A uniform thin rod AB of length L has linear mass density \mu (x)=a+\frac{bx}{L}      , where x is measured from A. If the CM of the   rod lies at adistance of   \left ( \frac{7}{12} L\right )from A, then a and b are related as :

 

  • Option 1)

    a=b

  • Option 2)

    a=2b

  • Option 3)

    2a=b

  • Option 4)

    3a=2b

 

Answers (1)

As we learnt

Centre of Mass of a continuous Distribution -

x_{cm}=\frac {\int xdm}{\int dm}, \; \;y_{cm}=\frac{\int ydm}{\int dm}, \;z_{cm}=\frac{\int zdm}{\int dm}

- wherein

dm is mass of small element. x, y, z are coordinates of dm part.

 

 As we know that 

\Rightarrow\ \;x_{C}=\frac{\int_{0}^{L}(\frac{a+bx}{L})xdx}{\int_{0}^{L}(\frac{a+bx}{L})dx}

x_{C}=\frac{\int_{0}^{L}(\frac{ax+bx^{2}}{L})dx}{\int_{0}^{L}(\frac{a+bx}{L})dx}=\frac{\frac{al^{2}}{2}+\frac{bL^{2}}{3}}{aL+\frac{bL}{2}}=\frac{L^{2}(\frac{a}{2}+\frac{b}{3})}{L(a+\frac{b}{2})}

Given x_{C}=\frac{7L}{12}

    \frac{7L}{12}=\frac{(\frac{a}{2}+\frac{b}{3})L}{(a+\frac{b}{2})}  After solving we get b = 2a.


Option 1)

a=b

This is an incorrect option.

Option 2)

a=2b

This is an incorrect option.

Option 3)

2a=b

This is the correct option.

Option 4)

3a=2b

This is an incorrect option.

Preparation Products

Knockout JEE Main April 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
Buy Now
Knockout JEE Main May 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-
Buy Now
Test Series JEE Main May 2021

Unlimited Chapter Wise Tests, Unlimited Subject Wise Tests, Unlimited Full Mock Tests, Get Personalized Performance Analysis Report,.

₹ 6999/- ₹ 2999/-
Buy Now
Knockout JEE Main May 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 14999/-
Buy Now
JEE Main Rank Booster 2021

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, 24x7 Doubt Chat Support,.

₹ 13999/- ₹ 6999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions