A vertically immersed surface is shown in the below figure. The distance of its centre of pressure from the water surface is <img src="/images/question-image/mechanical-engineering/hydraulics-and-fluid-mechanics-in-me/1664193121-a-vertically-immersed-surface-is-shown-in.jpg" title="Hydraulics and Fluid Mechanics in ME mcq question image" alt="Hydraulics and Fluid Mechanics in ME mcq question image">

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

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

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

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

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

The depth of centre of pressure (h) for a vertically immersed surface from the liquid surface is given by (where $${I_{\text{G}}}$$ = Moment of inertia of the immersed surface about horizontal axis through its centre of gravity, A = Area of immersed surface and x = Depth of centre of gravity of the immersed surface from the liquid surface)

$$\frac{{{I_{\text{G}}}}}{{{\text{A}}\overline {\text{x}} }} - \overline {\text{x}} $$

$$\frac{{{I_{\text{G}}}}}{{\overline {\text{x}} }} - {\text{A}}\overline {\text{x}} $$

$$\frac{{{\text{A}}\overline {\text{x}} }}{{{I_{\text{G}}}}} + \overline {\text{x}} $$

$$\frac{{{I_{\text{G}}}}}{{{\text{A}}\overline {\text{x}} }} + \overline {\text{x}} $$

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 block of mass m₁, placed on an inclined smooth plane is connected by a light string passing over a smooth pulley to mass m₂, which moves vertically downwards as shown in the below figure. The tension in the string is

$$\frac{{{{\text{m}}_1}}}{{{{\text{m}}_2}}}$$

$${{\text{m}}_1}{\text{g}}\sin \alpha $$

$$\frac{{{{\text{m}}_1}{{\text{m}}_2}}}{{{{\text{m}}_1} + {{\text{m}}_2}}}$$

$$\frac{{{{\text{m}}_1}{{\text{m}}_2}{\text{g}}\left( {1 + \sin \alpha } \right)}}{{{{\text{m}}_1} + {{\text{m}}_2}}}$$

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

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

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

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

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

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

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

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

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

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

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

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 body A of mass 6.6 kg which is lying on a horizontal platform 4.5 m from its edge is connected to the end of a light string whose other end is supporting a body of mass 3.2 kg as shown in below figure. If the friction between the platform and the body A is $$\frac{1}{3}$$, the acceleration is

0.5 m/sec2

0.75 m/sec2

1.00 m/sec2

1.25 m/sec2

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

The beam shown in below figure is supported by a hinge at ‘A’ and a roller at ‘B’. The reaction RA of the hinged support ‘A’ of the beam, is

10.8 t

10.6 t

10.4 t

10.2 t