When a force is applied on a wire of uniform cross-sectional area `3xx10^-6m^2` and length 4 m, the increase in length is 1mm. Energy stored in it will be `(Y=2xx10^(11)(N)/(m^2)`)

When a force is applied on a wire of uniform cross-sectional area 3×10 −6 m 2 and length 4m, the increase in length is 1 mm. Energy stored in it will be (Y=2×10 11 N/m 2 )

When a force is applied on a wire of uniform cross-sectional are 3 xx 10^(-6) m^(2) and length 4 m, the increase in length is 1 mm. Energy stored in it will be (Given, Y = 2 xx 10^(11) Nm^(-2) )

When a force of 100 N is applied on a wire of uniform cross-section as shown in figure then length of wire is increased by 1 mm. Energy stored in the wire will be

Calculate the amount of work done for a wire which area of cross section 10^-6m^2 and length 1.5m to increase the length 4 xx 10^-3 m. Young's modules of elasticity 2 xx 10^(11N//m^2) .

What is the energy stored per unit volume in a copper wire of uniform cross section and length 1.5 m, when it is stretched to a length of 1.51 m by a stress of 3 xx 10^(2) N//m^(2) ?

There is some change in length when a 33000 N tensile force is applied on a steel rod of area of cross-section 10^(-3)m^(2) . The change of elongation of the steel rod when heated is (Y = 3 xx 10^(11)N // m^(2) , alpha = 1.1 xx 10^(-5 //@)C)

A particular force (F) applied on a wire increases its length by 2 xx 10^(-3) m. To increases the wire's length by 4 xx 10^(-3) m, the applied force will be

A particular force (F) applied on a wire increases its length by 2 xx 10^(-3) m. To increases the wire's length by 4 xx 10^(-3) m, the applied force will be