(C) Let the carbon atom be at the origin $(0, 0)$ and the oxygen atom be placed at $(1.1, 0)$ on the $x$-axis.Given: $m_1 = 12 \ a.m.u.$ (carbon),$m_2

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

$0.63 \ \mathring{A}$ from the carbon atom

(C) Let the carbon atom be at the origin $(0, 0)$ and the oxygen atom be placed at $(1.1, 0)$ on the $x$-axis.Given: $m_1 = 12 \ a.m.u.$ (carbon),$m_2 = 16 \ a.m.u.$ (oxygen).Position vectors are $\vec{r}_1 = 0\hat{i}$ and $\vec{r}_2 = 1.1\hat{i}$.The center of mass $R_{cm}$ is given by:$R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$$R_{cm} = \frac{12(0) + 16(1.1)}{12 + 16}$$R_{cm} = \frac{17.6}{28} \ \mathring{A}$$R_{cm} \approx 0.63 \ \mathring{A}$ from the carbon atom.

Class 11 Physics System of Particles and Rotational Motion Centre of mass (Point Mass)

Medium

The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is $1.1 \ \mathring{A}$. Given,mass of carbon atom is $12 \ a.m.u.$ and mass of oxygen atom is $16 \ a.m.u.$,calculate the position of the center of mass of the carbon monoxide molecule.

A
$6.3 \ \mathring{A}$ from the carbon atom
B
$1 \ \mathring{A}$ from the oxygen atom
C
$0.63 \ \mathring{A}$ from the carbon atom
D
$0.12 \ \mathring{A}$ from the oxygen atom

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

System of Particles and Rotational Motion

Subtopic

Centre of mass (Point Mass)

Two particles of masses $2$ g and $4$ g are situated at the opposite ends, $A$ and $B$ of a wooden bar respectively. Let $l(AB) = 9$ cm. The center of mass of the system will be

DifficultMHT CET 2026

View Solution

Two point masses of $1.5 \ g$ and $2.5 \ g$ are placed at a distance of $16 \ cm$ from each other. If the center of mass is at a distance $x$ from the $1.5 \ g$ mass,then $x = \dots \ cm$.

Medium

View Solution

$A$ thin bar of length $L$ has a mass per unit length $\lambda$ that increases linearly with distance $x$ from one end. If its total mass is $M$ and its mass per unit length at the lighter end $(x=0)$ is $\lambda_0$,then the distance of the centre of mass from the lighter end is:

DifficultJEE MAIN 2014

View Solution

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.

MediumAP EAMCET 2021

View Solution

$A$ carpenter has constructed a toy as shown in the adjoining figure. If the density of the material of the sphere is $12$ times that of the cone,the position of the centre of mass of the toy is given by:

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

The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule is $1.1 \ \mathring{A}$. Given,mass of carbon atom is $12 \ a.m.u.$ and mass of oxygen atom is $16 \ a.m.u.$,calculate the position of the center of mass of the carbon monoxide molecule.

Similar Questions

Two particles of masses $2$ g and $4$ g are situated at the opposite ends, $A$ and $B$ of a wooden bar respectively. Let $l(AB) = 9$ cm. The center of mass of the system will be

Two point masses of $1.5 \ g$ and $2.5 \ g$ are placed at a distance of $16 \ cm$ from each other. If the center of mass is at a distance $x$ from the $1.5 \ g$ mass,then $x = \dots \ cm$.

$A$ thin bar of length $L$ has a mass per unit length $\lambda$ that increases linearly with distance $x$ from one end. If its total mass is $M$ and its mass per unit length at the lighter end $(x=0)$ is $\lambda_0$,then the distance of the centre of mass from the lighter end is:

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.

$A$ carpenter has constructed a toy as shown in the adjoining figure. If the density of the material of the sphere is $12$ times that of the cone,the position of the centre of mass of the toy is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module