A rod is of length `3` m and its mass acting per unit length is directly proportional to distance x from its one end. The centre of gravity of the rod from that end will be at

Options

A. 1.5 m
B. 2 m
C. 2.5 m
D. 3.0 m

Answer

B

Explanation

Here `rho=kx` where k is a constant mass of small element of dx length is dm=kx.dx
`x_"cm"=(intx.dm)/(intdm)=(int_0^3x(x dx))/(int_0^3x . dx)=([x^3/3]_0^3)/([x^2/2]_0^3)=(27/3)/(9/2)=2`