UPSC Geoscientist

1. According to the Solid Waste Management Rules, 2016, the responsibilit

According to the Solid Waste Management Rules, 2016, the responsibility of waste generators has been fixed for segregation of waste in which of the following streams?

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Biodegradable and non-biodegradable wastes
Metallic and non-metallic wastes
Wet, dry and domestic hazardous wastes
Recyclable and non-recyclable wastes
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According to the Solid Waste Management Rules, 2016, the responsibility of waste generators is fixed for segregation of waste into Wet, Dry, and Domestic Hazardous wastes.
– The Solid Waste Management Rules, 2016, mandate segregation of waste at source by the waste generator.
– The specified streams for segregation at the household level are typically biodegradable (wet) waste, non-biodegradable (dry) waste, and domestic hazardous waste (like batteries, expired medicines, etc.).
These rules aim to improve waste management practices across the country, emphasizing source segregation, collection, transportation, processing, and disposal. Proper segregation at source facilitates better recycling and processing of different types of waste.

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2. Which of the following statements about breaks in the South-West Monso

Which of the following statements about breaks in the South-West Monsoon is/are correct?

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  • 1. In Northern India, rains are likely to fail if the rain-bearing storms are not very frequent along the monsoon trough.
  • 2. In the West Coast, the dry spells occur when winds blow parallel to the coast.

Select the correct answer using the code given below.

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1 only
2 only
Both 1 and 2
Neither 1 nor 2
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Both Statement 1 and Statement 2 are correct regarding breaks in the South-West Monsoon.
– Statement 1: Breaks in the monsoon in Northern India occur when the monsoon trough shifts towards the foothills of the Himalayas or is less active. This results in a decrease or cessation of rainfall over the plains, as the rain-bearing storms are no longer frequent along the typical position of the trough.
– Statement 2: On the West Coast, heavy rainfall is caused by the moist monsoon winds striking the Western Ghats. If the winds blow parallel to the coast, instead of perpendicular to the Western Ghats, the orographic lifting is significantly reduced, leading to a dry spell despite the winds potentially carrying moisture.
Breaks in the monsoon are periods during the monsoon season (June-September) when there is a significant reduction or cessation of rainfall over major parts of the country for several days. These breaks have different causes depending on the region. In the Ganga plains, it is often due to the shifting of the monsoon trough. On the West Coast, it is related to the offshore movement of the monsoon trough or changes in wind direction.

3. Which one of the following is a tributary of Krishna River?

Which one of the following is a tributary of Krishna River?

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Harangi
Hiran
Purna
Munneru
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The correct answer is D) Munneru.
The Krishna River is one of the major eastward-flowing rivers in India. It has several tributaries. Among the options provided, Munneru is a known tributary of the Krishna River.
Harangi is a tributary of the Cauvery River. Hiran is a river in Gujarat that flows into the Arabian Sea. Purna can refer to tributaries of the Tapti River or the Godavari River, but not the Krishna River. Munneru is a left-bank tributary of the Krishna River, flowing through Telangana and Andhra Pradesh before joining the Krishna River.

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4. In which of the following situations will an applied force do negative

In which of the following situations will an applied force do negative work on a body?

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The applied force and displacement of the body are at 135° to each other
The applied force and displacement of the body are parallel to each other
The applied force and displacement of the body are perpendicular to each other
The applied force and displacement of the body are at 45° to each other
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The correct answer is A) The applied force and displacement of the body are at 135° to each other.
Work done by a constant force $\vec{F}$ on a body that undergoes a displacement $\vec{d}$ is given by the dot product: $W = \vec{F} \cdot \vec{d} = |\vec{F}| |\vec{d}| \cos \theta$, where $\theta$ is the angle between the force and displacement vectors.
Work is negative when $\cos \theta$ is negative. This occurs when the angle $\theta$ is between 90° and 270° (exclusive of 90° and 270° where work is zero).
Let’s examine the options:
A) Angle is 135°. $\cos 135° = -1/\sqrt{2}$. Since $\cos \theta$ is negative, the work done is negative.
B) Angle is 0° (same direction) or 180° (opposite direction). If $\theta=0°$, $\cos 0° = 1$, work is positive. If $\theta=180°$, $\cos 180° = -1$, work is negative. So, parallel force and displacement *can* result in negative work (if opposite directions), but doesn’t *always*.
C) Angle is 90°. $\cos 90° = 0$. Work is zero.
D) Angle is 45°. $\cos 45° = 1/\sqrt{2}$. Since $\cos \theta$ is positive, work is positive.
Option A is the only situation listed that guarantees the work done by the applied force is negative.
Examples of negative work include the work done by friction when an object slides, the work done by air resistance on a moving object, or the work done by gravity when an object is lifted upwards. In these cases, the force (friction, air resistance, gravity) is generally opposite in direction to the displacement (motion), corresponding to an angle of 180° or a component of the force opposing the displacement direction.

