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Question

A thin uniform bar of mass m and length 2 L is held at an angle 300 with the horizontal by means of two vertical inextensible strings, at each end as shown in the figure. If the string at the right end breaks, leaving the bar to swing, then immediately after string breaks,
156311_b1af105fbbbb40a587fbe85fb0138307.png

A
the tension in the string at the left end is T = mg11
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B
the tension in the string at the left end is T = 2mg11
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C
the angular acceleration of the bar a = 3311 gL
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D
the angular acceleration of the bar a = 2311 gL
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Solution

The correct options are
B the tension in the string at the left end is T = 2mg11
C the angular acceleration of the bar a = 3311 gL
Just after cutting the string the tension of right string will disappear and tension in left string will change. let us draw free body diagram of rod immediately after the string breaks.
Let ax and by be the linear accelerations of COM and α the
angular acceleration of the rod about COM as shown.
As no force in horizontal direction. Then Fx = 0
ax = 0 ......(i)
ay = Fym = mgTm ....(ii)
Applying torque equation about centre of mass
τ = Icmα
T.(Lcos300) = m(2L)212.α
α = τI = TLcos300m(2L)212
= 33T2mL (II)
Now after the string breaks, acceleration of point A in the vertical direction should be zero.
i.e.,
ay = Lαcos300 (iv)
Solving equations (i),(ii),(iii)and(iv)
T = 2mg11 and
a = 3311 gL
231533_156311_ans_487563d9038144e49abd5f8a020f70e7.png

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