For the figure shows, a rod of mass 10kg (of length 100cm) with some point masses tied to it at different positions. Find the distance of the point (from A) at which if the rod is picked over a knife edge, it will be in equilibrium about that knife
A
26.32cm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
28.72cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
30.43cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
32.50cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A26.32cm
Let distance from A at which the knife is pricked be xcm
For Rotational equilibrium balancing torques of masses about the knife's axis we get :
20x+15(x−20)=5(100−x−5)+7.5(100−x−20)
(Magnitude of Torque = Force × perpendicular distance )