UPSC CISF-AC-EXE Archives

21. Which one of the following elements forms compounds with pronounced co

Which one of the following elements forms compounds with pronounced covalent character?

Lithium (Li)

Sodium (Na)

Potassium (K)

Rubidium (Rb)

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Lithium (Li) forms compounds with the most pronounced covalent character among the given options.

The formation of covalent character in predominantly ionic compounds is explained by Fajans’ Rules. These rules state that a compound is more likely to have covalent character if:
1. The cation is small.
2. The cation has a high charge (not applicable here as all form +1 ions).
3. The anion is large (not directly comparable here as the anion is not specified, but assume it’s a common one like a halide).
4. The cation has a pseudo noble gas configuration (not applicable here as alkali metals form noble gas configuration ions).

The given elements are alkali metals: Lithium (Li), Sodium (Na), Potassium (K), and Rubidium (Rb). When they form compounds, they form +1 ions: Li⁺, Na⁺, K⁺, Rb⁺.
Their ionic radii increase down the group: Li⁺ < Na⁺ < K⁺ < Rb⁺. According to Fajans' Rules, smaller cations have higher polarizing power (the ability to distort the electron cloud of the anion), which leads to increased covalent character in the bond. Li⁺ is the smallest cation among Li⁺, Na⁺, K⁺, and Rb⁺. Therefore, Li⁺ has the highest polarizing power and forms compounds with anions that exhibit the most pronounced covalent character compared to the compounds of Na, K, and Rb with the same anion. For example, LiCl has significantly more covalent character than NaCl, KCl, or RbCl.

Lithium exhibits several properties that are anomalous compared to the other alkali metals, often showing similarities to magnesium (diagonal relationship). This pronounced covalent character is one such anomaly, explaining why compounds like LiCl are soluble in organic solvents and why Li₂CO₃ is less stable to heat than other alkali metal carbonates.

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22. Two reactants in a flask produce bubbles of gas that turn limewater mi

Two reactants in a flask produce bubbles of gas that turn limewater milky. The gas is

SO₂

NO₂

CO₂

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The gas that turns limewater milky is CO₂.

Limewater is a dilute aqueous solution of calcium hydroxide, Ca(OH)₂. It is a common test reagent for carbon dioxide (CO₂).
When carbon dioxide gas is bubbled through limewater, it reacts with calcium hydroxide to form a precipitate of calcium carbonate (CaCO₃), which is insoluble in water and makes the solution appear milky or cloudy.
The chemical reaction is:
Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l)

If excess carbon dioxide is bubbled through the milky solution, the calcium carbonate precipitate reacts further to form soluble calcium bicarbonate, Ca(HCO₃)₂, and the milky appearance disappears.
CaCO₃(s) + CO₂(g) + H₂O(l) → Ca(HCO₃)₂(aq)

Other gases listed:
– SO₂ (Sulfur dioxide) also reacts with limewater, but forms calcium sulfite (CaSO₃), which is also insoluble and causes turbidity. However, the reaction with CO₂ is the most common and characteristic test for CO₂.
– NO₂ (Nitrogen dioxide) is an acidic gas and would react with Ca(OH)₂, but it typically doesn’t produce a milky precipitate with limewater as characteristically as CO₂.
– CO (Carbon monoxide) is a neutral gas and does not react with limewater.

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Given the standard chemical tests, turning limewater milky is the definitive test for CO₂.

This reaction is commonly used in experiments to detect the production of carbon dioxide, such as in respiration or combustion.

23. The horizontal component of the earth’s magnetic field is zero at

The horizontal component of the earth’s magnetic field is zero at

magnetic equator

magnetic poles

South and North Poles

nowhere

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The horizontal component of the earth’s magnetic field is zero at magnetic poles.

The Earth’s magnetic field lines emerge from near the geographic South Pole (which is the North magnetic pole) and enter near the geographic North Pole (which is the South magnetic pole).
– At the magnetic poles (where a compass needle points vertically downwards or upwards), the magnetic field lines are essentially perpendicular to the Earth’s surface. Therefore, the magnetic field vector has only a vertical component, and the horizontal component is zero.
– At the magnetic equator, the magnetic field lines are approximately parallel to the Earth’s surface. Therefore, the magnetic field vector is primarily horizontal, and the vertical component is zero.
– The geographic poles (South and North Poles) are points on the Earth’s rotational axis and do not necessarily coincide with the magnetic poles. The horizontal component is generally not zero at the geographic poles unless they happen to coincide perfectly with the magnetic poles (which they do not).

