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ICSE Class 10 Physics 2023 Paper

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Question : 56 of 76
Marks: +1, -0
Calculate the distance covered by the Ultrasonic wave having a velocity of 1.5 kms−1\text{kms}^{-1} in 14 s14\ \text{s} , when it is received after reflection by the receiver of the SONAR.
Solution:  
Given : v=1.5 kms−1=1500 ms−1v=1.5\ \text{kms}^{-1}=1500\ \text{ms}^{-1}
  t=14 s.\;t=14\ \text{s}.
Let the distance between the source of ultra-sonic wave and the reflector be ' dd ' meter.
we know,
v  =  2dtv\;=\;\frac{2d}{t}
⇒    2d  =  v×t\Rightarrow\;\;2d\;=\;v\times t
⇒    d  =  v×t2  =  1500×142=10500 m\Rightarrow\;\;d\;=\;\frac{v\times t}{2}\;=\;\frac{1500\times14}{2}=10500\ \text{m}
Total distance covered =2d=2d
  =2×10500\;=2\times10500
  =21000 m\;=21000\ \text{m}
  =21 km\;=21\ \text{km}
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