Physics · Mechanical Properties of Matter

JEE Advanced 2019 — Paper 1 — Question 12

A block of weight 100 N is suspended by copper and steel wires of same cross sectional area 0.5 cm20.5 \mathrm{~cm}^{2} and, length 3 m\sqrt{3} \mathrm{~m} and 1 m , respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are 30∘30^{\circ} and 60∘60^{\circ}, respectively. If elongation in copper wire is (ΔℓC)\left(\Delta \ell_{C}\right) and elongation in steel wire is (ΔℓS)\left(\Delta \ell_{\mathrm{S}}\right), then the ratio ΔℓCΔℓS\frac{\Delta \ell_{\mathrm{C}}}{\Delta \ell_{\mathrm{S}}} is ____\_\_\_\_

[Young's modulus for copper and steel are 1×1011 N/m21 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} and 2×1011 N/m22 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}, respectively.]

Question figure

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

TSsin⁡30∘=TCsin⁡60∘\mathrm{T}_{\mathrm{S}} \sin 30^{\circ}=\mathrm{T}_{\mathrm{C}} \sin 60^{\circ}

ΔℓCΔℓS=TCℓCACYC(ASYSTSℓS)=2.00\frac{\Delta \ell_{\mathrm{C}}}{\Delta \ell_{\mathrm{S}}}=\frac{\mathrm{T}_{\mathrm{C}} \ell_{\mathrm{C}}}{\mathrm{A}_{\mathrm{C}} \mathrm{Y}_{\mathrm{C}}}\left(\frac{\mathrm{A}_{\mathrm{S}} \mathrm{Y}_{\mathrm{S}}}{\mathrm{T}_{\mathrm{S}} \ell_{\mathrm{S}}}\right)=2.00

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2019
Paper
Paper 1
Subject
Physics
Chapter
Mechanical Properties of Matter
Topic
Stress,Strain and Modulus of Elasticity