$A$ body of mass $100 \ g$ is tied to a spring of spring constant $8 \ N/m$,while the other end of the spring is fixed. If the body moves in a circular path on a smooth horizontal surface with a constant angular speed of $8 \ rad/s$,then the ratio of the extension in the spring to its natural length will be: (in $:1$)
- A$1$
- B$8$
- C$2$
- D$4$
Explore More
Similar Questions
The angular frequency of a block of mass $0.1 \ kg$ oscillating with the help of a spring of force constant $2.5 \ Nm^{-1}$ is
In the adjacent figure,if the inclined plane is smooth and the springs are identical,then the period of oscillation of this body is
Easy
View SolutionWhen a block of mass $m$ is suspended separately by two different springs having time periods $t_1$ and $t_2$,respectively. If the same mass is connected to the series combination of both springs,then its time period is given by:
Difficult
View SolutionTwo springs of force constants $2k$ and $k$ are connected to a mass $m$ as shown. The mass is displaced slightly to one side and released. The frequency of oscillation of the two springs-mass system is
$A$ $3 \ kg$ block is connected as shown in the figure. The spring constants of the two springs $K_1$ and $K_2$ are $50 \ Nm^{-1}$ and $150 \ Nm^{-1}$ respectively. The block is released from rest with the springs unstretched. The acceleration of the block in its lowest position is $(g=10 \ ms^{-2})$. (in $ms^{-2}$)
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo