Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

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 the figure. The tank is filled with water up to a height of 0.5 m. The specific gravity of the plank is 0.5. If the angle θ by the inclination of that the plank makes with the vertical in the equilibrium position (exclude the case θ = 0). Find the value of 1cos2⁡θ.

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 2.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Let y be the length of the plank inside water.          y=0.5cos⁡θLet A be the cross-sectional area of the plank. Then the buoyant force on it is Fb=Vρωg=(Av)ρωg Since the plank is in rotational equilibrium, so ∑τ→0=0 or mg×l2sin⁡θ−Fb×y2sin⁡θ=0 or mgl−Fb×y=0 or (Al×0.5)gl−(Ay)dvy=0 or 0.5l2=y2 or 0.5×(1)2=0.5cos⁡θ2 or cos2⁡θ=12
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring