Moment of inertia of uniform triangular plate about axis passing through sides AB,AC and BC are IP,IB and IH respectively and about an axis perpendicular to the plane and passing through point C is IC. Then
A
IH>IB>IC>IP
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B
IC>IP>IB>IH
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C
IP>IH>IB>IC
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D
none of these
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Solution
The correct option is BIC>IP>IB>IH
Moment of inertia is more when mass is
farther from the axis. In case of axis BC, mass
distribution is closest to it and in case of axis AB mass distribution is farthest. Hence IBC<IAC<IAB ⇒IP>IB>IH
Here I′A is moment of inertia of the plate about an axis perpendicular to it and passing through A. ⇒IC=I′A+m(y2−x2) IC=IP+IB+m(y2−x2)
It means IC>IP+IB
also IC>IP ∴IC>IP>IB>IH
Final Answer: (a)