AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Q.

A body of mass 200g is tied to a spring constant 12.5 N/m, while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad/s . Then the ratio of extension in the spring to its natural length will be :

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

a

1 : 2

b

2 : 5

c

1 : 1

d

2 : 3

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

detailed_solution_thumbnail

The required centripetal force is provided by the spring .

If 'l' is the length of the spring and the extension is given by 'e'

$\begin{array}{l} m (l + e) ω^{2} = Ke \\ \Rightarrow 200 \times 10^{- 3} (l + e) 25 = (\frac{25}{2}) e \\ \Rightarrow \frac{l + e}{e} = \frac{10}{4} = \frac{5}{2} \Rightarrow \frac{l}{e} = \frac{3}{2} \\ \Rightarrow \frac{e}{l} = \frac{2}{3} \end{array}$

Watch 3-min video & get full concept clarity

courses

No courses found

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

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

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!