Electric current – part (2)

Total Questions: 39

31. With reference to solar water pumps, consider the following statements: [LA.S. (Pre) 2020]

1. Solar power can be used for running surface pumps and not for submersible pumps.

2. Solar power can be used for running centrifugal pumps and not the ones with piston.

Which of the statement given above is/are correct?

(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither I nor 2

Correct Answer: (d) Neither I nor 2

Solution:Solar-powered pumps run on electricity generated by photovoltaic panels or the radiated thermal energy available from collected sunlight as opposed to grid electricity or diesel run water pumps. Solar power can be used for running surface pumps as well as for submersible pumps. Solar-powered submersible pumps support varied applications like field irrigation or supply potable water, while drawing water from underground water sources like bore-wells or sumps. Some solar submersible pumps use the same centrifugal mechanism as a surface pump, while others use a positive displacement mechanism. A solar-powered reciprocating pump is class of positive displacement pump with piston, which is used for various purposes such as, car washing, irrigation, colour spraying, large fountain, garden water pump etc. From the above explanation it is clear that both of the given statements are incorrect.

32. Which one of the following elements is used in solar cells? [Uttarakhand P.C.S. (Pre) 2012]

(a) Silicon
(b) Cerium
(c) Astatine
(d) Vanadium

Correct Answer: (a) Silicon

Solution:Silicon's Semiconductor Properties: Silicon is a semiconductor material, meaning its electrical conductivity lies between that of a conductor and an insulator. This property is crucial for the functioning of solar cells.

Cerium: Cerium is a rare earth metal with various applications, but it is not typically used in the semiconductor layer of solar cells.

Astatine: Astatine is a radioactive halogen.

Vanadium: Vanadium is a transition metal used in various alloys and chemical compounds.

33. A photoelectric cell is a device which : [69th B.P.S.C. (Pre) 2023]

(a) converts light energy into electric energy.
(b) converts electric energy into light energy.
(c) stores light energy.
(d) None of the above.

Correct Answer: (a) converts light energy into electric energy.

Solution:A photoelectric cell converts light energy into electric energy The photoelectric effect, where electricity is generated from a material when it absorbs light, was first discovered in 1887 by Heinrich Hertz. The photoelectric device can capture high-energy photons and generate a high-voltage output, while the photovoltaic cell can capture low-energy photons and generate a high-current output. This combination can increase the power output and efficiency of solar PV systems.

34. Which of the following photoelectric devices is most suitable for digital applications? [68th B.P.S.C. (Pre) 2022]

(a) Photodiode
(b) Photovoltaic cell
(c) Photoemitter
(d) More than one of the above
(e) None of the above

Correct Answer: (b) Photovoltaic cell

Solution:The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material. And photoelectric device means which uses photoelectric effect, in other words which produces electricity/electric signal from light. Photovoltaic cell is one such device. Photovoltaic cells or commonly known as solar cells have vast use in digital technology. For example, entire satellite communication depends on it, since this is the main source of power for it. Photoemitting or light emitting diodes emit light on flow of electricity te reverse of photoelectric effect. Similarly, photodiode changes its electrical properties on receiving light. It is like light controlled switch which allows flow of electricity on getting light but it does not generate electricity. Hence, option (b) is the correct answer.

35. Net Metering' is sometimes seen in the news in the context of promoting: [69th B.P.S.C. (Pre) 2023]

(a) the installation of CNG kits is motorcars.
(b) the installation of water meters in urban households.
(c) a billing mechanism for solar energy by consumers for the electricity they add to the grid.
(d) the use of piped natural gas in the kitchens of households.

Correct Answer: (c) a billing mechanism for solar energy by consumers for the electricity they add to the grid.

Solution:'Net metering' is a billing mechanism that credits solar energy system owners for the electricity they add to the grid. For example, if a residential customer has a PV system on their roof, it may generate more electricity than the home uses during daylight hours. If the home is net-metered, the electricity meter will run backwards to provide a credit against what electricity is consumed at night or other periods when the home's electricity use exceeds the system's output. Customers are only billed for their 'net' energy use.

