JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 4)
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 :
M
4M
M/2
2M
Explanation
Height of liquid rise in capillary tube $$h = {{2T\,\cos {\theta _c}} \over {\rho rg}}$$
$$ \Rightarrow h \propto {1 \over r}$$
When radius becomes double height become half
$$ \therefore $$ $${h^{'}} = {h \over 2}$$
Now, M = $$\pi $$r2h × $$\rho $$ and M' = $$\pi $$(2r)2 (h/2) × $$\rho $$ = 2M
$$ \Rightarrow h \propto {1 \over r}$$
When radius becomes double height become half
$$ \therefore $$ $${h^{'}} = {h \over 2}$$
Now, M = $$\pi $$r2h × $$\rho $$ and M' = $$\pi $$(2r)2 (h/2) × $$\rho $$ = 2M
Comments (0)
