Question

A. The resistance of a Im long
nichrome wire is 6 . If we reduce
the length of the wire to 70 cm.
what will its resistance be?​

Answer 1

We are assuming that the volume of the wire isn't changing

Its given that

L₁=100 cm

L₂=70 cm

R₁=6 ohm

R₂=R₂

We know that

R=ρL/A

where

ρ=resistivity of the material

L=lenght of the wire

A=area of the wire

Volume of the wire is same therefore

V₁=V₂

A₁L₁=A₂L₂

A₁/A₂=L₂/L₁

=7/10

or

A₂/A₁=10/7

R₁=ρL₁/A₁

R₂=ρL₂/A₂

dividing the above two equations

R₁/R₂=L₁/L₂*A₂/A₁

=10/7*10/7

=100/49

R₂=R₁*49/100

=6*49/100

=2.94 ohm

If volume isn't the same after reducing the lenght we can directly say that

R₁/R₂=L₁/L₂

Since A and ρ are constant therefore the ratio becomes independent of it.

R₂=R₁*L₂/L₁

=6*70/100

=4.2 ohm

Hope this helps

Answer 2

$\large{\mathfrak{\underline{\underline{\blue{Answer:-}}}}}$

${\mathfrak{\underline{\underline{\red{Step-By-Step-Explanation:-}}}}}$

$\large{\mathtt{\green{Given}}}$

⇒Volume doesn't change

Length of wire Before (L1) =100 cm

Length of wire after (L2) =70 cm

Resistance before (R1) =6 Ω

R1 = R2

_________________________

$\large{\mathtt{\green{Solution:-}}}$

We know that

R= ρL/A

where,

ρ(rho) = Resistivity (Which is a constant)

L= Lenght of the wire

A = Area of the wire

As we know,

Volume of the wire is same

⇒V1 = V2

And,

A1 L1 = A2 L2

A1 /A2 = L2 / L1

⇒7/10

Also,

A2 /A1 = 10/7

R1 = ρL1 /A1 ___________(1)

R2 = ρL2 /A2 ___________(2)

Divide eqn (1) by (2)

R1 /R2 = L1 /L2 × A2 /A1

= 10/7 × 10/7

= 100/49

As We know,

R₂ = R₁ × 49/100

= 6 × 49/100

=2.94 Ω

$\large{\star{\bf{\underline{\boxed{\orange{2.94 Ω }}}}}}$

________________________

And,

R1 /R2 = L1 /L2

So,

A and ρ are constant ,

R2 = R1 × L2 /L1

⇒ 6 × 70/100

⇒4.2 Ω

$\large{\star{\bf{\underline{\boxed{\orange{4.2 Ω }}}}}}$