Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A uniform rod of mass M is hanged with two strings P and Q such that rod is horizontal. A mass m is connected by a light ideal string from the end of rod as shown. Then the maximum value of mass m upto which the string P remains tight is [Length of rod is (a + b)]:

Text Solution
Verified by ExpertsThe correct answer is:
A
At maximum value of m, T P → 0
∴ T Q = (m + M) g … (1)
Now taking torque about centre of gravity of rod
T Q ×
= mg × 
or (m + M) [a – b] = m [a + b]
= 
=
⇒ m = 
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
A vessel containing water is given a constant acceleration a towards the right, along a straight ho…
A ship of mass \(3 \times 10^{7} \, \text{kg}\) initially at rest is pulled by a force of \(5 \time…
The mass of a body measured by a physical balance in a lift at rest is found to be m. If the lift i…
Three weights W, 2W and 3W are connected to identical springs suspended from a rigid horizontal rod…
When forces F_1, F_2, F_3 are acting on a particle of mass m such that F_2 and F_3 are mutually per…
The spring balance A reads 2 kg with a block m suspended from it. A balance B reads 5 kg when a bea…