A) Increases by 33%
B) Decreases by 33%
C) Increases by 66%
D) Decreases by 66%
Correct Answer: D
Solution :
[d] \[{{v}_{1}}=\frac{c}{{{\lambda }_{1}}}\] \[{{v}_{2}}=\frac{c}{{{\lambda }_{2}}}=\frac{c}{3{{\lambda }_{1}}}\] % change in frequency \[=\frac{{{v}_{2}}-{{V}_{1}}}{{{v}_{1}}}\times 100\] \[=\frac{\frac{c}{3{{\lambda }_{1}}}-\frac{c}{{{\lambda }_{1}}}}{\frac{c}{{{\lambda }_{1}}}}\times 100=\frac{-\frac{2c}{3{{\lambda }_{1}}}\times 100}{\frac{c}{{{\lambda }_{1}}}}\] \[=-66%\] %
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