Q2GATE 2024NAT

A $5 \mathrm{~m} \times 5 \mathrm{~m}$ closed tank of $10 \mathrm{~m}$ height contains water and oil, is connected to an overhead water reservoir as shown in the figure. Use $\gamma_{w}=10 \mathrm{kN} / \mathrm{m}^{3}$ and specific gravity of oil $=0.8$ . The total force (in $\mathrm{kN}$ ) due to pressure on the side $\mathrm{PQR}$ of the tank is ______ (rounded off to the nearest integer).

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📖 Explanation

\begin{aligned} P_{P}+(800)(10)(4)&=\left(10^{3}\right)(10)(10), \quad \text { where } P_{p} \text { is pressure at } P\\P_{P}+\left(32 \times 10^{3}\right)&=\left(100 \times 10^{3}\right)\\ P_{P}&=68 \times 10^{3} \mathrm{~Pa}\\ P_{P}&=68 \mathrm{kPa}\end{aligned}

Now, force on PQR, $F=\left[\frac{1}{2}(68+100) 4+\frac{1}{2}(100+160) 6\right] 5=(336+780) 5 =5580 \mathrm{kN}$

Q3GATE 2016NAT

A sector gate is provided on a spillway as shown in the figure. Assuming g = $10\: \: m/s^{2}$ the resultant force per meter length (expressed in kN/m) on the gate will be _______ .

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📖 Explanation

\begin{aligned} F_{x} &=\rho g \bar{h} A_{v} \\ &=\left(10^{3}\right)(10)\left(\frac{5}{2}\right)(5 \times 1)\\ &=125 \mathrm{kN}\text{ per unit width}\\ F_{y} &=\rho g \forall \\ \text { where } \forall &=\frac{\pi(5)^{2}}{6}-\left(\frac{1}{2} \times 5 \times 5 \cos 30^{\circ}\right)=2.264 \mathrm{m}^{3} \\ F_{y} &=(10)^{3}(10)(2.264)=22.64 \mathrm{kN} \end{aligned}

Resultant force per unit width,

\begin{aligned} F_{R}&=\sqrt{F_{x}^{2}+F_{y}^{2}}\\ &=\sqrt{125^{2}+22.64^{2}}=127.03 \mathrm{kN} \end{aligned}

Q4GATE 2015NAT

A triangular gate with a base width of 2 m and a height of 1.5 m lies in a vertical plane. The top vertex of the gate is 1.5 m below the surface of a tank which contains oil of specific gravity 0.8. Considering the density of water and acceleration due to gravity to be 1000 kg/ $m^{3}$ and 9.81 m/ $s^{2}$ , respectively, the hydrostatic force (in kN) exerted by the oil on the gate is__________.

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📖 Explanation

\begin{aligned} \bar{h} &=1.5+1.5\left(\frac{2}{3}\right)=2.5 \mathrm{m} \\ F_{R} &=\rho g \bar{h} A \\ &=(800)(9.81)(2.5)\left(\frac{1}{2} \times 2 \times 1.5\right) \\ &=29430 \mathrm{N}=29.43 \mathrm{kN} \end{aligned}

Q5GATE 2006MCQ

The necessary and sufficient condition for a surface to be called as a free surface is

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Q6GATE 2005MCQ

Cross-section of an object (having same section normal to the paper) submerged into a fluid consists of a square of sides 2 m and triangle as shown in the figure. The object is hinged at point P that is one meter below the fluid free surface. If the object is to be kept in the position as shown in the figure, the value of 'x' should be

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📖 Explanation

The vertical force on the surface bounded by square and triangle would be respectively, $F_{1}=4(S-1) \gamma_{w}$ for square edge and acting at 1 m from P $F_{2}=x(S-1) \gamma_{w}$ for inclined edge of triangle and acting at x / 3 from P Taking moments of both the forces about P, we get,

\begin{aligned} F_{ 1} \times 1 &=F_{2} \times \frac{x}{3} \\ \Rightarrow \quad 4(\mathrm{S}-1) \gamma_{w} &=x(S-1) \gamma_{w} \times \frac{x}{3} \\ \Rightarrow \quad x^{2} &=12 \\ \Rightarrow \quad x &=2 \sqrt{3} \mathrm{m} \end{aligned}

Q7GATE 2002MCQ

The force 'F' required at equilibrium on the semi-cylindrical gate shown below is

🚀 Solve in Practice Mode

📖 Explanation

Depth of water (H)=2 m Consider unit width, of cylinder,

\begin{aligned} F_{H} &=\rho g \bar{h} A_{h} \\ &=1000 \times 9.81 \times \left(\frac{2}{2}\right) \times (2 \times 1) \\ &=19.62 \mathrm{kN} / \mathrm{m} \text { width } \end{aligned}

Vertical component,

\begin{aligned} F_{v}&=\rho g(\text { volume })\\ &=1000 \times 9.81\left(\frac{\pi R^{2}}{2} \times 1\right) \\ &=15.41 \mathrm{kN} / \mathrm{m} \text { width } \end{aligned}

location of center of pressure of $F_{h}$ $h^{*}=\frac{2}{3} H=\frac{4}{3} m$ location of center of pressure of $F_{v}$ $\therefore \quad \bar{x}=\frac{4 R}{3 \pi}=\frac{4 \times 1}{3 \pi}=0.424 \mathrm{m}$ Moment about hinge,

\begin{aligned} \therefore \quad F_{H} \times \left(h^{*}-\bar{h}\right)&=F_{V} \times \bar{x}+F \times 1 \\ \therefore 19.62 \times \left(\frac{4}{3}-1\right)&=15.41 \times 0.424+F \times 1 \\ \therefore \quad F&=0 \mathrm{kN} \end{aligned}

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