The Earth’s magnetic poles are not fixed and drift over time. The angle between the magnetic north and geographic north is called the magnetic declination. The angle between the horizontal plane and the Earth’s magnetic field line is called the magnetic dip or inclination; the dip is 90 degrees at the magnetic poles and 0 degrees at the magnetic equator.

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24. Stars twinkle in the sky at night because

Stars twinkle in the sky at night because

refractive index of the atmosphere changes due to the change of temperature

stars emit light in the form of pulses

of interference of light coming from different stars

of diffraction of light

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Stars twinkle in the sky at night because refractive index of the atmosphere changes due to the change of temperature.

Twinkling of stars (scintillation) is caused by atmospheric refraction.
– Light from distant stars travels through the Earth’s atmosphere before reaching our eyes.
– The atmosphere is not uniform; it consists of layers with varying temperatures and densities.
– Variations in temperature and density cause variations in the refractive index of the air.
– As light from a star passes through these turbulent layers with changing refractive index, it undergoes continuous refraction in random directions.
– This causes fluctuations in the apparent position and brightness of the star as seen from Earth. These rapid fluctuations are perceived as twinkling.
– Planets, being much closer, appear as extended sources of light rather than point sources. The light from different parts of a planet’s disc undergoes similar but independent variations, which average out, so planets do not twinkle noticeably.

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Option A correctly identifies the cause: changes in atmospheric refractive index due to temperature variations (and hence density variations) lead to varying refraction of starlight.
Option B is incorrect; stars emit light continuously.
Option C is incorrect; twinkling is an effect on light from a single star due to atmospheric effects, not interference from different stars.
Option D is incorrect; while diffraction occurs, twinkling is primarily an effect of refraction due to atmospheric turbulence.

Atmospheric refraction is also responsible for phenomena like the apparent flattening of the sun at sunrise/sunset and the fact that we can see the sun just before it rises and just after it sets. The degree of twinkling is affected by atmospheric conditions (turbulence).

25. The minimum length of a plane mirror to see your full length image is

The minimum length of a plane mirror to see your full length image is

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one quarter of your height

one-third of your height

half of your height

equal to your height

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The minimum length of a plane mirror to see your full length image is half of your height.

This is a fundamental principle of reflection from a plane mirror. To see the full image of an object in a plane mirror, the minimum vertical extent of the mirror required is half the vertical extent of the object.

Consider a person standing in front of a mirror. Light rays from the top of their head reflect off the top part of the mirror and enter their eyes. Light rays from their feet reflect off the bottom part of the mirror and enter their eyes.
– The angle of incidence equals the angle of reflection.
– The law of reflection dictates that to see the top of your head, the top edge of the mirror must be halfway between the top of your head and your eye level.
– Similarly, to see your feet, the bottom edge of the mirror must be halfway between your feet and your eye level.
– The distance between these two points on the mirror is the minimum required length. This distance is (1/2 * distance from head to eye) + (1/2 * distance from eye to feet). Since (distance from head to eye) + (distance from eye to feet) equals the total height of the person, the required mirror length is half the person’s height.

The distance of the person from the mirror does not affect the *minimum length* required, although it does affect the *field of view*.

This principle is utilized in everyday life, for instance, when installing full-length mirrors. The mirror doesn’t need to be as tall as the person.

26. A Kelvin thermometer and a Fahrenheit thermometer both give the same r

A Kelvin thermometer and a Fahrenheit thermometer both give the same reading for a certain sample. The corresponding Celsius temperature is about

301 °C

614 °C

276 °C

273 °C

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The corresponding Celsius temperature is about 301 °C.

We are given that a Kelvin thermometer and a Fahrenheit thermometer give the same reading for a certain sample. Let this reading be x.
So, the temperature in Kelvin (K) is x, and the temperature in Fahrenheit (F) is x.
We need to find the corresponding temperature in Celsius (C).

