Question

Please give the proper solution with the whole process.

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Given ans:5:8.​

Answer 1

Answer:

Given:

The ratios of diameter, length and resistivity has been provided.

To find:

Ratio of resistance of the 2 wires.

Formulas used:

$R = \rho \times \frac{l}{a}$

where "ρ" is the resistivity, "l" is the length and "a" is the area

Now if we express this Equation in terms in diameter , we get

$R = \rho \times \frac{l}{( \frac{\pi {d}^{2} }{4}) }$

$= > R = \frac{4}{\pi} ( \rho \frac{l}{ {d}^{2} } )$

Calculation:

If we take ratio of all quantities :

$\frac{R1}{R2} = \frac{ \rho1}{ \rho2} \times \frac{l1}{l2} \times { (\frac{d2}{d1} )}^{2}$

$\frac{R1}{R2} = \frac{4}{9} \times \frac{5}{18} \times ( { \frac{3}{2} )}^{2}$

$\frac{R1}{R2} = \frac{5}{18}$

So the final answer is :

$\boxed{ ratio \: = \frac{R1}{R2} = \frac{5}{18} }$

Answer 2

Answer:

${\boxed{\sf{Answer=c)5\::\:18}}}$

Explanation:

$\bf{Given}\begin{cases}\sf{l_1 : l_2 = 5 : 18 } \\\sf{D_1 : D_2 = 2 : 3 } \\\sf {P_A : P_B = 4 : 9 }\\\sf {R_1 : R_2 =\:?}\end{cases}$

Now,

$\sf \frac{R_1}{R_2} = \frac{P_A}{P_B} \times \frac{l_1}{l_2} \times ({\frac{D_2}{D_1})}^{2}$

$\rightarrow\sf \frac{R_1}{R_2} = \frac{4}{9} \times \frac{5}{18} \times ({\frac{3}{2})}^{2}$

$\rightarrow\tt \frac{R_1}{R_2} = \frac{4}{9} \times \frac{5}{18}\times \frac{9}{4} \\\rightarrow\sf \frac{R_1}{R_2} = \frac{180}{648}$

$\rightarrow{\boxed{\tt{ \frac{R_1}{R_2} = \frac{5}{18}}}}$

$\therefore\bold{\underline{R_1 : R_2 = 5 : 18 }}$