A body of mass 500g moves along… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Mechanics

2023mcqeasy

A body of mass $500g$ moves along $x$ -axis such that it's velocity varies with displacement $x$ according to the relation $v=10\sqrt{x}m{s}^{-1}$ the force acting on the body is:

Official previous-year question

Held on 11 Apr 2023 · Verified 6 Jul 2026.

Options

A
$125N$
B
$25N$
C
$166N$
D
$5N$

Did you get this right?

Solution

It is given that

$v=10\sqrt{x}m{s}^{-1}$

Force is defined by $F=m\frac{dv}{dt}...(i)$

Multiplying and dividing equation (i) by $dx$

$F=m\frac{dv}{dt}\frac{dx}{dx} \Rightarrow F=m\frac{dx}{dt}\frac{dv}{dx}=mv\frac{dv}{dx}...(ii)$

Finding the values of the terms in equation (ii)

$\frac{dv}{dx}=\frac{10}{2\sqrt{x}}=\frac{5}{\sqrt{x}}$

$v\frac{dv}{dx}=a=10\sqrt{x}\times \frac{5}{\sqrt{x}}=50m{s}^{-2}$

Hence, the force is

$F=\frac{1}{2}\times 50=25N$ .

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Solution

It is given that

$v=10\sqrt{x}m{s}^{-1}$

Force is defined by $F=m\frac{dv}{dt}...(i)$

Multiplying and dividing equation (i) by $dx$

$F=m\frac{dv}{dt}\frac{dx}{dx} \Rightarrow F=m\frac{dx}{dt}\frac{dv}{dx}=mv\frac{dv}{dx}...(ii)$

Finding the values of the terms in equation (ii)

$\frac{dv}{dx}=\frac{10}{2\sqrt{x}}=\frac{5}{\sqrt{x}}$

$v\frac{dv}{dx}=a=10\sqrt{x}\times \frac{5}{\sqrt{x}}=50m{s}^{-2}$

Hence, the force is

$F=\frac{1}{2}\times 50=25N$ .

Options

Did you get this right?

Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics