Given below are two statements: one is labell | vedclass

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

Question

(B) According to the perpendicular axes theorem,for a planar body like a circular disc,$I_{z} = I_{x} + I_{y}$.The radius of gyration $K$ is defined b

Vedclass · Accepted Answer

$A$ is not correct but $R$ is correct.

Vedclass · Answer

(B) According to the perpendicular axes theorem,for a planar body like a circular disc,$I_{z} = I_{x} + I_{y}$.The radius of gyration $K$ is defined by the relation $I = MK^{2}$,where $M$ is the mass of the body.Substituting this into the theorem: $MK_{z}^{2} = MK_{x}^{2} + MK_{y}^{2}$,which simplifies to $K_{z}^{2} = K_{x}^{2} + K_{y}^{2}$.Since $K_{z}^{2} = K_{x}^{2} + K_{y}^{2}$,the radii of gyration about the three axes cannot be the same. Thus,Assertion $A$ is incorrect.Reason $R$ states that a rigid body has a fixed mass and shape,which is a correct definition of a rigid body in rotational mechanics. Therefore,$A$ is incorrect but $R$ is correct.

Similar Questions

Radius of gyration of a thin uniform circular disc about the axis passing through its centre and perpendicular to its plane is $K_c$. Radius of gyration of the same disc about a diameter of the disc is $K_d$. The ratio $K_d : K_c$ is

The moment of inertia of a big drop is $I$. If $8$ droplets are formed from the big drop,then the moment of inertia of a small droplet is:

The radius of gyration of a uniform rod of length $l$ about an axis passing through a point $\frac{l}{4}$ away from the centre of the rod and perpendicular to it is:

The ratios of the masses and radii of two circular rings are $1:2$ and $2:1$ respectively. What is the ratio of their moments of inertia?

For the adjoining diagram,the incorrect relation between $I_{1}, I_{2}$ and $I_{3}$ is ($I$ = moment of inertia).

Vedclass Test Series

Exam Paper Generator

Online Exam Module