Question

A resistor is of 4 ohm and the length of the conducting wire is doubled and the area of transverse cut is half than find the change in the resistance

Answer 1

Answer:

16

Explanation:

Let the initial length be 'l' and initial area 'A'

Initial Resistance = $\sf \rho \frac{l}{A} = 4$

New length = 2l (given)

New area = A/2 (given)

Hence, new resistance = $\sf \rho\frac{2l}{\frac{A}{2}}$

$\sf\implies \rho \frac{4l}{A}$

Now, given $\sf\rho \frac{l}{A} = 4$

$\sf\implies \rho \frac{4l}{A} = 4\rho \frac{l}{A}$

→ 4 × 4

→ 16 ohm.

$\rule{200}2$

★ Resistance of a conductor is directly proportional to length of the conductor.

★ Resistance of a conductor is directly proportional to the resistivity of the conductor (resistivity is also known as specific resistance).

★ Resistance of a conductor is inversely proportional to the area of cross section of conductor.

Answer 2

$\bold {\huge {Answer :-}}$

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Resistance of the conductor depends on the following factors:

1. Length of the conductor

2. Area of the cross-section of the conductor.

3. Material used as a conductor.

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✦ ✬Initially

➛Resistance (R) = 4 ohm

➛Length of wire (l) = l

➛Area of Cross section = A

Since we know that

$\Large\bold{\boxed{\boxed{R=\rho\frac{l}{A} }}}$

$\Large\bold{\Rightarrow\:4 =\rho\frac{l}{A}}$

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✦ ✬Finally

➛Length of wire (l) = 2l

➛Area of Cross section (A) = A/2

➛Resistance of wire (Rf) = ?

Again by putting the values we get

$\Large\bold{ R_f =\rho\frac{2l}{\frac{A}{2}}}$

$\Large\bold{\Rightarrow\:R_f =4\:\rho\frac{l}{A}}$

Now from initial

$\Large\bold{\Rightarrow\:R_f =4R}$

$\Large\bold{\Rightarrow\:R_f =16\:ohm}$

Hence, final resistance is 16 ohm