Question

Q-11 A copper wire 3mlong and 2mm'in cross-sectional area hange from
ceiling. What will be its elongation if 3kg mass is suspended from the lower end
Given Y= 1.25 x 10Nm?
(3)
essin for the orbital​

Answer 1

Answer:

hello good morning I am Siddharth Nagar Aurangabad Maharashtra India mobile

Answer 2

Answer:

A copper wire 3m long and 2 mm2 in cross-sectional area is hung from the ceiling. A 3 kg mass is suspended from the lower end of the wire if Y= 1.1 x 1011 N/m2& g = 10 m/s2 the elongation of the wire is $40.9*10^{-5}m$

Explanation:

• The Young's modulus (Y) is the ratio of longitudinal stress by corresponding strain

                   $Y= stress /strain$

• The stress is the force by area and the strain is the ratio of change in length divided by its original length

                  $stress=F/A$

                  $strain=l/L$

From the question,

the length of the copper wire $L=3m$

cross-sectional area $A=2mm^{2}=2*10^{-6}m^{2}$

here force is due to the weight of the body suspended at the end

weight $W=mg$

mass of the body $m=3kg$

acceleration due to gravity $g=10m/s^{2}$

⇒ weight $W=3*10=30N$

the stress is $=\frac{30}{2*10^{-6} } =15*10^{6} N/m^{2}$

and strain $l/3$

the young's modulus $Y=1.1*10^{11}$

put all the given values

⇒

the elongation $l=\frac{15*10^{6}*3 }{1.1*10^{11}}=40.9*10^{-5}m$