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Properties of Matter question

2020 · 9 Jan · Shift 2 · Q47
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Properties of Matter question

2020 · 9 Jan · Shift 2 · Q47

JEE MainPhysicsProperties of MatterMCQ+4 / −1
Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is 1 : 4, the ratio of their diameters is:
  1. A
    1 : 2
  2. B
    2 : 1
  3. C
    1:21:\sqrt 21:2​
  4. D
    2:1\sqrt 2 :12​:1
View written solutionFree

Correct answer: D

  1. Strain energy per unit volume for a stretched wire

For a wire under तनाव (stress) within elastic limit,

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

Using Hooke’s law,

strain=stressY\text{strain} = \frac{\text{stress}}{Y}strain=Ystress​

So,

= \frac{\text{stress}^2}{2Y}$$ Since both are **steel** wires, their Young’s modulus $Y$ is same. Hence, $$u \propto (\text{stress})^2$$ 2. **Stress in each wire** Stress is $$\text{stress} = \frac{F}{A}$$ Both wires carry the **same load**, so $F$ is same. Therefore, $$\text{stress} \propto \frac{1}{A}$$ Thus, $$u \propto \frac{1}{A^2}$$ 3. **Relating area to diameter** For a circular wire, $$A = \frac{\pi d^2}{4} \Rightarrow A \propto d^2$$ So, $$u \propto \frac{1}{d^4}$$ 4. **Using the given ratio** Given energy stored per unit volume ratio: $$u_1 : u_2 = 1:4$$ Using $u \propto \frac{1}{d^4}$, $$\frac{u_1}{u_2} = \frac{d_2^4}{d_1^4} = \frac{1}{4}$$ So, $$\left(\frac{d_2}{d_1}\right)^4 = \frac{1}{4}$$ Taking fourth root, $$\frac{d_2}{d_1} = \frac{1}{\sqrt{2}}$$ Hence, $$\frac{d_1}{d_2} = \sqrt{2}$$ Therefore, the ratio of diameters is $$d_1:d_2 = \sqrt{2}:1$$ 5. **Option check** - A: $1:2$ ❌ - B: $2:1$ ❌ - C: $1:\sqrt2$ ❌ - D: $\sqrt2:1$ ✅ So the correct option is **D**.
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