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 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 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

Calculate the length of an aluminium wire of diameter 0.4 mm required to have a resistance of 2 Omega at 20 ^@C . Resistivity of aluminium is 2.8 xx 10^-8Omegam .

An aluminium (respectively p = 2.2 xx10^8 Omega -m ) wire of a diameter 1.4 mm is used to make a 4Omega resister . The length of the wire is

Calculate the length of an aluminium wire of diameter 0.5 mm required to have a resistance of3 Omega at 20^(@)C .Rresisitvity of aluminium is 2.8xx10^(-8)Omegam