
A person places his ear at the end of a long steel pipe. He hears two distinct sounds at an interval of $0.5{\text{ s}}$ when another person hammers at the other end of the pipe. If the speed of the sound in metal and air are $3630m{s^{ - 1}}{\text{ and 330m}}{{\text{s}}^{ - 1}}$ respectively, then the distance between the two persons is
$
A.\;{\text{90}}{\text{.75m}} \\
{\text{B}}{\text{. 181}}{\text{.5m}} \\
{\text{C}}{\text{. 363m}} \\
{\text{D}}{\text{. 1650m}} \\
$
Answer
477.3k+ views
Hint: Here, use the concepts of the velocity/speed that is distance travelled upon the time taken to solve this question. Since, the distance between the two persons remains the same, equate the two equations, take correlation between the two terms and find the required unknown distance.
Complete step by step answer:
Let “d” be the distance between the two persons.
Velocity of sound in steel pipe/metal is $v{}_m = 3630m{s^{ - 1}}$
Velocity of sound in metal is $v{}_A = 330m{s^{ - 1}}$
Time taken by the steel pipe is $t = {t_1}$
Time taken by the air $t = {t_2} = t + 0.5$
Now, velocity
$
\Rightarrow {v_m} = \dfrac{d}{{{t_1}}} \\
\Rightarrow d = {v_m} \times t{\text{ }}......{\text{(1)}} \\
$
$
\Rightarrow {v_A} = \dfrac{d}{{{t_2}}} \\
\Rightarrow d = {v_A} \times (t + 0.5){\text{ }}......{\text{(2)}} \\
$
Since, the left hand side of both the equations $(1){\text{ and (2)}}$ are the same, equate the right hand side of the equations.
$\Rightarrow {v_m} \times t = {v_A} \times (t + 0.5){\text{ }}$
Place the known values-
$
\Rightarrow 3630 \times t = 330 \times (t + 0.5) \\
\Rightarrow 3630t = 330t + 330(0.5) \\
\Rightarrow 3630t - 330t = 165 \\
\Rightarrow 3300t = 165 \\
\Rightarrow t = \dfrac{{165}}{{3300}} \\
\therefore t = 0.05s \\
$
Place the value of “t” in equation$(1)$
$
\Rightarrow d = {v_m} \times t{\text{ }} \\
\Rightarrow d = 3630 \times 0.05 \\
\therefore d = 181.5m \\
$
Therefore, the required answer – the distance between the two person is $181.5m$
Hence, from the given multiple choices – the option B is the correct answer.
Note: Always check the given units and the units asked in the solutions. All the quantities should have the same system of units. There are three types of the system of units.
-MKS System (Metre Kilogram Second)
-CGS System (Centimetre Gram Second)
-System International (SI)
Also, remember the conversional relations among the system of units to make all the given units in the same format.
Complete step by step answer:
Let “d” be the distance between the two persons.
Velocity of sound in steel pipe/metal is $v{}_m = 3630m{s^{ - 1}}$
Velocity of sound in metal is $v{}_A = 330m{s^{ - 1}}$
Time taken by the steel pipe is $t = {t_1}$
Time taken by the air $t = {t_2} = t + 0.5$
Now, velocity
$
\Rightarrow {v_m} = \dfrac{d}{{{t_1}}} \\
\Rightarrow d = {v_m} \times t{\text{ }}......{\text{(1)}} \\
$
$
\Rightarrow {v_A} = \dfrac{d}{{{t_2}}} \\
\Rightarrow d = {v_A} \times (t + 0.5){\text{ }}......{\text{(2)}} \\
$
Since, the left hand side of both the equations $(1){\text{ and (2)}}$ are the same, equate the right hand side of the equations.
$\Rightarrow {v_m} \times t = {v_A} \times (t + 0.5){\text{ }}$
Place the known values-
$
\Rightarrow 3630 \times t = 330 \times (t + 0.5) \\
\Rightarrow 3630t = 330t + 330(0.5) \\
\Rightarrow 3630t - 330t = 165 \\
\Rightarrow 3300t = 165 \\
\Rightarrow t = \dfrac{{165}}{{3300}} \\
\therefore t = 0.05s \\
$
Place the value of “t” in equation$(1)$
$
\Rightarrow d = {v_m} \times t{\text{ }} \\
\Rightarrow d = 3630 \times 0.05 \\
\therefore d = 181.5m \\
$
Therefore, the required answer – the distance between the two person is $181.5m$
Hence, from the given multiple choices – the option B is the correct answer.
Note: Always check the given units and the units asked in the solutions. All the quantities should have the same system of units. There are three types of the system of units.
-MKS System (Metre Kilogram Second)
-CGS System (Centimetre Gram Second)
-System International (SI)
Also, remember the conversional relations among the system of units to make all the given units in the same format.
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