A resistance wire has a resistance R. Half of this wire is stretched to double its length and half is twisted to double its thickness, then its new resistance becomes

Question

A resistance wire has a resistance R. Half of this wire is stretched to double its length and half is twisted to double its thickness, then its new resistance becomes (A) (17 R /8) (B) (17 R /16) (C) (65 R /32) (D) (65 R /16)

Tardigrade Admin · Accepted Answer

Let the resistances of both segments be

R_{1}

and

R_{2}

then

R_{1} = \frac{ρ ℓ}{A ^{'}}

where

A^{'} ℓ = A \frac{ℓ}{2}

\Rightarrow A^{'} = \frac{A}{2}

so,

R_{1} = \frac{2 ρ ℓ}{A} = 2 R

R_{2} = \frac{ρ ℓ /4}{2 A} = \frac{R}{8}

Effective resistance

R_{e q} = R_{1} + R_{2} = \frac{17 R}{8}

Q. A resistance wire has a resistance R. Half of this wire is stretched to double its length and half is twisted to double its thickness, then its new resistance becomes

$\frac{17 R}{8}$

$\frac{17 R}{16}$

$\frac{65 R}{32}$

$\frac{65 R}{16}$

Q. A resistance wire has a resistance R. Half of this wire is stretched to double its length and half is twisted to double its thickness, then its new resistance becomes

817R​

1617R​

3265R​

1665R​

Let the resistances of both segments be R1​ and R2​ then R1​=A′ρℓ​ where A′ℓ=A2ℓ​ ⇒A′=2A​ so, R1​=A2ρℓ​=2R R2​=2Aρℓ/4​=8R​ Effective resistance Req​=R1​+R2​=817R​

$\frac{17 R}{8}$

$\frac{17 R}{16}$

$\frac{65 R}{32}$

$\frac{65 R}{16}$