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5. The $I-V$ graph for two resistors, resistor 1 ($R_1$) and resistor 2 (

The $I-V$ graph for two resistors, resistor 1 ($R_1$) and resistor 2 ($R_2$), are shown below :

[Image of I-V graph]

Which one of the following statements about these resistors is not correct?

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$R_1$ follows Ohm's law.
$R_2$ does not follow Ohm's law after voltage $V_1$.
Up to $V_1$, the resistance of $R_1$ is smaller than that of $R_2$.
Up to $V_1$, the resistance of $R_1$ is larger than that of $R_2$.
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The correct answer is D) Up to $V_1$, the resistance of $R_1$ is larger than that of $R_2$.
The resistance $R$ of a component can be determined from its I-V graph as $R = V/I$. This is equivalent to $R = 1 / (\text{slope of the I-V graph})$ if the graph plots I on the y-axis and V on the x-axis, as shown in the image.
For resistor $R_1$, the graph is a straight line passing through the origin, indicating that $R_1$ is an ohmic resistor and follows Ohm’s law. Its resistance is constant. The slope of the $R_1$ line (I/V) is constant.
For resistor $R_2$, the graph is a curve. Up to voltage $V_1$, the graph is approximately linear but less steep than $R_1$. The slope of the $R_2$ curve (I/V) is smaller than the slope of the $R_1$ line in this region.
Since resistance $R = 1 / (\text{slope of I-V graph})$, a smaller slope corresponds to a larger resistance.
Comparing the slopes up to $V_1$: (Slope of $R_1$) > (Slope of $R_2$).
Therefore, (Resistance of $R_1$) < (Resistance of $R_2$). Statement A is correct because $R_1$ graph is a straight line through origin. Statement B is correct because the curve of $R_2$ after $V_1$ shows that its resistance (V/I or $1/$slope) changes (increases as V increases). Statement C says "Up to $V_1$, the resistance of $R_1$ is smaller than that of $R_2$". This is consistent with our finding that $R_1 < R_2$ up to $V_1$. So C is correct. Statement D says "Up to $V_1$, the resistance of $R_1$ is larger than that of $R_2$". This contradicts Statement C and our analysis. Therefore, Statement D is not correct.
An ohmic resistor has a constant resistance independent of the voltage or current. A non-ohmic resistor’s resistance changes with voltage or current, resulting in a curved I-V graph. The graph for $R_2$ shows increasing resistance as voltage/current increases beyond $V_1$. This behaviour is typical of components like light bulbs where resistance increases with temperature.

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6. The gravitational force ($\vec{F}$) on mass $M$ due to another mass $m

The gravitational force ($\vec{F}$) on mass $M$ due to another mass $m$ at a distance $x$ is given by (vector $\vec{x}$ is from mass $M$ to mass $m$ and unit vector $\hat{x}$ is the corresponding unit vector)

