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λ = L/R ; R = L/λ = 8/1 = 8 ohm

(i) Initial di/dt = V/L = 100/8 = 12.5 A/s 

(ii) I_M = V/R = 100/8 = 12.5A

(iii) Here, i = 60% of 12.5 = 7.5 A

Now, i = I_m (1 −e^−t/λ)

∴ 7.5 = 12.5 (1 −e^−t/1);

t = 0.915 second

For growth ; 5 = I_m (1 −e ^−0.3/λ) ...(i)

For decay; 5 = I_m e ^−0.3/λ

Equating the two, we get, I_me ^−0.3/λ = (1 −e^{− 0.3/λ})I_m

or 2e^−0.3/λ = 1

∴ e ^−0.3/λ = 0.5 or λ = 0.4328 

Putting this value in (i), we get, 

5 = I_m = e ^0.3/0.4328 or I_m = 5 e ^{+ 0.3/0.4328} = 5 × 2 = 10 A.

Now, I_m = V/R

∴ 10 = 230/R or R = 230/10 = 23 Ω (approx.) 

As λ = L/R = 0.4328 ; L = 0.4328 × 23 = 9.95 H

At t = 0, i = 0, hence there is no iR drop and the applied voltage must equal the back e.m.f. in the coil. 

Hence, the voltage across L at t = 0 represents the applied voltage. 

At t = 0.025 second, voltage across L is 5 V, hence voltage across 

R = 25 − 5 = 20 V 

∴ iR = 20 V − at t = 0.025 second. 

Now i = I_m(I − e ^−t/λ) 

Here I_m = 25/R ampere, t = 0.025 second 

∴ i = 25/R(1 - e^-0.025/λ)

R x  25/R(1 - e^-0.0025/λ) = 20 or e^0.025/λ = 5

∴ 0.025/λ = 2.3log₁₀5 = 1.6077

∴ λ = 0.025/1.6077 

Now λ = L/R = 2/R 

∴ 2/R = 0.025/1.6077 

∴ R = 128.56 Ω

The voltage equation for an R-L circuit is

V = iR + L(di/dt) or L(di/dt) = V - iR or di/dt = 1/L(V - iR)

(i) when i = 0; di/dt = 1/L(V - 0 x R) = V/L = 80/20 = 4A/s

(ii) when i = 0.5A; di/dt = (80 - 0.5 x 80)/20 = 2A/s

(iii) when i = 1A; di/dt = (80 - 80 x1)/20 = 0.

In other words, the current has become steady at 1 ampere.

(a) Electric field intensity, E = 3 × 10³ î N/C
Magnitude of electric field intensity, |E| = 3 × 10³ N/C
Side of the square, s = 10 cm = 0.1 m
Area of the square, A = s² = 0.01 m²

The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field, θ = 0° Flux (ϕ) through the plane is given by the relation,

ϕ=|E| Acosθ
= 3 × 10³ × 0.01 × cos 0°
= 30 N m²/C

(b) Plane makes an angle of 60° with the x – axis. Hence, θ = 60°

Flux,ϕ =|E|
= 3 × 103 × 0.01 × cos 60°

 =30×1/2
= 15 N m²/C

(a) Net outward flux through the surface of the box,ϕ = 8.0 × 10³ N m²/C
For a body containing net charge q, flux is given by the relation, 

ϕ=q/ε_o

ε_o = Permittivity of free space
= 8.854 × 10⁻¹² N⁻¹C² m⁻²
q =ε_oϕ
= 8.854 × 10⁻¹² × 8.0 × 10³ C = 7.08 × 10⁻⁸ C = 0.07 μC
Therefore, the net charge inside the box is 0.07 μC.

(b) No

Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.

All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.

A. eddy currents

Correct Answer is: (b) 20 cm

X/R = AJ/JB for balance

Initially, 12/18 = AJ/100 - AJ' finally 12/8 = AJ'/100 - AJ or AJ' - AJ = 20 cm.

Correct option (A-q, s), (B-p,s), (C-r), (D-r)

Explanation:

(p) If current in increased, flux in the loop will increase in inside direction, then due to Lenz‟s law induced emf in the loop will be in anticlockwise direction. Due to this current, the current in the nearer side of loop to the wire will be in opposite direction to that of wire. 

Hence, there will be repulsion. 

(q) This situation is opposite to part (p) 

(r) If loop is moved away, then flux decreases and this becomes similar to part (q) 

(s) Similar to part (p)

1. Use less hot water 

2. Use the “off” switch

Correct option is (a) Stability of personnel

Use of energy-efficient vehicles and Compressed Natural Gas (CNG)

Correct answer is zipper

Radius of the ball (a) = 1mm = 1 × 10^-3 m 

Co-effecient of viscosity of liquid (η) = 0.2 Nsm^-2 

Speed of the ball (v) = 0.07 ms^-1 

According to Stoke’s law 

Viscous force F = 6 π η av 

= 6 × 3.14 × 1 × 10^-3 × 0.2 × 0.07

= 0.26376 × 10^-3 = 2.64 × 10^-4 N

Charminar: Eradication of plague

Correct option is A) seeking

The original length, change in temperature and nature of the material.

Cauliflower: Early: 45 × 30 cm Main: 60 × 45 cm

Cabbage: Early: 45 × 30 cm

Main: 60 × 45 cm

Browning: Browning is caused due to boron deficiency. In early stage, the water soaked areas appear on the stem and curd surface. As the plant grows, the stem becomes hollow with water soaked tissue covering the internal walls of the cavity. In advanced stage of deficiency, brown or pink coloured areas are seen on curd surface and therefore, it is also called brown rot or red rot or browning of the curd.

Control: The deficiency of boron may be corrected by applying borax. The quantity of borax depends on soil type, soil pH and the extent of deficiency. In acid soil, 10- 15 kg borax/ha is sufficient.

Whiptail: Whiptail disorder is caused due to deficiency of molybdenum. In young plants the deficiency symptoms are chlorosis of leaf margins and the whole leaves may turn white. The leaf blades do not develop properly. This condition is commonly known as 'Whiptail'. The deficiency of molybdenum generally occurs in acid soils when the soil pH is below 5.5.

Control: Lime application in acidic soils is done to increase the availability of molybdenum. The quantity of lime is determined by initially measuring the pH of the soil. Alternately, soil application of Sodium Molybdate (10-15 kg/ha) effectively controls the deficiency symptoms.

Buttoning: The development of small premature curds or buttons while the plants are young is known as buttoning. Several factors like poor nitrogen supply, planting of over-age seedlings, unfavorable climatic conditions and improper time of planting are reported to cause buttoning.

Control: Adequate supply of nitrogen and moisture for rapid vegetative growth of plant is considered important for preventing the occurrence of button plants.

Genetic mechanisms available in cauliflower, and cabbage:

Use of self-incompatible lines: The sporophytic system of self-incompatibility is used for hybrid seed production of cauliflower and cabbage. In general 3:1 self-incompatible and 1 self-compatible (pollinator) is followed to get sufficient amount of hybrid seeds.

Cytoplasmic Male Sterility: In recent times use of male sterility (CMS system) is getting more attention as an alternative to SI system due to its inherent advantage in the hybrid seed production of all cole crops.. Single cross hybrids are more uniform, while cost of the double cross hybrids is much cheaper than single cross hybrids.

Mechanism in solanaceous vegetables:

Male sterility:

Tomato: In hybrid seed production, male sterile lines can be used to cut done the expenditure incurred for emasculation operation. Stamenless and closed anther mutants can be used in hybrid seed production with success.

Chilli: Emasculation and hand pollination is most expensive method of hybrid seed production in chilli because of high labour cost and very low fruit set percentage. Therefore, genetic male sterility mechanism is more economical and can be exploited for hybrid seed production.

Sweet pepper: Both genic and cytoplasmic male sterility have been reported in Functional male sterility can be utilised in hybrid seed production.

Tomato:

Insect pest: Thrips; Aphids; Mites; Whitefly; Beetles; Fruit fly; Cluster caterpillars; Looper caterpillars; Potato moth; Heliothis (Helicoverpa).

Diseases: Tomato spotted wilt virus (TSWV); Tobacco mosaic virus (TMV) Bacterial spot; Damping-off; Powdery mildew; Tomato yellow leaf curl virus; Early blight; Bacterial wilt; Anthracnose; Fusarium wilt; Nematodes.

Brinjal:

Insect pest: Aphids; Thrips; Leafhoppers; Two-spotted mite; Beetles; Whitefly; Fruit and shoot borer.

Diseases: Damping-off; Tomato spotted wilt virus; Root rots; Tobacco mosaic virus; Tobamoviruses; Bacterial wilt; Nematodes.

Capsicum and Chilli:

Insect pest: Aphids; Thrips; Whitefly; Mites.

Diseases: Bacterial spot; Bacterial wilt; Anthracnose; Cercospora spot; Powdery mildew; Tobacco mosaic virus (TMV); Nematodes.

Potato: Thrips (thrips, Onion thrips); Aphids; Potato moth; Whitefly; Beetles; Looper caterpillars; Leafhoppers; Bugs; Potato moth.

Control measures:

1. Pre-sowing operation:

1. Deep summer ploughing: Helicoverpa, Spodoptera, Thrips, serpentine leaf miner and pinworm 

2. Soil solarization (with polythene sheet of 45 gauge (0.45 mm) thickness for three weeks before sowing): Helicoverpa, Spodoptera, Thrips, serpentine leaf miner and pinworm 

3. Apply Neem cake 250 kg/ha at the time of land preparation: Thrips and nematodes.

2. During nursery development: Raise Marigold (Tall African variety golden age bearing yellow and orange flowers) nursery 15-20 days before tomato nursery (as trap crop for Helicoverpa): Use nylon net of 40 gauge mesh to protect seedlings against whitefly infestation for leaf curl management.

3. Management in the main field: Transplant 20-25 days old tomato and 45-50 days old marigold simultaneously in the ratio of 16:1. Simultaneous flowering of both the crops ensures attraction of fruit borers to marigold flowers. A. Cultural methods B. Mechanical methods: Collection and destruction of eggs and early stages of larvae (Spodoptera): Handpick the older larvae during early stages of plant (Helicoverpa).

4. Physical methods: Use yellow/blue pan water / sticky traps @ 4-5 trap/acre Leaf miner, Thrips, Aphids 2 Use light trap @ 1/acre and operate between 6 pm and 10 pm Pinworm, Helicoverpa 3 Install pheromone traps @ 4-5/acre for monitoring Helicoverpa and 10-12 traps/acre for mass trapping of pinworm (replace the lures with fresh lures after every 2-3 weeks) Helicoverpa, Pinworm.

5. Chemical controll:

1. Fifteen days after planting spray imidacloprid 200 SL @ 0.4ml/l Whitefly, thrips, aphids 

2. Spray dicofol 18.5 EC (1.5 ml/l) Red spider mite 

3 Cyantraniliprole 10.26% OD @ 360 ml in 200 litre water/acre Thrips.

4 Spray indoxacarb 14.5% SC @ 0.8 ml/l of water Helicoverpa, Spodoptera and pin worm.

Correct option: d. Fruit borer

Correct option: d. Pusa Meghna

Urea is produced in the liver and is a metabolite (breakdown product) of amino acids. Ammonium ions are formed in the breakdown of amino acids. Some are used in the biosynthesis of nitrogen compounds. Excess ammonium ions are converted to urea.

(a) The Bhopal gas tragedy occurred on 3rd Dec. 1984 in which methyl isocyanate gas was released from a fertilizer manufacturing plant of Union Carbide causing death of approximately 2500 persons. Chernobyl disaster occurred on April 26, 1986, from an explosion at the chernobvl power station which released a huge radioactive cloud into the atmosphere in Ukrain.

Correct option: b. Brinjal

1. Ouch

2. Hey 

3. huh 

4. Wow 

5. bravo 

6. eureka 

7. Shoo 

8. Yippee 

9. Hurray 

10. Alas

Correct option is: (B) Blastopore

The parathyroids are four small glands embedded two in each posterior face of one thyroid lobe. The parathyroids secrete parathormone, a hormone that together with calcitonin and vitamin D regulates the calcium blood level.

(A) More than 10⁷

The testicles make androgenic hormones, the main of them being testosterone. The ovaries produce estrogen and progesterone.

Answer (c) RNA Polymerase III

First cellular form of life appeared on earth about 2000 million years ago.

Answer is (d) 6

Correct option (1) HCP =FCC > BCC > SC

Explanation:

packing HCP 74 % 

efficiency = FCC 74 % 

Sc = 52 % 

BCC = 68 %

The correct answer is Russia.

• Kudankulam Nuclear Power plant is situated in Tirunelveli district of Tamil Nadu.
• It is being developed by the Nuclear Power Corporation of India (NPCIL).
• Kudankulam Nuclear Power plant is the second power project in Tamil Nadu.
• The first phase of the construction was started in 2001 and the first two units were commissioned in 2013 and 2016 respectively.

• Russia collaborates with India to build the Kudankulam Nuclear Power Plant.
• On 20 November 1988, an inter-governmental agreement was signed between Rajiv Gandhi, and Mikhail Gorbachev, for the construction of two reactors of 2 GW.
• The power plant will have a combined capacity of 6000MW upon commissioning of its six units.
• The fuel to be used for the reactors at Kudankulam Nuclear Power plant is enriched Uranium(Uranium 235).

The correct answer is Russia.

Kudankulam Nuclear Power Plant: 

• It is the largest nuclear power plant in India.
• It is situated in the Tirunelveli district of Tamil Nadu.
• It was built in collaboration with Atomstroyexport, the Russian state company, and Nuclear Power Corporation of India Limited (NPCIL).
• It has a capacity of 6,000 MW of electricity.

The Arabian Sea lies to South-East of India.

(b) North-East Monsoons

The Arabian Sea lies to South-East of India.

The correct answer is Chennai - Ennore State Highway.

In news,

• The five-phase 133 km Peripheral Road for which land is being acquired will address the rapidly increasing road traffic demand in the Chennai Metropolitan Area.
• The project is expected to improve the connectivity in and around Chennai by formulating the radial-ring road network in collaboration with other ring roads such as Inner Ring Road, the Chennai Bypass, and Outer Ring Road to provide alternative routes for traffic as well as improve the road network.
• It will also provide direct access to Ennore Port and Kattupalli Port from industrial clusters located in suburban areas of Chennai Metropolitan Area to accelerate industrial and economic growth.​

• Six State Highway roads of about 452-km length have been upgraded as National Highways in Tamil Nadu in 2017.
  • The roads declared as NH are Thoppur Mettur-Bhavani-Erode (85 km).
  •  Salem-Thirupattur-Vaniyambadi (141.0 km),
  • Athur-Perambalur (55.4 km), T
  • irupur-Oddanchatram road (90.8),
  • Chengalpattu-Mahabalipuram (29 km) and
  • Kodaighat-Kodaikanal (52 km)
  • 125 km Vaniyambadi- Salem stretch was part of State highways until 2017 but It handed over to NH due to over traffic.

(d) Jharkhand