Question

A 20 kg load is suspended from the lower end
of a wire 10cm long and 1mm? in cross sectional
area. The upper half of the wire is made of
iron and the lower half with aluminium. The
total elongation in the wire is​

Answer 1

Answer:

The answer will be 0.064mm

Explanation:

According to the problem the mass of the load and the length of the wire is given.

As stated the upper section is made of iron,

Therefore the elongation of the iron wire , e1 = f(l/2)/ay1 = fl/2ay1= 20  x 10 x 0.1 /2 x10^(-6)x 20 x 10^10 =   20 / 4 x 10^5 = 0.05 mm

Therefore the elongation of the aluminium wire , e1 = f(l/2)/ay1 = fl/2ay1= 20  x 10 x 0.1 /2 x10^(-6)x 7 x 10^10 =   20 / 14 x 10^5 = 0.014mm

The total elongation of the wire = 0.05+0.014 = 0.064mm