Q41GATE 2009MCQ

The correct match of Column-I with Column-II is

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Q42GATE 2009MCQ

The reference pressure used in the determination of sound pressure level is

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📖 Explanation

The sound pressure of the faintest sound that can be heard by a normal healthy individual is about 20 $\mu Pa$ . Hence, this pressure is used as reference pressure in determination of sound pressure level.

Q43GATE 2009MCQ

Particulate matter (fly ash) carried in effluent gases from the furnaces burning fossil fuels are better removed by

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📖 Explanation

Electrostatic precipitator is generally used in thermal power plants and are used for removed of fly ash. Efficiency $\leq$ 99% ( $d_p$ ) Dia, of particulate removed $\geq \; 1 \mu$ Fabric filter : Highly efficient device $\eta \geq 99\%$ $d_p \leq \; 1\mu$

Q44GATE 2008MCQ

Two primary air pollutants are

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Q45GATE 2007MCQ

The dispersion of pollutants in atmosphere is maximum when

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Q46GATE 2007MCQ

Two electrostatic precipitators (ESPs) are in series. The fractional efficiencies of the upstream and downstream ESPs for size $d_{p}$ are 80% and 65%, respectively. What is the overall efficiency of the system for the same $d_{p}$ ?

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📖 Explanation

Given data: $\eta _{1}=80%;\; \; \eta _{2}=65%$ Since the efficiency of upstream ESP is 80%, only 20% of the particulate are not removed. These remaining particulates will face the downstream ESP whose efficiency is 65%. $\therefore$ Particulates removed by downstream, $ESP=20\times \frac{65}{100}=13\%$ The two ESPs are connected in series, therefore the overall efficiency = 80 + 13 = 93%.

Q47GATE 2006MCQ

The cumulative noise power distribution curve at a certain location is given below The value of $L_{40}$ is equal to

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📖 Explanation

\begin{aligned} \text{let,} L_{40}&=y \\ \text{Now,} \frac{100-0}{100-50}&=\frac{100-40}{y-50} \\ \therefore 2y-100 &=60 \\ \therefore 2y &=160 \\ \therefore y &=80\: \text{dB} \\ \therefore L_{40} &=80\: \text{dB} \end{aligned}

Q48GATE 2006MCQ

The mean indoor airborne Chloroform ( $CHCI_3$ ) concentration in a room was determine to be 0.4 $\mu g/m^{3}$ . Use the following data : T = 293 K, P = 1 atmosphere, R = $82.05\times 10^{-6} atm.m^{3}$ /mol-K, Atomic weight : C=12, H=1, Cl=35.5. This concentration expressed in parts per billion (volume basis, ppbv) is equal to

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📖 Explanation

The relation between $\mu g/m^{3}$ and ppm is, $1\mu g/m^{3}=\frac{1\: ppm \times \text{ molecular mass of pollutant \in g/mol } }{\text{L/mol of pollutant at given temp. and pressure} \left ( V_{2} \right )} \times 10^3 \; L/m^3$ By Avogadro's law, we known that,

\begin{aligned} \frac{P_{1}V_{1}}{T_{1}}&= \frac{P_{2}V_{2}}{T_{2}}\\ \text{where }\; P_{1} &=1\text{atmosphere;} \\ T_{1} &=0^{\circ}C=273 \text{K} \\ V_{1} &=\text{volume at }0^{\circ}C \text{at 1 atmosphere} \\ &=22.4\: L/mol \\ \text{Given}\: P_{2} &=1\: \text{atmosphere}; \\ T_{2} &=293\text{K} \\ \therefore \; \; V_{2} &=\frac{P_{1}V_{1}T_{2}}{P_{2}T_{1}} \\ V_{2}&=\frac{22.4\times 293}{273} \\ V_{2} &=24.04\; L/mol \\ \because \; \; \; 0.4\; \mu \mu g/m^{3}\: \text{of}CHCl_{3} &=\frac{CHCl_{3}\; \in\; ppm\times 119.5\times 10^{3}}{24.04} \end{aligned}

$\Rightarrow \; \; \; CHCl_{3}$ in ppm $=8.0469\times 10^{-5}$ $CHCl_{3}$ in ppbv $=8.0469\times 10^{-5}\times 10^{3} =0.08\; ppbv$