Question

The centre of gravity of a rod (of length $L$), whose linear mass density varies as the square of the distance from one end is at :

• A. $\frac{3L}{5}$
• B. $\frac{2L}{5}$
• C. $\frac{L}{3}$
• D. $\frac{3L}{4}$

Answer 1

The correct option is D $\frac{3L}{4}$

Le the linear density be $\rho =k{x}^2$

Centre of Gravity$=\frac{\int x\rho }{\int \rho }=\frac{{\int }_0^L{x}^{3}dx}{{\int }_0^L{x}^{2}dx}=\frac{3L}{4}$

So, Option (D)