JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 24)
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?
1.16 mm
1.36 mm
1.00 mm
0.90 mm
Explanation
$${{400} \over {{\pi \over 4}{d^2}}} = 379 \times {10^6}$$
$${d^2} = {{4 \times 400 \times {{10}^{ - 6}}} \over {\pi \times 379}} = 0.336 \times {10^{ - 6}} \times 4$$
$$d = 2\sqrt {0.336} \times {10^{ - 3}}M \simeq 1.16\,mm$$
$${d^2} = {{4 \times 400 \times {{10}^{ - 6}}} \over {\pi \times 379}} = 0.336 \times {10^{ - 6}} \times 4$$
$$d = 2\sqrt {0.336} \times {10^{ - 3}}M \simeq 1.16\,mm$$
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