Physics Elasticity Question 146
Question: A 5 m long aluminium wire ($ Y=7\times 10^{10}N/m^{2}) $ of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire $ (Y=12\times 10^{10}N/m^{2}) $ of the same length under the same weight, the diameter should now be, in mm.
[AMU 2000]
Options:
A) 1.75
B) 1.5
C) 2.5
D) 5.0
Show Answer
Answer:
Correct Answer: C
Solution:
$ l=\frac{FL}{\pi r^{2}Y}\Rightarrow r^{2}\propto \frac{1}{Y} $ (F,L and l are constant) $ \frac{r _{2}}{r _{1}}={{\left( \frac{Y _{1}}{Y _{2}} \right)}^{1/2}}={{\left( \frac{7\times 10^{10}}{12\times 10^{10}} \right)}^{1/2}} $
Therefore $ r _{2}=1.5\times {{\left( \frac{7}{12} \right)}^{1/2}} $ = 1.145 mm \ dia = 2.29 mm
BETA
