Question

A wire of resistance 200 Ω is melted and drawn into wire of half its original length. Its new resistance will be
(a) 30 Ω (b) 40 Ω (c) 50 Ω (d) 60 Ω

Answer 1

Answer:

50 ohms

Explanation:

when when wire is cut off at half of its original length then the resistance will drone one by fourth of its original value

Answer 2

$\underline{ \underline{ \bf{ question}}}$

A wire of resistance 200 Ω is melted and drawn into wire of half its original length. Its new resistance will be :

(a) 30 Ω

(b) 40 Ω

(c) 50 Ω ✔✔✔

(d) 60 Ω

$\underline{ \underline{ \bf{solution}}}$

We have,

➹ intital resistance ,R₁ = 200 Ω

➹ initial length of wire = l₁

➹ initial area of cross section = A₁

so,using ; resistance = ρ × l / A

( ρ is the resistivity, l is the length of conductor and A is the area of cross section of conductor)

we will get,

■ R₁ = ρ × l₁ / A₁

■ ρ = R₁ × A₁ / l₁ ...... eqn (1)

Since,

the length of wire is halved hence,

➹ new length, l₂ = l₁ / 2

and

➹ new area , A₂ = 2 A₁

➹ new resistance = R₂

so,

■ R₂ = ρ × l₂ / A₂

■ ρ = R₂ × A₂ / l₂ ........eqn (2)

According to eqn (1) and (2)

➷ R₁ × A₁ / l₁ = R₂ × A₂ / l₂

using the given information that

l₂ = l₁ / 2 and A₂ = 2 A₁ also, R₁ = 200 Ω

➷ 200 × A₁ / l₁ = R₂ × 2 A₁ / ( l₁ / 2 )

➷ 200 × A₁ / l₁ = R₂ × 4 A₁ / l₁

A₁ / l₁ will be eliminated being on both sides

so,

➷ R₂ = 200 / 4

➷ R₂ = 50 Ω

Hence,

the new resistance of the wire will be 50 ohms.