Q.

Two masses each equal to m are lying on X-axis at (- a,0) and (+ a, 0) respectively as shown in fig.They are connected by a light string. A force F is applied at the origin and along the Y-axis. As a result, the masses move towards each other What is the acceleration of each mass ? Assume the instantaneous position of the masses as (- x,0) and {x, 0) respectively

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

Fm⋅xa2−x2

b

F2m⋅xa2−x2

c

2Fm⋅xa2−x2

d

2Fm⋅a2−x2x

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

See figure.From figure F=2Tcos⁡θ or T=F/(2cos⁡θ)The force responsible for motion of masses on X-axis is Tsin⁡θ∴ ma=Tsin⁡θ=F2cos⁡θ×sin⁡θ =F2tan⁡θ=F2×OBOA=F2×x(a2−x2)So, a=F2m×x(a2−x2).
Watch 3-min video & get full concept clarity
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 Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
score_test_img

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 Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
whats app icon
Two masses each equal to m are lying on X-axis at (- a,0) and (+ a, 0) respectively as shown in fig.They are connected by a light string. A force F is applied at the origin and along the Y-axis. As a result, the masses move towards each other What is the acceleration of each mass ? Assume the instantaneous position of the masses as (- x,0) and {x, 0) respectively