Physics Elasticity Question 145
Question: Two similar wires under the same load yield elongation of 0.1 mm and 0.05 mm respectively. If the area of cross- section of the first wire is $ 4mm^{2}, $ then the area of cross section of the second wire is
[CPMT 2000; Pb. PET 2002]
Options:
A) $ 6mm^{2} $
B) $ 8mm^{2} $
C) $ 10,mm^{2} $
D)$ 12,mm^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ l=\frac{FL}{AY}\
Therefore l\propto \frac{1}{A} $ (F,L and Y are constant) $ \frac{A _{2}}{A _{1}}=\frac{l _{1}}{l _{2}}\Rightarrow A _{2}=A _{1}\left( \frac{0.1}{0.05} \right) $ = $ 2A _{1}=2\times 4=8mm^{2} $
BETA
