Logo
Jeeify
MCQEasyJEE 2023Young's Modulus, Bulk & Rigidity Modulus

JEE Physics 2023 Question with Solution

An aluminium rod with Young's modulus Y=7.0×1010 N/m2Y = 7.0 \times 10^{10} \, \text{N/m}^2 undergoes elastic strain of 0.04%0.04\%. The energy per unit volume stored in the rod in SI unit is:

  • A

    56005600

  • B

    28002800

  • C

    1120011200

  • D

    84008400

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Young's modulus is Y=7.0×1010 N/m2Y = 7.0 \times 10^{10} \, \text{N/m}^2 and elastic strain is 0.04%=0.00040.04\% = 0.0004.

Find: The energy per unit volume stored in the rod.

For elastic deformation, the energy per unit volume is

u = \frac{1}{2} \cdot \text{stress} \cdot \text{strain}$$ and $$\text{stress} = Y \cdot \text{strain}$$

Substituting stress into the energy density formula,

u=12⋅Y⋅(strain)2=12⋅7.0×1010⋅(0.0004)2\begin{aligned} u &= \frac{1}{2} \cdot Y \cdot (\text{strain})^2 \\ &= \frac{1}{2} \cdot 7.0 \times 10^{10} \cdot (0.0004)^2 \end{aligned}

Evaluating,

ν=5600 J/m3\nu = 5600 \, \text{J/m}^3

Therefore, the energy per unit volume stored in the rod is 5600 J/m35600 \, \text{J/m}^3. The correct option is A.

Stepwise Substitution

Given: Y=7.0×1010 N/m2Y = 7.0 \times 10^{10} \, \text{N/m}^2 and strain =0.04%= 0.04\%.

Find: Energy density in SI units.

First convert percentage strain into decimal form:

strain=0.04100=0.0004\text{strain} = \frac{0.04}{100} = 0.0004

Now use

u = \frac{1}{2} Y (\text{strain})^2$$ so

\nu = \frac{1}{2} \cdot 7.0 \times 10^{10} \cdot (0.0004)^2

Since

(0.0004)2=1.6×10−7(0.0004)^2 = 1.6 \times 10^{-7}

we get

ν=12⋅7.0×1010⋅1.6×10−7\nu = \frac{1}{2} \cdot 7.0 \times 10^{10} \cdot 1.6 \times 10^{-7}

Thus,

ν=5600 J/m3\nu = 5600 \, \text{J/m}^3

Therefore, the required energy per unit volume is 5600 J/m35600 \, \text{J/m}^3.

Common mistakes

  • Using u=12Y⋅strainu = \frac{1}{2}Y \cdot \text{strain} instead of u=12Y(strain)2u = \frac{1}{2}Y(\text{strain})^2 is incorrect because energy density depends on stress multiplied by strain. Always square the strain after substituting stress=Y⋅strain\text{stress} = Y \cdot \text{strain}.

  • Treating 0.04%0.04\% as 0.040.04 is wrong because percentage must be divided by 100100. Convert it correctly to 0.00040.0004 before substitution.

  • Confusing Young's modulus units with the final answer units leads to error. YY has units of N/m2\text{N/m}^2, but energy per unit volume is expressed as J/m3\text{J/m}^3, which is dimensionally equivalent.

Practice more Young's Modulus, Bulk & Rigidity Modulus questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Young's Modulus, Bulk & Rigidity Modulus

Related questions