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11. Which of the following tree species is/are found on Himalayas ? 1. O

Which of the following tree species is/are found on Himalayas ?

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1. Oak
2. Rhododendron
3. Rosewood

Select the correct answer using the code given below :

1 only

2 and 3

1 and 3

1 and 2

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The correct answer is D) 1 and 2.

The Himalayas exhibit a wide range of vegetation types depending on altitude and rainfall. Oak (Quercus species) trees are commonly found in the temperate forest zones of the Himalayas. Rhododendron species are also characteristic plants of the Himalayan region, particularly in the temperate to sub-alpine zones, known for their vibrant flowers. Rosewood (e.g., Dalbergia latifolia) is primarily a tropical hardwood species found in peninsular India, especially in the Western Ghats, and is not typical of the Himalayan flora.

Himalayan vegetation transitions from tropical deciduous forests at lower altitudes to temperate forests (including oaks, pines, deodars, firs, spruces), alpine meadows, and finally tundra/nival zones at the highest elevations. Rhododendrons are a prominent feature of the temperate and alpine flora.

12. Which one of the following is found in the innermost part of the Earth

Which one of the following is found in the innermost part of the Earth ?

Conrad discontinuity

Moho discontinuity

Guttenberg discontinuity

Lehmann discontinuity

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The correct answer is D) Lehmann discontinuity.

The Earth is structured in layers: crust, mantle, outer core, and inner core. Discontinuities are boundaries where seismic waves change speed due to changes in material properties. The Lehmann discontinuity is located at a depth of approximately 5150 km and separates the liquid outer core from the solid inner core. Since the inner core is the innermost part of the Earth, the Lehmann discontinuity is found at the boundary of this innermost region.

– The Conrad discontinuity is a theoretical or regional boundary within the continental crust.
– The Moho discontinuity (Mohorovičić discontinuity) is the boundary between the Earth’s crust and the mantle (around 35-70 km under continents, 5-10 km under oceans).
– The Guttenberg discontinuity (also known as the Wiechert–Gutenberg discontinuity or Core–mantle boundary) is the boundary between the Earth’s mantle and the outer core (around 2900 km depth).

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13. In the soil-forming regime, which one of the following occurs in a reg

In the soil-forming regime, which one of the following occurs in a region where evapotranspiration exceeds precipitation significantly ?

Calcification

Laterization

Podsolization

Gleization

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Soil-forming regimes are characteristic processes that occur in specific environmental conditions (primarily climate and vegetation) and lead to the development of different soil types. A region where evapotranspiration significantly exceeds precipitation is characterized by a dry climate, typical of arid or semi-arid environments.

– Calcification: Occurs in arid/semi-arid regions where precipitation is insufficient to leach soluble salts, particularly calcium carbonate, from the soil profile. Calcium accumulates in the upper horizons or forms a hardpan (caliche) due to upward capillary movement of water followed by evaporation.
– Laterization: Occurs in hot, humid climates with high rainfall, leading to intense leaching and accumulation of iron and aluminum oxides.
– Podsolization: Occurs in cool, humid climates, typically under coniferous forests, resulting in the leaching of iron, aluminum, and organic matter from upper horizons and their accumulation in lower horizons.
– Gleization: Occurs in waterlogged, anaerobic conditions, leading to the reduction of iron and characteristic grey/blue colours and mottles.

In regions where evapotranspiration exceeds precipitation, there is a net upward movement of water through capillary action in the soil profile. This water often carries dissolved minerals, including calcium. As the water evaporates at or near the surface, the dissolved minerals are left behind, leading to their accumulation. This process of calcium carbonate accumulation is known as calcification, which is characteristic of soils in dry climates.

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14. The Counter Insurgency and Jungle Warfare School of Indian Army is sit

The Counter Insurgency and Jungle Warfare School of Indian Army is situated at :

Dehradun

Vairengte

Gulmarg

Mhow

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The Counter Insurgency and Jungle Warfare School (CIJW School) is a premier training institution of the Indian Army dedicated to imparting training in counter-insurgency, counter-terrorism, and jungle warfare tactics. Its location is specific and well-known within India’s defense establishment.

The CIJW School is located in Vairengte, a town in the state of Mizoram in Northeast India. This region provides a suitable environment for jungle warfare training.

– Dehradun is home to the Indian Military Academy (IMA), which trains officers for the Indian Army.
– Gulmarg, in Jammu and Kashmir, is known for the High Altitude Warfare School (HAWS), which trains personnel in mountain and snow warfare.
– Mhow, in Madhya Pradesh, is the location of several Army institutions, including the Infantry School, Army War College, and MCTE (Military College of Telecommunication Engineering).

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15. Deendayal Port was earlier known as :

Deendayal Port was earlier known as :

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Paradip Port

Tuticorin Port

Kandla Port

Visakhapatnam Port

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Many ports in India have been renamed over the years, often after prominent national figures. Deendayal Port is a major port on the west coast of India, located in Gujarat.

The port formerly known as Kandla Port was renamed Deendayal Port in 2017 in honour of Pandit Deendayal Upadhyaya.

– Paradip Port: Major port in Odisha.
– Tuticorin Port: Major port in Tamil Nadu, officially renamed V.O. Chidambaranar Port.
– Kandla Port: Major port in Gujarat, renamed Deendayal Port. It is known for handling large volumes of cargo, especially petroleum, fertilizers, and food grains. It is one of the largest ports by cargo volume in India.
– Visakhapatnam Port: Major port in Andhra Pradesh.

16. Which of the following scheme(s) is/are included under Bharatmala Pari

Which of the following scheme(s) is/are included under Bharatmala Pariyojana ?

1. Develop the road connectivity to border areas
2. Development of coastal roads
3. Improvement in the efficiency of National Corridors

Select the correct answer using the code given below :

1 only

3 only

2 and 3 only

1, 2 and 3

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Bharatmala Pariyojana is an ambitious umbrella program of the Government of India for highways development, focusing on optimising the efficiency of freight and passenger movement across the country by bridging critical infrastructure gaps. It encompasses various components aimed at improving the overall road network.

The stated objectives/components of Bharatmala Pariyojana include:
1. Development of Economic Corridors (around 9,000 km).
2. Development of Inter Corridors and Feeder Routes (around 6,000 km).
3. National Corridor Efficiency Improvement (around 5,000 km, decongestion of existing corridors).
4. Border Road Development and International Connectivity Roads (around 3,400 km).
5. Coastal Road Development and Port Connectivity Roads (around 2,000 km).
6. Development of Greenfield Expressways (around 800 km).

All three points mentioned in the question (1. Develop the road connectivity to border areas, 2. Development of coastal roads, 3. Improvement in the efficiency of National Corridors) are explicit components or objectives of the Bharatmala Pariyojana as outlined by the Ministry of Road Transport and Highways, Government of India.

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17. What is the total resistance in the following circuit element ? [Image

What is the total resistance in the following circuit element ?
[Image of a circuit diagram with three resistors labelled R. Two are in parallel, and this combination is in series with the third resistor.]

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R/2

3R/2

2R/3

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The circuit element consists of three resistors, each with resistance R. Two of the resistors are connected in parallel, and this parallel combination is connected in series with the third resistor. To find the total resistance, we first calculate the equivalent resistance of the parallel part and then add it to the resistance of the series part.

– For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + …$
– For resistors in series, the equivalent resistance is the sum of individual resistances: $R_s = R_1 + R_2 + …$

Let the resistance of each resistor be R.
The two resistors in parallel have equivalent resistance $R_p$.
$\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}$.
$R_p = \frac{R}{2}$.
This parallel combination ($R_p$) is in series with the third resistor (R).
The total resistance $R_{total}$ is the sum of the series components:
$R_{total} = R_p + R = \frac{R}{2} + R = \frac{R + 2R}{2} = \frac{3R}{2}$.

18. In an electric circuit, a wire of resistance 10 Ω is used. If this wir

In an electric circuit, a wire of resistance 10 Ω is used. If this wire is stretched to a length double of its original value, the current in the circuit would become :

half of its original value.

double of its original value.

one-fourth of its original value.

four times of its original value.

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The resistance of a wire depends on its material (resistivity), length, and cross-sectional area. When a wire is stretched, its length increases, and its cross-sectional area decreases, while its volume remains constant. This change in dimensions affects the resistance. The current in the circuit is inversely proportional to the resistance (assuming constant voltage).

– Resistance R = $\rho \frac{L}{A}$, where $\rho$ is resistivity, L is length, and A is cross-sectional area.
– Volume of the wire V = A × L. When stretched, the volume remains constant: $A’L’ = AL$.
– Ohm’s Law: Current I = Voltage V / Resistance R.

