A wooden plank of length 1 m and uniform cross-section is hinged at one end to the bottom of a tank as shown in Figure. The tank is filled with water up to a height of 0.5 m. The specific gravity of the plank is 0.5. Find the angle (in degree) that the plank makes with the vertical in the equilibrium position (Exclude the case θ=0 ).
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
answer is 45.00.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Let y is the length of the plank inside water∴ y=0.5cosθLet A be the cross-sectional area of the plank, then buoyant force on itFb=Vρwg =(Ay)ρwgSince plank is in rotational equilibrium, so∑τ→o=0 ⇒mg×ℓ2sinθ−Fb×y2sinθ=0 ⇒mgℓ−Fb×y=0⇒(Aℓ×0.5)gℓ−(Ay)(1)y=0 ⇒0.5ℓ2=y2⇒0.5×(1)2=(0.5cosθ)2⇒cos2θ=12⇒cosθ=12⇒θ=45∘