Question

A copper wire of length 2.4 m and a steel wire of length 1.6 m, both of diameter 3 mm, are connected end to end. When stretched by a load, the net elongation found to be 0.7 mm. The load appled is (Y_("Copper") = 1.2 xx 10^(11) N m^(-2) , Y_("steel") = 2 xx 10^(11) N m^(-2))

Answer 1

Answer:

1.2×102N

1.8×102N

2.4×102N

3.2×102N

Answer :

B

Solution :

Let the load applied be W.

Since both the wires have the same tension (equal to the load W) and same area of the cross-section A, hence both wires have the same tensile stress.

As Stress =Young's modulus × Strain

∴WA=Yc×ΔLCLC=YS×ΔLSLS ....(i)

where the subscripts C and S refers to copper and steel respectively

∴ΔLCΔLS=LCLS×YSYC

Here,

For copper wire , LC=2.4m,YC=1.2×1011Nm−2

For steel wire, LS=16m

YS=2×1011Nm−2

ΔLCΔLS=(2.4m1.6m)×(2×1011Nm−2)(1.2×1011Nm−2)=52 ....(i)

Total elongation is

ΔLC+ΔLS=0.7mm=7×10−4m ....(ii)

Sloving Eq.(ii) and (iii), we get

ΔLC=5×10−4mandΔLs=2×10−4m

From Eq.(i)

W=A×YC×ΔLCLC

=π×(1.5×10−3m)2×1.2×1011Nm−2×5×10−4m2.4=1.8×102N

Explanation: