Question

a rod of length 3M has its mass per unit length directly proportional to the distance x from one of its in then its centre of gravity from that n will be at
​

Answer 1

centre of gravity from that one end will be at 2m

it is given that length of rod = 3m

and mass per unit length is directly proportional to from one of its end.

i.e., dm/dx ∝ x

⇒dm ∝ x dx

⇒dm = k xdx , where k is proportionality constant.

finding centre of mass of rod, x = $\frac{\int{x.dm}}{\int{dm}}$

= $\frac{\int\limits^3_0{x(xdx)}}{\int\limits^3_0{xdx}}$

= $\frac{\left[\frac{x^3}{3}\right]^3_0}{\left[\frac{x^2}{2}\right]^3_0}$

= 2m, from on of its end.

rod is symmetrical in shape, centre of gravity concides with its centre of mass. so, centre of gravity from that one end will be at 2m .

also read similar questions : Linear mass density of rod of length l is directly proportional to x cube where x is the distance from one end of rod.ce...

https://brainly.in/question/11436917

centre of gravity in a sphere

https://brainly.in/question/1943666

Answer 2

$\huge\bold\purple{Answer:-}$

A rod of length is 3m and its mass acting per unit length is directly proportional to distance x from one of its end then its centre of gravity from that end will be at (A) 1.5 m (C) 2.5 m ✅(B) 2 m (D) 3.0 m.