Moment of inertia of a hollow rectangular section as shown in the below figure about X-X axis, is <img src="/images/question-image/mechanical-engineering/engineering-mechanics/1528353467-6.jpg" title="Engineering Mechanics mcq question image" alt="Engineering Mechanics mcq question image">

$$\frac{{{\text{B}}{{\text{D}}^3}}}{{12}} - \frac{{{\text{b}}{{\text{d}}^3}}}{{12}}$$

$$\frac{{{\text{D}}{{\text{B}}^3}}}{{12}} - \frac{{{\text{d}}{{\text{b}}^3}}}{{12}}$$

$$\frac{{{\text{B}}{{\text{D}}^3}}}{{36}} - \frac{{{\text{b}}{{\text{d}}^3}}}{{36}}$$

$$\frac{{{\text{D}}{{\text{B}}^3}}}{{36}} - \frac{{{\text{d}}{{\text{b}}^3}}}{{36}}$$

Correct Answer: $$\frac{{{\text{B}}{{\text{D}}^3}}}{{12}} - \frac{{{\text{b}}{{\text{d}}^3}}}{{12}}$$

The moment of inertia of a uniform sphere of radius, r about an axis passing through its centre is given by $$\frac{2}{5}\left( {\frac{{4\pi }}{3}{r^5}\rho } \right).$$ A rigid sphere of uniform mass density $$\rho $$ and radius R has two smaller spheres of radii $$\frac{R}{2}$$ hollowed out of it as shown in the figure given below.

The moment of inertia of the resulting body about Y-axis is

$$\frac{{\pi \rho {R^5}}}{4}$$

$$\frac{{5\pi \rho {R^5}}}{{12}}$$

$$\frac{{7\pi \rho {R^5}}}{{12}}$$

$$\frac{{3\pi \rho {R^5}}}{4}$$

Moment of inertia of a hollow circular section, as shown in the below figure about an axis perpendicular to the section, is ________ than that about X-X axis.

Two times

Same

Half

None of these

Moment of inertia of a hollow circular section, as shown in the below figure about X-axis, is

$$\frac{\pi }{{16}}\left( {{{\text{D}}^2} - {{\text{d}}^2}} \right)$$

$$\frac{\pi }{{16}}\left( {{{\text{D}}^3} - {{\text{d}}^3}} \right)$$

$$\frac{\pi }{{32}}\left( {{{\text{D}}^4} - {{\text{d}}^4}} \right)$$

$$\frac{\pi }{{64}}\left( {{{\text{D}}^4} - {{\text{d}}^4}} \right)$$

The moment of inertia of the shaded portion of the area shown in below figure about the X-axis, is

229.34 cm4

329.34 cm4

429.34 cm4

529.34 cm4

A particle of mass m is attached to a thin uniform rod of length a and mass 4m. The distance of the particle from the centre of mass of the rod is $$\frac{a}{2}.$$

The moment of inertia of the combination about an axis passing through a normal to the rod is

$$\frac{{64}}{{48}}m{a^2}$$

$$\frac{{91}}{{48}}m{a^2}$$

$$\frac{{27}}{{48}}m{a^2}$$

$$\frac{{51}}{{48}}m{a^2}$$

The moment of the force ‘P’ about ‘O’ as shown in the below figure is

P × OA

P × OB

P × OC

P × AC

A vertically immersed surface is shown in the below figure. The distance of its centre of pressure from the water surface is

$$\frac{{{\text{b}}{{\text{d}}^2}}}{{12}} + \overline {\text{x}} $$

$$\frac{{{{\text{d}}^2}}}{{12\overline {\text{x}} }} + \overline {\text{x}} $$

$$\frac{{{{\text{b}}^2}}}{{12}} + \overline {\text{x}} $$

$$\frac{{{{\text{d}}^2}}}{{12}} + \overline {\text{x}} $$

According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. $${I_{\text{P}}}$$) is given by (where, A = Area of the section, $${I_{\text{G}}}$$ = Moment of inertia of the section about an axis passing through its C.G. and h = Distance between C.G. and the parallel axis.)

$${I_{\text{P}}} = {I_{\text{G}}} + {\text{A}}{{\text{h}}^2}$$

$${I_{\text{P}}} = {I_{\text{G}}} - {\text{A}}{{\text{h}}^2}$$

$${I_{\text{P}}} = \frac{{{I_{\text{G}}}}}{{{\text{A}}{{\text{h}}^2}}}$$

$${I_{\text{P}}} = \frac{{{\text{A}}{{\text{h}}^2}}}{{{I_{\text{G}}}}}$$

A circular hole of 50 mm diameter is cut out from a circular disc of 100 mm diameter as shown in the below figure. The center of gravity of the section will lie

In the shaded area

In the hole

At ‘O’

None of these

A torsional system with discs of moment of inertia $${I_1}$$ and $${I_2}$$ as shown in the below figure, is gear driven such that the ratio of speed of shaft B to shaft A is 'G'. Neglecting the inertia of gears, the equivalent inertia of disc on shaft B at the speed of shaft A is equal to

$${\text{G}} \times {I_2}$$

$${{\text{G}}^2} \times {I_2}$$

$$\frac{{{I_2}}}{{\text{G}}}$$

$$\frac{{{I_2}}}{{{{\text{G}}^2}}}$$