62yes that method is right riddle.. we have to do exactly what you have done .. :)
3okie so let it be at angle @ wit vertical......
the length of rod inside liquid = h/cos@ ................
therefore the center of buoyancey of rod is at h/2cos@ from pivot .............
F(hsin@/2cos@) - Ïlgsin@(l/2) = 0 ...........
σ/Ï(hg/cos@)(hsin@/2cos@) - lgsin@(l/2) = 0
2(h2g/2cos2@) - g(l2/2) = 0
0.25*2 = cos2@ .........
1/2 = cos2@ ....
@ = 45 .................
cheers!!!!!!!!!!!!!!!!!!!
1This thread is going be to ended by riddle who will throw a riddle to riddle up ur mind !!
Three eyes have I, all in a row;
when the red one opens, all freeze.
145°
equating torque about the hinge
length of rod in side water = 0.5 secθ
Force due to water = 1* 0.5secθ*A *g
wt of rod = ( .5* 1*A *g)
torque = F * length * sin angle...
(1* 0.5secθ*A *g )* ( 0.5secθ/2)*sinθ =( .5* 1*A *g) * (1/2)*sinθ
secθ = √2
θ = 45°
11BUT SIR IN HCV
I THINK THERE IS A MISTAKE IN THE SOLUTION
HE HAVE ALSO EQUATED THE TORQUES BUT THERE
HE DIDN'T TOOK R ANS THE PERPENDICULAR
THATS Y I ASKD
62mani.. the expression is rxF
r is radius vector and F is the force
in one condition it is equal to -mg j
in another condition it is in the upward direction due to bouyant force..
11@RIDDDLE
BHAI PATA HAI AAP MAZAKIA HO
PAR ITNA MAZAK MAT KARO
THODA ELOBORATE KAR KE BATAO!!!!!!!!!!!
11I HAVE A DOUBT
I NEVER ASKED 4 THE ANSWER
PROVE UR ANSWER
I ASKED 4 THE PROCEDURE
I HAVE A SOLUTION BUT
IT SEEMED TO BE FISHY
11Options pls
Gut feeling is 0 and a bit of reasoning
May be wrong
145°
equating torque about the hinge
F * sin angle * length ...
(1* 0.5secθ )*sinθ * ( 0.5secθ/2) = .5* 1 *sinθ * 1/2
secθ = √2
θ = 45°
3has been done bfore [similar 1] .....
have to go for my workout now will do later [1][1][1][1][1][1][1]
