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 · 8 Apr · Shift 1 · Q61
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 · 8 Apr · Shift 1 · Q61

Properties of Matter question

2023 · 8 Apr · Shift 1 · Q61

JEE MainPhysicsProperties of MatterMCQ+4 / −1
An aluminium rod with Young's modulus Y=7.0×1010 N/m2Y=7.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}Y=7.0×1010 N/m2 undergoes elastic strain of 0.04%0.04 \%0.04%. The energy per unit volume stored in the rod in SI unit is:
  1. A
    5600
  2. B
    2800
  3. C
    11200
  4. D
    8400
View written solutionFree

Correct answer: A

  1. Given data
  • Young's modulus: Y=7.0×1010 N/m2Y = 7.0 \times 10^{10}\ \text{N/m}^2Y=7.0×1010 N/m2
  • Strain: 0.04%=0.04100=4×10−40.04\% = \frac{0.04}{100} = 4 \times 10^{-4}0.04%=1000.04​=4×10−4
  1. Formula for elastic energy per unit volume

The strain energy density is u=12(stress)(strain)u = \frac{1}{2}(\text{stress})(\text{strain})u=21​(stress)(strain)

Using Hooke's law, stress=Y×strain\text{stress} = Y \times \text{strain}stress=Y×strain

So, u=12Y(strain)2u = \frac{1}{2}Y(\text{strain})^2u=21​Y(strain)2

  1. Substitute the values

u=12(7.0×1010)(4×10−4)2u = \frac{1}{2}(7.0 \times 10^{10})(4 \times 10^{-4})^2u=21​(7.0×1010)(4×10−4)2

First compute the square: (4×10−4)2=16×10−8=1.6×10−7(4 \times 10^{-4})^2 = 16 \times 10^{-8} = 1.6 \times 10^{-7}(4×10−4)2=16×10−8=1.6×10−7

Now, u=12(7.0×1010)(1.6×10−7)u = \frac{1}{2}(7.0 \times 10^{10})(1.6 \times 10^{-7})u=21​(7.0×1010)(1.6×10−7)

Multiply: 7.0×1.6=11.27.0 \times 1.6 = 11.27.0×1.6=11.2 1010×10−7=10310^{10} \times 10^{-7} = 10^31010×10−7=103

Thus, u=12(11.2×103)=5.6×103u = \frac{1}{2}(11.2 \times 10^3) = 5.6 \times 10^3u=21​(11.2×103)=5.6×103

u=5600 J/m3u = 5600\ \text{J/m}^3u=5600 J/m3

  1. Match with options
  • A: 5600
  • B: 2800
  • C: 11200
  • D: 8400

So the correct option is A.

PreviousNext

More from Properties of Matter

  • An air bubble of diameter 6 mm rises steadily through a solution of density 1750 kg/m3 at the rate of 0.35 cm/s. The co-efficient of viscosity of the solution (neglect density of…2023 · Numerical
  • A hydraulic automobile lift is designed to lift vehicles of mass 5000 kg. The area of cross section of the cylinder carrying the load is 250 cm2. The maximum pressure the smaller piston would have to bear is […2023 · MCQ
  • Given below are two statements: Statement I : Pressure in a reservoir of water is same at all points at the same level of water. Statement II : The pressure applied to enclosed water is transmitted in all directions equally. In the light…2023 · MCQ
  • Two wires each of radius 0.2 cm and negligible mass, one made of steel and the other made of brass are loaded as shown in the figure. The elongation of the steel wire is ​× 10 −6 m. [Young's modulus for… Includes diagram2023 · Numerical
  • Young's moduli of the material of wires A and B are in the ratio of 1:4, while its area of cross sections are in the ratio of 1:3. If the same amount of load is applied to both the wires, the amount of elongation produced in the…2023 · MCQ
  • Figure below shows a liquid being pushed out of the tube by a piston having area of cross section 2.0 cm2. The area of cross section at the outlet is 10 mm2. If the piston is pushed at a speed of 4 cm s−1… Includes diagram2023 · Numerical
  • The length of a wire becomes l1​ and l2​ when 100 N and 120 N tensions are applied respectively. If 10 l2​=11 l1​, the natural length of wire will be x1​ l1​. Here the value of x is ​…2023 · Numerical
  • Eight equal drops of water are falling through air with a steady speed of 10 cm/s. If the drops coalesce, the new velocity is:-2023 · MCQ