MikaStars39/nano-eval huggingface dataset:
An infinitely long straight line is uniformly charged, with a charge per unit length of λ. Under the influence of its electric field, a particle with mass $m$ and charge $q$ performs uniform circular motion around it as its axis. What are the particle's speed $\upsilon$ and period $T$?

Options:
A. $$
{\sqrt{\frac{q m} {2 \pi\varepsilon_{0} \lambda}}},~2 \pi r \sqrt{\frac{2 \varepsilon_{0} q} {-m \lambda}}
$$
B. $$
{\sqrt{\frac{\varepsilon_{0} \lambda} {2 \pi q m}}},~2 r \pi \sqrt{\frac{-2 m \varepsilon_{0}} {q \lambda}}
$$
C. $$
{\sqrt{\frac{\lambda q} {2 \pi m \varepsilon_{0}}}},~2 \pi r \sqrt{\frac{-2 \pi m} {q \lambda \varepsilon_{0}}}
$$
D. $$
{\sqrt{\frac{-q \lambda} {2 \varepsilon_{0} \pi m}}},~2 r \pi \sqrt{\frac{\varepsilon_{0} m} {-q 2 \lambda}}
$$
E. $$
{\sqrt{\frac{-q \lambda} {4 \varepsilon_{0} \pi m}}},~2 r \pi \sqrt{\frac{\varepsilon_{0} m} {-2 q \lambda}}
$$
F. $$
{\sqrt{\frac{-\pi q} {2 \varepsilon_{0} m \lambda}}},~r \pi \sqrt{\frac{2 \pi m} {-q \lambda}}
$$
G. $$
{\sqrt{\frac{q \lambda} {2 \pi\varepsilon_{0} m}}},~2 \pi r \sqrt{\frac{-2 \pi\varepsilon_{0} m} {q \lambda}}
$$
H. $$
{\sqrt{\frac{-q \pi} {2 \lambda\varepsilon_{0} m}}},~r \sqrt{\frac{4 \pi\varepsilon_{0} m} {-q \lambda}}
$$
I. $$
{\sqrt{\frac{-q \lambda} {2 \pi\varepsilon_{0} m}}},~2 \pi r \sqrt{\frac{2 \pi\varepsilon_{0} m} {-q \lambda}}
$$
J. $$
{\sqrt{\frac{-\lambda q} {4 \pi\varepsilon_{0} m}}},~4 \pi r \sqrt{\frac{2 \varepsilon_{0} m} {q \lambda}}
$$
Answer: I
A metal pipe with an outer diameter of 75 mm and an inner diameter of 55 mm is used to transport a certain hot fluid. Known data: the inner wall temperature of the metal pipe is 120°C, the outer wall temperature is 115°C, and the heat dissipation rate per meter of pipe length is 4545 W/m. To reduce heat loss, an additional layer of asbestos is applied, with a thermal conductivity of 0.15 W/m·°C, and the outer wall temperature of the asbestos layer is 10°C, while the heat dissipation rate per meter of pipe length is reduced to 3.87% of the original. Determine the thickness of the asbestos layer and the temperature at the contact surface between the metal pipe and the asbestos layer is ().

Options:
A. 170.10mm,115.6°C
B. 167.88mm,119.8°C
C. 164.45mm,116.2°C
D. 167.50mm,120.5°C
E. 166.34mm,121.1°C
F. 166.98mm,122.7°C
G. 168.77mm,123.4°C
H. 163.80mm,124.8°C
I. 169.22mm,118.3°C
J. 165.90mm,117.9°C
Answer: B
Calculate the free settling velocity of spherical quartz particles with a density of $2650 \, \text{kg/m}^{3}$ and a diameter of $50 \, \mu\text{m}$ in air at 20°C under standard atmospheric pressure is ().

Options:
A. 0.3m/s
B. 1.5m/s
C. 0.8m/s
D. 0.9m/s
E. 0.2m/s
F. 1.1m/s
G. 0.6m/s
H. 0.5m/s
I. 0.4m/s
J. 0.7m/s
Answer: E
A certain fluid undergoes turbulent flow in a smooth straight circular pipe. If the pipe length and diameter remain unchanged while the flow rate doubles, the friction factor $\lambda$ in both cases can be calculated using the Blasius formula. What is the multiple of the original energy loss due to flow resistance?

Options:
A. 5.684 times
B. 1.562 times
C. 2.112 times
D. 3.364 times
E. 5.239 times
F. 2.901 times
G. 3.976 times
H. 6.101 times
I. 4.432 times
J. 4.789 times
Answer: D
The total airflow is 20000 $ft^{3}/sec$ (0.072 $lbf/ft^{3}$) passing through two propellers with a diameter of 6.5 ft, which are installed on an aircraft flying at 150 $mile/hr$ through still air. Find the total thrust.

Options:
A. $$
=4 1 9 0 \mathrm{~ l b f ~}
$$
B. $$
=9 4 5 0 \mathrm{~ l b f ~}
$$
C. $$
=8 3 0 0 \mathrm{~ l b f ~}
$$
D. $$
=1 0 5 6 0 \mathrm{~ l b f ~}
$$
E. $$
=7 2 8 0 \mathrm{~ l b f ~}
$$
F. $$
=2 3 7 0 \mathrm{~ l b f ~}
$$
G. $$
=8 5 2 0 \mathrm{~ l b f ~}
$$
H. $$
=9 9 9 0 \mathrm{~ l b f ~}
$$
I. $$
=5 6 7 0 \mathrm{~ l b f ~}
$$
J. $$
=6 2 4 0 \mathrm{~ l b f ~}
$$
Answer: E
There is a rod subjected to a tensile load, with a known load $F ( \mu_{F}, \ \sigma_{F} ) = F ( 80,000, \ 1,200 ) \ \mathrm{N}$. The diameter of the rod is $d ( \mu_{d}, \ \sigma_{d} ) = d ( 40, \ 0.8 ) \ \operatorname{mm}$, the length of the rod $l = l ( \mu_{l}, \ \sigma_{l} ) = l ( 6,000, \ 60 ) \ \mathrm{mm}$, and the elastic modulus of the material is $E ( \mu_{E}, \ \sigma_{E} ) = E ( 21 \times 10^{4}, \ 3,150 ) \ \mathrm{N/mm}^{2}$. Find the elongation $\delta$ of the rod within the range of elastic deformation.

Options:
A. $$
( 2. 3 0 \pm\! 0. 2 3 1 0 ) \operatorname{m m}
$$
B. $$
( 1. 8 8 \pm\! 0. 2 3 3 1 ) \operatorname{m m}
$$
C. $$
( 1. 6 5 \pm\! 0. 2 5 9 2 ) \operatorname{m m}
$$
D. $$
( 1. 7 9 \pm\! 0. 2 4 7 3 ) \operatorname{m m}
$$
E. $$
( 1. 8 3 \pm\! 0. 2 5 4 4 ) \operatorname{m m}
$$
F. $$
( 2. 0 1 \pm\! 0. 2 7 3 5 ) \operatorname{m m}
$$
G. $$
( 1. 9 2 \pm\! 0. 2 6 8 1 ) \operatorname{m m}
$$
H. $$
( 2. 5 0 \pm\! 0. 2 3 9 9 ) \operatorname{m m}
$$
I. $$
( 1. 5 0 \pm\! 0. 2 5 6 7 ) \operatorname{m m}
$$
J. $$
( 2. 7 5 \pm\! 0. 3 5 7 8 ) \operatorname{m m}
$$
Answer: E
The parameters of the turbine at Jiugongshan Power Station are as follows: flow rate $Q = 8 \, \mathrm{m}^{3}/\mathrm{s}$, generator output $P_{\mathrm{o}} = 12500 \, \mathrm{kW}$, measured pressure at the snail casing inlet $\frac{p_{1}}{\gamma} = 165 \, \mathrm{m}$, snail casing inlet diameter $d_{1} = 1.1 \, \mathrm{m}$, velocity at the draft tube exit $\upsilon_{5} = 1.5 \, \mathrm{m/s}$, and the height from the center of the snail casing inlet cross-section to the tail water level is $z_{1} = 5.5 \, \mathrm{m}$. Determine the working head of the turbine.

Options:
A. $$
1 8 5. 5 \mathrm{m}
$$
B. $$
1 7 4. 2 \mathrm{m}
$$
C. $$
1 6 2. 0 \mathrm{m}
$$
D. $$
1 8 8. 7 \mathrm{m}
$$
E. $$
1 5 7. 4 \mathrm{m}
$$
F. $$
1 6 5. 8 \mathrm{m}
$$
G. $$
1 6 9. 3 \mathrm{m}
$$
H. $$
1 8 2. 6 \mathrm{m}
$$
I. $$
1 9 0. 8 \mathrm{m}
$$
J. $$
1 7 1. 9 \mathrm{m}
$$
Answer: B
To ensure that the confidence interval of the measurement at a 95% confidence level does not exceed $\bar{x} \pm S$, how many parallel determinations should be conducted at minimum?

Options:
A. 20
B. 12
C. 3
D. 8
E. 10
F. 9
G. 15
H. 5
I. 6
J. 7
Answer: J
Determination of Nitrogen Content in a Certain Drug. Weigh a sample of 0.6464g, dissolve it and transfer it to a 100ml volumetric flask to make it up to volume. Take 25.00ml, add an excess amount of NaOH solution, and introduce the generated NH₃ into 40.00ml of 0.05100mol/L H₂SO₄ standard solution. Titrate the remaining $\mathbf{H}_{2} \mathbf{S O}_{4}$ with 0.09600mol/L NaOH, consuming 17.00ml to the endpoint. What is the nitrogen content in the calculated sample?

