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 <br><img src="/images/question-image/mechanical-engineering/engineering-mechanics/1528353276-9.jpg" title="Engineering Mechanics mcq question image" alt="Engineering Mechanics mcq question image">

In the shaded area

In the hole

At ‘O’

None of these

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 center of gravity of a trapezium with parallel sides 'a' and 'b' lies at a distance of 'y' from the base 'b', as shown in the below figure. The value of 'y' is

$${\text{h}}\left( {\frac{{2{\text{a}} + {\text{b}}}}{{{\text{a}} + {\text{b}}}}} \right)$$

$$\frac{{\text{h}}}{2}\left( {\frac{{2{\text{a}} + {\text{b}}}}{{{\text{a}} + {\text{b}}}}} \right)$$

$$\frac{{\text{h}}}{3}\left( {\frac{{2{\text{a}} + {\text{b}}}}{{{\text{a}} + {\text{b}}}}} \right)$$

$$\frac{{\text{h}}}{3}\left( {\frac{{{\text{a}} + {\text{b}}}}{{2{\text{a}} + {\text{b}}}}} \right)$$

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 r<sub>1</sub>, where 0 1 0

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 centre of gravity of the trapezium as shown in below figure from the side is at a distance of

$$\frac{{\text{h}}}{3} \times \frac{{{\text{b}} + {\text{2a}}}}{{{\text{b}} + {\text{a}}}}$$

$$\frac{{\text{h}}}{3} \times \frac{{2{\text{b}} + {\text{a}}}}{{{\text{b}} + {\text{a}}}}$$

$$\frac{{\text{h}}}{2} \times \frac{{{\text{b}} + {\text{2a}}}}{{{\text{b}} + {\text{a}}}}$$

$$\frac{{\text{h}}}{2} \times \frac{{2{\text{b}} + {\text{a}}}}{{{\text{b}} + {\text{a}}}}$$

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 weight of 100 kg is supported by a string whose ends are attached to pegs ‘A’ and ‘B’ at the same level shown in below figure. The tension in the string is

50 kg

750 kg

100 kg

120 kg

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

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

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

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

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

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

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

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

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

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