Manipal Entrance Test 2015 Physics Paper

Length of rod inside the water = 1 sec θ = sec θ
Upthrust, F=(

2

2

)(sec‌θ)(

1

500

)(1000)(10)
⇒F=20‌sec‌θ
Weight of rod, w = 20 × 10 = 20 N
For rotational equilibrium of rod, net torque about O should be zero.
∴F(

sec‌θ

2

)(sin‌θ)=w=(1‌sin‌θ)
⇒

20

2

sec2θ=20
⇒θ=45°
⇒F=20‌sec‌45°
F=20√2N
For vertical equilibrium of rod, force exerted by the hinge on the rod will be (20√2−20) N downwards i.e. 8.3 N downwards