Verified Solution • 2025–26 Session moderate

Q6.2: NCERT Physics - Part 1 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 6.2 Solution: In the HCl molecule, the separation between the nuclei of the…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often write the CM equation with a negative term for the heavier mass (e.g., $m_H(1.27-x)+m_{Cl}(-x)=0$), mixing up directions and cancelling the total mass denominator. Instead, place the origin at one nucleus, use only positive distances, and apply $m_H(1.27-x)=m_{Cl}x$ (or $x=\frac{m_H\,1.27}{m_H+m_{Cl}}$) to avoid sign mistakes.

In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 � (1 � = 10___ m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Step-by-Step Verified Solution

**Answer:** The centre of mass lies at a distance of 0.035 � from the chlorine atom. **Step-by-Step Explanation:** Let the distance of CM from the chlorine atom be xx. Then its distance from the hydrogen atom is (1.27−x)(1.27 - x) �. Taking the centre of mass at the origin and using the definition of CM:
x=mH(1.27−x)+mCl(−x)mH+mCl=0x = \frac{m_H(1.27 - x) + m_{Cl}(-x)}{m_H + m_{Cl}} = 0
With mCl=35.5mHm_{Cl} = 35.5 m_H:
1.27−x=35.5x  ⟹  x=1.2736.5=0.035 �1.27 - x = 35.5x \implies x = \frac{1.27}{36.5} = 0.035 \text{ �}
[citation:33]

All Questions in this Chapter

Q6.1 Q6.2 Q6.3 Q6.4 Q6.5 Q6.6 Q6.7 Q6.8 Q6.9 Q6.10 Q6.11 Q6.12 Q6.13 Q6.14 Q6.15 Q6.16 Q6.17

Related Learning Resources & Exam Preparation

Chapter Practice Bank
Practice all topic-wise questions from SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
Reference Textbook Solutions
Explore complete chapter exercises for NCERT Physics - Part 1
Competitive PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring