Question

ANTHE 2018 - CLASS X

PHYSICS

An ultrasound equipment installed at a height of 80m above sea water sends out ultrasound that returns from the sea bed and is detected after 5.4s. If the speed of ultrasound through sea water is 1500 m/s and that through air is 320 m/s, then the depth of the sea at the place is:-

(1) 4900 m
(2) 2290 m
(3) 3675 m
(4) 2450 m

Answer 1

Hello vidhi jiiii

Time taken from transmission to reception is given as total time

5.4= time required to travel in water + time required to travel in air

time = Distance /Speed

And the distance above sea level is given to be 80 m

[Tex]5.4= \frac{(80*2)}{320}+\frac{(d*2)}{1500}[/tex]

d is the depth of water

[Tex]2.7=\frac{80}{320}+\frac{d}{1500}[/tex]

[Tex]d=2.7-\frac{80}{320}*1500[/tex]

[Tex]d=2.45*1500[/tex]

[Tex]d=3675 [/tex]

Option 3rd is correct one

@skb

Answer 2

This question can be divided  in two parts.

First  time  taken in travelling  in Air,

Second  time  taken in travelling in sea.

1. When sound goes in downward  direction:-

(i) To find  time  taken to travel in air

As given that  ultrasound  equipment  installed at  height of 80m above sea level. It  means that distance  between sea  surface  and  installation position is 80 m. This  distance  is traveled by  sound  in air with the speed  of  320 m/s as  given.

Distance  = 80m

Speed = 320 m

We  know that,

Time  = distance\speed

T₁ = 80/320

T₁ = 8/32

T₁ = 1/4 second.

(i) To find  time  taken to travel in sea

Let the depth of the sea  be  x. Now, sound  has to travel x distance with the  speed  of 1500m/s .

distance = x

speed = 1500 m/s

Time = x/1500

T₂ = x/1500 second

Total Tiime  taken in travelling A to C = T₁ + T₂

2 When sound goes in upward  direction (C to A):-

Clearly, time  taken in travelling  A to C will be  equal to time  taken in travelling C to A

Hence, Time  Taken in travellinf C to A = T₁ + T₂

Now,

Sound goes A to C , comes C to A ,

Total time in this journey = 2 × (T₁ + T₂)

Now, As per given in question, Total time  in this journey = 5.4 second

$2(T_1+T_2)=5.4\\\\2(\frac{1}{4}+\frac{x}{1500})=5.4\\\;\\\frac{2}{2}(\frac{1}{2}+\frac{x}{750})=5.4\\\;\\\frac{1}{2}+\frac{x}{750}=5.4\\\;\\0.5+\frac{x}{750}=5.4\\\;\\\frac{x}{750}=5.4-0.5\\\;\\\frac{x}{750}=4.9\\\;\\x=750\times4.9\\\;\\x=3675\;m$

Hence option 3) 3675m is correct