36. With reference to solar power production in India, consider the following statements: [I.A.S. (Pre) 2018]

1. India is the third largest in the world in the manufacture of silicon wafers used in photovoltaic units.

2. The solar power tariffs are determined by the Solar Energy Corporation of India.

Which of the statements given above is/are correct?

(a) 1 only
(b) 2 only
(c) both 1 and 2
(d) Neither 1 nor 2

Correct Answer: (d) Neither 1 nor 2

Solution:As per IC Insights Global Wafer Capacity 2021-2025 report, Taiwan led the world with 21.4% of global silicon wafer sigui capacity installed in that country. In second place was South altis Korea, which accounted for 20.4% of global silicon wafer incapacity. Japan and China were at the third and fourth place respectively, India is not among the leading manufacturers of silicon wafers in the world. The solar power tariffs in India are determined by the Central Electricity Regulatory Commission (CERC) and State Electricity Regulatory Consmissions (SERCs) on the hosts of competitive bidding Solar Energy Corporation of India (SECT) has a Power trading license, but it does not determine solar power tariffs.

37. Which of the following statements is not correct regarding wind power capacity in India? [M.P. P.C.S. (Pre) 2022]

(a) Wind power capacity in India has significantly.
(b) Gujarat has around 22% of the total wind power capacity of the country.
(c) Tamil Nadu's wind power capacity was about 24% of the country in the year 2021.
(d) Madhya Pradesh has third rank in the country in wind.

Correct Answer: (d) Madhya Pradesh has third rank in the country in wind.

Solution:

Madhya Pradesh has seventh rank in the country in wind power installed capacity, As per the data of 2021, other three statements are correct. However, Gujarat has surpassed Tamil Nadu in total wind power installed capacity in 2023. As on 31 March, 2024, total wind power installed capacity in India is 45886.51 MW in which Gujarat has around 25.5% share (11722.72 MW) while Tamil Nadu has around 23.1% share (10603.54 MW).

38. Two wires have their length, diameter and resistivities all in the ratio of 1:2 if the resistance of the thinner wire is 10 ohm the resistance of the thicker wire is : [I.A.S (pre) 2001]

(a) 5 ohm
(b) 10 ohm
(c) 20 ohm
(d) 40 ohm

Correct Answer: (b) 10 ohm

Solution:we know that

R=(ρ×l)/a

R= resistance

ρ= resistivity

l=length

a=area

let the ρ₁,l₁ and r₁ is resistance length and radius of the first wire respectively then

as per question

resistivity of second wire ρ₂=2ρ₁

length of the second wire l₂=2l₁

radius of the second wire r₂=2r₁

⇒R₁=(ρ₁×l₁)/a₁=(ρ₁×l₁)/πr₁²

R₂=(ρ₂×l₂)/a₂ =(2ρ₁×2l₁)/π(r₁)²

=4(ρ₁×l₁)/4πr₁²=(ρ₁×l₁)/πr₁²

⇒R₁/R₂=1

Since the resistance of thin wire R₁=10Ω

therefore resistance of the thicker wire R₂=10Ω.

39. Small drops of the same size are charged to V volts each If n such drops coalesce to form a single large drop its potential will be : [U.P LOWER Sub (MAINS) 2015]

(a) (n^2/3 )V
(b) n⁵
(c) n×V
(d) N⁻¹

Correct Answer: (a) (n^2/3 )V

Solution:Let the charge on every small drop=q

radius of every small drops =r

charge on the large drop (Q)=nq

radius of large drop =R

∴ the total volume of n drops=volume of the large drops

∴ ( n) 4/3πr³ = 4/3πR³

⇒ R = (n^1/3) r

V = kq/r and V' = kQ/R , where k is constant.

now, V'/V= Qr/qR

=nqr/q(n^1/3) r

⇒ V'/V= n^(1-1/3)

⇒ So potential of large drop V' = (n^2/3)V

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Electric current – part (2)