The magnitude of normal reaction exerted by the inclined plane on the cylinder is:

50 N

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75 N

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100 N

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Solution

The correct option is A 75 N
FBD of rod and cylinder is shown in the given figure.
Net torque on rod about hinge 'O' $=$ 0
$∴ N_{1} \times L = m g \times \frac{L}{2} o r N_{1} = \frac{m g}{2} = \frac{5 g}{2} = 25 N$
Net torque on cylinder about its centre C is zero.
$∴ f_{1} R = f_{2} R o r f_{1} = f_{2}$
Net torque on cylinder about hinge O is zero.
$∴ N_{2} \times L = N_{1} \times L + m g L o r N_{2} = \frac{3 m g}{2} = \frac{3 \times 5 g}{2} = 75 N$

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