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)]:
A
B
C
D
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 = 
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