QJEEM21Ph95fd: JEE Main 2021 Physics - Class 12 JEEM21Ph95fd Solution: A conducting wire of length 'l', area of cross-section A and…

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QJEEM21Ph95fd: JEE Main 2021 Physics - Class 12 JEEM21Ph95fd Solution: A conducting wire of length 'l', area of cross-section A and…

A conducting wire of length 'l', area of cross-section A and electric resistivity

ρ

is connected between the terminals of a battery. A potential difference V is developed between its ends, causing an electric current.

If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be :

Multiple Choice Options

A 4VAρl
B 34VAρl
C 14VAρl
D 14ρlVA

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Accepted Answer

We know that

R = ρ \frac{l}{A}

Now, new length :

l^{'} = 2 l

new area of cross section :

A^{'} = A / 2

∴

New resistance :

R^{'} = ρ . \frac{2 l}{A / 2}

\Rightarrow R^{'} = 4 \frac{ρ l}{A}

\Rightarrow R^{'} = 4 R

∴

Resultant current :

I = \frac{V}{4 R}

\Rightarrow

I = \frac{1}{4} \frac{V A}{ρ l}

QJEEM21Ph95fd: JEE Main 2021 Physics - Class 12 JEEM21Ph95fd Solution: A conducting wire of length 'l', area of cross-section A and…

Multiple Choice Options

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