Numerical Problems Based on Indirect Methods for Long Distances for Class 11 Physics

Given total time taken for the echo $t = 200\text{ s}$ and the speed of sound in water $v = 1.450\text{ km s}^{-1}$.

Since this is an echo, the wave travels to the enemy submarine and back. The total distance traveled is $2d$, where $d$ is the distance to the submarine.

Given distance $d = 6.3 \times 10^{10}\text{ m}$. The total time taken for the signal to travel to the planet and return is $t = 7\text{ minutes}$.

First, convert the time to seconds:

Using the echo formula ($2d = v \times t$), we rearrange for speed ($v$):

Given distance to the rock $d = 1595\text{ m}$ and speed $v = 1450\text{ m s}^{-1}$.

The time taken for the signal to travel to the rock and echo back is found using $2d = v \times t$:

Let the height of the hill be $h$ and the original horizontal distance from the point to the base of the hill be $x$.

From the first observation point ($\theta_1 = 30^\circ$):

After walking $1\text{ km}$ towards the hill, the new distance is $(x – 1)$. From the second observation point ($\theta_2 = 45^\circ$):

Equating the two expressions for $x$:

Rationalizing the denominator:

Given the base-line $b = R_e = 6.4 \times 10^6\text{ m}$.

The parallax angle $\theta = 57’$ (57 minutes). We must convert this angle into radians.

Using the parallax formula $D = \frac{b}{\theta}$, the distance to the moon ($D$) is: