JEE Main 2019PhysicsMechanical Properties of FluidsEasyMCQ

JEE Main 2019 — Mechanical Properties of Fluids Question with Solution

JEE Main 2019 (10 Apr Shift 2)

Question

A submarine experiences a pressure of 5.05×106 Pa at a depth of d1 in a sea. When it goes further to a depth of  d2 , it experiences a pressure of  8.08×106 Pa . Then d2-d1 is approximately (density of water =103 kg/m3 and acceleration due to gravity  =10 ms-2 ):

Choose an option

Show full solutionCorrect option: C
Correct answer
C300 m

Step-by-step explanation

The pressure at a depth d is p=p0+ρgd, here p0 is the atmospheric pressure.
∴p1=p0+ρgd1
and p2=p0+ρgd2
The difference in pressure is p2-p1=ρgd2-d1

Putting the values in the above equation
8.08×106-5.05×106=103×10d2-d1
d2-d1=3.03×106104
=303 m 

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About this question

This is a previous-year question from JEE Main 2019, covering the Mechanical Properties of Fluids chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.