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NEET PHYSICSMechanical Properties of SolidsMedium

Question

Two wires are made of the same material and have the same volume. The first wire has a cross-sectional area AA and the second wire has a cross-sectional area 3A3A. If the length of the first wire is increased by Δl\Delta l on applying a force FF, how much force is needed to stretch the second wire by the same amount?

A

9F

B

6F

C

4F

D

F

Step-by-Step Solution

  1. Young's Modulus Formula: Young's modulus (YY) is a property of the material and is defined as stress divided by strain: Y=F/AΔl/L    F=YAΔlLY = \frac{F/A}{\Delta l/L} \implies F = \frac{YA\Delta l}{L} .
  2. Constraint (Same Volume): The volume (VV) of a wire is V=A×LV = A \times L. Since both wires have the same volume, A1L1=A2L2A_1 L_1 = A_2 L_2. Given A1=AA_1 = A and A2=3AA_2 = 3A, we have: AL1=3AL2    L2=L13A \cdot L_1 = 3A \cdot L_2 \implies L_2 = \frac{L_1}{3}
  3. Comparison of Forces: We need to find F2F_2 given that the extension Δl\Delta l and material (YY) are the same for both. F1=YA1ΔlL1=FF_1 = \frac{Y A_1 \Delta l}{L_1} = F F2=YA2ΔlL2F_2 = \frac{Y A_2 \Delta l}{L_2}
  4. Substitution: Substitute A2=3AA_2 = 3A and L2=L1/3L_2 = L_1/3 into the equation for F2F_2: F2=Y(3A)Δl(L1/3)=3×3×YAΔlL1=9(YAΔlL1)F_2 = \frac{Y (3A) \Delta l}{(L_1/3)} = 3 \times 3 \times \frac{Y A \Delta l}{L_1} = 9 \left( \frac{Y A \Delta l}{L_1} \right) F2=9FF_2 = 9 F

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Mechanical Properties of Solids. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMechanical Properties of Solidsmaterialvolumecrosssectionalsecondcrosssectional

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