The figure shows two cases. In the first case,a spring (spring constant $K$) is pulled by two equal and opposite forces $F$ at both ends. In the second case,it is pulled by a force $F$ at one end while the other end is fixed. The extensions $(x)$ in the springs will be:

  • A
    In both cases $x = \frac{2F}{K}$
  • B
    In both cases $x = \frac{F}{K}$
  • C
    In the first case $x = \frac{2F}{K}$,in the second case $x = \frac{F}{K}$
  • D
    In the first case $x = \frac{F}{K}$,in the second case $x = \frac{2F}{K}$

Explore More

Similar Questions

$A$ spring of force constant $k$ is cut into two pieces such that one piece is three times the length of the other. The longer piece will have a force constant of

$A$ spring of length $l$ has a force constant $k$. When a weight $W$ is attached to it,the extension produced is $x$. If the spring is cut into two equal parts and these parts are connected in parallel to support the same weight $W$,what will be the new extension?

Difficult
View Solution

If a spring of stiffness $k$ is cut into two parts $A$ and $B$ of length $l_{A}: l_{B}=2: 3$,then the stiffness of spring $A$ is given by

The force required to stretch a spring varies with the distance as shown in the figure. If the experiment is performed with the above spring of half length,the line $OA$ will

If a spring is extended to length $l,$ then according to Hooke's law,

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo