JEE Main 2019 — Physics solution

JEE Main2019Physics

The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit?

A
0.90 mm
B
1.16 mm
C
1.00 mm
D
1.36 mm

Solution

$S t r ess = \frac{400}{π r ^{2}} \leq 379 \times 1 0^{6} N / m^{2} \Rightarrow r^{2} \geq \frac{400}{379 \times 10 ^{6} π} 2 r \geq 1.15 mm$

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