UPSC CDS-1

1. Which one of the following pairs of animals is warm-blooded?

Which one of the following pairs of animals is warm-blooded?

Crocodile and Ostrich
Hagfish and Dogfish
Tortoise and Ostrich
Peacock and Camel
This question was previously asked in
UPSC CDS-1 – 2019
Download PDFAttempt Online
Warm-blooded (endothermic) animals are those that maintain a stable internal body temperature regardless of the external environment. Birds (like Peacock and Ostrich) and mammals (like Camel) are warm-blooded. In option D, both Peacock (a bird) and Camel (a mammal) are warm-blooded animals.
Birds and Mammals are warm-blooded (endothermic).
Crocodiles and Tortoises are reptiles, which are cold-blooded (ectothermic). Hagfish and Dogfish are fish, which are also cold-blooded.

2. Which one of the following has a bilateral symmetry in its body

Which one of the following has a bilateral symmetry in its body organization?

Asterias
Sea anemone
Nereis
Echinus
This question was previously asked in
UPSC CDS-1 – 2019
Download PDFAttempt Online
Nereis is a marine polychaete worm belonging to the phylum Annelida. Annelids typically exhibit bilateral symmetry, where the body can be divided into two mirror-image halves along a single sagittal plane.
Bilateral symmetry divides an organism into two symmetrical halves (left and right).
Asterias (starfish) and Echinus (sea urchin) are echinoderms, which typically exhibit radial symmetry (specifically, pentamerous radial symmetry in adults). Sea anemones belong to the phylum Cnidaria and exhibit radial symmetry.

3. Which one of the following organisms is dependent on saprophytic mode

Which one of the following organisms is dependent on saprophytic mode of nutrition?

Agaricus
Ulothrix
Riccia
Cladophora
This question was previously asked in
UPSC CDS-1 – 2019
Download PDFAttempt Online
Agaricus is a genus of fungi, commonly known as mushrooms. Fungi are heterotrophic organisms that obtain nutrients by absorption. Many fungi, including Agaricus, are saprophytic, meaning they feed on dead and decaying organic matter.
Saprophytic nutrition involves obtaining nutrients from dead organic material. Fungi are often saprophytic.
Ulothrix and Cladophora are green algae, which are photosynthetic and therefore autotrophic. Riccia is a liverwort, a type of bryophyte, which is also photosynthetic and autotrophic.

4. Which one of the following cell organelles mainly functions as storeho

Which one of the following cell organelles mainly functions as storehouse of digestive enzymes?

Desmosome
Ribosome
Lysosome
Vacuoles
This question was previously asked in
UPSC CDS-1 – 2019
Download PDFAttempt Online
Lysosomes are membrane-bound organelles containing a variety of hydrolytic enzymes (like proteases, lipases, nucleases, glycosidases) that are active at acidic pH. Their main function is the intracellular digestion of macromolecules, worn-out organelles, and pathogens. Thus, they serve as the principal storehouse of digestive enzymes within the cell.
Lysosomes are responsible for cellular digestion and waste breakdown.
Ribosomes are involved in protein synthesis. Desmosomes are structures that provide strong adhesion between cells. Vacuoles in animal cells are small and involved in storage or transport, while in plant cells they are large and maintain turgor pressure and store water, nutrients, and waste; they can also contain some digestive enzymes in plant cells, but lysosomes are the primary organelle for this function in many eukaryotic cells.

5. Which one of the following functions is not carried out by smooth en

Which one of the following functions is not carried out by smooth endoplasmic reticulum?