Let the original length be L and the original cross-sectional area be A. The original resistance is $R = \rho \frac{L}{A}$.
The wire is stretched to a new length $L’ = 2L$.
Since the volume remains constant, $A’L’ = AL$.
$A'(2L) = AL \Rightarrow A’ = \frac{A}{2}$.
The new resistance is $R’ = \rho \frac{L’}{A’} = \rho \frac{2L}{A/2} = \rho \frac{2L \times 2}{A} = 4 \rho \frac{L}{A}$.
So, $R’ = 4R$. The resistance becomes four times the original resistance.
Assuming the voltage V across the circuit element (or the voltage source) is constant:
Original current $I = \frac{V}{R}$.
New current $I’ = \frac{V}{R’} = \frac{V}{4R} = \frac{1}{4} \left(\frac{V}{R}\right) = \frac{1}{4}I$.
The current in the circuit would become one-fourth of its original value.

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19. Which one of the following is an example of Second Class Lever ?

Which one of the following is an example of Second Class Lever ?

A pair of scissors

A bottle opener

A cricket bat

A bow and arrow

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Levers are classified into three types based on the relative positions of the fulcrum (pivot point), the load (resistance), and the effort (applied force).

– Class 1 Lever: Fulcrum is between the effort and the load (e.g., see-saw, scissors, pliers).
– Class 2 Lever: Load is between the fulcrum and the effort (e.g., wheelbarrow, bottle opener, nutcracker). Class 2 levers always provide mechanical advantage (>1).
– Class 3 Lever: Effort is between the fulcrum and the load (e.g., fishing rod, forceps, cricket bat, tweezers). Class 3 levers always have mechanical advantage (<1), prioritizing range of motion or speed over force multiplication.

– A pair of scissors consists of two Class 1 levers joined at the pivot (fulcrum).
– A bottle opener uses the lip of the bottle cap as the fulcrum. The opener is used to lift the cap (load) by applying force (effort) at the end of the handle. The load (cap) is lifted between the fulcrum and the effort. Hence, it is a Class 2 lever.
– A cricket bat has the player’s hands applying effort between the pivot (often lower hand near the handle end) and the point of impact with the ball (load). Hence, it is a Class 3 lever.
– A bow and arrow is not typically classified as a simple lever system.

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20. A mass is attached to a spring that hangs vertically. The extension pr

A mass is attached to a spring that hangs vertically. The extension produced in the spring is 6 cm on Earth. The acceleration due to gravity on the surface of the Moon is one-sixth of its value on the surface of the Earth. The extension of the spring on the Moon would be :

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6 cm

1 cm

0 cm

36 cm

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According to Hooke’s Law, the extension of a spring is directly proportional to the force applied to it, provided the elastic limit is not exceeded. The force stretching the vertical spring is the weight of the attached mass. The weight of the mass is proportional to the acceleration due to gravity.

– Hooke’s Law: Force (F) = spring constant (k) × extension (x), or F = kx.
– Weight (W) = mass (m) × acceleration due to gravity (g).
– On Earth: $W_{Earth} = mg_{Earth} = kx_{Earth}$.
– On the Moon: $W_{Moon} = mg_{Moon} = kx_{Moon}$.

Given: Extension on Earth $x_{Earth} = 6$ cm.
Acceleration due to gravity on the Moon $g_{Moon} = \frac{1}{6} g_{Earth}$.
From Hooke’s Law on Earth: $mg_{Earth} = k \times 6$ cm. So, $\frac{mg_{Earth}}{k} = 6$ cm.
On the Moon: $mg_{Moon} = kx_{Moon}$.
Substitute $g_{Moon}$: $m \left(\frac{g_{Earth}}{6}\right) = kx_{Moon}$.
$\frac{1}{6} (mg_{Earth}) = kx_{Moon}$.
Substitute $mg_{Earth} = k \times 6$ cm: $\frac{1}{6} (k \times 6 \text{ cm}) = kx_{Moon}$.
$k \times 1 \text{ cm} = kx_{Moon}$.
$x_{Moon} = 1$ cm.
The extension is directly proportional to the weight, and thus directly proportional to gravity. Since gravity on the Moon is one-sixth of Earth’s, the extension will also be one-sixth.