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If the linear density of rod of length 3m varies as (lambda) = 2+x then the distance of centre of gravity of the rod is: (a) 7/3m (b)12/7m (c)10/7m (d)9/7m Pls provide with the complete workout of solution and also with proper explanation..
If the linear density of rod of length 3m varies as (lambda) = 2+x then the distance of centre of gravity of the rod is: (a) 7/3m  (b)12/7m  (c)10/7m  (d)9/7m Pls provide with the complete workout of solution and also with proper explanation..


4 years ago

## Answers : (3)

Vikas TU
14149 Points
							total mass = density * lengththat isdm = lamda * dxdm = (x+2)dxintegrating both sides,we get,m = x^2/2 + 2x   = l^2/2 + 2lput l = 3 meter.m = 9/2 + 6 = 21/2 m that is approx. 10.5 m

4 years ago
Raman Mishra
67 Points
							Here it is given that $\lambda =x+2$Considering an element dx of the rod, its mass dm will be: $dm=\lambda dx=(x+2)dx$If xcg denotes the centre of gravity of the rod, then:             $x_{cg}=\frac{\int_{0}^{L}xdm}{\int_{0}^{L}dm}$     =>     $x_{cg}=\frac{\int_{0}^{L}x\lambda dx}{\int_{0}^{L}\lambda dx}$Substituting the value of $\lambda$, we get:           $x_{cg}=\frac{\int_{0}^{L}x(x+2)dx}{\int_{0}^{L}(x+2)dx}$Solving the above integral and further putting the value of L=3m, we get          ${\color{DarkRed}x_{cg}= \frac{12}{7}m }$  Ans.

4 years ago
616 Points
							Dear student,Please find the attached solution to your problem below. Given, λ = dm/dx = x + 2hence, dm = (x + 2)dxLet, xcg denote the center of gravity of rod.Now, xcg = ∫xdm / ∫dm = ∫x(x + 2)dx / ∫(x+2)dx              [ both integral varying from 0 to L, where L = 3m]Hence, xcg = (x3/3 + 2x2/2) / (x2/2 + 2x)or, xcg = (33/3 + (3)2) / (32/2 + 2(3))           = 36/21 = 12/7 m Hope it helps.Thanks and regards,Kushagra

6 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions