A) 1.47
B) 1.42
C) 1.50
D) 1.45
Correct Answer: B
Solution :
| [b] |
| \[\because \frac{{{n}_{1}}}{{{\gamma }_{1}}-1}+\frac{{{n}_{2}}}{{{\gamma }_{2}}-1}=\frac{{{n}_{1}}+{{n}_{2}}}{{{\gamma }_{m}}-1}\] |
| \[\Rightarrow \frac{3}{\frac{4}{3}-1}+\frac{2}{\frac{5}{3}-1}=\frac{5}{\gamma m-1}\] |
| \[\Rightarrow \frac{9}{1}+\frac{2\times 3}{2}=\frac{5}{{{\gamma }_{m}}-1}\] |
| \[\Rightarrow {{\gamma }_{m}}-1=\frac{5}{12}\] |
| \[\Rightarrow {{\gamma }_{m}}=\frac{17}{12}=1.42\] |
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