each other. A weight w kg suspended from the midpoint of the wire. The strain produced in the wire, (if the vertical distance through which the midpoint of the wire moves down x < l ) will be (A) (x2/l2) (B) (2 x2/l2) (C) (x2/2 l2) (D) (x/2 l)

Tardigrade Admin · Accepted Answer

From figure the increase in length Δ l=(P R + R Q)- PQ = 2PR - PQ =2(l2 + x2)1 / 2-2l=2l(1 + (x2/l2))1 / 2-2l =2l[1 + (1/2) (x2/l2)]-2l ( By Binomial theorem) =x2/l ∴ Strain= Δ l/2l=x2/2l2 <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-jnxb8cp8a5q1.png" />

$\frac{x ^{2}}{l ^{2}}$

$\frac{2 x ^{2}}{l ^{2}}$

$\frac{x ^{2}}{2 l ^{2}}$

$\frac{x}{2 l}$

From figure the increase in length $Δ l = (PR + RQ) - PQ$
= 2PR - PQ
$= 2 (l^{2} + x^{2})^{1/2} - 2 l = 2 l (1 + \frac{x ^{2}}{l ^{2}})^{1/2} - 2 l$
$= 2 l [1 + \frac{1}{2} \frac{x ^{2}}{l ^{2}}] - 2 l$ ( By Binomial theorem)
$= x^{2} / l$
$∴ St r ain = Δ l /2 l = x^{2} /2 l^{2}$

l2x2​

l22x2​

2l2x2​

2lx​

From figure the increase in length Δl=(PR+RQ)−PQ = 2PR - PQ =2(l2+x2)1/2−2l=2l(1+l2x2​)1/2−2l =2l[1+21​l2x2​]−2l ( By Binomial theorem) =x2/l ∴Strain=Δl/2l=x2/2l2

$\frac{x ^{2}}{l ^{2}}$

$\frac{2 x ^{2}}{l ^{2}}$

$\frac{x ^{2}}{2 l ^{2}}$

$\frac{x}{2 l}$

From figure the increase in length $Δ l = (PR + RQ) - PQ$
= 2PR - PQ
$= 2 (l^{2} + x^{2})^{1/2} - 2 l = 2 l (1 + \frac{x ^{2}}{l ^{2}})^{1/2} - 2 l$
$= 2 l [1 + \frac{1}{2} \frac{x ^{2}}{l ^{2}}] - 2 l$ ( By Binomial theorem)
$= x^{2} / l$
$∴ St r ain = Δ l /2 l = x^{2} /2 l^{2}$