Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

60 of 130

Question 60 of 130

The position vector of the center of mass r⃗cm{\vec{r}}_{cm} of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:

Show answer and solution

Answer: (A) r⃗cm=138Lx^+58Ly^{\vec{r}}_{cm}=\dfrac{13}{8}L\hat{x}+\dfrac{5}{8}L\hat{y}

The position vector of center of mass r⃗cm{\vec{r}}_{cm} is given as
r⃗cm=Xcmx^+Ycmy^{\vec{r}}_{cm}={X}_{cm}\hat{x}+{Y}_{cm}\hat{y}
Where, Xcm={X}_{cm}=x−x-coordinate of center of mass and Ycm=y−{Y}_{cm}=y-coordinate of center of mass
r⃗cm=(m1x1+m2x2+m3x3m1+m2+m3)x^+(m1y1+m2y2+m3y3m1+m2+m3)y^{\vec{r}}_{cm}=\left(\dfrac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}+{m}_{3}{x}_{3}}{{m}_{1}+{m}_{2}+{m}_{3}}\right)\hat{x}+\left(\dfrac{{m}_{1}{y}_{1}+{m}_{2}{y}_{2}+{m}_{3}{y}_{3}}{{m}_{1}+{m}_{2}+{m}_{3}}\right)\hat{y}
r⃗cm=[(2m)×L+(m)×2L+m(5L2)2m+m+m]x^+[(2m)×L+(m)×L2+m×(0)2m+m+m]y^{\vec{r}}_{cm}=\left[\dfrac{\left(2m\right)\times L+\left(m\right)\times 2L+m\left(\dfrac{5L}{2}\right)}{2m+m+m}\right]\hat{x}+\left[\dfrac{\left(2m\right)\times L+\left(m\right)\times \dfrac{L}{2}+m\times \left(0\right)}{2m+m+m}\right]\hat{y}

r⃗cm=13L8x^+5L8y^{\vec{r}}_{cm}=\dfrac{13L}{8}\hat{x}+\dfrac{5L}{8}\hat{y}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
Revise Rotational Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Center of Mass, Momentum & Collision (Centre of mass of continuous mass distribution), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.