The conversion formulas between these scales are:
1. Fahrenheit to Celsius: C = (F – 32) * 5/9
2. Kelvin to Celsius: C = K – 273.15 (or often approximated as C = K – 273)

Let’s use the exact conversion K = C + 273.15 and F = (9/5)C + 32.
Since K = x and F = x, we have:
x = C + 273.15 (Equation 1)
x = (9/5)C + 32 (Equation 2)

Equating the right sides of Equation 1 and Equation 2:
C + 273.15 = (9/5)C + 32

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Rearrange the terms to solve for C:
273.15 – 32 = (9/5)C – C
241.15 = (9/5 – 5/5)C
241.15 = (4/5)C

C = (241.15 * 5) / 4
C = 1205.75 / 4
C = 301.4375

Rounding to the nearest whole number or considering the options, the corresponding Celsius temperature is about 301 °C.

The question asks for “about” the Celsius temperature, indicating an approximation is acceptable. Using the approximation C = K – 273 would yield C = 301.4375 – 0.15 ≈ 301.2875, still close to 301 °C. The unusual point where Kelvin and Fahrenheit scales read the same is 301.4375 in both K and °F.

27. Which one of the following pairs of quantities has no length in their

Which one of the following pairs of quantities has no length in their dimension?

Surface tension and angular momentum

Surface tension and strain

Angular momentum and mass density

Pressure gradient and angle

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The correct answer is Surface tension and strain.

We need to find the pair of quantities whose dimensions do not include length ([L]). Let’s analyze the dimensions of each quantity:
– **Surface tension:** Defined as force per unit length. Dimension is [Force]/[Length] = [MLT⁻²]/[L] = [MT⁻²]. It has no length dimension.
– **Angular momentum:** Defined as the product of moment of inertia and angular velocity (Iω) or cross product of position and linear momentum (r x p). Dimension is [ML²] * [T⁻¹] = [ML²T⁻¹]. It has a length dimension.
– **Strain:** Defined as the ratio of change in dimension to the original dimension (e.g., change in length / original length). Dimension is [L]/[L] = [Dimensionless]. It has no length dimension.
– **Mass density:** Defined as mass per unit volume. Dimension is [M]/[L³] = [ML⁻³]. It has a length dimension.
– **Pressure gradient:** Defined as change in pressure per unit distance. Dimension is [Pressure]/[Length]. Pressure is [Force]/[Area] = [MLT⁻²]/[L²] = [ML⁻²T⁻²]. So, pressure gradient is [ML⁻²T⁻²]/[L] = [ML⁻³T⁻²]. It has a length dimension.
– **Angle:** Defined as the ratio of arc length to radius. Dimension is [L]/[L] = [Dimensionless]. It has no length dimension.

Now let’s check the pairs:
A) Surface tension ([MT⁻²]) and angular momentum ([ML²T⁻¹]). Angular momentum has length.
B) Surface tension ([MT⁻²]) and strain ([Dimensionless]). Neither has length.
C) Angular momentum ([ML²T⁻¹]) and mass density ([ML⁻³]). Both have length.
D) Pressure gradient ([ML⁻³T⁻²]) and angle ([Dimensionless]). Pressure gradient has length.

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The pair with no length dimension is Surface tension and Strain.

Understanding the dimensional formulas of common physical quantities is crucial for solving such problems. Dimensional analysis helps verify the consistency of equations and understand the fundamental nature of physical quantities.

28. The transportation cost charged by a shipping company is proportional

The transportation cost charged by a shipping company is proportional to the square root of the distance and proportional to the volume of the parcel. If the distance is increased to 4 times, how much volume of the parcel can be transported with the same cost?

25%

50%

66%

75%

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The correct answer is 50%.

The transportation cost (C) is proportional to the square root of the distance (D) and proportional to the volume (V). This can be written as C ∝ √D * V, or C = k * √D * V, where k is a constant.

We are given that the cost remains the same (C₁ = C₂) while the distance is increased to 4 times (D₂ = 4D₁). We need to find the new volume (V₂) in terms of the original volume (V₁).

Using the formula:
C₁ = k * √D₁ * V₁
C₂ = k * √D₂ * V₂

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Since C₁ = C₂, we have:
k * √D₁ * V₁ = k * √D₂ * V₂
√D₁ * V₁ = √D₂ * V₂

Substitute D₂ = 4D₁:
√D₁ * V₁ = √(4D₁) * V₂
√D₁ * V₁ = 2√D₁ * V₂

Assuming D₁ > 0, we can divide both sides by √D₁:
V₁ = 2 * V₂

Solving for V₂:
V₂ = V₁ / 2

This means the new volume V₂ is half of the original volume V₁, which is 50%.

