Q2GATE 2024NAT

The return period of a large earthquake for a given region is 200 years. Assuming that earthquake occurrence follows Poisson's distribution, the probability that it will be exceeded at least once in 50 years is ______ $\%$ . (rounded off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \Rightarrow \quad \text { Risk } & =1-\left[\frac{\lambda^{r} \cdot e^{-\lambda}}{r !}\right]\$ \\ & =1-\left[\frac{\lambda^{0} \cdot e^{-\lambda}}{0 !}\right]\$ \\ \lambda & =n p=50 \times \frac{1}{200}=0.25 \\ \text { Now, } \quad \text { Risk } & =1-\left[e^{-0.25}\right]\$ \\ & =0.2211 \text { or } 22.11 \% \end{aligned}

Q3GATE 2024NAT

A $2 \mathrm{~m} \times 2 \mathrm{~m}$ tank of $3 \mathrm{~m}$ height has inflow, outflow and stirring mechanisms. Initially, the tank was half filled with fresh water. At $t=0$ , an inflow of salt a solution of concentration $5 \mathrm{~g} / \mathrm{m}^{3}$ at the rate of 2 litre/s and an outflow of the well stirred mixture at the rate of 1 litre/s are initiated. This process can be modeled using the following differential equation. $\frac{d m}{d t}+\frac{m}{6000+t}=0.01$ where $m$ is the mass (grams) of salt at time $t$ (seconds). The mass of the salt (in grams) in the tank at $75 \%$ of its capacity is ______(rounded off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

Total capacity of tank $=2 \times 2 \times 3=12 \mathrm{~m}^{3}$ At $50 \%$ capacity, $\text { Volume }\left(V_{50}\right)=0.5 \times 12=6 \mathrm{~m}^{3}$ At $75 \%$ capacity, $\text { Volume }\left(V_{75}\right)=0.75 \times 12 \mathrm{~m}^{3}=9 \mathrm{~m}^{3}$ Now, increase in volume $\Delta V=V_{75}-V_{50}=(9-6) \mathrm{m}^{3}=3 \mathrm{~m}^{3}$ Also, $\text { Inflow }(I)=2 \mathrm{l} / \mathrm{s}$ Outflow $(Q)=1 \mathrm{l} / \mathrm{s}$ Rate of change storage,

\begin{aligned} \frac{\Delta S}{\Delta t} & =I-Q=2-1=1 \mathrm{l} / \mathrm{s} \\ & =\frac{3000 l}{1 \mathrm{l} / \mathrm{s}}=3000 \mathrm{sec} \end{aligned}

Now, $\frac{d m}{d t}+\frac{m}{6000+t}=0.01$ $\Rightarrow \quad \frac{d m}{d t}+\frac{m}{6000+3000}=0.01$ $\Rightarrow \quad \frac{d m}{d t}+\frac{m}{9000}=0.01$ $\Rightarrow \quad \frac{d m}{0.01-\frac{m}{9000}}=d t$ Integrating both sides $\Rightarrow \quad \int_{m_{0}=0}^{m} \frac{d m}{0.01-\frac{m}{9000}}=\int_{t=0}^{t=3000} d t$ $\Rightarrow \quad \frac{\left[\ln \left(0.01-\frac{m}{9000}\right)\right]_{0}^{m}}{-\frac{1}{9000}}=(3000-0)$ $\Rightarrow \quad \ln \left(0.01-\frac{m}{9000}\right)-\ln 0.01=-\frac{1}{9000} \times 3000$

\begin{array}{rlrl} \Rightarrow & & \ln \left(0.01-\frac{m}{9000}\right) & =-4.938 \\ \Rightarrow & 0.01-\frac{m}{9000} & =7.165 \times 10^{-3} \\ \Rightarrow & m & =25.5 \mathrm{gm} \end{array}

Q4GATE 2023NAT

A catchment may be idealized as a circle of radius $30 \mathrm{~km}$ . There are five rain gauges, one at the center of the catchment and four on the boundary (equi-spaced), as shown in the figure (not to scale). The annual rainfall recorded at these gauges in a particular year are given below.

\begin{array}{|c|c|c|c|c|c|} \hline Gauge & \mathrm{G}_{1} & \mathrm{G}_{2} & \mathrm{G}_{3} & \mathrm{G}_{4} & \mathrm{G}_{5} \\ \hline Rainfall (mm) & 910 & 930 & 925 & 895 & 905 \\ \hline \end{array}

Using the Thiessen polygon method, what is the average rainfall (in $\mathrm{mm}$ , rounded off to two decimal places) over the catchment in that year ? ___

🚀 Solve in Practice Mode

📖 Explanation

Radius of basin $=30 \mathrm{~km}$ We know that in Theissen polygon method, the polygons are obtained by joining the perpendicular bisectors of triangles formed, when various rain gauge stations are joined. The Theissen polygon in the given basin can be drawn as: Area under influence of station GI $=30 \times 30=900 \mathrm{~km}^{2}$ Area under influence of stations G2, G3, G4, G5 $=\frac{\pi \times 30^{2}-900}{4}$ $=481.86 \mathrm{~km}^{2}$ mean rainfall

=\frac{\begin{array}{c}910 \times 900+481.86 \times 930+ \\ 481.86 \times 925+481.86 \times 895+ \\ 481.86 \times 905\end{array}}{900+481.86 \times 4}

$=912.56 \mathrm{~mm}$

Q5GATE 2021NAT

A tube-well of 20 cm diameter fully penetrates a horizontal, homogeneous and isotropic confined aquifer of infinite horizontal extent. The aquifer is of 30 m uniform thickness. A steady pumping at the rate of 40 litres/s from the well for a long time results in a steady drawdown of 4 m at the well face. The subsurface flow to the well due to pumping is steady, horizontal and Darcian and the radius of influence of the well is 245 m. The hydraulic conductivity of the aquifer (in m/day,round off to integer) is ______________

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} 40 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s} &=\frac{2 \pi k \times 30 \mathrm{~m} \times 4 \mathrm{~m}}{\ln \left(\frac{245}{0.1}\right)} \\ k &=4.14 \times 10^{-4} \mathrm{~m} / \mathrm{s} \\ \text{or, } \qquad \qquad \qquad \qquad k &=35.77 \mathrm{~m} / \mathrm{d} \\ &\approx 36 \mathrm{~m} / \mathrm{d} (\text{ round off to nearest integer}) \end{aligned}

Q6GATE 2021NAT

Two reservoirs are connected through a homogeneous and isotropic aquifer having hydraulic conductivity (K) of 25 m/day and effective porosity $(\eta)$ of 0.3 as shown in the figure (not to scale). Ground water is flowing in the aquifer at the steady state. If water in Reservoir 1 is contaminated then the time (in days, round off to one decimal place) taken by the contaminated water to reach to Reservoir 2 will be _____________

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} k&=25 \mathrm{~m} / \mathrm{d} \\ n&=0.3 \\ i&=\frac{20 \mathrm{~m}}{2000 \mathrm{~m}}=0.01 \\ V&=k i \\ V_{s}&=\frac{V}{n} \\ t&=\frac{l}{V_{s}} \\ t&=2400 \mathrm{days} \end{aligned}

Q7GATE 2019MCQ

A catchment may be idealised as a rectangle. There are three rain gauges located inside the catchment at arbitrary locations. The average precipitation over the catchment is estimated by two methods: (i) Arithmetic mean ( $P_A$ ), and (ii) Thiessen polygon ( $P_T$ ). Which one of the followingstatements is correct?

🚀 Solve in Practice Mode

📖 Explanation

The result from Thiessen polygon method is more accurate than arithmetic mean method. But there is no any close relationship between values obtained by Thiessen polygon method and Arithmetic mean method. Therefore, there is no any relation between $P_A$ and $P_T$ .

Q8GATE 2018NAT

Rainfall depth over a watershed is monitored through six number of well distributed rain gauges. Gauged data are given below The Thiessen mean value (in mm, up to one decimal place) of the rainfall is ______

🚀 Solve in Practice Mode

📖 Explanation

Thiessen mean value of rainfall; $P_{avg}=\frac{\sum_{i=1}^{6}P_{i}A_{i}}{\sum_{i=1}^{6}A_{i}}$ $P_{avg}=\frac{470\times 95+465\times 100+435\times 98+525\times 80+480\times 85+510\times 92}{95+100+98+80+85+92} =479.09 mm$

Q9GATE 2017NAT

A catchment is idealized as a 25 km x 25 km square. It has five rain gauges, one at each corner and one at the center, as shown in the figure. During a month, the precipitation at these gauges is measured as $G_{1}$ = 300 mm, $G_{2}$ = 285 mm, $G_{3}$ = 272mm, $G_{4}$ = 290mm and $G_{5}$ = 288mm. The average precipitation (in mm, up to one decimal place) over the catchment during this month by using the Thiessen polygon method is ______

🚀 Solve in Practice Mode

📖 Explanation

Let sides of square as, a=25 km Now area of the polygon is calculated as,

\begin{aligned} A_{1}&=A_{2}=A_{3}=A_{4} \\ &=\frac{1}{2}\times \frac{a}{2}\times \frac{a}{2} \\ &=\frac{a^{2}}{8}=\frac{25^{2}}{8}=78.125 \\ A_{5}&=\text{Area of square of sides } \frac{\sqrt{2}a}{2} \\ &=\frac{a}{\sqrt{2}} =\frac{a^{2}}{2}=\frac{25^{2}}{2}=312.5\, km^{2} \\ \therefore & \text{Mean precipitation,}\\ \bar{P}&=\frac{G_{1}A_{1}+G_{2}A_{2}+G_{3}A_{3}+G_{4}A_{4}+G_{5}A_{5}}{A_{1}+A_{2}+A_{3}+A_{4}+A_{5}} \\ \therefore \;\;\bar{P}&=\frac{\left ( 300+255+272+290 \right )\times 78.125+288\times 312.5}{\left ( 4\times 78.125 \right )+312.5} \\ &= 287.375 mm\end{aligned}

Q10GATE 2015NAT

In a catchment, there are four rain-gauge stations, P, Q, R, and S. Normal annual precipitation values at these stations are 780 mm, 850 mm, 920 mm, and 980 mm respectively. In the year 2013, stations Q, R, and S, were operative but P was not. Using the normal ratio method, the precipitation at station P for the year 2013 has been established as 860 mm. If the observed precipitation at station Q and R for the year 2013 were 930 mm and 1010 mm, respectively; what was the observed precipitation (in mm) at station S for that year?

🚀 Solve in Practice Mode

📖 Explanation

Let $P_{p}=$ Precipitation in 2013 at station P $N_{p}=$ Noramal annual precipitation at station P