A) 20m/s
B) 10m/s
C) 30m/s
D) 33m/s
Correct Answer: C
Solution :
[c] When source approaches observer \[{{f}_{1}}={{f}_{{}^\circ }}\frac{V}{V-{{V}_{sourec}}}\] \[{{f}_{1}}={{f}_{{}^\circ }}\frac{330}{330-{{V}_{sourec}}}....(1)\] When source recedes from observer \[{{f}_{1}}={{f}_{{}^\circ }}\frac{V}{V+{{V}_{sourec}}}\] \[{{f}_{1}}={{f}_{{}^\circ }}\frac{330}{330-{{V}_{sourec}}}....(2)\] \[\frac{{{f}_{1}}}{{{f}_{2}}}=\frac{6}{5}=\frac{330+{{V}_{source}}}{330-{{V}_{source}}}\] \[6\,[330-{{V}_{source}}]\,=5\,[330+{{V}_{source}}]\] \[1980-6{{V}_{source}}=1650+5{{V}_{source}}\] \[330=11{{V}_{source}}\] \[{{V}_{source}}=30\,m/s\]You need to login to perform this action.
You will be redirected in
3 sec