Gravitation Question 221
Question: Two exactly similar wires of steel and copper are stretched by equal forces. If the difference in their elongations is 0.5 cm, the elongation ($ [l] $ ) of each wireis$ [Y_{s}(steel)=2.0\times 10^{11}N/m^{2}] $ $ [Y_{c}(copper)=1.2\times 10^{11}N/m^{2}] $
Options:
A) $ [l_{s}=0.75,cm,,l_{c}=1.25,cm] $
B)$ [l_{s}=1.25,cm,,l_{c}=0.75,cm] $
C)$ [l_{s}=0.25,cm,,l_{c}=0.75,cm] $
D)$ [l_{s}=0.75,cm,,l_{c}=0.25,cm] $
Show Answer
Answer:
Correct Answer: A
Solution:
-
$ [l\propto \frac{1}{Y}\Rightarrow \frac{Y_{s}}{Y_{c}}=\frac{l_{c}}{l_{s}}\Rightarrow \frac{l_{c}}{l_{s}}=\frac{2\times 10^{11}}{1.2\times 10^{11}}=\frac{5}{3}] $ (i) Also $ [l_{c}-l_{s}=0.5] $
(ii) On solving (i) and (ii), we get $ [l_{c}=0.5] $ cm and $ [l_{s}=0.75] $ cm
