question_answer

The length of an elastic spring is a meters when a force of \[4\text{ }N\]is applied, and b meters when the \[\text{5 }N\] force is applied. Then the length of the spring when the \[9N\] force is applied is

A)  \[a+b\]         

B)  \[~9b-9a\]

C)  \[5b-4a\]       

D)  \[4a-5b\]

Correct Answer: C

Solution :

From Hookes law, restoring force F is \[F=kl\] where k is spring constant. When L is original length of spring, and k the spring constant, then                 \[L+\left( \frac{5}{R} \right)=b\]  Also,         \[L+\left( \frac{4}{k} \right)=a\] \[\therefore \] \[\frac{5}{k}-\frac{4}{k}=b-a\] \[\Rightarrow \] \[k=\frac{1}{b-a}\] \[\therefore \] \[L=b-\frac{5}{k}\] \[\Rightarrow \] \[L=b-5(b-a)=5a-4b\] When tension is 9 N. Length of spring \[=L+\frac{9}{K}\] Length of spring \[=(5a-4b)+9(b-a)\] Length of spring \[=5b-4a\]