A) \[{{\alpha }_{1}}{{l}_{1}}={{\alpha }_{2}}{{l}_{2}}\]
B) \[{{\alpha }_{1}}{{l}_{2}}={{\alpha }_{2}}{{l}_{1}}\]
C) \[\alpha _{1}^{2}{{l}_{2}}=\alpha _{2}^{2}{{l}_{1}}\]
D) \[{{\alpha }_{1}}l_{2}^{2}={{\alpha }_{2}}l_{1}^{2}\]
Correct Answer: A
Solution :
| Linear expansion coefficient |
| \[=\frac{\text{change in length}}{\text{original length }\times \text{ rise temperature}}\] |
| or \[\alpha =\frac{\Delta l}{lt}\] |
| or \[\Delta l=l\,\alpha \,t\] |
| For brass rod, \[\Delta {{l}_{1}}={{l}_{1}}{{\alpha }_{1}}t\] |
| For steel rod, \[\Delta {{l}_{2}}={{l}_{2}}{{\alpha }_{2}}t\] |
| Since, \[{{l}_{2}}-{{l}_{1}}=\] constant (give) |
| So, \[\Delta {{l}_{2}}-\Delta {{l}_{1}}=0\] |
| or \[\Delta {{l}_{2}}=\Delta {{l}_{1}}\] |
| \[\therefore \] \[{{l}_{2}}{{\alpha }_{2}}t={{l}_{1}}{{\alpha }_{1}}t\] |
| As \[t\ne 0,\] hence |
| \[{{l}_{2}}{{\alpha }_{2}}={{l}_{1}}{{\alpha }_{1}}\] |
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