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Question: A splash is heard after 3.12s of a stone is dropped into a well 45m deep. The speed of sound in air ...

A splash is heard after 3.12s of a stone is dropped into a well 45m deep. The speed of sound in air is: [g=10ms2g = 10m{s^{ - 2}}]
(A) 330m/s330m/s
(B) 375m/s375m/s
(C) 340m/s340m/s
(D) 346m/s346m/s

Explanation

Solution

We need to calculate the time taken by the stone to reach the well at first and deduct that amount from the total time, which is 3.12s3.12s. Depth of the well is given, therefore, this remaining time was taken by sound to reach the observer, and hence we can find the

Complete step by step solution:
In the question it is given that the splash is heard after 3.12s3.12s, this means that the stone reached the well and then the sound travelled back to the observer, this took 3.12s3.12s.
The following values are given to us:
Time after which the splash is heard = 3.12s$$$$(t)
Depth of the well =45m = 45m
g =10ms2 = 10m{s^{ - 2}}
Let t1{t_1} be the time taken by the stone to reach the water.
And hh be the depth of the well =45m = 45m
We know, the three equations of motion, from there let us consider the equation which helps us to calculate the distance travelled from initial velocity, acceleration and time taken.
h=ut+12gt12h = ut + \dfrac{1}{2}g{t_1}^2
Since the stone is thrown, we consider the initial velocity to be zero.
U=0\therefore U = 0
So, h=12gt12h = \dfrac{1}{2}g{t_1}^2
Now, rearranging the equation, we get:
t1=2hg{t_1} = \sqrt {\dfrac{{2h}}{g}}
Putting the values, we obtain:
t1=2×4510s{t_1} = \sqrt {\dfrac{{2 \times 45}}{{10}}} s
On solving, we get:
t1=3s{t_1} = 3s
Total time taken by the splash to reach the observer is =3.12s = 3.12s, therefore, time taken for the sound to travel to the observer is: t2=tt1{t_2} = t - {t_1}
Therefore, t2=(3.123)s=0.12s{t_2} = (3.12 - 3)s = 0.12s
We know. Distance travelled by the sound is depth of the well, which is =45m = 45m
Thus, we can calculate the speed of sound, as we know:
\Rightarrow Speed=Distance travelledTime TakenSpeed = \dfrac{{{\text{Distance travelled}}}}{{{\text{Time Taken}}}}
Putting the values, we obtain:
\Rightarrow Speed=450.12Speed = \dfrac{{45}}{{0.12}}
Thus, we finally arrive at:
\Rightarrow Speed=375m/sSpeed = 375m/s
This is the required solution.

Therefore, option (B) is correct.

Note: SI unit of speed is m/sm/s, so if we have distances given in kilometre and time given I hours, we must convert them to meters and seconds respectively, and not write km/hrkm/hr, as the unit of speed. Initial velocity of the stone is taken to be zero, as the velocity remains the same, with respect to the reference frame.