The elastic potential energy stored in a steel wire of length ``20 m` stretched through `2 cm` is `80 j`. The cross sectional area of the wire is ___mm^2.
(Given , y =2.0times 10^(11)Nm^(-2))

A steel wire of 4.0 m in length is stretched through 2.00 mm. The cross-sectional area of the wire is 2.0mm^(2) if young's modulus of steel is 2.0xx10^(11)N//m^(2) find (i) the energy density of wire (ii) the elastic potential energy stored in the wire.

A steel wire of 4.0 m is stretched through 2.0 mm. The cross - sectional area of the wire is 2.0 mm^2. If young's modulus of steel is 2.0 xx10^(11) Nm^(-2) find (i) the energy density of the wire, (ii) the elastic potential energy stored in the wire.

A steel wire 4.0 m in length is stretched through 2.0 mm. The cross-sectional area of the wire is 2.0 mm^(2) . If young's modulus of steel is 2.0 xx 10^(11) Nm^(-2) , then find the energy density of wire.

A steel wire 4.0m in length is stretched through 2.0mm .The cross -sectional area of the wire is 2.0 mm^(2) .If young's modulus of steel is 2.0xx10^(11) N//m^(2) (a) the enrgy density of wire, (b) the elastic potential energy stored in the wire.

A steel wire of length 2.0 m/s is stretched through 2.0 mm. The cross sectional area of the wire is 4.0 mm^2. Calculate the elastic potential energy stored in the wire in the stretched condition. Young modulus of steel =2.0x10^11Nm^-2

A steel wire of length 3.0m is stretched through 3.0 mm. the cross-sectional area of the wire is 5.0mm^(2) . Calculate the elastic potential energy stroed in the wire in the stretched condition. Young's modulus of steel is 2.0 xx 10^(11) Nm^(-2) .

Figure shows the graph of elastic potential energy (U) stored versus extension, for a steel wire (Y = 2xx10^(11) Pa) of volume 200 c c . If area of cross- section A and original length L, then

The work per unit volume to stretch the length by 1% of a wire with cross sectional area of 1mm^2 will be. [Y=9xx10^(11)(N)/(m^2) ]

A wire of length 50 cm and cross sectional area of 1 sq. mm is extended by 1mm . The required work will be (Y=2xx10^(10) Nm^(-2))