An earthquake generates both transverse S and | vedclass

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(C) Let the distance of the epicenter of the earthquake from the point of observation be $d$.Speed of $S$-wave, $v_S = 4.5 \,km/s$.Speed of $P$-wave,

Vedclass · Accepted Answer

$2160$

Vedclass · Answer

(C) Let the distance of the epicenter of the earthquake from the point of observation be $d$.Speed of $S$-wave, $v_S = 4.5 \,km/s$.Speed of $P$-wave, $v_P = 8.0 \,km/s$.Since the distance $d$ is the same for both waves, we have $d = v_P t_P = v_S t_S$.Thus, $8.0 t_P = 4.5 t_S$, which gives $t_P = \frac{4.5}{8.0} t_S \quad \dots (i)$.The first $P$-wave arrives $3.5 \,min$ earlier than the first $S$-wave, so $t_S - t_P = 3.5 \,min = 3.5 	imes 60 \,s = 210 \,s \quad \dots (ii)$.Substituting $(i)$ into $(ii)$:$t_S - \frac{4.5}{8.0} t_S = 210$$\frac{8.0 t_S - 4.5 t_S}{8.0} = 210$$\frac{3.5 t_S}{8.0} = 210$$t_S = \frac{210 	imes 8.0}{3.5} = 480 \,s$.Now, calculating the distance $d = v_S t_S = 4.5 \,km/s 	imes 480 \,s = 2160 \,km$.

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