Match Column-$I$ with Column-$II$.
Column-$I$Column-$II$
$(1)$ $\frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}$$(a)$ Reduced mass of a two-particle system
$(2)$ $\frac{{{r_1} + {r_2}}}{2}$$(b)$ Position vector of the center of mass for a system of two equal masses

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A) For $(1)$, the reduced mass $\mu$ of a system of two particles with masses $m_1$ and $m_2$ is defined as $\mu = \frac{m_1 m_2}{m_1 + m_2}$. Thus, $(1)$ matches with $(a)$.
For $(2)$, the center of mass $R_{cm}$ of a system of two particles with masses $m_1$ and $m_2$ at positions $r_1$ and $r_2$ is given by $R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$. If $m_1 = m_2 = m$, then $R_{cm} = \frac{m(r_1 + r_2)}{2m} = \frac{r_1 + r_2}{2}$. Thus, $(2)$ matches with $(b)$.
The correct matching is $(1-a, 2-b)$.

Explore More

Similar Questions

Four identical spheres, each of radius $10 \,cm$ and equal mass $1 \,kg$, are placed on a horizontal surface touching each other such that their centers are located at the vertices of a square of side $20 \,cm$. What is the distance of their center of mass from the center of any sphere?

The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is $1.1 Å$. Given,mass of carbon atom is $12 amu$ and mass of oxygen atom is $16 amu$. Calculate the position of the centre of mass of the carbon monoxide molecule.

Four particles of masses $m_1 = 2m$,$m_2 = 4m$,$m_3 = m$,and $m_4$ are placed at the four corners of a square of side $a$. Let the corners be $(0,0)$,$(a,0)$,$(a,a)$,and $(0,a)$ respectively. What should be the value of $m_4$ so that the centre of mass of the system is at the centre of the square,i.e.,at $(\frac{a}{2}, \frac{a}{2})$?

Difficult
View Solution

Four masses are arranged along a circle of radius $1 \ m$ as shown in the figure. The center of mass of this system of masses is at

Three masses of $2\,kg$,$4\,kg$,and $4\,kg$ are placed at the three points $(1, 0, 0)$,$(1, 1, 0)$,and $(0, 1, 0)$ respectively. The position vector of its center of mass is:

Difficult
View Solution

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