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Question

A uniform metre stick of mass 250 gm is suspended from the ceiling through two vertical strings of equal lengths. A small object of mass 50 gm is placed on the stick at a distance of 70 cm from the left end. Find the tension in both the strings. Take g=10 m/s2.


A
T1=1.15 N,T2=2.15 N
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B
T1=1.06 N,T2=1.14 N
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C
T1=1.40 N,T2=1.60 N
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D
T1=1.72 N,T2=1.28 N
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Solution

The correct option is C T1=1.40 N,T2=1.60 N

Let C be the centre of mass of the stick.
For translation equilibrium,
T1+T2=(m1+m2)g
T1+T2=0.3g=3 ...(i)

For rotational equilibrium, let the net torque be zero about centre C:
τnet=0 ...(ii)
Taking anticlockwise sense of rotation for torque as +ve
& τ=F×r
Force m2g passes through reference point C, hence its torque is zero about point C.
Here, AC=CB=50 cm=0.50 m
m1=0.05 kg, m2=0.25 kg
Substituting the torques with proper signs in Eq (ii),
τ=0
(T1×0.5)+0(m1g×0.20)+(T2×0.50)=0
(T1×0.50)(0.05×10×0.20)+(T2×0.50)=0
T2T12=0.1
T2T1=0.20 ...(iii)

Adding Eq (i) & (iii),
2T2=3.20
T2=1.60 N
and T1=31.6=1.40 N

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