QJEEM21Ph9c5a: JEE Main 2021 Physics - Class 11 JEEM21Ph9c5a Solution: The normal reaction 'N' for a vehicle of 800 kg mass,…

Question

Verified Solution • 2025–26 Session moderate

The normal reaction 'N' for a vehicle of 800 kg mass, negotiating a turn on a 30

^\circ

banked road at maximum possible speed without skidding is ____________

\times

10³ kg m/s². [Given cos30

^\circ

= 0.87,

\mu

_s = 0.2]

Accepted Answer

The given situation can be represented as

JEE Main 2021 (Online) 20th July Morning Shift Physics - Circular Motion Question 47 English Explanation

Equating forces perpendicular to the inclined plane,

N = mg\cos 30^\circ + {{m{v^2}} \over R}\sin 30^\circ

\Rightarrow N - mg\cos 30^\circ = {{m{v^2}} \over R}\sin 30^\circ

.... (i)

Equating forces along the inclined plane,

mg\sin 30^\circ + {\mu _s}N = {{m{v^2}} \over R}\cos 30^\circ

.... (ii)

On dividing Eq. (i) by Eq. (ii), we get

{{N - mg\cos 30^\circ } \over {mg\sin 30^\circ + {\mu _s}N}} = \tan 30^\circ

[

\because

\cos 30^\circ = {{\sqrt 3 } \over 2}

and

\sin 30^\circ = {1 \over 2}

]

\Rightarrow

{{N - mg(\sqrt3/2) } \over {mg(1/2) + (0.2)N}} = {1 \over {\sqrt 3 }}

N\sqrt 3 - mg{{\sqrt 3 } \over 2}.\sqrt 3 = {{mg} \over 2} + 0.2N

\Rightarrow (\sqrt 3 - 0.2)N = mg{{(1 + 3)} \over 2} = 2mg

\Rightarrow N = {{2mg} \over {\sqrt 3 - 0.2}} = {{2 \times 800 \times 10} \over {1.532}}

= 10.44 \times {10^3}V

Therefore,

N = 10.2 \times {10^3}

kg-m/s²

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