Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Properties of Matter question

2023 · 13 Apr · Shift 1 · Q72
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Properties of Matter
  5. /2023 · 13 Apr · Shift 1 · Q72

Properties of Matter question

2023 · 13 Apr · Shift 1 · Q72

JEE MainPhysicsProperties of MatterNumerical+4 / −1
The elastic potential energy stored in a steel wire of length 20 m20 \mathrm{~m}20 m stretched through 2 cm2 \mathrm{~cm}2 cm is 80 J80 \mathrm{~J}80 J. The cross sectional area of the wire is ‾mm2\underline{\hspace{2cm}}\mathrm{mm}^{2}​mm2. (\left(\right.( Given, y=2.0×1011Nm−2)\left.y=2.0 \times 10^{11} \mathrm{Nm}^{-2}\right)y=2.0×1011Nm−2)
Numerical answer
View written solutionFree

Correct answer: 40

  1. Given data
  • Length of wire: L=20 mL = 20\,\text{m}L=20m
  • Extension: ΔL=2 cm=0.02 m\Delta L = 2\,\text{cm} = 0.02\,\text{m}ΔL=2cm=0.02m
  • Elastic potential energy stored: U=80 JU = 80\,\text{J}U=80J
  • Young's modulus: Y=2.0×1011 N/m2Y = 2.0 \times 10^{11}\,\text{N/m}^2Y=2.0×1011N/m2

We need to find the cross-sectional area AAA.

  1. Formula for elastic potential energy in a stretched wire

For a wire obeying Hooke’s law,

U=12F ΔLU = \frac{1}{2}F\,\Delta LU=21​FΔL

Also,

Y=F/AΔL/L=FLAΔLY = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L}Y=ΔL/LF/A​=AΔLFL​

So,

F=YAΔLLF = \frac{YA\Delta L}{L}F=LYAΔL​

Substitute this into the energy formula:

U=12(YAΔLL)ΔLU = \frac{1}{2}\left(\frac{YA\Delta L}{L}\right)\Delta LU=21​(LYAΔL​)ΔL U=12⋅YA(ΔL)2LU = \frac{1}{2}\cdot \frac{YA(\Delta L)^2}{L}U=21​⋅LYA(ΔL)2​

Hence,

A=2ULY(ΔL)2A = \frac{2UL}{Y(\Delta L)^2}A=Y(ΔL)22UL​
  1. Substitute the values
A=2×80×20(2.0×1011)(0.02)2A = \frac{2 \times 80 \times 20}{(2.0 \times 10^{11})(0.02)^2}A=(2.0×1011)(0.02)22×80×20​

First calculate (0.02)2(0.02)^2(0.02)2:

(0.02)2=4×10−4(0.02)^2 = 4 \times 10^{-4}(0.02)2=4×10−4

So,

A=32002.0×1011×4×10−4A = \frac{3200}{2.0 \times 10^{11} \times 4 \times 10^{-4}}A=2.0×1011×4×10−43200​ A=32008×107A = \frac{3200}{8 \times 10^{7}}A=8×1073200​ A=4×10−5 m2A = 4 \times 10^{-5}\,\text{m}^2A=4×10−5m2
  1. Convert to mm2^22

Since

1 m2=106 mm21\,\text{m}^2 = 10^6\,\text{mm}^21m2=106mm2

therefore,

A=4×10−5×106=40 mm2A = 4 \times 10^{-5} \times 10^6 = 40\,\text{mm}^2A=4×10−5×106=40mm2
  1. Final answer
40\boxed{40}40​

The cross-sectional area of the wire is 40 mm240\,\text{mm}^240mm2.

PreviousNext

More from Properties of Matter

  • Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A : A spherical body of radius (5±0.1)mm having a particular density is falling through a…2023 · MCQ
  • A wire of length ' L' and radius 'r' is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by ' l'. Another wire of same material of length '2 L' and radius '2r'…2023 · MCQ
  • There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.075 Nm−1 and density 1000 kg m−3 at a depth of 10 cm below the free surface. The amount by which the…2023 · Numerical
  • A 100 m long wire having cross-sectional area 6.25×10−4 m2 and Young's modulus is 1010 Nm−2 is subjected to a load of 250 N, then the elongation in the wire will be :2023 · MCQ
  • Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Steel is used in the construction of buildings and bridges. Reason R : Steel is more elastic and its elastic limit is high.…2023 · MCQ
  • The frequency (u) of an oscillating liquid drop may depend upon radius (r) of the drop, density (ρ) of liquid and the surface tension (s) of the liquid as u=raρbsc. The values of a, b and c respectively are2023 · MCQ
  • A Spherical ball of radius 1mm and density 10.5 g/cc is dropped in glycerine of coefficient of viscosity 9.8 poise and density 1.5 g/cc. Viscous force on the ball when it attains constant velocity is 3696×10−x N. The value of x…2023 · Numerical
  • Match List I with List II Choose the correct answer from the options given below : Includes table2023 · MCQ