A brass rod of length L and cross section area 2 cm2 is attached end to end to a steel rod of length 2m and cross sectional area 1 cm2. The compound rod is subjected to equal and opposite tensile force of magnitude 5 × 104 N at its ends. If the elongation of the two rods are equal. Then the length of brass rod is [Ybrass = 1011 N/m2, Ysteel = 2 × 1011 N/m2]

Engineering

Physics

Stress Strain and Hookes Law New

Question

A brass rod of length L and cross section area 2 cm² is attached end to end to a steel rod of length 2m and cross sectional area 1 cm². The compound rod is subjected to equal and opposite tensile force of magnitude 5 × 10⁴ N at its ends. If the elongation of the two rods are equal. Then the length of brass rod is [Y_brass = 10¹¹ N/m², Y_steel = 2 × 10¹¹ N/m²]

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

When rods are in series, the force F is the same in both. The elongation ΔL is given by ΔL = (F L)/(Y A). Given ΔL_brass = ΔL_steel.

Therefore: $\frac{F L_{b}}{Y_{b} A_{b}} = \frac{F L_{s}}{Y_{s} A_{s}}$

Cancel F and substitute the given values: Y_b = 1×10¹¹, Y_s = 2×10¹¹, A_b = 2×10^-4 m², A_s = 1×10^-4 m², L_s = 2 m.

$\frac{L_{b}}{(1 \times 10^{11}) (2 \times 10^{- 4})} = \frac{2}{(2 \times 10^{11}) (1 \times 10^{- 4})}$

Solving this equation gives L_b = 2 m.

Final Answer: 2 m