A long metal rod of length l and relative density ∂ is held vertically with its lower end just touching the surface of water.The speed of the rod when it just sinks in water is given by
a)√(2 g l)
b)√(2 g l ∂)
c)√(2 g l (1 - 12∂))
d)√(2 g l (2∂ - 1))
the ans given is c
1∂ = \frac{\rho _{m}}{\rho _{w}}
mg - Upthrust = ma
\Rightarrow \pi r^{2}l\rho _{m}g-\pi r^{2}l\rho _{w}g= \pi r^{2}l\rho _{m}a
\Rightarrow \pi r^{2}lg(\rho _{m}-\rho _{w})=\pi r^{2}l\rho _{m}a
\Rightarrow g\left(\frac{\rho _{m}}{\rho _{m}} -\frac{\rho _{w}}{\rho _{m}}\right)=a
g(1 - 1∂) = a
BY using v2 = u2 + 2as
since initially rod at rest so u = 0
v^{2}=2aS \Rightarrow v= \sqrt{2al}
Substituting value of 'a' we get
\Rightarrow v= \sqrt{2lg\left(1-\frac{1}{\frac{\rho _{m}}{\rho _{w}}} \right)} here ∂ = \frac{\rho _{m}}{\rho _{w}}
THIS IS WAHT I GOT.....
106Let at any isnt of time t, length of rod immersed in water = x.
So upthrust = Axdg (d= density of water, Ï = density of rod)
So writing force equation,
AlÏg - Axdg = AlÏa
=> (lÏ - xd)g = lÏa
∫adx = ∫vdv
=> ∫(1- xdlÏ)gdx = v2/2 (limits 0 to l)
=> 2g(l-ld2Ï) = v2
=> v = \sqrt{2gl(1-\frac{1}{2\partial })}
1seems like i made a mistake taking acceleration constant.
and using equation v2 = u2 + 2as..SRY for the mistake
[2][2] shuld have integrated as asish did.......
