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  • The density of a non-uniform rod of length 1 m is given by \rho (x)=a(1+bx^{2}) where a and b are constants and o\leq x\leq 1.. The center of mass of the rod will be at

The density of a non-uniform rod of length 1 m is given by \rho (x)=a(1+bx^{2}) where a and b are constants and o\leq x\leq 1.. The center of mass of the rod will be at

(a)\frac{3(2+b)}{4(3+b)}

(b)\frac{4(2+b)}{3(3+b)}

(c)\frac{3(3+b)}{4(2+b)}

(d)\frac{4(3+b)}{3(2+b)}

Answers (1)

The correct answer is the option(a)\frac{3(2+b)}{4(3+b)}

The density of non-uniform rod is given by     

\rho (x)=a(1+bx^{2})

So if b = 0, the density of the rod would be uniform and its centre of mass would lie at the middle point of the length of the rod, i.e.) 0.5

Putting b = 0 for all the options, we get 0.5 only for A.

Hence A is the correct option.

Posted by

infoexpert23

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