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Properties of Matter question

2016 · 9 Apr · Shift 1 · Q51
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Properties of Matter question

2016 · 9 Apr · Shift 1 · Q51

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A uniformly tapering conical wire is made from a material of Young’s modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values R and 3 R, respectively. The upper end of the wire is fixed to a rigid support and a mass M is suspended from its lower end. The equilibrium extended length, of this wire, would equal :
  1. A
    L (1+29MgπYR2)\left( {1 + {2 \over 9}{{Mg} \over {\pi Y{R^2}}}} \right)(1+92​πYR2Mg​)
  2. B
    L (1+13MgπYR2)\left( {1 + {1 \over 3}{{Mg} \over {\pi Y{R^2}}}} \right)(1+31​πYR2Mg​)
  3. C
    L (1+19MgπYR2)\left( {1 + {1 \over 9}{{Mg} \over {\pi Y{R^2}}}} \right)(1+91​πYR2Mg​)
  4. D
    L (1+23MgπYR2)\left( {1 + {2 \over 3}{{Mg} \over {\pi Y{R^2}}}} \right)(1+32​πYR2Mg​)
View written solutionFree

Correct answer: B

  1. Given data
  • Length of conical wire =L= L=L
  • Young's modulus =Y= Y=Y
  • Radius at upper end =R= R=R
  • Radius at lower end =3R= 3R=3R
  • Load suspended =Mg= Mg=Mg

We need the extended length of the wire in equilibrium.

So first we find the extension ΔL\Delta LΔL, then add it to LLL.

  1. Radius as a function of distance

Let xxx be the distance measured from the upper end. Since the wire tapers uniformly, radius varies linearly from RRR at x=0x=0x=0 to 3R3R3R at x=Lx=Lx=L.

Hence,

r(x)=R+(3R−R)Lx=R(1+2xL)r(x)=R+\frac{(3R-R)}{L}x=R\left(1+\frac{2x}{L}\right)r(x)=R+L(3R−R)​x=R(1+L2x​)

So area of cross-section at xxx is

A(x)=πr2(x)=πR2(1+2xL)2A(x)=\pi r^2(x)=\pi R^2\left(1+\frac{2x}{L}\right)^2A(x)=πr2(x)=πR2(1+L2x​)2
  1. Small extension of an element

For a small element dxdxdx, the tensile force throughout the wire is the same and equal to

F=MgF=MgF=Mg

The extension of element dxdxdx is

d(ΔL)=F dxYA(x)d(\Delta L)=\frac{F\,dx}{Y A(x)}d(ΔL)=YA(x)Fdx​

Therefore,

d(ΔL)=Mg dxYπR2(1+2xL)2d(\Delta L)=\frac{Mg\,dx}{Y\pi R^2\left(1+\frac{2x}{L}\right)^2}d(ΔL)=YπR2(1+L2x​)2Mgdx​

Integrating from x=0x=0x=0 to x=Lx=Lx=L:

ΔL=MgπYR2∫0Ldx(1+2xL)2\Delta L=\frac{Mg}{\pi YR^2}\int_0^L \frac{dx}{\left(1+\frac{2x}{L}\right)^2}ΔL=πYR2Mg​∫0L​(1+L2x​)2dx​
  1. Evaluate the integral

Let

u=1+2xLu = 1+\frac{2x}{L}u=1+L2x​

Then

du=2Ldx⇒dx=L2dudu=\frac{2}{L}dx \quad \Rightarrow \quad dx=\frac{L}{2}dudu=L2​dx⇒dx=2L​du

When x=0x=0x=0, u=1u=1u=1. When x=Lx=Lx=L, u=3u=3u=3.

So,

ΔL=MgπYR2⋅L2∫13u−2 du\Delta L=\frac{Mg}{\pi YR^2}\cdot \frac{L}{2}\int_1^3 u^{-2}\,duΔL=πYR2Mg​⋅2L​∫13​u−2du

Now,

∫u−2du=−u−1\int u^{-2}du=-u^{-1}∫u−2du=−u−1

Thus,

∫13u−2du=[−1u]13=−13+1=23\int_1^3 u^{-2}du=\left[-\frac{1}{u}\right]_1^3=-\frac13+1=\frac23∫13​u−2du=[−u1​]13​=−31​+1=32​

Hence,

ΔL=MgπYR2⋅L2⋅23\Delta L=\frac{Mg}{\pi YR^2}\cdot \frac{L}{2}\cdot \frac23ΔL=πYR2Mg​⋅2L​⋅32​ ΔL=LMg3πYR2\Delta L=\frac{LMg}{3\pi YR^2}ΔL=3πYR2LMg​
  1. Extended length

Therefore the equilibrium extended length is

L+ΔL=L+LMg3πYR2L+\Delta L=L+\frac{LMg}{3\pi YR^2}L+ΔL=L+3πYR2LMg​ =L(1+13MgπYR2)= L\left(1+\frac{1}{3}\frac{Mg}{\pi YR^2}\right)=L(1+31​πYR2Mg​)
  1. Matching with options

This matches:

Option B

L(1+13MgπYR2)L\left(1+\frac{1}{3}\frac{Mg}{\pi YR^2}\right)L(1+31​πYR2Mg​)
  1. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

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