Options:
A. 19.84%
B. 24.65%
C. 20.11%
D. 21.22%
E. 23.05%
F. 18.29%
G. 22.93%
H. 20.89%
I. 19.57%
J. 22.47%
Answer: D
Which of the following statements about the value of $\alpha_{\mathrm{Y ( H )}}$ is correct ( ).

Options:
A. The $\alpha_{\mathrm{Y ( H )}}$ value is unaffected by temperature variations in the solution.
B. The $\alpha_{\mathrm{Y ( H )}}$ value is inversely related to the dilution of the solvent.
C. The $\alpha_{\mathrm{Y ( H )}}$ value is constant across all pH levels in the solution.
D. The $\alpha_{\mathrm{Y ( H )}}$ value decreases with the decreasing pressure of the gas.
E. The $\alpha_{\mathrm{Y ( H )}}$ value remains constant regardless of the solution's acidity.
F. The $\alpha_{\mathrm{Y ( H )}}$ value decreases with the rising salinity of the mixture.
G. The $\alpha_{\mathrm{Y ( H )}}$ value increases with the rising acidity of the solution.
H. The $\alpha_{\mathrm{Y ( H )}}$ value is independent of any chemical reaction in progress.
I. The $\alpha_{\mathrm{Y ( H )}}$ value is solely dependent on the molecular weight of solute.
J. The $\alpha_{\mathrm{Y ( H )}}$ value decreases with the decreasing acidity of the solvent.
Answer: G
Weigh 0.1173 g of NaCl standard reagent, dissolve it and add 30.00 ml of AgNO₃ standard solution. The excess Ag⁺ requires 3.20 ml of NH₄SCN standard solution for titration to the endpoint. It is known that 20.00 ml of AgNO₃ standard solution completely reacts with 21.00 ml of NH₄SCN standard solution. What is the concentrations of the AgNO₃ and NH₄SCN solutions?

Options:
A. $$C_{\mathrm{AgNO}_{3}}=0.1100(\mathrm{~mol/L})\\C_{\mathrm{NH_{4}SCN}} = 0. 1150 ( \mathrm{mol/L})$$
B. $$C_{\mathrm{AgNO}_{3}}=0.1080(\mathrm{~mol/L})\\C_{\mathrm{NH_{4}SCN}} = 0. 1130 ( \mathrm{mol/L})$$
C. $$C_{\mathrm{AgNO}_{3}}=0.1125(\mathrm{~mol/L})\\C_{\mathrm{NH_{4}SCN}} = 0. 1071 ( \mathrm{mol/L})$$
D. $$C_{\mathrm{AgNO}_{3}}=0.1050(\mathrm{~mol/L})\\C_{\mathrm{NH_{4}SCN}} = 0. 1120 ( \mathrm{mol/L})$$
E. $$C_{\mathrm{AgNO}_{3}}=0.1150(\mathrm{~mol/L})\\C_{\mathrm{NH_{4}SCN}} = 0. 1100 ( \mathrm{mol/L})$$
F. $$C_{\mathrm{AgNO}_{3}}=0.1200(\mathrm{~mol/L})\\C_{\mathrm{NH_{4}SCN}} = 0. 1000 ( \mathrm{mol/L})$$
G. $$C_{\mathrm{AgNO}_{3}}=0.1180(\mathrm{~mol/L})\\C_{\mathrm{NH_{4}SCN}} = 0. 1090 ( \mathrm{mol/L})$$
H. $$C_{\mathrm{AgNO}_{3}}=0.1000(\mathrm{~mol/L})\\C_{\mathrm{NH_{4}SCN}} = 0. 1200 ( \mathrm{mol/L})$$
I. $$C_{\mathrm{AgNO}_{3}}=0.1300(\mathrm{~mol/L})\\C_{\mathrm{NH_{4}SCN}} = 0. 1080 ( \mathrm{mol/L})$$
J. $$C_{\mathrm{AgNO}_{3}}=0.1250(\mathrm{~mol/L})\\C_{\mathrm{NH_{4}SCN}} = 0. 1050 ( \mathrm{mol/L})$$
Answer: C
Analysis of the sample of mirabilite $\mathrm{Na_{2} SO_{4} \cdot10 H_{2} O}$ estimates its purity to be approximately 90\%. How many grams of the sample should be weighed for each portion when using the $\mathrm{BaSO_{4}}$ gravimetric method?

Options:
A. $0.6g$
B. $0.47g$
C. $1.5g$
D. $1.0g$
E. $1.1g$
F. $1.8g$
G. $1.2g$
H. $0.8g$
I. $1.3g$
J. $0.5g$
Answer: H
The absorption peak of the complex $\mathbf{M} \mathbf{R}_{2}$ is at 480 nm. When the ligand is in more than fivefold excess, the absorbance is only related to the total concentration of metal ions and follows Beer's Law. Neither the metal ions nor the ligand absorb at 480 nm. Given a solution containing $M^{2+}$ at 0.000230 mol/L and $R$ at 0.00860 mol/L, its absorbance at 480 nm, measured with a 1 cm cuvette, is 0.690. Another solution containing $M^{2+}$ at 0.000230 mol/L and $R$ at 0.000500 mol/L has an absorbance of 0.540 under the same conditions. What is the stability constant of the complex?

Options:
A. $$
2. 0 5 \times1 0^{5}
$$
B. $$
2. 9 9 \times1 0^{6}
$$
C. $$
6. 1 2 \times1 0^{5}
$$
D. $$
1. 4 3 \times1 0^{3}
$$
E. $$
5. 5 1 \times1 0^{7}
$$
F. $$
3. 9 8 \times1 0^{4}
$$
G. $$
1. 2 6 \times1 0^{6}
$$
H. $$
4. 3 1 \times1 0^{8}
$$
I. $$
1. 8 4 \times1 0^{8}
$$
J. $$
8. 7 3 \times1 0^{9}
$$
Answer: I
The theoretical plate number of a certain liquid chromatography column is 4000. When separating a certain chiral drug, the retention times of the two enantiomers are 8.7 and 9.5 minutes, respectively. What is the resolution between them?

Options:
A. 1.24
B. 0.98
C. 0.53
D. 1.73
E. 1.09
F. 2.05
G. 0.78
H. 1.58
I. 1.39
J. 1.84
Answer: I
To determine the content of drug A in a sample using gas chromatography, B was used as the internal standard. An accurate weight of 5.456g of the sample and 0.2537g of the internal standard were taken, mixed thoroughly, and injected into the system. The peak areas for the drug and the internal standard were measured as 156mm$^{2}$ and 143mm$^{2}$, respectively. Separately, 0.2941g of standard A and 0.2673g of standard B were accurately weighed, diluted to a certain volume, mixed thoroughly, and analyzed under the same conditions as the sample. The measured peak areas were 54.5mm$^{2}$ and 46.6mm$^{2}$, respectively. What is the percentage content of A in the sample?

Options:
A. $$
5. 3 1 \%
$$
B. $$
7. 1 8 \%
$$
C. $$
9. 4 3 \%
$$
D. $$
6. 0 9 \%
$$
E. $$
4. 7 7 \%
$$
F. $$
2. 5 9 \%
$$
G. $$
5. 8 2 \%
$$
H. $$
3. 6 4 \%
$$
I. $$
8. 2 0 \%
$$
J. $$
6. 9 5 \%
$$
Answer: E
A certain drug was analyzed using high-performance capillary electrophoresis, and the measured migration time was 5.76 minutes. It is known that the total length of the capillary is $64 \, \mathrm{cm}$, the effective capillary length $L_{\mathrm{d}}$ is 55 cm, and the applied voltage is 25 kV. The electroosmotic mobility of this system is $3.42 \times 10^{-4}$. What is the electrophoretic mobility of this drug?

Options:
A. $\mu_{ap}=4. 0 7 \times1 0^{-4} \ ( \mathrm{c m}^{2} / V \cdot s )$
$\mu_{\mathrm{e p}}=\, 6. \, 5 \, \times\, 1 0^{-5} \, ( \, \mathrm{c m}^{2} / V \cdot s )$
B. $\mu_{ap}=9. 1 7 \times1 0^{-4} \ ( \mathrm{c m}^{2} / V \cdot s )$
$\mu_{\mathrm{e p}}=\, 5. \, 5 \, \times\, 1 0^{-5} \, ( \, \mathrm{c m}^{2} / V \cdot s )$
C. $\mu_{ap}=2. 2 7 \times1 0^{-4} \ ( \mathrm{c m}^{2} / V \cdot s )$
$\mu_{\mathrm{e p}}=\, 1. \, 5 \, \times\, 1 0^{-5} \, ( \, \mathrm{c m}^{2} / V \cdot s )$
D. $\mu_{ap}=7. 0 7 \times1 0^{-4} \ ( \mathrm{c m}^{2} / V \cdot s )$
$\mu_{\mathrm{e p}}=\, 8. \, 5 \, \times\, 1 0^{-5} \, ( \, \mathrm{c m}^{2} / V \cdot s )$
E. $\mu_{ap}=4. 0 6 \times2 0^{-4} \ ( \mathrm{c m}^{2} / V \cdot s )$
$\mu_{\mathrm{e p}}=\, 7. \, 0 \, \times\, 1 0^{-4} \, ( \, \mathrm{c m}^{2} / V \cdot s )$
F. $\mu_{ap}=6. 5 3 \times1 0^{-4} \ ( \mathrm{c m}^{2} / V \cdot s )$
$\mu_{\mathrm{e p}}=\, 3. \, 5 \, \times\, 1 0^{-5} \, ( \, \mathrm{c m}^{2} / V \cdot s )$
G. $\mu_{ap}=3. 8 7 \times1 0^{-4} \ ( \mathrm{c m}^{2} / V \cdot s )$
$\mu_{\mathrm{e p}}=\, 4. \, 5 \, \times\, 1 0^{-5} \, ( \, \mathrm{c m}^{2} / V \cdot s )$
H. $\mu_{ap}=5. 4 7 \times1 0^{-4} \ ( \mathrm{c m}^{2} / V \cdot s )$
$\mu_{\mathrm{e p}}=\, 8. \, 2 \, \times\, 1 0^{-5} \, ( \, \mathrm{c m}^{2} / V \cdot s )$
I. $\mu_{ap}=5. 9 7 \times1 0^{-4} \ ( \mathrm{c m}^{2} / V \cdot s )$
$\mu_{\mathrm{e p}}=\, 9. \, 3 \, \times\, 1 0^{-5} \, ( \, \mathrm{c m}^{2} / V \cdot s )$
J. $\mu_{ap}=4. 0 6 \times1 0^{-4} \ ( \mathrm{c m}^{2} / V \cdot s )$
$\mu_{\mathrm{e p}}=\, 5. \, 8 \, \times\, 1 0^{-5} \, ( \, \mathrm{c m}^{2} / V \cdot s )$
Answer: A
U - tube manometer containing mercury was used to find the negative pressure in the pipe, containing water. The right limb was open to the atmosphere. Find the vacuum pressure in the pipe, if the difference of mercury level in the two limbs was 100 mm and height of water in the left limb was 40 mm below is ().

Options:
A. -12.45kPa
B. -15.25kPa
C. -11.80kPa
D. -15.00kPa
E. -14.50kPa
F. -13.90kPa
G. -13.50kPa
H. -12.60kPa
I. -13.73kPa
J. -14.00kPa
Answer: I
If the variation range of the flow rate is 3 times, what is the range of head variation when measuring the flow using a rectangular weir?

Options:
A. 1.60
B. 1.75
C. 1.40
D. 1.65
E. 1.45
F. 1.35
G. 1.80
H. 1.70
I. 1.50
J. 1.55
Answer: J
Circular non-pressure concrete pipe, with a bed slope of $i=0.003$, diameter $d=2 \ \mathrm{m}$, and roughness coefficient $n=0.014$. When the water depth is $h_{0}=1.0 \ \mathrm{m}$, try to calculate the average flow velocity and the flow rate through the section of the channel is ().

Options:
A. $2.0 5 {m / s}$,$8.20m^{3}/s$
B. $2.4 6 {m / s}$,$7.72m^{3}/s$
C. $1.8 7 {m / s}$,$5.30m^{3}/s$
D. $3.1 2 {m / s}$,$8.45m^{3}/s$
E. $2.6 2 {m / s}$,$8.90m^{3}/s$
F. $2.8 9 {m / s}$,$6.50m^{3}/s$
G. $3.5 8 {m / s}$,$9.60m^{3}/s$
H. $2.1 4 {m / s}$,$7.90m^{3}/s$
I. $2.3 1 {m / s}$,$7.00m^{3}/s$
J. $3.0 4 {m / s}$,$7.50m^{3}/s$
Answer: B
There is a rectangular cross-section channel made of masonry block stones. Given the channel length $l = 500 \mathrm{m}$, bottom width $b = 3.2 \mathrm{m}$, water depth in the channel $h_0 = 1.6 \mathrm{m}$, roughness coefficient $n = 0.025$, and discharge (or volumetric flow rate) through $q_{\mathrm{V}} = 6 \mathrm{m}^3 / \mathrm{s}$, try to determine the friction head loss along the channel length is ().

Options:
A. 1.245m
B. 0.489m
C. 1.098m
D. 0.734m
E. 0.578m
F. 0.812m
G. 0.923m
H. 1.032m
I. 0.675m
J. 0.359m
Answer: E
To determine the resistance of the full-sized ship, a towing experiment is conducted in the towing tank using a ship model made to a length scale ratio of $\lambda_{l}=30$. If the wave-making resistance in the model experiment is $F_{\mathrm{m}}=1.47 \mathrm{N}$, what is the corresponding wave-making resistance $F_{\mathrm{p}}$ of the actual ship?

Options:
A. 100.12N
B. 61.73N
C. 51.89N
D. 82.45N
E. 15.63N
F. 39.69kN
G. 72.11N
H. 94.37N
I. 43.22N
J. 29.47N
Answer: F
The kinematic viscosity of the oil is $0.43 \mathrm{cm}^{2} / \mathrm{s}$, which is used in flows where both viscous resistance and gravity are active. Given that the geometric scale ratio of the model is $\lambda_{1}=5$, find the required kinematic viscosity for the model liquid is ().

Options:
A. $$
v_{\mathrm{m}}=0. 0 3 2 4 \mathrm{c m}^{2} / \mathrm{s}
$$
B. $$
v_{\mathrm{m}}=0. 0 3 8 5 \mathrm{c m}^{2} / \mathrm{s}
$$
C. $$
v_{\mathrm{m}}=0. 0 5 1 2 \mathrm{c m}^{2} / \mathrm{s}
$$
D. $$
v_{\mathrm{m}}=0. 0 2 9 6 \mathrm{c m}^{2} / \mathrm{s}
$$
E. $$
v_{\mathrm{m}}=0. 0 3 5 7 \mathrm{c m}^{2} / \mathrm{s}
$$
F. $$
v_{\mathrm{m}}=0. 0 4 2 1 \mathrm{c m}^{2} / \mathrm{s}
$$
G. $$
v_{\mathrm{m}}=0. 0 3 9 2 \mathrm{c m}^{2} / \mathrm{s}
$$
H. $$
v_{\mathrm{m}}=0. 0 4 0 1 \mathrm{c m}^{2} / \mathrm{s}
$$
I. $$
v_{\mathrm{m}}=0. 0 4 3 8 \mathrm{c m}^{2} / \mathrm{s}
$$
J. $$
v_{\mathrm{m}}=0. 0 2 8 8 \mathrm{c m}^{2} / \mathrm{s}
$$
Answer: B
There is a circular pipe through which water is flowing at $\nu=0.013 \text{ cm}^{2}/\text{s}$. The measured flow rate is $40 \text{ cm}^{3}/\text{s}$. Over a pipe section 25m long, the head loss is measured to be $2 \text{ cm}$. Try to determine the inner diameter $d$ of the circular pipe is ().

Options:
A. 1.95cm
B. 1.89cm
C. 2.75cm
D. 3.00cm
E. 2.05cm
F. 2.67cm
G. 2.28cm
H. 2.33cm
I. 2.50cm
J. 3.15cm
Answer: G
Arrange two wells, 1 and 2, along the groundwater seepage direction, spaced $500 \mathrm{m}$ apart. The measured water surface elevation of well 1 is 13m, with a bottom elevation of 10m; well 2 has a water surface elevation of 12.75m and a bottom elevation of 8.75m. The soil seepage coefficient is $k = 0.05 \mathrm{cm/s}$. Calculate the unit width seepage flow rate of the groundwater is ().

Options:
A. $7. 4 5 \times1 0^{-7} {\mathrm{m}}^{3} / ( {\mathrm{s}} \cdot{\mathrm{m}} )$
B. $9. 3 2 \times1 0^{-7} {\mathrm{m}}^{3} / ( {\mathrm{s}} \cdot{\mathrm{m}} )$
C. $6. 5 8 \times1 0^{-7} {\mathrm{m}}^{3} / ( {\mathrm{s}} \cdot{\mathrm{m}} )$
D. $6. 3 7 \times1 0^{-7} {\mathrm{m}}^{3} / ( {\mathrm{s}} \cdot{\mathrm{m}} )$
E. $8. 6 7 \times1 0^{-7} {\mathrm{m}}^{3} / ( {\mathrm{s}} \cdot{\mathrm{m}} )$
F. $9. 0 2 \times1 0^{-7} {\mathrm{m}}^{3} / ( {\mathrm{s}} \cdot{\mathrm{m}} )$
G. $8. 6 9 \times1 0^{-7} {\mathrm{m}}^{3} / ( {\mathrm{s}} \cdot{\mathrm{m}} )$
H. $8. 9 1 \times1 0^{-7} {\mathrm{m}}^{3} / ( {\mathrm{s}} \cdot{\mathrm{m}} )$
I. $7. 8 5 \times1 0^{-7} {\mathrm{m}}^{3} / ( {\mathrm{s}} \cdot{\mathrm{m}} )$
J. $7. 9 4 \times1 0^{-7} {\mathrm{m}}^{3} / ( {\mathrm{s}} \cdot{\mathrm{m}} )$
Answer: G
Assume that an antenna radiates conically only within the range of $\theta = 0^\circ$ to $45^\circ$ with the z-axis, and radiates uniformly. Find the maximum directivity coefficient of the antenna.

