Three copper wires of length and cross-sectional area are l-A- 2l-A2 and l2- 2A and their resistance are R1 - R2 and R3 respectively- The increasing order of resistances is-

Given, l_1=l ; l_2=2l ; l_3=l2 A_1=A ; A_2=A2 ; A_3=2A Resistance of a wire is, R= ρ lA ⇒ R_1 = ρ lA=R ⇒ R_2 = ρ 2lA2=4(ρ lA)=4R ⇒ R_3 = ρ l2(2 ...

You visited us 5 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Series and Parallel Combination of Resistors

Three copper ...

Question

Three copper wires of length and cross-sectional area are $(l, A)$ , $(2 l, \frac{A}{2})$ and $(\frac{l}{2}, 2 A)$ and their resistance are $R_{1}, R_{2}$ and $R_{3}$ respectively. The increasing order of resistances is-

R_{1} > R_{2} > R_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R_{3} > R_{2} > R_{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R_{2} > R_{3} > R_{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R_{3} < R_{1} < R_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $R_{3} < R_{1} < R_{2}$
Given, $l_{1} = l; l_{2} = 2 l; l_{3} = \frac{l}{2}$

$A_{1} = A; A_{2} = \frac{A}{2}; A_{3} = 2 A$

Resistance of a wire is,

$R = \frac{ρ l}{A}$

$\Rightarrow R_{1} = \frac{ρ l}{A} = R$

$\Rightarrow R_{2} = \frac{ρ 2 l}{\frac{A}{2}} = 4 (\frac{ρ l}{A}) = 4 R$

$\Rightarrow R_{3} = \frac{ρ l}{2 (2 A)} = \frac{1}{4} (\frac{ρ l}{A}) = \frac{R}{4}$

$∴ R_{3} < R_{1} < R_{2}$

 Hence, $(D)$ is the correct answer.

Suggest Corrections

Similar questions

Q. Three copper wires of lengths and cross sectional areas are

(l, A) (2 l, \frac{A}{2})

and

(\frac{l}{2}, 2 A)

. Resistance is minimum in

The lengths and cross-sectional areas of four copper wires P, Q, R and S are respectively (l/2,2A),(2l,A/2),(2l,2A) and (l/2, A/2). The wire which has the maximum resistance is

The lengths and cross-sectional areas of four copper wires A, B, C and D are respectively $(\frac{l}{2}, 2 A), (2 l, \frac{A}{2})$ and $(\frac{l}{2}, \frac{A}{2})$ . The wire which has the maximum resistance is

Q. A wire of a given material having length

l

and area of cross-section

A

has a resistance

8 Ω

. Determine the resistance of another wire of same material having length

2 l

and area of cross-section

\frac{A}{2}

Q. A given piece of wire of length l, cross sectional area A and resistance R is stretched uniformly to a wire of length 2l. The new resistance will be

Join BYJU'S Learning Program

Three copper wires of length and cross-sectional area are (l,A), (2l,A2) and (l2,2A) and their resistance are R1,R2 and R3 respectively. The increasing order of resistances is-

The correct option is D R3<R1<R2Given, l1=l ; l2=2l ; l3=l2 A1=A ; A2=A2 ; A3=2A Resistance of a wire is, R=ρlA ⇒ R1=ρlA=R ⇒ R2=ρ2lA2=4(ρlA)=4R ⇒ R3=ρl2(2A)=14(ρlA)=R4 ∴ R3<R1<R2  Hence, (D) is the correct answer.

Three copper wires of length and cross-sectional area are $(l, A)$ , $(2 l, \frac{A}{2})$ and $(\frac{l}{2}, 2 A)$ and their resistance are $R_{1}, R_{2}$ and $R_{3}$ respectively. The increasing order of resistances is-