Properties Of Matter Ques 38
- 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?
(2019 Main, 10 April II)
(a) $0.90 mm$
(b) $1.00 mm$
(c) $1.16 mm$
(d) $1.36 mm$
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Answer:
Correct Answer: 38.(c)
Solution:
- Let $d _{\text {min }}=$ minimum diameter of brass.
Then, stress in brass rod is given by
$$ \sigma=\frac{F}{A}=\frac{4 F}{\pi d _{\min }^{2}} \quad\left[\because A=\frac{\pi d^{2}}{4}\right] $$
For stress not to exceed elastic limit, we have
$\sigma \leq 379 MPa$
$\Rightarrow \quad \frac{4 F}{\pi d^{2} \min } \leq 379 \times 10^{6}$
Here,
$$ F=400 N $$
$\therefore \quad d _{\text {min }}^{2}=\frac{1600}{\pi \times 379 \times 10^{6}}$
$\Rightarrow \quad d _{\text {min }}=1.16 \times 10^{-3} m=1.16 mm$
BETA