This type of problem involves understanding and applying direct and inverse proportionality relationships described in the problem statement. Keeping the cost constant requires adjusting the volume inversely proportionally to the square root of the distance.

29. A man rows at 6 km/hr in still water and with the same effort he rows

A man rows at 6 km/hr in still water and with the same effort he rows 4.5 km/hr against the stream of a river. What is the speed if he rows (with the same effort) along the stream of the river?

7½ km/hr

7 km/hr

6½ km/hr

6¼ km/hr

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Let v be the speed of the man in still water and u be the speed of the river stream.
Speed of man in still water, v = 6 km/hr.

Speed against the stream (upstream speed) = Speed in still water – Speed of stream = v – u.
We are given that the speed against the stream is 4.5 km/hr.
v – u = 4.5
Substitute the value of v:
6 – u = 4.5
u = 6 – 4.5 = 1.5 km/hr.
The speed of the river stream is 1.5 km/hr.

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We need to find the speed if he rows along the stream (downstream speed) with the same effort. The effort remaining the same means his speed relative to the water is still v.
Speed along the stream (downstream speed) = Speed in still water + Speed of stream = v + u.
Substitute the values of v and u:
Downstream speed = 6 + 1.5 = 7.5 km/hr.

The question asks for the speed in one of the given options. 7.5 km/hr is equivalent to 7 and a half km/hr.
7.5 = 7 + 0.5 = 7 + 1/2 = 7½ km/hr.

– Speed upstream = Speed in still water – Speed of stream.
– Speed downstream = Speed in still water + Speed of stream.
– The effort remaining the same implies the speed of the man relative to the water (speed in still water) is constant.

If upstream speed and downstream speed are given, the speed in still water and speed of the stream can be found:
Speed in still water = (Downstream speed + Upstream speed) / 2
Speed of stream = (Downstream speed – Upstream speed) / 2
In this case, we found v=6 and u=1.5.
Upstream speed = 6 – 1.5 = 4.5 km/hr (Given).
Downstream speed = 6 + 1.5 = 7.5 km/hr (Calculated).
Check using the formulas:
Speed in still water = (7.5 + 4.5) / 2 = 12 / 2 = 6 km/hr (Matches given v).
Speed of stream = (7.5 – 4.5) / 2 = 3 / 2 = 1.5 km/hr (Matches calculated u).

30. The diameter of a wheel is 1.26 m. How far will it travel in 500

The diameter of a wheel is 1.26 m. How far will it travel in 500 revolutions?

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1492 m

1980 m

2530 m

2880 m

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The distance a wheel travels in one revolution is equal to its circumference.
The diameter of the wheel is given as 1.26 m.
The circumference of a circle is given by the formula $C = \pi \times d$, where d is the diameter.
Circumference = $\pi \times 1.26$ meters.

The wheel makes 500 revolutions.
The total distance traveled is the distance per revolution multiplied by the number of revolutions.
Total distance = Circumference $\times$ Number of revolutions
Total distance = $(\pi \times 1.26) \times 500$ meters.

We can use the approximation $\pi \approx \frac{22}{7}$ or $\pi \approx 3.14$. Since the diameter 1.26 is divisible by 7 ($1.26 = 0.18 \times 7$), using $\pi \approx \frac{22}{7}$ will likely give a precise answer or one of the options.
$1.26 / 7 = 0.18$.
Total distance = $\frac{22}{7} \times 1.26 \times 500$
Total distance = $22 \times (1.26 / 7) \times 500$
Total distance = $22 \times 0.18 \times 500$
Total distance = $22 \times (0.18 \times 500)$
$0.18 \times 500 = 18 \times 5 = 90$.
Total distance = $22 \times 90$.
Total distance = $1980$ meters.

– The distance covered by a wheel in one revolution is equal to its circumference.
– Circumference of a circle = $\pi \times \text{diameter}$ or $2 \times \pi \times \text{radius}$.
– Total distance = Circumference $\times$ Number of revolutions.

Using $\pi \approx 3.14$:
Circumference $\approx 3.14 \times 1.26 \approx 3.9564$ m.
Total distance $\approx 3.9564 \times 500 \approx 1978.2$ m.
This is close to 1980 m, suggesting that 1980 m is the correct answer and the calculation with 22/7 was intended. The value 1.26 is specifically chosen to be a multiple of 0.07, making the use of $\pi = 22/7$ convenient.