The Earth has a radius of $6400 \ km$. The inner core of $1000 \ km$ radius is solid. Outside it,there is a region from $1000 \ km$ to a radius of $3500 \ km$ which is in a molten state. Then again from $3500 \ km$ to $6400 \ km$,the Earth is solid. Only longitudinal $(P)$ waves can travel inside a liquid. Assume that the $P$ wave has a speed of $8 \ kms^{-1}$ in solid parts and of $5 \ kms^{-1}$ in liquid parts of the Earth. An earthquake occurs at some place close to the surface of the Earth. Calculate the time after which it will be recorded in a seismometer at a diametrically opposite point on the Earth if the wave travels along the diameter.

(N/A) The total time taken $t$ is the sum of the time taken to travel through each distinct layer along the diameter.
The diameter consists of three regions: the inner solid core,the molten middle region,and the outer solid crust.
The total distance for each region along the diameter is twice its radial thickness.
$1$. Inner solid core ($0$ to $1000 \ km$): Distance $d_1 = 2 \times 1000 \ km = 2000 \ km$,Speed $v_1 = 8 \ kms^{-1}$.
Time $t_1 = \frac{d_1}{v_1} = \frac{2000}{8} = 250 \ s$.
$2$. Molten region ($1000 \ km$ to $3500 \ km$): Distance $d_2 = 2 \times (3500 - 1000) \ km = 5000 \ km$,Speed $v_2 = 5 \ kms^{-1}$.
Time $t_2 = \frac{d_2}{v_2} = \frac{5000}{5} = 1000 \ s$.
$3$. Outer solid crust ($3500 \ km$ to $6400 \ km$): Distance $d_3 = 2 \times (6400 - 3500) \ km = 5800 \ km$,Speed $v_3 = 8 \ kms^{-1}$.
Time $t_3 = \frac{d_3}{v_3} = \frac{5800}{8} = 725 \ s$.
Total time $t = t_1 + t_2 + t_3 = 250 + 1000 + 725 = 1975 \ s$.