JEE Main & Advanced
JEE Main Paper (Held on 12-4-2019 Afternoon)
question_answer
A uniform cylindrical rod of length L and radius r, is made from a material whose Young's modulus of Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equals to : [JEE Main 12-4-2019 Afternoon]
A)\[F/(3\pi {{r}^{2}}YT)\]
B)\[3F/(\pi {{r}^{2}}YT)\]
C)\[6F/(\pi {{r}^{2}}YT)\]
D)\[9F/(\pi {{r}^{2}}YT)\]
Correct Answer:
B
Solution :
\[\therefore \] Length of cylinder remains unchanged so\[{{\left( \frac{F}{A} \right)}_{Compressive}}={{\left( \frac{F}{A} \right)}_{Thermal}}\] \[\frac{F}{\pi {{r}^{2}}}=Y\alpha T\] (\[\alpha \]is linear coefficient of expansion) \[\therefore \]\[\alpha =\frac{F}{YT\pi {{r}^{2}}}\] \[\therefore \]The coefficient of volume expansion\[\gamma =3\alpha \] \[\therefore \]\[\gamma =3\frac{F}{YT\pi {{r}^{2}}}\]