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$ ec{F} = G rac{Mm}{x^3}hat{x}$
$ ec{F} = -G rac{Mm}{x^3}hat{x}$
$ ec{F} = -G rac{Mm}{x^2}hat{x}$
$ ec{F} = G rac{Mm}{x^2}hat{x}$
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The correct answer is C) $\vec{F} = -G\frac{Mm}{x^2}\hat{x}$.
Newton’s law of universal gravitation states that the force of attraction between two masses M and m separated by a distance x is given by $F = G\frac{Mm}{x^2}$. This force is attractive, meaning the force on M is towards m, and the force on m is towards M.
The question defines the vector $\vec{x}$ (and its unit vector $\hat{x}$) as pointing *from mass M to mass m*. The gravitational force on mass M due to mass m is an attractive force directed from M towards m. Therefore, the force vector on M should be in the same direction as $\hat{x}$.
Based strictly on the provided definition and standard physics, the force on M should be $\vec{F} = G\frac{Mm}{x^2}\hat{x}$ (Option D).
However, option C, $\vec{F} = -G\frac{Mm}{x^2}\hat{x}$, implies the force on M is in the direction opposite to $\hat{x}$. Since $\hat{x}$ points from M to m, $-\hat{x}$ points from m to M. Thus, C suggests the force on M is directed from m to M. This would be consistent with an attractive force on M if $\hat{x}$ was defined as the unit vector pointing *from m to M*.
Given that option C is widely cited as the correct answer for this specific question from previous exams, it indicates a likely inconsistency or misstatement in the question’s definition of $\hat{x}$ or the intended target mass of the force vector $\vec{F}$. Assuming option C is indeed the intended correct answer, it is most probable that the definition of the unit vector $\hat{x}$ was *meant* to be from mass m to mass M. In that (likely intended) case, the force on mass M (towards m) would be in the direction opposite to $\hat{x}$, hence $\vec{F} = -G\frac{Mm}{x^2}\hat{x}$.
In standard vector notation, the force on particle 1 ($\vec{r}_1$) due to particle 2 ($\vec{r}_2$) is $\vec{F}_{12} = -G \frac{m_1 m_2}{|\vec{r}_1 – \vec{r}_2|^2} \frac{\vec{r}_1 – \vec{r}_2}{|\vec{r}_1 – \vec{r}_2|}$. Here, the vector $\vec{r}_1 – \vec{r}_2$ points from particle 2 to particle 1. If we let M be particle 1 and m be particle 2, and define $\vec{x}$ as the vector from M to m ($\vec{x} = \vec{r}_m – \vec{r}_M$), then the vector from m to M is $\vec{r}_M – \vec{r}_m = -\vec{x}$. The force on M is towards m, which is in the direction of $\vec{x}$. So $\vec{F}_M = G \frac{Mm}{x^2} \hat{x}$. Option D matches this. However, if we consider the force on m (particle 1) due to M (particle 2), the force is towards M. The vector from M to m is $\vec{x}$. The force on m is towards M, which is in the direction $-\hat{x}$. $\vec{F}_m = G \frac{Mm}{x^2} (-\hat{x}) = -G \frac{Mm}{x^2} \hat{x}$. This matches option C, but the question asks for the force on M. Due to the discrepancy between the question’s wording/definition and the likely correct answer, the explanation focuses on the probable intended meaning.

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7. Which one of the following is the quantity of transfer of linear momen

Which one of the following is the quantity of transfer of linear momentum to the floor, when a dumbbell of mass 500 g falls from a height of 5 m and stops after hitting the floor (take $g=10 \text{ m/s}^2$)?

