A) 0.005/°C
B) 0.0005/°C
C) 0.002/°C
D) 0.0002/°C
Correct Answer: A
Solution :
\[\frac{{{h}_{1}}}{{{h}_{2}}}=\frac{{{\rho }_{1}}}{{{\rho }_{2}}}=\frac{(1+\gamma \,{{\theta }_{1}})}{(1+\gamma \,{{\theta }_{2}})}\] \[\left[ \because \,\rho =\frac{{{\rho }_{0}}}{(1+\gamma \theta )} \right]\] Þ \[\frac{50}{60}=\frac{1+\gamma \times 50}{1+\gamma \times 100}\] Þ \[\gamma =0.005/{}^\circ C\]
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