question_answer

The length of an elastic spring is a metres when a force of 4 N is applied, and b metres when the 5N force is applied. Then the length of the spring when the 9 N 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\[=5b-4a\]