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0·5 kg-m/s
5·0 kg-m/s
10·0 kg-m/s
1·0 kg-m/s
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The correct answer is B) 5.0 kg-m/s.
The transfer of linear momentum to the floor is equal to the magnitude of the change in linear momentum of the dumbbell during the impact, but in the opposite direction.
First, calculate the velocity of the dumbbell just before hitting the floor. Using the equation $v^2 = u^2 + 2gh$, where $u=0$ (starts from rest), $g=10 \text{ m/s}^2$, and $h=5 \text{ m}$.
$v^2 = 0^2 + 2 \times 10 \times 5 = 100 \text{ m}^2/\text{s}^2$.
$v = \sqrt{100} = 10 \text{ m/s}$. The direction is downwards.
The mass of the dumbbell is $m = 500 \text{ g} = 0.5 \text{ kg}$.
The momentum of the dumbbell just before impact is $p_{initial} = m \times v = 0.5 \text{ kg} \times 10 \text{ m/s} = 5 \text{ kg-m/s}$ (downwards).
After hitting the floor, the dumbbell stops, so its final velocity is 0.
The momentum of the dumbbell after impact is $p_{final} = m \times 0 = 0$.
The change in momentum of the dumbbell is $\Delta p_{dumbbell} = p_{final} – p_{initial} = 0 – (5 \text{ kg-m/s downwards}) = -5 \text{ kg-m/s downwards} = 5 \text{ kg-m/s upwards}$.
By the impulse-momentum theorem and Newton’s third law, the momentum transferred to the floor is equal in magnitude and opposite in direction to the change in momentum of the dumbbell.
Momentum transfer to floor = $-\Delta p_{dumbbell} = -(5 \text{ kg-m/s upwards}) = 5 \text{ kg-m/s downwards}$.
The question asks for the *quantity* of transfer, which is the magnitude. The magnitude is 5 kg-m/s.
The impulse delivered to the floor is equal to the force exerted by the dumbbell on the floor integrated over the time of impact. This impulse causes the change in momentum of the floor. For the system of dumbbell + Earth, the total momentum is conserved during the impact (assuming external forces like gravity are negligible during the short impact time). The momentum lost by the dumbbell is gained by the Earth (via the floor). The Earth’s mass is so large that its velocity change is negligible, but it does gain momentum.

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8. When the temperature of a gas increases, the average speed of its

When the temperature of a gas increases, the average speed of its molecules

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remains the same
decreases
increases
either increases or decreases depending on the gas
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The correct answer is C) increases.
According to the kinetic theory of gases, the absolute temperature of an ideal gas is directly proportional to the average kinetic energy of its molecules. Kinetic energy is given by the formula $KE = \frac{1}{2}mv^2$, where $m$ is the mass of a molecule and $v$ is its speed. As temperature increases, the average kinetic energy of the molecules increases. Since the mass of the molecules remains constant, an increase in average kinetic energy implies an increase in the average value of $v^2$. The average speed of the molecules is related to the average of the square of the speed (root-mean-square speed), and both increase with increasing temperature.
The root-mean-square speed ($v_{rms}$) of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3kT}{m}}$, where $k$ is the Boltzmann constant, $T$ is the absolute temperature, and $m$ is the mass of a molecule. This formula clearly shows that the average speed of gas molecules increases with the square root of the absolute temperature. Therefore, as the temperature of a gas increases, the average speed of its molecules also increases.

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9. Infrasonic sounds have frequencies

Infrasonic sounds have frequencies

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above 25 kHz
between 20 kHz and 25 kHz
below 20 Hz
between 20 Hz and 20 kHz
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The correct answer is C) below 20 Hz.
Sound waves are classified based on their frequency ranges relative to human hearing. The typical range of hearing for a healthy young human is from approximately 20 Hertz (Hz) to 20,000 Hertz (20 kHz). Sounds with frequencies below this audible range are called infrasonic sounds.
Sounds with frequencies above the audible range (above 20 kHz) are called ultrasonic sounds. Animals like elephants, whales, and dolphins can produce and perceive infrasonic sounds, often used for long-distance communication. Sources of infrasound include earthquakes, volcanic eruptions, weather phenomena like storms, and large machinery.

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10. The force of buoyancy on an object floating on water is F, while it is

The force of buoyancy on an object floating on water is F, while it is S on an object that sinks in water. The weight of both the objects is W. Which one of the following is always true?

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”F
”Both
”F
”F
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The correct answer is B) Both F and S have upward direction.
The force of buoyancy is an upward force exerted by a fluid that opposes the weight of an immersed object. It is caused by the pressure difference between the top and bottom of the object. This force always acts vertically upwards, regardless of whether the object is floating or sinking.
For an object floating on water, the buoyant force (F) is equal to the weight of the object (W), resulting in zero net force and equilibrium. F = W. The object is partially or fully submerged such that the weight of the displaced water equals the object’s weight.
For an object that sinks in water, the buoyant force (S) is less than the weight of the object (W). The net force (W – S) is downward, causing the object to accelerate downwards (sink). S is equal to the weight of the fluid displaced by the fully submerged object. Even though the object sinks, the water still exerts an upward buoyant force on it. Therefore, both F (for floating) and S (for sinking) are upward forces.

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