Am elastic material has a length of 36cm when a load of 40N is hung on it and a length of 45cm when a load of 60N is hung on it. The original length of the string is
12cm
20cm
18cm
15cm
Correct answer is C
For Hooke's law F = Ke
\( \implies F/e = K \\
f1/e1=f2/e2 \\
\text{Let the original length} = t_0 \\
\text{therefore} e1 = (36 - t_0)cm ; e2 = (46 - t_0)cm \\
\text{if f1} = 40N \text{f2} = 60N \\
\text{Then } \frac{40}{36 - t_0} = \frac{60}{45 - t_0} \\
\implies 40(45 - t_0) = 60(36 - t_0) \\
\text{therefore } 1800 - 40t_0 \\
2160 - 60t_0 \\
60t_0 - 40t_0 = 2160 - 1800 \\
20t_0 = 360 \\
t_0 = 18cm \)