If ' M ' is the mass of water that rises in a capillary tube of radius ' r '. then mass of water which will rise in a capillary tube of radius ' 2 r' is :

Question

If ' M ' is the mass of water that rises in a capillary tube of radius ' r '. then mass of water which will rise in a capillary tube of radius ' 2 r' is : (A) 4 M (B) M (C) 2 M (D) M/2

Tardigrade Admin · Accepted Answer

Height of liquid rise in capillary tube

h = \frac{2 T c o s θ _{C}}{ρ r g}

\Rightarrow h \propto \frac{1}{r}

when radius becomes double height become half

∴ h^{'} = \frac{h}{2}

Now,

M = π r^{2} h \times ρ

and

M^{'} = π (2 r)^{2} (h /2) \times ρ = 2 M

Q. If ' M ' is the mass of water that rises in a capillary tube of radius ' r '. then mass of water which will rise in a capillary tube of radius ' 2r' is :

4 M

M

2 M

M/2

Height of liquid rise in capillary tube h=ρrg2TcosθC​​ ⇒h∝r1​ when radius becomes double height become half ∴h′=2h​ Now, M=πr2h×ρ and M′=π(2r)2(h/2)×ρ=2M

Q. If ' $M$ ' is the mass of water that rises in a capillary tube of radius ' $r$ '. then mass of water which will rise in a capillary tube of radius ' $2 r$ ' is :

Height of liquid rise in capillary tube
$h = \frac{2 T c o s θ _{C}}{ρ r g}$
$\Rightarrow h \propto \frac{1}{r}$
when radius becomes double height become half
$∴ h^{'} = \frac{h}{2}$
Now, $M = π r^{2} h \times ρ$
and $M^{'} = π (2 r)^{2} (h /2) \times ρ = 2 M$