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 · 11 Apr · Shift 2 · Q64
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 · 11 Apr · Shift 2 · Q64

Properties of Matter question

2023 · 11 Apr · Shift 2 · Q64

JEE MainPhysicsProperties of MatterNumerical+4 / −1
A wire of density 8×103 kg/m38 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}8×103 kg/m3 is stretched between two clamps 0.5 m0.5 \mathrm{~m}0.5 m apart. The extension developed in the wire is 3.2×10−4 m3.2 \times 10^{-4} \mathrm{~m}3.2×10−4 m. If Y=8×1010 N/m2Y=8 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}Y=8×1010 N/m2, the fundamental frequency of vibration in the wire will be ‾Hz\underline{\hspace{2cm}}\mathrm{Hz}​Hz.
Numerical answer
View written solutionFree

Correct answer: 80

  1. Given data
  • Density of wire: ρ=8×103 kg/m3\rho = 8 \times 10^{3}\,\text{kg/m}^3ρ=8×103kg/m3
  • Length between clamps: L=0.5 mL = 0.5\,\text{m}L=0.5m
  • Extension: ΔL=3.2×10−4 m\Delta L = 3.2 \times 10^{-4}\,\text{m}ΔL=3.2×10−4m
  • Young's modulus: Y=8×1010 N/m2Y = 8 \times 10^{10}\,\text{N/m}^2Y=8×1010N/m2

We need the fundamental frequency of the stretched wire.

  1. Use relation between stress and strain

For a wire,

Y=stressstrain=T/AΔL/LY = \frac{\text{stress}}{\text{strain}} = \frac{T/A}{\Delta L/L}Y=strainstress​=ΔL/LT/A​

where TTT is tension and AAA is cross-sectional area.

So,

TA=YΔLL\frac{T}{A} = Y\frac{\Delta L}{L}AT​=YLΔL​

Hence,

T=AYΔLLT = AY\frac{\Delta L}{L}T=AYLΔL​
  1. Linear mass density of wire

If volume density is ρ\rhoρ, then linear density is

μ=ρA\mu = \rho Aμ=ρA
  1. Wave speed on stretched string

The speed of transverse wave is

v=Tμv = \sqrt{\frac{T}{\mu}}v=μT​​

Substitute T=AYΔLLT = AY\frac{\Delta L}{L}T=AYLΔL​ and μ=ρA\mu = \rho Aμ=ρA:

v=AY(ΔL/L)ρA=YΔLρLv = \sqrt{\frac{AY(\Delta L/L)}{\rho A}} = \sqrt{\frac{Y\Delta L}{\rho L}}v=ρAAY(ΔL/L)​​=ρLYΔL​​

Now put values:

v=(8×1010)(3.2×10−4)(8×103)(0.5)v = \sqrt{\frac{(8\times 10^{10})(3.2\times 10^{-4})}{(8\times 10^{3})(0.5)}}v=(8×103)(0.5)(8×1010)(3.2×10−4)​​

Simplify numerator:

8×1010⋅3.2×10−4=25.6×1068\times 10^{10} \cdot 3.2\times 10^{-4} = 25.6\times 10^68×1010⋅3.2×10−4=25.6×106

Simplify denominator:

8×103⋅0.5=4×1038\times 10^3 \cdot 0.5 = 4\times 10^38×103⋅0.5=4×103

Thus,

v=25.6×1064×103=6.4×103=6400=80 m/sv = \sqrt{\frac{25.6\times 10^6}{4\times 10^3}} = \sqrt{6.4\times 10^3} = \sqrt{6400} = 80\,\text{m/s}v=4×10325.6×106​​=6.4×103​=6400​=80m/s
  1. Fundamental frequency

For a string fixed at both ends,

f1=v2Lf_1 = \frac{v}{2L}f1​=2Lv​

So,

f1=802×0.5=801=80 Hzf_1 = \frac{80}{2\times 0.5} = \frac{80}{1} = 80\,\text{Hz}f1​=2×0.580​=180​=80Hz
  1. Final answer
80\boxed{80}80​

This matches the stored correct answer.

PreviousNext

More from Properties of Matter

  • Glycerin of density 1.25×103 kg m−3 is flowing through the conical section of pipe The area of cross-section of the pipe at its ends are 10 cm2 and 5 cm2 and pressure drop…2023 · Numerical
  • The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross - section. Cross sectional areas at A is 1.5 cm2, and B is 25 mm2, if the speed of liquid… Includes diagram2023 · MCQ
  • Under isothermal condition, the pressure of a gas is given by P=a V−3, where a is a constant and V is the volume of the gas. The bulk modulus at constant temperature is equal to2023 · MCQ
  • The elastic potential energy stored in a steel wire of length 20 m stretched through 2 cm is 80 J. The cross sectional area of the wire is ​mm2. (…2023 · Numerical
  • 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