A) 1.98 m
B) 2.04 m
C) 2.00 m
D) 1.99 m
Correct Answer: B
Solution :
Frequency \[f=\frac{9}{3}=3\,Hz\] \[f=v\left[ \frac{1}{{{\lambda }_{1}}}-\frac{1}{{{\lambda }_{2}}} \right]\] \[3=300\left[ \frac{1}{2}-\frac{1}{{{\lambda }_{2}}} \right]\] \[\frac{1}{100}=\frac{1}{2}-\frac{1}{{{\lambda }_{2}}}\] \[\frac{1}{{{\lambda }_{2}}}=\frac{1}{2}-\frac{1}{100}\] \[\frac{1}{{{\lambda }_{2}}}=\frac{50-1}{100}=\frac{49}{100}\] \[{{\lambda }_{2}}=\frac{100}{49}=2.04\,m\]You need to login to perform this action.
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