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

Last updated date: 13th Jun 2024
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Hint: 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 \[3.12s\]. 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 \[3.12s\], this means that the stone reached the well and then the sound travelled back to the observer, this took \[3.12s\].
The following values are given to us:
Time after which the splash is heard \[ = 3.12s\]\[(t)\]
Depth of the well \[ = 45m\]
g \[ = 10m{s^{ - 2}}\]
Let \[{t_1}\] be the time taken by the stone to reach the water.
And \[h\] be the depth of the well \[ = 45m\]
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.
\[h = ut + \dfrac{1}{2}g{t_1}^2\]
Since the stone is thrown, we consider the initial velocity to be zero.
\[\therefore U = 0\]
So, \[h = \dfrac{1}{2}g{t_1}^2\]
Now, rearranging the equation, we get:
\[{t_1} = \sqrt {\dfrac{{2h}}{g}} \]
Putting the values, we obtain:
\[{t_1} = \sqrt {\dfrac{{2 \times 45}}{{10}}} s\]
On solving, we get:
\[{t_1} = 3s\]
Total time taken by the splash to reach the observer is \[ = 3.12s\], therefore, time taken for the sound to travel to the observer is: \[{t_2} = t - {t_1}\]
Therefore, \[{t_2} = (3.12 - 3)s = 0.12s\]
We know. Distance travelled by the sound is depth of the well, which is \[ = 45m\]
Thus, we can calculate the speed of sound, as we know:
$\Rightarrow$ \[Speed = \dfrac{{{\text{Distance travelled}}}}{{{\text{Time Taken}}}}\]
Putting the values, we obtain:
$\Rightarrow$ \[Speed = \dfrac{{45}}{{0.12}}\]
Thus, we finally arrive at:
$\Rightarrow$ \[Speed = 375m/s\]
This is the required solution.

Therefore, option (B) is correct.

Note: SI unit of speed is \[m/s\], 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 \[km/hr\], 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.