A particle is projected horizontally from the top of a tower with a velocity $v_{0}$ . If $v$ be its velocity at any instant, then the radius of curvature of the path of the particle at that instant is directly proportional to :

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\frac{1}{v}$

$v^{3}$

$v^{2}$

$v$

answer is A.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

As we know : $a_{c} = \frac{v^{2}}{R}$ (Centripetal acceleration)

From figure, $g \sin θ = \frac{V^{2}}{R}$

$\Rightarrow g \cdot \frac{v_{0}}{V} = \frac{v^{2}}{R} (∵ \sin θ_{i} = \frac{v_{0}}{V}) \Rightarrow R \propto v^{3}$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test