Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A stone dropped from the top of a tower of height  200 m high splashes into the water of a pound near the base of the tower. When is the splash heard at the top? Given that the speed of sound

A stone dropped from the top of a tower of height  200 m high splashes into the water of a pound near the base of the tower. When is the splash heard at the top? Given that the speed of sound in air is  332 m / s ,(g=9.8 m /s2)

Option: 1

1.634


Option: 2

5.267


Option: 3

2.622


Option: 4

3.465


Answers (1)

best_answer


Here,  h=200 m \,\,\,\, g=9.8 m/s^{2} and                                         

velocity of sound   V=332 m/s

Let  t_1  be the time taken by the stone to reach the surface of the pound

Then using  S=u t+\frac{1}{2} a t^2 \,\,\,\,\quad h=0+\frac{1}{2} \,g \,t_1^2

{ or }\,\,\, t_1 =\sqrt{\frac{2 h}{g}} \\

\therefore t_1 =\sqrt{\frac{2 \times 200}{9.8}}=\sqrt{\frac{2 \times 2 \times 10^2}{98}}=\frac{\sqrt{2 \times 10^2}}{\sqrt{49}} \\

=\frac{\sqrt{2} \times 10}{7}=\frac{1.414 \times 10}{7}=\frac{1414 \times 10^{-2}}{7}

=202 \times 10^{-2}=2.02 \mathrm\,\,{sec} \text {. }

Also, if  t_2 is the time taken by the sound to reach at a height  h  then

t_2=\frac{h}{v}=\frac{200}{332}=0.602 \,\,\mathrm{sec}

\therefore Total time after which sound of splash is heard 

=t_{1}+t_{2}=2.02+0.602=2.622

Posted by

Kuldeep Maurya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: NIT Rourkela cut-offs for CSE, chemical engineering drop; last years' closing ranks
anam-khan

JEE Main 2025 Live: NTA exam city slip at jeemain.nta.nic.in soon; session 1 schedule
anam-khan

JEE Main 2025 schedule out for session 1; paper 1 from January 22

Latest Question