Question

two identical wires having different resistivity in ratio 1:3 the resistance of first wire is 20 ohm find the resistance of second resistance

​

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\sf\implies Two \ wires \ have \ resistivity \ in \ ratio \ 1 : 3. \\\sf\implies Resistance \ of \ first \ wire \ is 20\Omega. \\\sf\implies The\ two \ wires \ are \ identical.$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\sf \implies The \ resistance \ of \ second \ wire.$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Let :- }}}$

$\sf\implies Resistance\ of \ Second \ wire \ be \ R_2. \\\sf\implies Ratio \ of \ resistivity \ be \ 1x : 3x.$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

Since the two wires are identical they will have same area and cross- sectional area. The ratio of resistivity is 1: 3. By formula we know that resistance is $\sf\rho\dfrac{l}{a}$ where $\bf \rho$ is the resistivity of the wire , l is the length of the wire & a is the cross-sectional area.

$\sf :\implies R_1 : R_2 = \rho_1\dfrac{l}{a} : \rho_2\dfrac{l}{a} \\\\\sf:\implies 20\Omega : R_2 = \rho_1 : \rho_2 \\\\\sf:\implies \dfrac{20\Omega}{R_2}=\dfrac{1}{3} \\\\\sf\implies R_2 = ( 20 \times 3 ) \Omega \\\\\boxed{\red{\sf \implies Resistance_{2nd\ wire }= 60\Omega }}$

$\boxed{\green{\bf \pink{\dag}\:\: Hence \ the \ resistance \ of \ second \ wire \ is 60\Omega. }}$

$\rule{200}2$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: More \ to \ know :- }}}$

When the lenghth of the conductor is varied , then its area also changes , e.g. When we stretch a wire , it becomes thin , its lenght increases but its cross - sectional area decreases. During this transformation the Volume remains constant .

➳ $\underline{\pink{\bf When \ lenght \ is \ increased :- }}$

When we increase lenghth $\sf L_1$ to $\sf L_2$ then its resistance becomes :-

$\qquad\boxed{\red{\bf R_{new}= \bigg\lgroup\dfrac{L_2}{L_1}\bigg\rgroup ^2 R_{initial}}}$

➳ $\underline{\pink{\bf When \ Area \ is \ increased :- }}$

When we increase area $\sf A_1$ to $\sf A_2$ then its resistance becomes :-

$\qquad\boxed{\red{\bf R_{new}= \bigg\lgroup\dfrac{A_1}{A_2}\bigg\rgroup ^2 R_{initial}}}$

➳ $\underline{\pink{\bf When \ radius\ is \ increased :- }}$

When we increase radius $\sf r_1$ to $\sf r_2$ then its resistance becomes :-

$\qquad\boxed{\red{\bf R_{new}= \bigg\lgroup\dfrac{r_1}{r_2}\bigg\rgroup ^4R_{initial}}}$