Question

A force F is applied on the wire of radius r and length L and change in the length of wire is l. If the same force F is applied on the wire of the same material and radius 2r and length 2L. Then the change in length of the other wire is

Answer 1

$\huge\fcolorbox{red}{pink}{Answer}$

➡ Given:

• For wire - 1

✏ Force applied = F

✏ Radius of cross section = r

✏ Length = L

✏ Elongation = l

• For wire - 2

✏ Force applied = F

✏ Radius of cross section = 2r

✏ Length = 2L

➡ To Find:

✏ Elongation in wire - 2

➡ Concept:

✏ Here, both wires are made up from same material therefore young modulus of both wires will be same.

➡ Formula:

✏ Young modulus of material is given by

$\: \: \: \: \: \: \: \: \: \: \: \: \dag \: \underline{ \boxed{ \rm{ \pink{ \bold{Y = \frac{F \times L}{A \times \triangle{L}}}}}}} \: \dag$

➡ Terms:

✏ F denotes applied force

✏ A denotes area of cross section

✏ L denotes length

✏ ΔL denotes elongation

➡ Calculation:

$\implies \rm \: Y_1 = Y_2 \\ \\ \therefore \rm \: \frac{F_1 \times L_1}{A_1 \times \triangle{L_1}} = \frac{F_2 \times L_2}{A_2 \times \triangle{L_2}} \\ \\ \therefore \rm \: \frac{F \times L}{\pi {r}^{2} \times l} = \frac{F \times 2L}{\pi( {2r)}^{2} \times \triangle{L_2}} \\ \\ \therefore \rm \: l = \frac{4 \triangle{L_2}}{2} \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \huge{ \orange{ \triangle{L_2} = \frac{l}{2} }}}}}} \: \purple{ \clubsuit}$

Answer 2

Solution

To finD

Extension in the new wire

We know that,

$\sf Stress \propto Strain \\ \\ \dashrightarrow \sf Y = \dfrac{Stress}{Strain} \\ \\ \dashrightarrow \boxed{\boxed{\sf Y = \dfrac{F \times L}{A \times e}}}$

Case I

• Length of the Wire (L)

• Radius of the wire (r)

• Extension in the wire (l)

Putting the values,we get :

$\sf Y = \dfrac{F \times L}{\pi r^2 \times l}-----------(1)$

Case II

• Length of the Wire (2L)

• Radius of the Wire (2r)

• Extension in the wire = ?

Let the extension in the wire be e

Putting the values,we get :

$\sf Y = \dfrac{F \times 2L }{\pi (2r)^2 \times e} \\ \\ \longrightarrow \ \sf Y = \dfrac{1}{2} \times\dfrac{F \times e}{\pi r^2 \times L}-----------(2)$

Here,

• Same force is applied on the two wires

• The wires are of same material,their Young's Modulus is equal

From relations (1) and (2),we can write :

$\dashrightarrow \sf l = \dfrac{1}{2} \times e \\ \\ \large{ \dashrightarrow \ \boxed{\boxed{\sf l : e = 2 : 1}}}$

Thus,the extension in the new wire in 1/2 of the extension in the first wire