Moment of inertia of a hollow circular section, as shown in the below figure about X-axis, is <img src="/images/question-image/mechanical-engineering/engineering-mechanics/1528352873-5.jpg" title="Engineering Mechanics mcq question image" alt="Engineering Mechanics mcq question image">

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

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

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

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

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

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 rectangular section as shown in the below figure about X-X axis, is

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

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

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

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

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

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 mass m is constrained to move on a horizontal frictionless surface. It is set in circular motion with radius r₀ and angular speed ω₀ by an applied force $$\overrightarrow {\bf{F}} $$ communicated through an inextensible thread that passesthrough a hole on the surface as shown in figure given below. Then, this force is suddenly doubled.

The magnitude of the radial velocity of the mass

increases till mass falls into hole

decreases till mass falls into hole

remains constant

becomes zero at radius r1, where 0 1 0

A square hole is made in a circular lamina, the diagonal of the square is equal to the radius of the circle as shown in below figure the shift in the centre of gravity is

$$\frac{{{\text{r}}\left( {\pi - 0.75} \right)}}{{\left( {\pi - 0.5} \right)}}$$

$$\frac{{{\text{r}}\left( {\pi - 0.25} \right)}}{{\left( {\pi - 0.75} \right)}}$$

$$\frac{{{\text{r}}\left( {\pi - 0.5} \right)}}{{\left( {\pi - 0.75} \right)}}$$

$$\frac{{{\text{r}}\left( {\pi - 0.5} \right)}}{{\left( {\pi - 0.25} \right)}}$$

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

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

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

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

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

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

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

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

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

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