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 1213mg
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C
Decreases in magnitude and becomes 517mg
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D
Decrease in magnitude and becomes 6mg13mg
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Solution
The correct option is C Decreases in magnitude and becomes 517mg
mg−T=ma…(1)
Tl=Iα=[ml212+m(2l)212]α
T=512mlα…(2)
Since acceleration of point P can not be vertical a=lα…(3)