An 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
A
12cm
B
20cm
C
18cm
D
15cm
Verified Correct Answer: C.
Physical Principles & Numerical Solution
For Hooke's law F = Ke
\( \implies \frac{F}{e} = K\)
\(\frac{f_1}{e_1} = \frac{f_2}{e_2}\)
\(\text{Let the original length} = t_0 \)
\(\text{therefore}\) e\(_1\) = (36 - t_0)\)cm ; e\(_2 = (46 - t_0)\)cm
\(f_1\) = 40N, \(f_2\) = 60N
\(\frac{40}{36 - t_0} = \frac{60}{45 - t_0}\)
\(\implies 40(45 - t_0) = 60(36 - t_0)\)
\(\text{therefore } 1800 - 40t_0\) = 2160 - 60\(t_0\)
60\(t_0\) - 40\(t_0\) = 2160 - 1800
20\(t_0\) = 360
\(t_0\) = 18cm
Source: Myschool Past Questions Archive (Record #41442)