Question

Please help me with this question....

Answer 1

Answer:-

$\displaystyle\sf {\underline {\underline {4)\ R=\dfrac {\rho}{2\pi l}\log_e\left (\dfrac {b}{a}\right)}}}$

Solution:-

We know the relation between resistance and resistivity.

$\displaystyle\longrightarrow\sf {R=\dfrac {\rho L}{A}}$

where,

• $\displaystyle\sf {R=}$ resistance.

• $\displaystyle\sf {\rho=}$ resistivity.

• $\displaystyle\sf {L=}$ length of the tube.

• $\displaystyle\sf {A=}$ cross sectional area of the tube.

But beware, here,

★ the curved surface area of the tube is considered as the cross sectional area, not the circular area at the edge of the tube.

★ since the curved surface area is taken into consideration, the width of the elementary particle ($\displaystyle\sf {dr\!}\!$) should be considered as the length, not the length of the tube $\displaystyle\sf {l.}$

The cross sectional area of the elementary particle of radius $\displaystyle\sf {r}$ with width $\displaystyle\sf {dr}$ (as shown in the fig.) is,

$\displaystyle\longrightarrow\sf {dA=2\pi rl}$

So it's resistance is,

$\displaystyle\longrightarrow\sf {dR=\dfrac {\rho\ dr}{dA}}$

$\displaystyle\longrightarrow\sf {dR=\dfrac {\rho\ dr}{2\pi rl}}$

Hence the resistance of the cylindrical tube between the inner and the outer surfaces is,

$\displaystyle\longrightarrow\sf {R=\int\limits_a^b\dfrac {\rho\ dr}{2\pi rl}}$

$\displaystyle\longrightarrow\sf {R=\dfrac {\rho}{2\pi l}\int\limits_a^b\dfrac {1}{r}\ dr}$

$\displaystyle\longrightarrow\sf {R=\dfrac {\rho}{2\pi l}\big[\log_e(r)\big]_a^b}$

$\displaystyle\longrightarrow\sf {R=\dfrac {\rho}{2\pi l}\big[\log_e(b)-\log_e(a)\big]}$

$\displaystyle\longrightarrow\sf {\underline {\underline {R=\dfrac {\rho}{2\pi l}\log_e\left (\dfrac {b}{a}\right)}}}$