RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • question_answer Two spheres of radii in the ratio 1 : 2 and densities in the ratio 2 : 1 and of same specific                 heat, are heated to same temperature and                 left in the same surrounding. Their rate of cooling will be in the ratio :

    A)  2 : 1                      

    B)         1 : 1

    C)  1 : 2                      

    D)         1 : 4

    Correct Answer: B

    Solution :

    The formula for rate of cooling is given by \[=\frac{mc}{t}\] As, mass = volume density Mass of sphere \[=\frac{4}{3}\pi {{r}^{3}}\times \rho ,\] where p is density Mass per unit area\[\frac{\frac{4}{3}\pi {{r}^{3}}\times \rho }{4\pi {{r}^{2}}}=\frac{1}{3}rp\] Hence, rate of cooling per unit area must be proportional to \[rp\] (here r is the radius of sphere and \[\rho \] is the density. Hence, ratio of rate of cooling for two spheres is \[=\frac{{{r}_{1}}\rho }{{{r}_{2}}{{\rho }_{2}}}\] where\[{{r}_{1}}:{{r}_{2}}\,=1:2\]and\[{{\rho }_{1}}:{{\rho }_{2}}=2:1\]                                                       \[=\frac{{{r}_{1}}{{\rho }_{1}}}{{{r}_{2}}{{\rho }_{2}}}\]


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