Q156 / 180 · COMEDK 2022 · Physics
A force $F$ applied on the wire of radius $r$ and length $L$ and change in the length of the wire is l. If the same force $F$ is applied on the wire of the same material and radius $4 r$ and length $4 l$, then change in length of the other wire is,
- Correct answerA$\dfrac{l}{4}$
- B$2 l$
- C$\dfrac{l}{2}$
- D$4 l$
Solution
Young's modulus $Y$ is defined as $Y = \dfrac{F L}{A l}$, where $A = \pi r^2$ is the cross-sectional area of the wire.
For the first wire, $Y = \dfrac{F L}{(\pi r^2) l}$.
For the second wire, let the change in length be $l'$. The material is the same, so $Y$ remains constant. The force is $F$, the radius is $4r$, and the length is $4L$.
$Y = \dfrac{F (4L)}{\pi (4r)^2 l'} = \dfrac{4 F L}{16 \pi r^2 l'} = \dfrac{F L}{4 \pi r^2 l'}$.
Equating the expressions for $Y$:
$\dfrac{F L}{\pi r^2 l} = \dfrac{F L}{4 \pi r^2 l'}$
$\dfrac{1}{l} = \dfrac{1}{4 l'}$
$l' = \dfrac{l}{4}$.
Answer: $\dfrac{l}{4}$
For the first wire, $Y = \dfrac{F L}{(\pi r^2) l}$.
For the second wire, let the change in length be $l'$. The material is the same, so $Y$ remains constant. The force is $F$, the radius is $4r$, and the length is $4L$.
$Y = \dfrac{F (4L)}{\pi (4r)^2 l'} = \dfrac{4 F L}{16 \pi r^2 l'} = \dfrac{F L}{4 \pi r^2 l'}$.
Equating the expressions for $Y$:
$\dfrac{F L}{\pi r^2 l} = \dfrac{F L}{4 \pi r^2 l'}$
$\dfrac{1}{l} = \dfrac{1}{4 l'}$
$l' = \dfrac{l}{4}$.
Answer: $\dfrac{l}{4}$
Hence, the correct answer is option: 'A'.