Question

two wires of same material and having length in the ratio 2 : 3 are connected in series the potential difference across the wires are 4.2 volt 3.6 volt respectively the ratio of their radii​

Answer 1

Let I is the current through wires which are connected in series combination.

so, resistance on first wire, $R_1$ = $\frac{V_1}{I}$

= 4.2/I .....(1)

resistance on 2nd wire, $R_2$=$\frac{V_2}{I}$

= 3.6/I ..... (2)

using formula of resistance, R = pl/A

where , p is resistivity , l is length and A is cross sectional area.

as, both wires are made of same material so, resistivities of both are same.

cross section area of first wire, $A_1=\pi r_1^2$

cross sectional area of 2nd wire, $A_2=\pi r_2^2$

now, $R_1=p\frac{2l}{\pi r_1^2}$

and $R_2=p\frac{3l}{\pi r_2^2}$

from equations (1) and (2),

$\frac{R_1}{R_2}=\frac{4.2}{3.6}=\frac{2\pi r_2^2}{3\pi r_1^2}$

or, $\frac{7}{6}=\frac{2r_2^2}{3r_2^2}$

or, $\frac{7}{4}=\frac{r_2^2}{r_1^2}$

hence, $\frac{r_1}{r_2}=\sqrt{\frac{4}{7}}$

so, answer is 2 : √7

Answer 2

Answer:

Explanation:

Let I is the current through wires which are connected in series combination.

so, resistance on first wire,

= 4.2/I .....(1)

resistance on 2nd wire,

= 3.6/I ..... (2)

using formula of resistance, R = pl/A

where , p is resistivity , l is length and A is cross sectional area.

as, both wires are made of same material so, resistivities of both are same.

cross section area of first wire,

cross sectional area of 2nd wire,

now,

and

from equations (1) and (2),

or,

hence,

so, answer is 2 : √7