Question

A stone is dropped into a $200\text{m}$ deep well, if the well is dry, after how much time the sound of stone hitting the bottom will be heared.
(speed of sound in air $=340\text{m/s}$ and $g=10{\text{m/s}}^2$)

• A. $6.91\text{s}$
  
• B. $8.52\text{s}$
  
• C. $11.43\text{s}$
  
• D. $2.51\text{s}$
  

Answer 1

The correct option is A $6.91\text{s}$


Depth of the well, $h=200\text{m}$
Speed of sound, $v=340\text{m/s}$
Let time taken by stone to reach the bottom of the well be $t_1$, so, we have
$s=u{t}_1+\frac{1}{2}a{t}_1^2$
$\Rightarrow 200=0+\frac{1}{2}\times 10\times t_1^2$$[\because s=h,u=0\text{and}a=g$]
$\Rightarrow t_1=2\sqrt{10}=6.32\text{s}$
Now, time taken by sound to reach the mouth of the well
$t_2=\frac{h}{v}$
$t_2=\frac{200}{340}=\frac{10}{17}\text{s}=0.59\text{s}$
Time after which sound is heard is
$t=t_1+t_2=6.91\text{s}$