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 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 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)

A uniform wire of length 10m and diameter 0.8 mm stretched 5 mm under a certain force . If the change in diameter is 10 xx 10^(-8) m , calculate the Poissson's ratio.

If the strain in a wire is not more than 1//1000 and Y =2xx10^(11)N//m^(2) Diameter of wire is 1 mm. The maximum weight hung from the wire is

A metallic wire of diameter 0.3 mm and length 3m is stretched by hanging a weight of 2 kg . If the elongation produced is 2 mm

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