Options:
A. $$
5. 9 3 ( \mathrm{d B} )
$$
B. $$
6. 5 8 ( \mathrm{d B} )
$$
C. $$
6. 3 4 ( \mathrm{d B} )
$$
D. $$
9. 0 2 ( \mathrm{d B} )
$$
E. $$
7. 8 7 ( \mathrm{d B} )
$$
F. $$
5. 4 9 ( \mathrm{d B} )
$$
G. $$
7. 5 0 ( \mathrm{d B} )
$$
H. $$
8. 3 4 ( \mathrm{d B} )
$$
I. $$
6. 7 1 ( \mathrm{d B} )
$$
J. $$
7. 2 5 ( \mathrm{d B} )
$$
Answer: H
The characteristic impedance of a uniform, lossless transmission line is $Z_{0} = 100 \Omega$. The first voltage wave node measured along the transmission line, moving away from the terminal towards the power source, is at $z_{\operatorname*{minl}}^{\prime} = 2.5 \mathrm{cm}$. The voltage at the wave node is $|U|_{\operatorname*{min}} = 1 \mathrm{V}$, and the voltage standing wave ratio (VSWR) is $\rho = 2$. The phase wavelength is $\lambda_{\mathrm{p}} = 20 \mathrm{cm}$. Find: the load impedance $Z_{L}$.

Options:
A. $$
1 0 5-\mathrm{j}6 5 ( \Omega)
$$
B. $$
1 0 0+\mathrm{j}5 0 ( \Omega)
$$
C. $$
7 5-\mathrm{j}5 0 ( \Omega)
$$
D. $$
9 0+\mathrm{j}4 0 ( \Omega)
$$
E. $$
8 0-\mathrm{j} 6 0 ( \Omega)
$$
F. $$
5 0-\mathrm{j}8 0 ( \Omega)
$$
G. $$
1 2 0+\mathrm{j}2 5 ( \Omega)
$$
H. $$
6 0+\mathrm{j}7 0 ( \Omega)
$$
I. $$
8 5-\mathrm{j}3 5 ( \Omega)
$$
J. $$
7 0+\mathrm{j}3 0 ( \Omega)
$$
Answer: E
Suppose that two stations $1.5 km$ apart are communicating with each other. Each station uses a half-wave oscillator as an antenna and operates at a frequency of $300 MHz$. If the power transmitted by one station is $100 W$, what is the maximum receiving power of the other station?

Options:
A. $$
1. 2 5 \mu{\bf W}
$$
B. $$
0. 5 4 \mu{\bf W}
$$
C. $$
0. 7 6 \mu{\bf W}
$$
D. $$
1. 1 8 \mu{\bf W}
$$
E. $$
0. 6 5 \mu{\bf W}
$$
F. $$
0. 8 1 \mu{\bf W}
$$
G. $$
0. 7 9 \mu{\bf W}
$$
H. $$
1. 0 2 \mu{\bf W}
$$
I. $$
0. 9 0 \mu{\bf W}
$$
J. $$
1. 0 0 \mu{\bf W}
$$
Answer: C
It is known that the directional coefficient of the current element is $D=1.76 \mathrm{dB}$, and that of the half-wave dipole is $D=2.15 \mathrm{dB}$. If the electric fields of the two are equal at equal distances in the maximum direction, calculate the ratio of their radiation powers.

Options:
A. 1.0933
B. 1.0759
C. 1.0899
D. 1.1264
E. 1.1125
F. 1.0815
G. 1.0628
H. 1.0872
I. 1.0967
J. 1.1050
Answer: A
What is the radiation resistance of an electric fundamental oscillator of length $\frac{\lambda_{0}} {10}$ ($\lambda_{0}$ is the free space wavelength)?

Options:
A. $$
8. 3 \Omega
$$
B. $$
2. 9 \Omega
$$
C. $$
0. 7 \Omega
$$
D. $$
3. 4 \Omega
$$
E. $$
9. 1 \Omega
$$
F. $$
6. 5 \Omega
$$
G. $$
4. 2 \Omega
$$
H. $$
7. 9 \Omega
$$
I. $$
5. 6 \Omega
$$
J. $$
1. 8 \Omega
$$
Answer: H
There is a microstrip line with characteristic impedance $Z_{o} = 50 \, \Omega$ , the terminal is connected to a matching load, and the other end is connected to a device with impedance of 6.4n through a 1/4 wavelength impedance transformer. Given that the working wavelength $\lambda = 10 \, \mathrm{cm}$, the thickness of the microstrip substrate $h = 0.8 \, \mathrm{mm}$, and the relative dielectric constant of the substrate medium $\varepsilon_{r} = 9$. Calculate the width $W$ and the length $l_{o}$ of the microstrip forming the transformer is____.

Options:
A. $$3.5(mm),9.5(mm)$$
B. $$3.9(mm),9.4(mm)$$
C. $$4.4(mm),8.7(mm)$$
D. $$3.8(mm),8.6(mm)$$
E. $$4.1(mm),8.8(mm)$$
F. $$4.3(mm),9.3(mm)$$
G. $$4.2(mm),9.0(mm)$$
H. $$3.7(mm),8.9(mm)$$
I. $$3.6(mm),9.1(mm)$$
J. $$4.0(mm),9.2(mm)$$
Answer: F
The $S$ parameters of a certain microwave transistor at $10GHz$ relative to a $50\Omega$ reference impedance are given by:
$$
\mathbf{S} {=} \left[ \begin{matrix} {{0. \, 4 5 \, \mathrm{e}^{\mathrm{j} 1 5 0^{\mathrm{o}}}}} & {{0. \, 0 1 \mathrm{e}^{-\mathrm{j} 1 0^{\mathrm{o}}}}} \\ {{2. \, 0 5 \, \mathrm{e}^{\mathrm{j} 1 0^{\mathrm{o}}}}} & {{0. \, 4 \mathrm{e}^{-\mathrm{j} 1 5 0^{\mathrm{o}}}}} \\ \end{matrix} \right]
$$
The source impedance is $20\Omega$, and the load impedance is $30\Omega$. Calculate the available power gain $G_{\mathrm{A}}$, transducer power gain $G_{\mathrm{T}}$, and power gain $G$ .

Options:
A. $$
\approx8. 0 1
$$
$$
\approx7, 9
$$
$$
\approx8. 2 4
$$
B. $$
\approx2. 5 4
$$
$$
\approx2, 4
$$
$$
\approx2. 7 1
$$
C. $$
\approx7. 0 5
$$
$$
\approx6, 9
$$
$$
\approx7. 3 8
$$
D. $$
\approx7. 5 9
$$
$$
\approx7, 4
$$
$$
\approx7. 8 2
$$
E. $$
\approx6. 2 0
$$
$$
\approx6, 1
$$
$$
\approx6. 3 5
$$
F. $$
\approx3. 7 8
$$
$$
\approx3, 6
$$
$$
\approx3. 9 2
$$
G. $$
\approx5. 8 6
$$
$$
\approx5, 5
$$
$$
\approx5. 9 4
$$
H. $$
\approx4. 3 4
$$
$$
\approx4, 2
$$
$$
\approx4. 5 8
$$
I. $$
\approx9. 1 0
$$
$$
\approx8, 9
$$
$$
\approx9. 3 5
$$
J. $$
\approx1. 0 5
$$
$$
\approx0, 9
$$
$$
\approx1. 2 8
$$
Answer: G
Given the stripline dimensions $b = 2 \mathrm{mm}, t = 0.1 \mathrm{mm}, W = 1.4 \mathrm{mm}$, with a dielectric constant $\varepsilon_{r} = 2.1$, determine the characteristic impedance $Z_{0}$ of the stripline and the maximum signal frequency permitted for $TEM$ mode transmission.

Options:
A. $$
5 3. 0 ( \Omega)
$$
$$
5 9. 8 5 ( \mathrm{G H z} )
$$
B. $$
5 0. 7 ( \Omega)
$$
$$
5 0. 2 5 ( \mathrm{G H z} )
$$
C. $$
5 2. 4 ( \Omega)
$$
$$
5 6. 7 5 ( \mathrm{G H z} )
$$
D. $$
5 4. 2 ( \Omega)
$$
$$
5 8. 6 0 ( \mathrm{G H z} )
$$
E. $$
4 9. 6 ( \Omega)
$$
$$
6 1. 1 0 ( \mathrm{G H z} )
$$
F. $$
5 5. 3 ( \Omega)
$$
$$
6 0. 2 0 ( \mathrm{G H z} )
$$
G. $$
5 0. 3 ( \Omega)
$$
$$
5 7. 3 5 ( \mathrm{G H z} )
$$
H. $$
5 5. 8 ( \Omega)
$$
$$
5 4. 4 5 ( \mathrm{G H z} )
$$
I. $$
4 8. 9 ( \Omega)
$$
$$
5 5. 9 0 ( \mathrm{G H z} )
$$
J. $$
5 1. 1 ( \Omega)
$$
$$
5 1. 7 5 ( \mathrm{G H z} )
$$
Answer: J
The propagation constant of a transmission line is $\gamma=0.0091+\mathrm{j} 0.8 \, \text{km}^{-1}$, and the characteristic impedance is $Z_{0} = 500 e^{-j 12^{\circ}} \, \Omega$. Suppose the input signal voltage at the start of the transmission line is $u(z, t) \mid_{z=0} = 10 \sin(10^{5} t) \,\mathrm{V}$, and the terminal is connected to a matched load. Determine the voltage $E u (z, t)$ and current $i(z, t)$ along the line. If the transmission line length is $160 \,\mathrm{km}$, calculate the time required for the above signal to be transmitted from the starting end to the terminal.

Options:
A. 1.77(ms)
B. 2.34(ms)
C. 2.90(ms)
D. 1.50(ms)
E. 1.28(ms)
F. 3.18(ms)
G. 1.05(ms)
H. 0.75(ms)
I. 3.45(ms)
J. 0.92(ms)
Answer: E
Design a pyramid horn as a feed source, with an operating wavelength of $\lambda=3 \, \, \, \mathrm{cm}$. If the width between the two main planes hosting -10dB level is required to be 72°, the horn must be matched with the structure of the standard waveguide $a \times b = 2.3\, \mathrm{cm} \times 1.0\, \mathrm{cm}$, and the horn must be limited. The maximum phase deviation on the aperture does not exceed $\pi/8$. Find the diameter and length of the horn.

Options:
A. 224.8mm, 195.2mm
B. 233.0mm, 178.8mm
C. 250.1mm, 190.7mm
D. 222.7mm, 184.4mm
E. 218.4mm, 192.1mm
F. 211.9mm, 180.3mm
G. 215.3mm, 182.5mm
H. 230.5mm, 175.6mm
I. 205.6mm, 185.0mm
J. 240.0mm, 199.9mm
Answer: D
If the velocity distribution within a flat plate turbulent boundary layer is ${\frac{u} {V_{\infty}}}=\left( {\frac{y} {\delta}} \right)^{\frac{1} {9}}$, and there is $\lambda=0.185 ( R e_{\delta} )^{-0.2}$, where $V_{\infty}$ is the free stream velocity, $\delta$ is the boundary layer thickness, and $\lambda$ is the friction loss coefficient, and $R e_{\delta}={\frac{V_{\infty} \delta} {\nu}}$. What is the formula for calculating the boundary layer thickness?

Options:
A. $$
\delta=0. 2 3 9 9 x ( R e_{y} )^{-\frac{1} {5}}
$$
B. $$
\delta=0. 4 7 5 2 x ( R e_{z} )^{-\frac{1} {8}}
$$
C. $$
\delta=0. 6 0 2 8 x ( R e_{n} )^{-\frac{1} {3}}
$$
D. $$
\delta=0. 7 3 9 5 x ( R e_{b} )^{-\frac{1} {7}}
$$
E. $$
\delta=0. 3 8 6 9 x ( R e_{l} )^{-\frac{1} {9}}
$$
F. $$
\delta=0. 5 0 3 4 x ( R e_{x} )^{-\frac{1} {7}}
$$
G. $$
\delta=0. 4 4 2 3 x ( R e_{t} )^{-\frac{1} {4}}
$$
H. $$
\delta=0. 1 2 4 7 x ( R e_{m} )^{-\frac{1} {4}}
$$
I. $$
\delta=0. 4 0 6 1 x ( R e_{x} )^{-\frac{1} {6}}
$$
J. $$
\delta=0. 5 1 8 3 x ( R e_{k} )^{-\frac{1} {2}}
$$
Answer: I
What is the water flow rate $q_{V}$ when water at 15°C flows through a riveted steel pipe with a diameter of $d=300 \ \mathrm{mm}$, given that the absolute roughness $\Delta=3\ \mathrm{mm}$, and the head loss over a pipe length of $l=300\ \mathrm{m}$ is $h_{1}=6\ \mathrm{m}$?

Options:
A. $$
0. 1 0 9 7 {{\mathrm{m}}^{3}} / {\mathrm{s}}
$$
B. $$
0. 1 2 5 9 {{\mathrm{m}}^{3}} / {\mathrm{s}}
$$
C. $$
0. 1 1 9 8 {{\mathrm{m}}^{3}} / {\mathrm{s}}
$$
D. $$
0. 1 2 4 5 {{\mathrm{m}}^{3}} / {\mathrm{s}}
$$
E. $$
0. 1 0 3 6 {{\mathrm{m}}^{3}} / {\mathrm{s}}
$$
F. $$
0. 1 4 8 9 {{\mathrm{m}}^{3}} / {\mathrm{s}}
$$
G. $$
0. 1 5 2 4 {{\mathrm{m}}^{3}} / {\mathrm{s}}
$$
H. $$
0. 1 3 6 7 {{\mathrm{m}}^{3}} / {\mathrm{s}}
$$
I. $$
0. 1 6 1 4 {{\mathrm{m}}^{3}} / {\mathrm{s}}
$$
J. $$
0. 1 3 4 4 {{\mathrm{m}}^{3}} / {\mathrm{s}}
$$
Answer: D
The airflow with stagnation parameters $p_{0}=10.35 \times 10^{5} \, \mathrm{Pa}, \; \; T_{0}=350 \, \mathrm{K}$ enters a converging nozzle with an exit circular cross-section diameter of $d = 15 \, \mathrm{mm}$. What is the mass flow rate of the nozzle when the external back pressure $\rho_{b}$ is $7 \times 10^{5} \, \mathrm{Pa}$ and $5 \times 10^{5} \, \mathrm{Pa}$ respectively?

Options:
A. $$
0. 4 0 0 1 \mathrm{k g} / \mathrm{s}
$$
B. $$
0. 3 7 8 2 \mathrm{k g} / \mathrm{s}
$$
C. $$
0. 4 2 3 4 \mathrm{k g} / \mathrm{s}
$$
D. $$
0. 3 6 8 4 \mathrm{k g} / \mathrm{s}
$$
E. $$
0. 3 9 5 1 \mathrm{k g} / \mathrm{s}
$$
F. $$
0. 4 0 1 5 \mathrm{k g} / \mathrm{s}
$$
G. $$
0. 4 1 2 9 \mathrm{k g} / \mathrm{s}
$$
H. $$
0. 3 8 9 0 \mathrm{k g} / \mathrm{s}
$$
I. $$
0. 3 7 5 8 \mathrm{k g} / \mathrm{s}
$$
J. $$
0. 3 9 9 7 \mathrm{k g} / \mathrm{s}
$$
Answer: E
Randomly place 2 red balls and 2 white balls into two boxes, A and B, so that each box contains 2 balls. Each time, take one ball from each box, swap them, and then put them back. Let $X(0)$ denote the number of red balls in box A at the start. $X(n) (n \geq 1)$ represents the number of red balls in box A after $n$ swaps. { $X(n) \,, n=0 \,, 1 \,, 2 \,, \cdots$ } is a homogeneous Markov chain. Try to write out the state transition probability matrix is ().

Options:
A. $$
P=\left[ \begin{array} {c c c} {{{0}}} & {{{1}}} & {{{0}}} \\ {{{\frac{1} {4}}}} & {{{\frac{1} {2}}}} & {{{\frac{1} {4}}}} \\ {{{0}}} & {{{1}}} & {{{0}}} \\ \end{array} \right]
$$
B. $$
P=\left[ \begin{array} {c c c} {{{0}}} & {{{0}}} & {{{1}}} \\ {{{\frac{1} {2}}}} & {{{0}}} & {{{\frac{1} {2}}}} \\ {{{1}}} & {{{0}}} & {{{0}}} \\ \end{array} \right]
$$
C. $$
P=\left[ \begin{array} {c c c} {{{0}}} & {{{\frac{1} {4}}}} & {{{\frac{3} {4}}}} \\ {{{\frac{1} {4}}}} & {{{\frac{1} {2}}}} & {{{\frac{1} {4}}}} \\ {{{1}}} & {{{0}}} & {{{0}}} \\ \end{array} \right]
$$
D. $$
P=\left[ \begin{array} {c c c} {{{\frac{1} {2}}}} & {{{\frac{1} {2}}}} & {{{0}}} \\ {{{1}}} & {{{0}}} & {{{0}}} \\ {{{0}}} & {{{\frac{1} {2}}}} & {{{\frac{1} {2}}}} \\ \end{array} \right]
$$
E. $$
P=\left[ \begin{array} {c c c} {{{0}}} & {{{1}}} & {{{0}}} \\ {{{1}}} & {{{0}}} & {{{0}}} \\ {{{\frac{1} {2}}}} & {{{\frac{1} {2}}}} & {{{0}}} \\ \end{array} \right]
$$
F. $$
P=\left[ \begin{array} {c c c} {{{0}}} & {{{\frac{1} {2}}}} & {{{\frac{1} {2}}}} \\ {{{\frac{1} {4}}}} & {{{\frac{1} {2}}}} & {{{\frac{1} {4}}}} \\ {{{0}}} & {{{0}}} & {{{1}}} \\ \end{array} \right]
$$
G. $$
P=\left[ \begin{array} {c c c} {{{0}}} & {{{0}}} & {{{1}}} \\ {{{\frac{1} {4}}}} & {{{\frac{1} {4}}}} & {{{\frac{1} {2}}}} \\ {{{1}}} & {{{0}}} & {{{0}}} \\ \end{array} \right]
$$
H. $$
P=\left[ \begin{array} {c c c} {{{1}}} & {{{0}}} & {{{0}}} \\ {{{\frac{1} {3}}}} & {{{\frac{1} {3}}}} & {{{\frac{1} {3}}}} \\ {{{0}}} & {{{0}}} & {{{1}}} \\ \end{array} \right]
$$
I. $$
P=\left[ \begin{array} {c c c} {{{1}}} & {{{0}}} & {{{0}}} \\ {{{\frac{1} {2}}}} & {{{0}}} & {{{\frac{1} {2}}}} \\ {{{1}}} & {{{0}}} & {{{0}}} \\ \end{array} \right]
$$
J. $$
P=\left[ \begin{array} {c c c} {{{0}}} & {{{1}}} & {{{0}}} \\ {{{\frac{1} {3}}}} & {{{\frac{1} {3}}}} & {{{\frac{1} {3}}}} \\ {{{0}}} & {{{0}}} & {{{1}}} \\ \end{array} \right]
$$
Answer: A
If customers arrive according to a Poisson process at an average rate of 2 per minute, find the probability that a total of 3 customers arrive in the time intervals $[1, 3)$ and $[3, 5)$ is ().

Options:
A. $$
\frac{2 5 6} {3} \mathrm{e}^{-8}
$$
B. $$
\frac{2 5 7} {3} \mathrm{e}^{-8}
$$
C. $$
\frac{1 5 6} {3} \mathrm{e}^{-9}
$$
D. $$
\frac{1 4 9} {3} \mathrm{e}^{-6}
$$
E. $$
\frac{3 5 6} {3} \mathrm{e}^{-7}
$$
F. $$
\frac{2 5 6} {4} \mathrm{e}^{-9}
$$
G. $$
\frac{3 6 9} {4} \mathrm{e}^{-8}
$$
H. $$
\frac{2 5 6} {5} \mathrm{e}^{-8}
$$
I. $$
\frac{2 4 6} {3} \mathrm{e}^{-7}
$$
J. $$
\frac{2 7 6} {3} \mathrm{e}^{-8}
$$
Answer: A
Given that $X(t)$ is a zero-mean stationary Gaussian process with an autocorrelation function $R_{X}(\tau)$, and that $X(t)$ passes through a square-law detector resulting in an output process $Y(t) = X^2(t)$, find the mean $m_{z}$ and variance $\sigma_{Z}^{2}$ of $Y(t)$ is ().

Options:
A. $$
m_{Y}=E [ Y ( t ) ]=E [ X ( t ) Y ( t ) ]=R_{Y} ( 1 )
$$
$$
\sigma_{Y}^{2}=R_{Z} ( 1 )-m_{Z}^{2}=R_{X}^{2} ( 1 )
$$
B. $$
m_{Y}=E [ Y ( t ) ]=E [ X ( -t ) ]=0
$$
$$
\sigma_{Y}^{2}=R_{Y} ( 0 )-m_{Y}^{2}=0
$$
C. $$
m_{Y}=E [ Y ( t ) ]=E [ X^{3} ( t ) ]=0
$$
$$
\sigma_{Y}^{2}=R_{Y} ( 0 )-m_{Y}^{2}=R_{X} ( 0 )
$$
D. $$
m_{Y}=E [ Y ( t ) ]=E [ X^{2} ( t ) ]=0
$$
$$
\sigma_{Y}^{2}=R_{Y} ( 1 )-m_{Y}^{2}=2 R_{X} ( 1 )
$$
E. $$
m_{Y}=E [ Y ( t ) ]=E [ X^{2} ( t ) ]=R_{X} ( 1 )
$$
$$
\sigma_{Y}^{2}=R_{Y} ( 0 )-m_{Y}^{2}=R_{X}^{2} ( 0 )
$$
F. $$
m_{Y}=E [ Y ( t ) ]=E [ X^{2} ( 2t ) ]=R_{X} ( 0 )
$$
$$
\sigma_{Y}^{2}=R_{Y} ( 0 )-m_{Y}^{2}=R_{X}^{2} ( 0 )
$$
G. $$
m_{Y}=E [ Y ( t ) ]=\operatorname{Var}[ X ( t ) ]=1
$$
$$
\sigma_{Y}^{2}=R_{Y} ( 1 )-m_{Y}^{2}=3 R_{X} ( 0 )
$$
H. $$
m_{Y}=E [ Y ( t ) ]=2 E [ X^{2} ( t ) ]=2 R_{X} ( 0 )
$$
$$
\sigma_{Y}^{2}=R_{Y} ( 0 )-m_{Y}^{2}=4 R_{X}^{2} ( 0 )
$$
I. $$
m_{Y}=E [ Y ( t ) ]=E [ X^{2} ( t ) ]=R_{X} ( 0 )
$$
$$
\sigma_{Y}^{2}=R_{Y} ( 0 )-m_{Y}^{2}=2 R_{X}^{2} ( 0 )
$$
J. $$
m_{Y}=E [ Y ( t ) ]=0
$$
$$
\sigma_{Y}^{2}=R_{Y} ( 0 )-m_{Y}^{2}=2 R_{X} ( 0 )
$$
```
Answer: I
The voltage $X(t)$ across the diode is a stationary Gaussian process with zero mean and autocorrelation function $R_{X}(\tau) = c \mathrm{e}^{-a |\tau|}$. Find the mean of the diode current $Y(t) = I e^{\mathrm{ar}(t)}$ is ().

Options:
A. $$
\begin{aligned} {{m_{Y} ( t )}} & {{} {{}=E \langle I \mathrm{e}^{-\mathrm{e} X ( t )} \rangle}} \\ \end{aligned}
$$
B. $$
\begin{aligned} {{m_{Y} ( t )}} & {{} {{}=E \langle I \tan(\mathrm{e} X ( t )) \rangle}} \\ \end{aligned}
$$
C. $$
\begin{aligned} {{m_{Y} ( t )}} & {{} {{}=E \langle I \sqrt{\mathrm{e} X ( t )} \rangle}} \\ \end{aligned}
$$
D. $$
\begin{aligned} {{m_{Y} ( t )}} & {{} {{}=E \langle I \cos(\mathrm{e} X ( t )) \rangle}} \\ \end{aligned}
$$
E. $$
\begin{aligned} {{m_{Y} ( t )}} & {{} {{}=E \langle I \arctan(\mathrm{e} X ( t )) \rangle}} \\ \end{aligned}
$$
F. $$
\begin{aligned} {{m_{Y} ( t )}} & {{} {{}=E \langle I \ln(\mathrm{e} X ( t )) \rangle}} \\ \end{aligned}
$$
G. $$
\begin{aligned} {{m_{Y} ( t )}} & {{} {{}=E \langle I |\mathrm{e} X ( t )| \rangle}} \\ \end{aligned}
$$
H. $$
\begin{aligned} {{m_{Y} ( t )}} & {{} {{}=E \langle I \mathrm{e}^{\mathrm{e} X ( t )} \rangle}} \\ \end{aligned}
$$
I. $$
\begin{aligned} {{m_{Y} ( t )}} & {{} {{}=E \langle I \sin(\mathrm{e} X ( t )) \rangle}} \\ \end{aligned}
$$
J. $$
\begin{aligned} {{m_{Y} ( t )}} & {{} {{}=E \langle I \mathrm{e}^{\mathrm{sin} X ( t )} \rangle}} \\ \end{aligned}
$$
Answer: H
The composite Poisson process is defined as
$$\left|Y_{i}=\sum_{n=1}^{x_{i}}\xi_{n},\quad t\geqslant0\right|$$
Assume $\lambda=5,\xi_n$ follows the distribution:
$$p(x)=\begin{cases}\frac{1}{1000},&1000\leqslant x\leqslant2000\\\\0,&\text{otherwise}\end{cases}$$
If points follow the distribution:
$$P(\xi_n<x)=\begin{cases}1-\mathrm{e}^{-\mu t},&x\geqslant0\\0,&x<0\end{cases}$$
find the mean $E(Y_i)$, variance var$(Y_i)$, and characteristic function $\phi_Y(v)$ is ().

Options:
A. $$
\begin{array} {l} {{{E ( Y_{i} )=\frac{5 t} {\mu}}}} \\ {{{\mathrm{v a r} ( Y_{i} )=\frac{1 0 t} {\mu^{2}}}}} \\ {{{\phi_{\mathrm{r}} ( v )=\operatorname{e x p} \left( \frac{\mathrm{j} 5 v t} {\mu-\mathrm{j} v} \right)}}} \\ \end{array}
$$
B. $$
\begin{array} {l} {{{E ( Y_{i} )=\frac{3 t} {\mu}}}} \\ {{{\mathrm{v a r} ( Y_{i} )=\frac{9 t} {\mu^{2}}}}} \\ {{{\phi_{\mathrm{r}} ( v )=\operatorname{e x p} \left( \frac{\mathrm{j} 4 v t} {\mu-\mathrm{j} v} \right)}}} \\ \end{array}
$$
C. $$
\begin{array} {l} {{{E ( Y_{i} )=\frac{7 t} {\mu}}}} \\ {{{\mathrm{v a r} ( Y_{i} )=\frac{12 t} {\mu^{2}}}}} \\ {{{\phi_{\mathrm{r}} ( v )=\operatorname{e x p} \left( \frac{\mathrm{j} 7 v t} {\mu-\mathrm{j} v} \right)}}} \\ \end{array}
$$
D. $$
\begin{array} {l} {{{E ( Y_{i} )=\frac{6 t} {\mu}}}} \\ {{{\mathrm{v a r} ( Y_{i} )=\frac{8 t} {\mu^{2}}}}} \\ {{{\phi_{\mathrm{r}} ( v )=\operatorname{e x p} \left( \frac{\mathrm{j} 6 v t} {\mu-\mathrm{j} v} \right)}}} \\ \end{array}
$$
E. $$
\begin{array} {l} {{{E ( Y_{i} )=\frac{2 t} {\mu}}}} \\ {{{\mathrm{v a r} ( Y_{i} )=\frac{4 t} {\mu^{2}}}}} \\ {{{\phi_{\mathrm{r}} ( v )=\operatorname{e x p} \left( \frac{\mathrm{j} 2 v t} {\mu-\mathrm{j} v} \right)}}} \\ \end{array}
$$
F. $$
\begin{array} {l} {{{E ( Y_{i} )=\frac{9 t} {\mu}}}} \\ {{{\mathrm{v a r} ( Y_{i} )=\frac{6 t} {\mu^{2}}}}} \\ {{{\phi_{\mathrm{r}} ( v )=\operatorname{e x p} \left( \frac{\mathrm{j} 9 v t} {\mu-\mathrm{j} v} \right)}}} \\ \end{array}
$$
G. $$
\begin{array} {l} {{{E ( Y_{i} )=\frac{4 t} {\mu}}}} \\ {{{\mathrm{v a r} ( Y_{i} )=\frac{8 t} {\mu^{2}}}}} \\ {{{\phi_{\mathrm{r}} ( v )=\operatorname{e x p} \left( \frac{\mathrm{j} 5 v t} {\mu-\mathrm{j} 2v} \right)}}} \\ \end{array}
$$
H. $$
\begin{array} {l} {{{E ( Y_{i} )=\frac{8 t} {\mu}}}} \\ {{{\mathrm{v a r} ( Y_{i} )=\frac{6 t} {\mu^{2}}}}} \\ {{{\phi_{\mathrm{r}} ( v )=\operatorname{e x p} \left( \frac{\mathrm{j} 8 v t} {\mu-\mathrm{j} v} \right)}}} \\ \end{array}
$$
I. $$
\begin{array} {l} {{{E ( Y_{i} )=\frac{5 t} {\mu}}}} \\ {{{\mathrm{v a r} ( Y_{i} )=\frac{5 t} {\mu^{2}}}}} \\ {{{\phi_{\mathrm{r}} ( v )=\operatorname{e x p} \left( \frac{\mathrm{j} 3 v t} {\mu-\mathrm{j} v} \right)}}} \\ \end{array}
$$
J. $$
\begin{array} {l} {{{E ( Y_{i} )=\frac{5 t} {\lambda}}}} \\ {{{\mathrm{v a r} ( Y_{i} )=\frac{10 t} {\mu^{3}}}}} \\ {{{\phi_{\mathrm{r}} ( v )=\operatorname{e x p} \left( \frac{\mathrm{j} 5 v t} {\lambda-\mathrm{j} v} \right)}}} \\ \end{array}
$$
Answer: A
A single-phase voltage regulation device is described, with an input power supply of single-phase industrial frequency $220 \mathrm{V}$ and a load that comprises a series of resistive and inductive components, where $R=1 \Omega$ and $L=2 \mathrm{mH}$. The device combines a transformer (assumed to be an ideal transformer) with a single-phase voltage regulation circuit based on thyristors, requiring the load output power to be adjustable within the range of $0-10 \mathrm{kW}$. Calculate the following: when $\alpha=\pi/2$, determine the root mean square (RMS) value of the thyristor current, the conduction angle of the thyristor, and the power factor on the supply side is().

Options:
A. $$43.6A,210°,0.490$$
B. $$43.6A,200°,0.422$$
C. $$43.6A,120°,0.522$$
D. $$42.0A,120°,0.673$$
E. $$46.2A,120°,0.600$$
F. $$41.5A,120°,0.325$$
G. $$44.0A,150°,0.500$$
H. $$43.6A,90°,0.553$$
I. $$43.6A,60°,0.621$$
J. $$45.1A,180°,0.501$$
Answer: C
When the water inflow at the shaft sinking face exceeds $50 \, \mathrm{m}^{3} / \mathrm{h}$, the recommended drainage method is ____.

Options:
A. Mechanical piston pumping directly from the excavation site
B. Submersible pump used in conjunction with horizontal pump for drainage
C. Vertical borehole drilling with immediate surface discharge
D. Gravity drainage through an inclined channel to the surface
E. Automated weir control supported by manual bucket removal
F. Hydraulic pressure balancing with controlled overflow release
G. Bladder tank retention until gradual evaporation is achieved
H. Continuous siphon operation using external vacuum assistance
I. Sequential evaporation combined with periodic suction pumping
J. Direct surface release via strategically placed sump portals
Answer: B
Based on the classification of the surrounding rock in roadways supported by anchor spraying, type II rock characteristics are ____.

Options:
A. Requires frequent support and reinforcement adjustments.
B. With reinforcement, large fragments are stabilized.
C. Minimizes crack propagation through self-stabilization.
D. Inclined layers securing stability with less intervention.
E. Compressive strength under tension without failure.
F. Constant settling observed even with applied supports.
G. Unstable layers requiring constant reinforcement renewal.
H. Without support, small pieces may fall off.
I. High elasticity, reducing likelihood of shear deformation.
J. Prone to collapse under high-pressure without bolting.
Answer: H
Using Cu $K_{a} (\lambda=0.154056 \ \mathrm{nm})$, the X-ray diffraction pattern of an OCr18Ni9Ti austenitic stainless steel sample that underwent solution nitriding and was quenched in water after being pre-cooled to 1050°C was measured. The $2\theta$ angles corresponding to the diffraction peaks of austenite are 43.28°, 50.48°, 74.10°, 89.90°, and 95.12°. If $a=0.36165 \ \mathrm{nm}$, calculate the Miller indices corresponding to these diffraction peaks.

Options:
A. (222),(220),(200),(311),(111)
B. (222),(311),(200),(220),(111)
C. (220),(200),(311),(111),(222)
D. (311),(200),(222),(111),(220)
E. (111),(220),(311),(200),(222)
F. (311),(222),(111),(200),(220)
G. (220),(111),(222),(311),(200)
H. (200),(111),(311),(220),(222)
I. (111),(200),(220),(311),(222)
J. (200),(311),(220),(111),(222)
Answer: I
A $\mathrm{C u}-{30}\%\mathrm{Z n}$ brass cold-rolled sheet undergoes 25% cold deformation, reducing its thickness to $1 \, \mathrm{cm}$. Then, the thickness is further reduced to $0.6 \, \mathrm{cm}$. Calculate the total engineering strain and true linear strain.

Options:
A. $$
\approx-5 9 \%
$$
$$
\approx-7 9 \%
$$
B. $$
\approx-5 2 \%
$$
$$
\approx-8 2 \%
$$
C. $$
\approx-6 5 \%
$$
$$
\approx-7 8 \%
$$
D. $$
\approx-5 4 \%
$$
$$
\approx-8 7 \%
$$
E. $$
\approx-5 5 \%
$$
$$
\approx-8 0 \%
$$
F. $$
\approx-5 0 \%
$$
$$
\approx-9 0 \%
$$
G. $$
\approx-6 8 \%
$$
$$
\approx-7 3 \%
$$
H. $$
\approx-6 0 \%
$$
$$
\approx-7 0 \%
$$
I. $$
\approx-5 3 \%
$$
$$
\approx-8 5 \%
$$
J. $$
\approx-5 8 \%
$$
$$
\approx-7 5 \%
$$
Answer: E
After a single crystal metal is cold-pressed to 80% deformation and held at 20°C for $7$ days, its properties recover to a certain extent. If it requires holding at 100°C to achieve the same extent of recovery, it needs to be maintained for $50 \operatorname*{min}$. Find the recovery activation energy of this metal.

Options:
A. $$
\approx 6. 3 2 \times 1 0^{4} ( \mathrm{J / m o l} )
$$
B. $$
\approx 5. 7 5 \times 1 0^{4} ( \mathrm{J / m o l} )
$$
C. $$
\approx 7. 2 1 \times 1 0^{4} ( \mathrm{J / m o l} )
$$
D. $$
\approx 5. 8 9 \times 1 0^{4} ( \mathrm{J / m o l} )
$$
E. $$
\approx 7. 4 8 \times 1 0^{4} ( \mathrm{J / m o l} )
$$
F. $$
\approx 6. 5 0 \times 1 0^{4} ( \mathrm{J / m o l} )
$$
G. $$
\approx 6. 9 7 \times 1 0^{4} ( \mathrm{J / m o l} )
$$
H. $$
\approx 7. 0 3 \times 1 0^{4} ( \mathrm{J / m o l} )
$$
I. $$
\approx6. 0 4 \times1 0^{4} ( \mathrm{J / m o l} )
$$
J. $$
\approx 5. 6 6 \times 1 0^{4} ( \mathrm{J / m o l} )
$$
Answer: I
In the known ceramic material $\mathrm{Al}_{2} \mathrm{O}_{3}-\mathrm{ZrO}_{2}$, the $w_{\mathrm{ZrO}_{2}}$ is 42.6%. Find the mass percentage of $\mathrm{Al}$, $\mathrm{Zr}$ and $\mathrm{O}$ in the composite structure.

Options:
A. ${34.6} \%$,
${30.9} \%$,
${29.8} \%$
B. ${29.1} \%$,
${37.2} \%$,
${33.6} \%$
C. ${31.2} \%$,
${35.9} \%$,
${34.7} \%$
D. ${35.4} \%$,
${29.9} \%$,
${36.1} \%$
E. ${30.4} \%$,
${31.5} \%$,
${38.1} \%$
F. ${27.3} \%$,
${34.4} \%$,
${31.7} \%$
G. ${33.8} \%$,
${32.9} \%$,
${35.2} \%$
H. ${36.5} \%$,
${28.8} \%$,
${37.3} \%$
I. ${28.5} \%$,
${36.4} \%$,
${32.8} \%$
J. ${32.1} \%$,
${33.5} \%$,
${30.7} \%$
Answer: E
Assume there is a $100 \mathrm{g}$ $96\%\operatorname{Al}-4\%\operatorname{Cu}$ alloy that reaches equilibrium at $620^\circ\mathrm{C}$, forming an $\alpha$ phase and a liquid phase $\mathbf{L}_{o}$. Then, the alloy is rapidly cooled to $550^\circ\mathrm{C}$, preventing the original solid from participating in the reaction, and the liquid phase remains. What is the composition of this liquid?

Options:
A. $$
\begin{array} {c} {{{\mathrm{L_{2} : 0. 9 0 A l-0. 1 0 C u ;}}}} \\ {{{\alpha: 0. 7 5 0 \mathrm{~ A l-0. 2 5 0 ~ C u}}}} \\ \end{array}
$$
B. $$
\begin{array} {c} {{{\mathrm{L_{2} : 0. 8 2 A l-0. 1 8 C u ;}}}} \\ {{{\alpha: 0. 8 7 0 \mathrm{~ A l-0. 1 3 0 ~ C u}}}} \\ \end{array}
$$
C. $$
\begin{array} {c} {{{\mathrm{L_{2} : 0. 6 7 A l-0. 3 3 C u ;}}}} \\ {{{\alpha: 0. 9 4 4 \mathrm{~ A l-0. 0 5 6 ~ C u}}}} \\ \end{array}
$$
D. $$
\begin{array} {c} {{{\mathrm{L_{2} : 0. 7 5 A l-0. 2 5 C u ;}}}} \\ {{{\alpha: 0. 8 8 0 \mathrm{~ A l-0. 1 2 0 ~ C u}}}} \\ \end{array}
$$
E. $$
\begin{array} {c} {{{\mathrm{L_{2} : 0. 6 0 A l-0. 4 0 C u ;}}}} \\ {{{\alpha: 0. 9 2 0 \mathrm{~ A l-0. 0 8 0 ~ C u}}}} \\ \end{array}
$$
F. $$
\begin{array} {c} {{{\mathrm{L_{2} : 0. 9 5 A l-0. 0 5 C u ;}}}} \\ {{{\alpha: 0. 8 3 0 \mathrm{~ A l-0. 1 7 0 ~ C u}}}} \\ \end{array}
$$
G. $$
\begin{array} {c} {{{\mathrm{L_{2} : 0. 7 2 A l-0. 2 8 C u ;}}}} \\ {{{\alpha: 0. 9 1 0 \mathrm{~ A l-0. 0 9 0 ~ C u}}}} \\ \end{array}
$$
H. $$
\begin{array} {c} {{{\mathrm{L_{2} : 0. 6 5 A l-0. 3 5 C u ;}}}} \\ {{{\alpha: 0. 9 5 0 \mathrm{~ A l-0. 0 5 0 ~ C u}}}} \\ \end{array}
$$
I. $$
\begin{array} {c} {{{\mathrm{L_{2} : 0. 8 0 A l-0. 2 0 C u ;}}}} \\ {{{\alpha: 0. 9 0 0 \mathrm{~ A l-0. 1 0 0 ~ C u}}}} \\ \end{array}
$$
J. $$
\begin{array} {c} {{{\mathrm{L_{2} : 0. 8 5 A l-0. 1 5 C u ;}}}} \\ {{{\alpha: 0. 9 0 5 \mathrm{~ A l-0. 0 9 5 ~ C u}}}} \\ \end{array}
$$
Answer: C
Heat the austenitic stainless steel cold-rolled plate to 900°C and 1,100°C respectively, maintain for 10 hours, and measure the average grain sizes $\overline{{{{D}}}}_{900}=0.05 \ \mathrm{nm}, \overline{{{{D}}}}_{1100}=0.15 \ \mathrm{nm}$. Assuming $\overline{{{D}}}_{i} \gg \overline{{{D}}}_{0}$, calculate the activation energy for grain boundary migration is ().

Options:
A. $$
\approx 1 1 8 \mathrm{~ k J / m o l ~}
$$
B. $$
\approx 1 9 0 \mathrm{~ k J / m o l ~}
$$
C. $$
\approx 1 3 5 \mathrm{~ k J / m o l ~}
$$
D. $$
\approx 1 4 0 \mathrm{~ k J / m o l ~}
$$
E. $$
\approx 1 6 0 \mathrm{~ k J / m o l ~}
$$
F. $$
\approx 1 2 0 \mathrm{~ k J / m o l ~}
$$
G. $$
\approx 1 5 5 \mathrm{~ k J / m o l ~}
$$
H. $$
\approx 1 4 7 \mathrm{~ k J / m o l ~}
$$
I. $$
\approx 1 8 5 \mathrm{~ k J / m o l ~}
$$
J. $$
\approx 1 7 2 \mathrm{~ k J / m o l ~}
$$
Answer: H
A $PNP$ type germanium transistor with a junction operating at room temperature is given. The doping concentrations of the emitter, base, and collector regions are $10^{17}\mathrm{cm^{-3}}$, $10^{15}\mathrm{cm^{-3}}$, and $2 \times 10^{16} \mathrm{cm^{-3}}$, respectively. The base width is $W=2 \times 10^{-3} \mathrm{cm}$, and the cross-sectional area is $S=5 \times 10^{-3} \mathrm{cm}^{2}$. The mobilities are given as $$\mu_{pb} = 1800 \mathrm{cm}^{2} / \mathrm{V} \cdot \mathrm{s}, \quad \mu_{ne} = 1900 \mathrm{cm}^{2} / \mathrm{V} \cdot \mathrm{s}, \quad \mu_{nc} = 3500 \mathrm{cm}^{2} / \mathrm{V} \cdot \mathrm{s},$$ and the diffusion lengths are $L_{pb} = L_{ne} = L_{nc} = 0.1 \mathrm{mm}$. Calculate the common-base current gain $\alpha$.

Options:
A. $$
= 0. 9 2 1 0
$$
B. $$
= 0. 8 8 8 8
$$
C. $$
= 0. 7 5 6 8
$$
D. $$
= 0. 6 5 7 2
$$
E. $$
= 0. 9 7 7 9
$$
F. $$
= 0. 9 6 1 4
$$
G. $$
= 0. 6 6 9 9
$$
H. $$
= 0. 7 9 9 8
$$
I. $$
= 0. 8 3 4 5
$$
J. $$
= 0. 9 0 2 3
$$
Answer: E
A bipolar amplifier has a relationship of ${\frac{f_{p 2}} {f_{p 1}}}=10$, with the dominant pole frequency at 1MHz and a low-frequency gain of $K(0)$. When a negative feedback is applied with $K(0)k_{f}=0.8$, calculate the two new poles of the feedback amplifier.

Options:
A. $$
=-3 1. 6 \times1 0^{6} \operatorname{or}-5 0. 6 \times1 0^{6} \operatorname{rad} / {\operatorname{s}}
$$
B. $$
=-1 6. 8 \times1 0^{6} \operatorname{or}-4 7. 9 \times1 0^{6} \operatorname{rad} / {\operatorname{s}}
$$
C. $$
=-1 1. 4 \times1 0^{6} \operatorname{or}-6 0. 9 \times1 0^{6} \operatorname{rad} / {\operatorname{s}}
$$
D. $$
=-1 5. 2 \times1 0^{6} \operatorname{or}-5 7. 4 \times1 0^{6} \operatorname{rad} / {\operatorname{s}}
$$
E. $$
=-1 3. 6 \times1 0^{6} \operatorname{or}-4 9. 2 \times1 0^{6} \operatorname{rad} / {\operatorname{s}}
$$
F. $$
=-2 5. 0 \times1 0^{6} \operatorname{or}-5 5. 0 \times1 0^{6} \operatorname{rad} / {\operatorname{s}}
$$
G. $$
=-2 2. 8 \times1 0^{6} \operatorname{or}-5 8. 0 \times1 0^{6} \operatorname{rad} / {\operatorname{s}}
$$
H. $$
=-2 0. 5 \times1 0^{6} \operatorname{or}-4 8. 8 \times1 0^{6} \operatorname{rad} / {\operatorname{s}}
$$
I. $$
=-1 2. 6 \times1 0^{6} \operatorname{or}-5 6. 6 \times1 0^{6} \operatorname{rad} / {\operatorname{s}}
$$
J. $$
=-1 4. 0 \times1 0^{6} \operatorname{or}-4 6. 4 \times1 0^{6} \operatorname{rad} / {\operatorname{s}}
$$
Answer: I