Question

One end of a horizontal thick copper wire of length 2L and radius 2R is weded to an end of another horizontal thin copper wire of length L and radius R .When the arrangement is stretched by applying forces at two ends , the ratio of the elongation in the thin wire to that in the thick wire is

Answer 1

The ratio of the elongation in the thin wire to that in the thick wire is:

• One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R . The arrangement is stretched by applying forces at two ends.

• Change in length, Δl=$\frac{FL}{YA}$

        Here F=Force applied

        L= Length of the wire

        Y= Young's Modulus of the wire

        A=Cross Sectional Area of the wire

• As same force is applied, $F_1 = F_2$

• Δl is proportional to L and A. Y and F are constants.

• $\frac{dl_1}{dl_2} = \frac{l_1}{l_2} * \frac{A_2}{A_1}$

        ⇒$\frac{dl_1}{dl_2} = \frac{l_1}{l_2} * \frac{R_2^2}{R_1^2}$

• Putting values of $l_1, l_2, R_1 and R_2$ , we get $\frac{dl_2}{dl_1} = 2$

• Hence ratio of the elongation in the thin wire to that in the thick wire = 2 : 1