Physics · Mechanical Properties of Matter

JEE Main 2024 — 5 April, Shift 1 — Question 53

The density and breaking stress of a wire are 6×6 \times 104 kg/m310^{4} \mathrm{~kg} / \mathrm{m}^{3} and 1.2×108 N/m21.2 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2} respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity is 1rd3\frac{1^{\mathrm{rd}}}{3} of the value on the surface of earth. The maximum length of the wire with breaking is \qquad m (take, g=\mathrm{g}= 10 m/s2)\left.10 \mathrm{~m} / \mathrm{s}^{2}\right)

Answer: 600

Numerical answer — enter this value.

Step-by-step solution

figure

T=mg\mathrm{T}=\mathrm{mg}

σ=TA=mgA\sigma=\frac{\mathrm{T}}{\mathrm{A}}=\frac{\mathrm{mg}}{\mathrm{A}}

(σAℓ)gA\frac{(\sigma \mathrm{A} \ell) \mathrm{g}}{\mathrm{A}}

⇒ℓ=σρg=1.2×108×36×104×10=600\Rightarrow \ell=\frac{\sigma}{\rho g}=\frac{1.2 \times 10^{8} \times 3}{6 \times 10^{4} \times 10}=600

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Physics
Chapter
Mechanical Properties of Matter
Topic
Stress,Strain and Modulus of Elasticity
The density and breaking stress of a wire are 6 × 10 4 kg / m 3 and… | JEE Main 2024 PYQ with Solution · DhiX AI