JEE Main & Advanced Physics Transmission of Heat Question Bank Conduction

  • question_answer
    A cylindrical rod having temperature \[{{T}_{1}}\] and \[{{T}_{2}}\] at its ends. The rate of flow of heat is \[{{Q}_{1}}\]cal/sec. If all the linear dimensions are doubled keeping temperature constant then rate of flow of heat \[{{Q}_{2}}\]will be                        [CBSE PMT 2001]                        

    A)            \[4{{Q}_{1}}\]                      

    B)            \[2{{Q}_{1}}\]

    C)            \[\frac{{{Q}_{1}}}{4}\]      

    D)            \[\frac{{{Q}_{1}}}{2}\]

    Correct Answer: B

    Solution :

                       Rate of heat flow \[\left( \frac{Q}{t} \right)=\frac{k\pi {{r}^{2}}({{\theta }_{1}}-{{\theta }_{2}})}{L}\propto \frac{{{r}^{2}}}{L}\]            \ \[\frac{{{Q}_{1}}}{{{Q}_{2}}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}\left( \frac{{{l}_{2}}}{{{l}_{1}}} \right)={{\left( \frac{1}{2} \right)}^{2}}\times \left( \frac{2}{1} \right)=\frac{1}{2}\]Þ \[{{Q}_{2}}=2{{Q}_{1}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner