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Grade: 11
        
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..
3 years ago

Answers : (2)

Vikas TU
11143 Points
							
total mass = density * length
that is
dm = lamda * dx
dm = (x+2)dx
integrating both sides,
we get,
m = x^2/2 + 2x
   = l^2/2 + 2l
put l = 3 meter.
m = 9/2 + 6 = 21/2 m that is approx. 10.5 m
3 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.
            
3 years ago
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