Ncert Exercises & Solutions

The density of a non-uniform rod of length 1m is given by $\rho ( x) = a \left( 1 + bx^{2} \right)$ where, $a$, and $b$ are constants and $0 \leq x \leq 1$. The centre of mass of the rod will be at:

1.	$\dfrac{3(2+b)}{4(3+b)}$	2.	$\dfrac{4(2+b)}{3(3+b)}$
3.	$\dfrac{3(3+b)}{4(2+b)}$	4.	$\dfrac{4(3+b)}{3(2+b)}$

(a) Hint: Apply the concept of centre of mass.

Step 1: Find the position of centre of mass.

$\begin{array}{l} Density is given as ρ (x) = a (1 + {bx}^{2}) \\ where a and b are constants and 0 \leq x \leq 1 \\ Let b \to 0, in this case \\ ρ (x) = a = constant \end{array}$

Hence. center of mass will be at x 0.5m. (middle of the rod)
Putting, b = Oin all the options, only (a) gives 0.5.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions

1.	\(\dfrac{3(2+b)}{4(3+b)}\)	2.	\(\dfrac{4(2+b)}{3(3+b)}\)
3.	\(\dfrac{3(3+b)}{4(2+b)}\)	4.	\(\dfrac{4(3+b)}{3(2+b)}\)