A) 2000 m/sec
B) 200 m/sec
C) 20 m/sec
D) 2 m/sec
Correct Answer: C
Solution :
\[N'=\frac{v-{{v}_{0}}}{v-{{v}_{s}}}N\] Given, \[N'=\frac{9}{8}N\] \[v'=340\,\,m/s\] \[\xrightarrow{\begin{smallmatrix} sound \\ \,\,\,\,(V) \end{smallmatrix}}\] \[\xrightarrow[\begin{smallmatrix} Source\,\,train \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({{V}_{S}}) \end{smallmatrix}]{}\] \[\xleftarrow[\begin{smallmatrix} Observer\,\,Train \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({{V}_{O}}) \end{smallmatrix}]{}\] Let speed of each of the trains be\[v'\] \[\therefore \] \[\frac{9}{8}N=\frac{340-(-v')}{340-v'}N\] (Replacing \[{{v}_{0}}\] by \[-v'\] for the observer train, as it moves in a direction opposite to that of sound) \[\Rightarrow \] \[\frac{9}{8}=\frac{340+v'}{340-v'}\] \[\Rightarrow \] \[9(340-v')=8(340+v')\] \[\Rightarrow \] \[v'=20\,\,m/s\]
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