Question

A metal wire of length 2.5 m and area of cross section 1.5 × 10⁻⁶ m² is stretched through 2 mm. If its Young's modulus is 1.25 × 10¹¹ N m⁻², find the tension in the wire.

Answer 1

Hii dear,

◆ Answer-
F = 150 N

◆ Explanation-
# Given-
L = 2.5 m
A = 1.5×10^-6 m^2
x = 2 mm = 2×10^-3 m
Y = 1.25×10^11 N/m^2

# Solution-
Tension produced in the wire is given by-
F = YAx / L
F = 1.25×10^11 × 1.5×10^-6 × 2×10^-3 / 2.5
F = 150 N

Therefore, tension produced in the wire is 150 N.

Hope that was useful..

Answer 2

Answer:

A metal wire of length 2.5 m and area of cross-section 1.5 × 10⁻⁶ m² is stretched through 2 mm. If Young's modulus is 1.25 × 10¹¹ N m⁻², the tension in the wire will be 150 N.

Explanation:

Young's modulus of a material (Y) is given by the expression,

Y = $\frac{FL}{Al}$ ....(1)

where

F - Stretching force

L - Initial length of the wire

l -  Extended length

A - Area

Given,

L     = 2.5 m

l      =  2 mm  = $2\times 10^{-3}$ m

Young's modulus (Y)  =  $1.25\times 10^{11}$ $N/m^{2}$

Area (A) = $1.5\times 10^{-6}$ $m^{2}$

From equation (1),

The stretching force or the tension on the wire will be,

F  =  $\frac{YAL}{l}$

Substituting the values to the above equation

 F  =  $\frac{1.25\times10^{11}\times1.5\times10^{-6} \times1.5 }{2\times10^{-3} }$

     =  150 N

Ans :

Tension on the wire  = 150 N