Question

A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of (a) the stresses developed in the two wires and (b) the strains developed. Y of steel = 2 × 1011 N m−2. Y of copper = 1.3 × 10 11 N m−2.

Answer 1

20:13 is Required Ratio.

Explanation:

Given:

Young's modulus of steel = $2 \times 10^1^1 N m^-^2$

Young's modulus of copper = $1.3 \times 10^1^1 N m^-^2$

Both wires are of equal length and equal cross-sectional area. Also, equal tension is applied on them.

As per the question:

$L_s_t_e_e_l = L_C_u$

$A_s_t_e_e_l = A_C_u$

$F_C_u = F_S_t_e_e_l$  

Here: $L_s_t_e_e_l & L_C_u$ denote the lengths of steel and copper wires, respectively.

         $A_s_t_e_e_l & A_C_u$ denote the cross-sectional areas of steel and copper wires, respectively.

        $F_C_u = F_S_t_e_e_l$ denote the tension of steel and cooper wires, respectively.

(a$\frac {Stress\ of\ Cu}{Stress\ of\ Steel} = \frac {F_C_u \times A_S_t_e_e_l}{A_C_u \times F_S_t_e_e_l} = 1$

(b)

$\frac {Strain\ of\ Cu}{Strain\ of\ steel} = \frac {\Delta LSteel}{LSteel}$

$\Delta L_C_uL_C_u = \frac {F_S_t_e_e_l L_S_t_e_e_l A_C_u Y_c_u}{A_S_t_e_e_l Y_S_t_e_e_l F_C_u L_c_u}$

$(Using \frac {\Delta L}{L} = \frac {F}{AY}$

⇒$\frac {Strain\ of\ Cu}{Strain\ of\ steel}$ = $\frac {Y_C_u}{Y_S_t_e_e_l}$ = $\frac {1.3 \times 10^1^1}{2\times 10^1^1}$

⇒$\frac {Strain\ of\ Cu}{Strain\ of\ steel} = \frac {13}{20}$

⇒$\frac {Strain\ of\ steel}{Strain\ of\ Cu} = \frac {20}{13}$

Hence, The required ratio is 20 : 13.

Answer 2

Answer:

Explanation:how can he take strain of copper is directly proportional to youngs modulas of copper