Question

When $0.4\text{V}$ is applied to the ends of mercury column contained in a thin glass tube $X,5\text{A}$ current flows. Same mercury is poured into a glass tube $Y$ which has diameter one third of the tube $X$ and same voltage is applied to it. Then

• A. the ratio of resistance of mercury in the tube $X$ to tube $Y$ is $1/81$.
• B. the ratio of resistance of mercury in the tube $X$ to tube $Y$ is $1/9$.
• C. current in the tube $Y$ is $5/81\text{A}$.
• D. current in the tube $Y$ is $5/9\text{A}$.

Answer 1

Answer: (A), (C)

The correct options are
A the ratio of resistance of mercury in the tube $X$ to tube $Y$ is $1/81$.
C current in the tube $Y$ is $5/81\text{A}$.

For glass Tube $X$ :-
Resistance $\left[R_1\right]=\frac{\rho l_1}{A_1}$
For glass tube $Y$ :-
Resistance $\left[R_2\right]=\frac{\rho l_2}{A_2}$
$\frac{A_2}{A_1}=\frac{\frac{\pi }{4}{d}_2^2}{\frac{\pi }{4}{d}_1^2}={\left[\frac{d_1}{3d_1}\right]}^2$
$A_2=\frac{A_1}{9}$
As Mercury volume is same in both $X$ and $Y$ tube.
$l_1{A}_1=l_2{A}_2$
$l_1{A}_1=l_2\frac{A_1}{9}$
$l_2=9l_1$
The ratio of resistance:
$\frac{R_1}{R_2}=\frac{\frac{\rho l_1}{A_1}}{\frac{\rho l_2}{A_2}}=\frac{l_1{A}_2}{l_2{A}_1}$
$=[\frac{1}{9}\times \frac{1}{9}]=\frac{1}{81}$
given same voltage is applied to both $X$ and $Y$ Tube.
$V=iR=$ same
$i_1{R}_1=i_2{R}_2$
$i_2=\frac{i_1{R}_1}{R_2}$
$=\left[5\right]\left[\frac{1}{81}\right]$
$=\frac{5}{81}\text{A}$
option A and C are correct.