Physics Elasticity Question 101
Question: An aluminum rod (Young’s modulus $ =7\times 10^{9},N/m^{2}) $ has a breaking strain of 0.2%. The minimum cross-sectional area of the rod in order to support a load of $ 10^{4} $ Newton’s is
[MP PMT 1991]
Options:
A) $ 1\times {{10}^{-2}},m^{2} $
B) $ 1.4\times {{10}^{-3}},m^{2} $
C) $ 3.5\times {{10}^{-3}},m^{2} $
D) $ 7.1\times {{10}^{-4}},m^{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ Y=\frac{F/A}{r}\Rightarrow A=\frac{F}{Y\times r} $ = $ \frac{10^{4}}{7\times 10^{9}\times 0.002} $ = $ \frac{1}{14}\times {{10}^{-2}} $
$ =7.1\times {{10}^{-4}}m^{2} $
BETA
