A) zero
B) 2.5%
C) 5%
D) 10%
Correct Answer: D
Solution :
\[\frac{n}{n}=\left[ \frac{v+{{v}_{o}}}{v} \right]\] \[\frac{n}{n}=\left[ \frac{330+33}{330} \right]\] \[n[1.1]\,\,n\] Apparent increase in frequency \[=1.1\,\,n-n\] \[=0.1\times n\] \[=\frac{0.1\times n\times 100}{n}=10%\]You need to login to perform this action.
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