Question

Question 10.
Two exactly similar wires of steel and copper are stretched by equal forces. If the total elongation is 1 cm, find by how much each wire is elongated. Given Y for steel = 20 × 1011 dyne cm-2, Y for copper = 12 × 1011 dyne cm-2.

Answer 1

Question:-

Two exactly similar wires of steel and copper are stretched by equal forces. If the total elongation is 1 cm, find by how much each wire is elongated. Given Y for steel = 20 × 1011 dyne cm-2, Y for copper = 12 × 1011 dyne cm-2.

Answer:-

0.375cm

Explanation:-

Let $\sf \: Δl_s \: and \: Δl_c$ be the elongation produced in steel and copper wires respectively.

$\tt \: L_s, L_c be \: their \: respective \: lengths, \\ \tt \: L_s = L_c (∵ wires \: are \: similar)\\ \tt \: Y_s = 20 × 10 {}^{11} dyne \: cm {}^{2} \\ \tt \: = 12 × 10^11 \: dyne \: cm {}^{2} \\ \tt \: Δl_s + Δl_c = 1 \: cm$

• A = area of cross-section of each wire.
• F = equal force applied.

∴ Using the relation

$\tt \: Y= \frac{ F}{A} \times \frac{ L}{ ∆l} \\ \\ \tt \: ∆l_c= \frac{ F}{A} \times \frac{ L}{Y_c}⇢a) \\ \\ \tt \: l_s= \frac{FL_S}{ AY_S} = \frac{ F}{A} \frac{ L_C}{ Y_C}⇢b)$

Please refer to the attachment (:

$\huge \bold {{@QianNiu}}$