If the radius of curvature of the path of two particles of same mass are in the ratio 3 : 4, then in order to have constant centripetal force, their velocities will be in the ratio of

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
If the radius of curvature of the path of two particles of same mass are in the ratio 3 : 4, then in order to have constant centripetal force, their velocities will be in the ratio of [2024]

1 $\sqrt{3} : 2$

2 $1 : \sqrt{3}$

3 $\sqrt{3} : 1$

4 $2 : \sqrt{3}$

Ans.
(1)

Given $m_{1} = m_{2}$ and $\frac{r_{1}}{r_{2}} = \frac{3}{4}$

As centripetal force $F = \frac{m v^{2}}{r}$

To have constant centripetal force, $F_{1} = F_{2}$

$\frac{m_{1} v_{1}^{2}}{r_{1}} = \frac{m_{2} v_{2}^{2}}{r_{2}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{r_{1}}{r_{2}}} = \frac{\sqrt{3}}{2}$

Want Us to Email you About Special Offers & Updates?