Two rings of radius R and nR having different | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The moment of inertia $I$ of a ring of mass $M$ and radius $r$ about an axis passing through its centre and perpendicular to its plane is $I = Mr^

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The moment of inertia $I$ of a ring of mass $M$ and radius $r$ about an axis passing through its centre and perpendicular to its plane is $I = Mr^2$.Since the rings are made of the same wire,their linear mass density $\lambda$ is constant.The mass of a ring is given by $M = \lambda 	imes 	ext{circumference} = \lambda (2\pi r)$.For the first ring: $M_1 = \lambda (2\pi R)$ and $r_1 = R$. Thus,$I_1 = M_1 R^2 = \lambda (2\pi R) R^2 = 2\pi \lambda R^3$.For the second ring: $M_2 = \lambda (2\pi nR)$ and $r_2 = nR$. Thus,$I_2 = M_2 (nR)^2 = \lambda (2\pi nR) (nR)^2 = 2\pi \lambda n^3 R^3$.The ratio of the moments of inertia is given as $\frac{I_1}{I_2} = \frac{1}{8}$.Substituting the expressions: $\frac{2\pi \lambda R^3}{2\pi \lambda n^3 R^3} = \frac{1}{n^3} = \frac{1}{8}$.Therefore,$n^3 = 8$,which gives $n = 2$.

Two rings of radius $R$ and $nR$ having different masses and made up of the same wire have the ratio of moment of inertia about an axis passing through the centre as $1 : 8$. The value of $n$ is

Similar Questions

The moment of inertia of a ring about a diameter is

If a wheel rotating with a constant angular speed experiences a break during its motion,what happens to its radius of gyration?

Two solid spheres $A$ and $B$ each of radius $R$ are made of materials of densities $\rho_A$ and $\rho_B$ respectively. Their moments of inertia about a diameter are $I_A$ and $I_B$ respectively. The value of $\frac{I_A}{I_B}$ is

Three identical rods each of mass $M$ and length $L$ are joined to form a symbol $H$. The moment of inertia of the system about one of the sides of $H$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module