Question

A wire of given material having length ‘l’ and area of cross section ‘A’ has a

resistance of 16 Ω. What would be the new resistance of another wire of the same

material having length ‘l/2’ and area of cross section ‘2A’?​

Answer 1

Answer:Wire of uniform cross-section has a resistance of R. Let the diameter be d,  the length of the wire be l cross-sectional area be A.

Now the resistance of a wire can be expressed as:

R=ρ

A

l

​

Where resistivity is ρ.

Now a second wire is twice as long and of half cross section.

Let final length be l

1

​

, diameter be d

1

​

 and

cross section Area be A

1

​

.

A

1

​

=

2

A

​

R

1

​

=ρ

A

1

​

l

1

​

​

Putting the value of l

1

​

=2l and A

1

​

=

2

A

​

We have, resistance of the new second wire, R

2

​

=ρ

A

4l

​

=4×R

1

​

=4×8=32Ω

Explanation:

Answer 2

$\huge\underline{\underline{\color{navy}{αηsωεя —☆}}}$

ғᴏʀ ғɪʀsᴛ ᴡɪʀᴇ,

$\sf{R₁ = ρ \frac{l}{A} = 4 \: Ω }$

ɴᴏᴡ ғᴏʀ sᴇᴄᴏɴᴅ ᴡɪʀᴇ,

$\sf{R₂ = ρ \frac{ \frac{l}{2} }{2A} = \frac{1}{4} \times ρ \times \frac{l}{A} }$

$\sf{R₂ = \frac{1}{4}R₁ }$

$\sf\underline{\underline{\bold{∴ R₂ = 1 \: Ω}}}$

The resistance of the new wire is 1 Ω.

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