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 :

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 '2r' is : (A) 4M (B) M (C) 2M (D) (M/2)

Tardigrade Admin · Accepted Answer

Height of liquid rise in capillary tube h =

\frac{2 T cos θ _{c}}{ρ r g}

\Rightarrow

h ∝

\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

Option (3)

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 :

$4 M$

$M$

$2 M$

$\frac{M}{2}$

Height of liquid rise in capillary tube h = $\frac{2 T cos θ _{c}}{ρ r g}$
$\Rightarrow$ h ∝ $\frac{1}{r}$
when radius becomes double height become half
$∴ h^{'} = \frac{h}{2}$
Now, M = $π r^{2} h \times ρ$
and M' = $π (2 r^{2}) (h /2) \times ρ = 2 M$
Option (3)

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 :

4M

M

2M

2M​

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(2r2)(h/2)×ρ=2M Option (3)

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 :

$4 M$

$M$

$2 M$

$\frac{M}{2}$

Height of liquid rise in capillary tube h = $\frac{2 T cos θ _{c}}{ρ r g}$
$\Rightarrow$ h ∝ $\frac{1}{r}$
when radius becomes double height become half
$∴ h^{'} = \frac{h}{2}$
Now, M = $π r^{2} h \times ρ$
and M' = $π (2 r^{2}) (h /2) \times ρ = 2 M$
Option (3)