Question

⭐Ⓗⓔⓛⓛⓞ ⭐

✨A wire is stretched so that its length becomes 6/5 times its original length. If its original resistance is 25 ohm, find its new resistance and Իḙṧ!ṧт!ṽ!тʏ?

❗❗ ṽḙԻ!ḟ!իժ ᾰℵṧωḙԻ Իḙǭṳ!իḙժ ❗❗​

Answer 1

$\huge\green{Required}$ $\huge\red{Answer!}$

R = p L/A

R = 25 ohm

Volume of Material in the conductor = LA

Volume of the conductor Remains The Same

Volume = (Area × Length)

→ A' L'

→ A' = AL/L'

After Streching The Wire :-

L' = $\frac{6L}{5}$

A' = $\frac{5AL}{6L}$

→ $\frac{5A}{6}$

Resistance of streched wire = R'

→ pL'/A'

→ p(6L /5) × (6 / (5A)

R' = 36pL / 25A

R' = 36R / 25

Now Resistance :-

→ (36×25 / 25)

→ 36 Ohm's

Thanking You ✊

Answer 2

Heya mate!!

R = 25 ohm

Volume of Material in the conductor = LA

Volume of the conductor Remains The Same

Volume = (Area × Length)

→ A' L'

→ A' = AL/L'

After Streching The Wire :-

L' = \frac{6L}{5}

5

6L

A' = \frac{5AL}{6L}

6L

5AL

→ \frac{5A}{6}

6

5A

Resistance of streched wire = R'

→ pL'/A'

→ p(6L /5) × (6 / (5A)

R' = 36pL / 25A

R' = 36R / 25

Now Resistance :-

→ (36×25 / 25)

→ 36 Ohm's

Hope it helps you dear✌️✌️