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Properties of Matter question

2022 · 27 Jul · Shift 2 · Q47
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Properties of Matter question

2022 · 27 Jul · Shift 2 · Q47

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A steel wire of length 3.2 m(Ys=2.0×1011 Nm−2)3.2 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{s}}=2.0 \times 10^{11} \,\mathrm{Nm}^{-2}\right)3.2 m(Ys​=2.0×1011Nm−2) and a copper wire of length 4.4 m(Yc=1.1×1011 Nm−2)4.4 \mathrm{~m}\left(\mathrm{Y}_{\mathrm{c}}=1.1 \times 10^{11} \,\mathrm{Nm}^{-2}\right)4.4 m(Yc​=1.1×1011Nm−2), both of radius 1.4 mm1.4 \mathrm{~mm}1.4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm1.4 \mathrm{~mm}1.4 mm. The load applied, in Newton, will be: (\quad\left(\right.( Given π=227\pi=\frac{22}{7}π=722​)
  1. A
    360
  2. B
    180
  3. C
    1080
  4. D
    154
View written solutionFree

Correct answer: D

  1. Concept used

When two wires are connected in series (end to end), the same load FFF acts through both wires.

Elongation of a wire:

ΔL=FLAY\Delta L = \frac{FL}{AY}ΔL=AYFL​

where:

  • FFF = load
  • LLL = length
  • AAA = cross-sectional area
  • YYY = Young’s modulus

So total elongation is:

ΔLtotal=FLsAYs+FLcAYc\Delta L_{\text{total}} = \frac{F L_s}{A Y_s} + \frac{F L_c}{A Y_c}ΔLtotal​=AYs​FLs​​+AYc​FLc​​
  1. Given data

For steel:

Ls=3.2 m,Ys=2.0×1011 N/m2L_s = 3.2\,\text{m}, \qquad Y_s = 2.0\times 10^{11}\,\text{N/m}^2Ls​=3.2m,Ys​=2.0×1011N/m2

For copper:

Lc=4.4 m,Yc=1.1×1011 N/m2L_c = 4.4\,\text{m}, \qquad Y_c = 1.1\times 10^{11}\,\text{N/m}^2Lc​=4.4m,Yc​=1.1×1011N/m2

Radius of each wire:

r=1.4 mm=1.4×10−3 mr = 1.4\,\text{mm} = 1.4\times 10^{-3}\,\text{m}r=1.4mm=1.4×10−3m

Total elongation:

ΔL=1.4 mm=1.4×10−3 m\Delta L = 1.4\,\text{mm} = 1.4\times 10^{-3}\,\text{m}ΔL=1.4mm=1.4×10−3m

Also,

π=227\pi = \frac{22}{7}π=722​
  1. Cross-sectional area
A=πr2=227(1.4×10−3)2A = \pi r^2 = \frac{22}{7}(1.4\times 10^{-3})^2A=πr2=722​(1.4×10−3)2

Now,

(1.4)2=1.96(1.4)^2 = 1.96(1.4)2=1.96

so

A=227×1.96×10−6A = \frac{22}{7}\times 1.96 \times 10^{-6}A=722​×1.96×10−6

Since

227×1.96=6.16\frac{22}{7}\times 1.96 = 6.16722​×1.96=6.16

we get

A=6.16×10−6 m2A = 6.16\times 10^{-6}\,\text{m}^2A=6.16×10−6m2
  1. Set up total elongation equation
1.4×10−3=F(3.2(6.16×10−6)(2.0×1011)+4.4(6.16×10−6)(1.1×1011))1.4\times 10^{-3} = F\left(\frac{3.2}{(6.16\times 10^{-6})(2.0\times 10^{11})} + \frac{4.4}{(6.16\times 10^{-6})(1.1\times 10^{11})}\right)1.4×10−3=F((6.16×10−6)(2.0×1011)3.2​+(6.16×10−6)(1.1×1011)4.4​)

Factor out AAA:

ΔL=FA(LsYs+LcYc)\Delta L = \frac{F}{A}\left(\frac{L_s}{Y_s} + \frac{L_c}{Y_c}\right)ΔL=AF​(Ys​Ls​​+Yc​Lc​​)

Thus,

F=ΔL ALsYs+LcYcF = \frac{\Delta L\,A}{\frac{L_s}{Y_s} + \frac{L_c}{Y_c}}F=Ys​Ls​​+Yc​Lc​​ΔLA​
  1. Compute the denominator
LsYs=3.22.0×1011=1.6×10−11\frac{L_s}{Y_s} = \frac{3.2}{2.0\times 10^{11}} = 1.6\times 10^{-11}Ys​Ls​​=2.0×10113.2​=1.6×10−11 LcYc=4.41.1×1011=4.0×10−11\frac{L_c}{Y_c} = \frac{4.4}{1.1\times 10^{11}} = 4.0\times 10^{-11}Yc​Lc​​=1.1×10114.4​=4.0×10−11

So,

LsYs+LcYc=5.6×10−11\frac{L_s}{Y_s} + \frac{L_c}{Y_c} = 5.6\times 10^{-11}Ys​Ls​​+Yc​Lc​​=5.6×10−11
  1. Compute the load
F=(1.4×10−3)(6.16×10−6)5.6×10−11F = \frac{(1.4\times 10^{-3})(6.16\times 10^{-6})}{5.6\times 10^{-11}}F=5.6×10−11(1.4×10−3)(6.16×10−6)​

First multiply numerator:

1.4×6.16=8.6241.4\times 6.16 = 8.6241.4×6.16=8.624

so

F=8.624×10−95.6×10−11F = \frac{8.624\times 10^{-9}}{5.6\times 10^{-11}}F=5.6×10−118.624×10−9​ F=8.6245.6×102=1.54×102=154 NF = \frac{8.624}{5.6}\times 10^2 = 1.54\times 10^2 = 154\,\text{N}F=5.68.624​×102=1.54×102=154N
  1. Option check

The correct option is:

154 N\boxed{154\,\text{N}}154N​

which corresponds to Option D.

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