HELP CRACK DA SOLUTION!!! : PHY

I had thot it wud b sumthing related 2 centrifugal force = mR₁w² and mr2w^2 [2]

okie ill give it a try stay online .. gimme 2-3 min...

take a particle sumwhere inbeteen on the surface of the liquid ,,,,,,,,,

the forces actin are.......

g downward....

w²x away frm axis .....[takin particle x m frm axis] ........

now if i take the tan of the angle between the net accn and vertical ..[perp to net accn will be tangent to surface which is parabolic].........

w²x/g = tan @ = dy/dx ...

w²x²/2g = y ......

y is height and x is distance frm axis........

so if we take distances as r1,r2 we can see ratio of heights will be r1²/r2²

hope it was useful!!!!

telme if u didnt understand!!!!!!!

lovely solution DUDE, thnx yaar!!

wait i'll post ek aur questn HA hA

solution :

40 (c) Kspring = (1/6)MVo²

gimme 2-3 mins again nice prob.[1]

lets divide spring into various segments of length dx and let a segment be at distance x frm fixed end...........

now clearly velocity of each segment is x/L(v0) [ assuming a linear velocity gradient]

dm = dxM/L [ASSUMING UNIFORM DENSITY[ .........

∫1/2dmv² = net KE of spriing ...

∫1/2(dx/L * M)(x/L*V0)² ..

=1/6MV0²

KE OF BLOCK IS CLEARLY 1/2mV0²

now

COME

1/2mV0² + 1/6MV0^2 = 1/2KX² ...

SOLVIN I THINK ULL GET C ONLY!!!!

SORRY FOR LATE REPLY WAZ WACHIN GAJINI.........[3]

YEP U JUST LUK AT THE BEHAVIOR OF VELOCITY ..........AT THE END ITS SAME AS TH BLOCK..V0..AT THE FIXED END ITS ZERO... THEN U CAn formulate it

hmmm pt. but then it cud also be v=v₀sin(wt) or sumthin like dat.... ????

no point arguin coz it must b a convention.......

so can u tell me 4 sure dat v assume uniform vel grad. wen its givn???

YEP. IM QUITE SURE... KAHAN SE ""wt"" AA GAYA v is a funtion of x .......it is assumed unlees informed dat it linear variation

k k.. thnx.... dat was a typo