Answer
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Hint: As we know that the sound can travel through all the medium solid, liquid, and gas. But the speed with which it will travel through the different mediums will be different. So, we have to calculate the time for each case separately, that is time for the sound to travel through air and the time for the sound to travel through the aluminium rod.
Complete step-by-step answer:
So, the time can be calculated by the basic formula relating time, distance and speed.
That is,
$speed = \dfrac{{distance}}{{time}}$
$v = \dfrac{d}{t}$
$ \Rightarrow t = \dfrac{d}{v}$
So, for the sound to travel in the air,
As we know that the speed of the sound is air= ${v_a} = 340$m/s
And the distance between both the children will be same in both the cases, so let that distance =$d$
So now time for the sound to travel in the air will be,
${t_a} = \dfrac{d}{{{v_a}}}$
$ \Rightarrow {t_a} = \dfrac{d}{{340}}$------equation (1)
Now similarly for the sound to travel through the aluminium rod,
So as we also know that the value of the speed of the sound in the aluminium= ${v_{Al}}$=$6420$m/s
So now the time for the sound to travel in the aluminium will be,
${t_{Al}} = \dfrac{d}{{{v_{Al}}}}$
$ \Rightarrow {t_{Al}} = \dfrac{d}{{6420}}$--------equation (2)
Now we have got the time equation for both the cases so to get the ratio of times taken by the sound wave in air and in aluminium to reach the second child, we will divide the equations,
So, on dividing equation (1) by equation (2), we get
$\dfrac{{{t_a}}}{{{t_{Al}}}} = \dfrac{{\dfrac{d}{{340}}}}{{\dfrac{d}{{6420}}}}$
$ \Rightarrow \dfrac{{{t_a}}}{{{t_{Al}}}} = \dfrac{{6420}}{{340}}$
$ \Rightarrow \dfrac{{{t_a}}}{{{t_{Al}}}} = 18.88$
Hence, the ratio of times taken by the sound wave in air and in aluminium is \[18.88\].
Note: If we compare the speed of the sound in all the three states of matter, we find that the speed of the sound in the solids is highest and speed of sound in the air is lowest. So, if we arrange the speed of the sound in all the three states of matter in the decreasing order then it will be as follows,
$solid > liquid > gas.$
Complete step-by-step answer:
So, the time can be calculated by the basic formula relating time, distance and speed.
That is,
$speed = \dfrac{{distance}}{{time}}$
$v = \dfrac{d}{t}$
$ \Rightarrow t = \dfrac{d}{v}$
So, for the sound to travel in the air,
As we know that the speed of the sound is air= ${v_a} = 340$m/s
And the distance between both the children will be same in both the cases, so let that distance =$d$
So now time for the sound to travel in the air will be,
${t_a} = \dfrac{d}{{{v_a}}}$
$ \Rightarrow {t_a} = \dfrac{d}{{340}}$------equation (1)
Now similarly for the sound to travel through the aluminium rod,
So as we also know that the value of the speed of the sound in the aluminium= ${v_{Al}}$=$6420$m/s
So now the time for the sound to travel in the aluminium will be,
${t_{Al}} = \dfrac{d}{{{v_{Al}}}}$
$ \Rightarrow {t_{Al}} = \dfrac{d}{{6420}}$--------equation (2)
Now we have got the time equation for both the cases so to get the ratio of times taken by the sound wave in air and in aluminium to reach the second child, we will divide the equations,
So, on dividing equation (1) by equation (2), we get
$\dfrac{{{t_a}}}{{{t_{Al}}}} = \dfrac{{\dfrac{d}{{340}}}}{{\dfrac{d}{{6420}}}}$
$ \Rightarrow \dfrac{{{t_a}}}{{{t_{Al}}}} = \dfrac{{6420}}{{340}}$
$ \Rightarrow \dfrac{{{t_a}}}{{{t_{Al}}}} = 18.88$
Hence, the ratio of times taken by the sound wave in air and in aluminium is \[18.88\].
Note: If we compare the speed of the sound in all the three states of matter, we find that the speed of the sound in the solids is highest and speed of sound in the air is lowest. So, if we arrange the speed of the sound in all the three states of matter in the decreasing order then it will be as follows,
$solid > liquid > gas.$
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