A 5 aluminium wire `(Y = 7 xx 10^(10) Nm^(-2)`) of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire `(Y = 12 xx 10^(10) Nm^(-2)`) of the same length under the same weight, the diameter (in mm) should be

A 5 m long aluminium wire (Y=7xx10^10(N)/(m^2)) of diameter 3 mm supprts a 40 kg mass. In order to have the same elongation in a copper wire (Y=12xx10^10(N)/(m^2)) of the same length under the same weight, the diameter should now, in mm

A 5 m long aluminium wire (Y=7xx10^10(N)/(m^2) ) of diameter 3 mm supprts a 40 kg mass. In order to have the same elongation in a copper wire (Y=12xx10^10(N)/(m^2) ) of the same length under the same weight, the diameter should now, in mm

A 4m long aluminium wire whose diameter is 3 mm is used to support a mass of 50 kg. What will be elongation of the wire ? Y for aluminium is 7xx10^(10) Nm6(-2)

An iron wire of length 4m and diameter 2mm is loaded with a weight of 8kg. If the Young's modulus 'Y' for iron is 2xx10^11 Nm^(-2) then the increase in length of the wire is

A force of 6 xx 10^(6) Nm^(-2) required for breaking a material. The denisty rho of the material is 3 xx 10^(3) kg m^(-3) . If the wire is to break under its own weight, then the length of the wire made of that material should be (Given, g = 10 ms^(-2) )

When a copper wire of area of cross section 1 mm^(2) is stretched by a force of 10N. Then the work done per unit volume in stretching the wire is (Y=1.2xx10^(11)Nm^(-2))

A steel wire of length 4 m and diameter 5 mm is stretched by kg-wt. Find the increase in its length if the Young's modulus of steel wire is 2.4 xx 10^(12) dyn e cm^(-2)