Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The linear mass density of a thin rod AB of length L varies from A to B as <img alt="\lambda (x)=\lambda _{0}\left ( 1+\frac{x}{L} \right )," src="http://entrancecorner.codecogs.com/gif.latex?%5Clambda%20%28x%29%3D%5Clambda%20_%7B0%7D%5Cleft%20%28%201&

The linear mass density of a thin rod AB of length L varies from A to B as \lambda (x)=\lambda _{0}\left ( 1+\frac{x}{L} \right ), where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is :  
Option: 1 \frac{5}{12}ML^{2}
 
Option: 2 \frac{7}{18}ML^{2}
Option: 3 \frac{2}{5}ML^{2}  
Option: 4 \frac{3}{7}ML^{2}

Answers (1)

best_answer

\begin{aligned} &I=\int r^{2} d m=\int x^{2} \lambda d x\\ &I=\int_{0}^{L} x^{2} \lambda_{0}\left(1+\frac{x}{L}\right) d x\\ &I=\lambda_{0} \int_{0}^{1}\left(x^{2}+\frac{x^{3}}{L}\right) d x\\ &I=\lambda\left[\frac{L^{3}}{3}+\frac{L^{3}}{4}\right]\\ &I=\frac{7 L^{3} \lambda_{0}}{12} \dots(i)\\ &M=\int_{0}^{L} \lambda d x=\int_{0}^{L} \lambda_{0}\left(1+\frac{x}{L}\right) d x\\ &M=\lambda_{0}\left(L+\frac{L}{2}\right)=\lambda_{0} \frac{3 L}{2}\\ &\frac{2}{3} M=\left(\lambda_{0} L\right) \dots(ii) \end{aligned}

\begin{aligned} &\text { From (i) \& (ii) }\\ &I=\frac{7}{12}\left(\frac{2}{3} M\right) L^{2}=\frac{7 M L^{2}}{18} \end{aligned}

 

Posted by

Deependra Verma

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

Amrita BTech Admission 2025: JEE Main-based registration deadline extended to August 30

Latest Question