MCQ
When a metal wire is stretched by a load, the fractional change in its volume ∆V/V is proportional to
Options
\[\frac{∆ \text{l}}{\text{ l }}\]
\[\left( \frac{∆ \text{ l }}{\text{ l }} \right)^2\]
\[\sqrt{∆ \text{ l / l}}\]
none of these
Advertisement Remove all ads
Solution
\[\frac{∆ \text{l}}{\text{ l }}\]
\[\text{ C . S . A . = A }\]
\[\text{ Length = l}\]
\[\text{ Volume of the wire V = Al }\]
\[\text{ Assuming no lateral strain when longitudinal strain occurs: }\]
\[\text{ Increase in volume: ∆ V = A ∆ l }\]
\[ \Rightarrow \frac{∆ V}{V} = \frac{A ∆ \text{ l }}{\text{ Al }} = \frac{∆ \text{ l }}{\text{ l }}\]
\[\text{ So,} \frac{∆ V}{V} \text{ is directly proportional to } \frac{∆ \text{l}}{\text{ l }}\] .
Concept: Stress and Strain
Is there an error in this question or solution?
Advertisement Remove all ads
APPEARS IN
Advertisement Remove all ads
Advertisement Remove all ads