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Question

A uniform rectangular plate of mass m & side 2l & l is suspended by two strings as shown in figure.

As soon as one of the strings is burnt tension in remaining string becomes

A
Remains same i.e. mg2
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B
Increases in magnitude and becomes 1213 mg
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C
Decreases in magnitude and becomes 517 mg
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D
Decrease in magnitude and becomes 6mg13 mg
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Solution

The correct option is C Decreases in magnitude and becomes 517 mg


mgT=ma (1)

Tl=Iα=[ml212+m(2l)212]α

T=512mlα (2)

Since acceleration of point P can not be vertical
a=lα (3)

Solving (1) (2) (3) we get

T=517mg & a=12g17

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