Question

plz solve it as soon as possible​

Answer 1

Required Answer:-

As we have to find the total resistance of the conducting wires in series but it should be in form of $\rho$, l and A. Hence, we will use the formula:

$\large{ \boxed{ \sf{R = \rho\frac{l}{A}}}}$

The two wires are of same material, so the Resistivity ($\rho$) is same in both case. Now we have to find the resistance for both considering their different lengths and cross sectional area.

Resistance of P:

• Resistivity = $\rho$
• Length = l
• Cross-sectional area = A

Then by formula,

$\rm{Rp = \rho\dfrac{l}{A}}$

Resistance of Q:

• Resistivity = $\rho$
• Length = 2L
• Cross-sectional area = A / 2

Then by Formula,

$\rm{Rq = \rho\dfrac{2l}{ \dfrac{A}{2} }}$

$\rm{Rq = 4\rho\dfrac{l}{A}}$

Total Resistance in series is calculated by the algebraic sum of the resistors connected end to end.

Hence, Total Resistance:

= Rp + Rq

$\rm{ = \rho\dfrac{l}{A} + 4\rho\dfrac{l}{A}}$

$\rm{ = 5 \rho\dfrac{l}{A}}$

Therefore the total resistance of the combination of wires is $\rm{5 \rho\dfrac{l}{A}}$. Option (A)

Answer 2

Answer:

SoluTion :-

Firstly let's find resistance of p

$\sf R \: = ρ \dfrac{I}{A}$

Resistivity = ρ

Length = l

Cross-sectional area = A

$\sf rp = ρ \dfrac{l}{a}$

Resistance of Q

Resistivity = ρ

Length = l

Cross-sectional area = A/2

$\sf Rq = \rho \: \dfrac{ \dfrac{21}{A} }{2}$

$\sf \: Rq = 4 \rho \: \dfrac{l}{A}$

Total Resistance :-

$\sf \: Rp + Rq$

$\sf \: \rho \: \dfrac{l}{A} + 4 \rho \: \dfrac{l}{A}$

$\sf \: = 5 \rho \: \dfrac{l}{a}$

Therefore,

$\bigodot \sf Option \: a$