Transport of materials
Synthesis of lipid
Synthesis of protein
Synthesis of steroid hormone
This question was previously asked in
UPSC CDS-1 – 2019
Download PDFAttempt Online
The correct option is C.
The endoplasmic reticulum (ER) is involved in various cellular functions. The rough endoplasmic reticulum (RER) is characterized by the presence of ribosomes on its surface and is primarily responsible for the synthesis and modification of proteins destined for secretion, insertion into membranes, or delivery to other organelles. The smooth endoplasmic reticulum (SER) lacks ribosomes and is involved in lipid synthesis, steroid hormone synthesis, detoxification, and calcium storage. Protein synthesis is mainly carried out by ribosomes, either free in the cytoplasm or attached to the RER. Therefore, synthesis of protein is not a primary function of the smooth endoplasmic reticulum.
Transport of materials (lipids, proteins) through vesicles budding from the ER is a function of both RER and SER. Steroid hormones are lipids derived from cholesterol, and their synthesis occurs in the SER.

6. If two vectors $\vec{A}$ and $\vec{B}$ are at an angle $\theta \n

If two vectors $\vec{A}$ and $\vec{B}$ are at an angle $\theta \neq 0^\circ$, then

”<tex>$|ec{A}|+|ec{B}|=|ec{A}+ec{B}|$</tex>”
”<tex>$|ec{A}|+|ec{B}|>|ec{A}+ec{B}|$</tex>”
”<tex>$|ec{A}|+|ec{B}|<|ec{A}+ec{B}|$</tex>”
”<tex>$|ec{A}|+|ec{B}|=|ec{A}-ec{B}|$</tex>”
This question was previously asked in
UPSC CDS-1 – 2019
Download PDFAttempt Online
The correct option is B.
For any two vectors $\vec{A}$ and $\vec{B}$, the triangle inequality states that the magnitude of their sum is less than or equal to the sum of their magnitudes: $|\vec{A}+\vec{B}| \le |\vec{A}|+|\vec{B}|$. Equality holds only when the vectors are collinear and point in the same direction, which corresponds to the angle between them $\theta = 0^\circ$. If $\theta \neq 0^\circ$, the vectors form two sides of a triangle, and the sum vector forms the third side. In a triangle, the length of any side is strictly less than the sum of the lengths of the other two sides. Therefore, if $\theta \neq 0^\circ$, $|\vec{A}+\vec{B}| < |\vec{A}|+|\vec{B}|$, or equivalently, $|\vec{A}|+|\vec{B}| > |\vec{A}+\vec{B}|$.
The magnitude of the sum of two vectors is given by $|\vec{A}+\vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta}$. Since $|\vec{A}|+|\vec{B}|$ is always positive, we can compare their squares: $(|\vec{A}|+|\vec{B}|)^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|$. Comparing this with $|\vec{A}+\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta$, we see that $(|\vec{A}|+|\vec{B}|)^2 > |\vec{A}+\vec{B}|^2$ when $\cos\theta < 1$, which is true for $\theta \neq 0^\circ$.

7. An electron and a proton starting from rest get accelerated through po

An electron and a proton starting from rest get accelerated through potential difference of 100 kV. The final speeds of the electron and the proton are V$_e$ and V$_p$ respectively. Which one of the following relations is correct?

”V<tex>$_e$</tex>
”V<tex>$_e$</tex>
”V<tex>$_e$</tex>
”Cannot
This question was previously asked in
UPSC CDS-1 – 2019
Download PDFAttempt Online
The correct option is A.
When a charged particle is accelerated from rest through a potential difference V, its kinetic energy is equal to the work done on it by the electric field, which is |qV|. So, (1/2)mv^2 = |qV|. The speed is given by v = sqrt(2|q|V/m). The electron and the proton have the same magnitude of charge (|q| = e) and are accelerated through the same potential difference V. Thus, their kinetic energies gained are equal: KE_e = KE_p = eV.
(1/2)m_e * V_e^2 = (1/2)m_p * V_p^2
m_e * V_e^2 = m_p * V_p^2
V_e / V_p = sqrt(m_p / m_e)
Since the mass of the electron (m_e) is significantly less than the mass of the proton (m_p), the ratio m_p / m_e is greater than 1. Therefore, V_e / V_p > 1, which means V_e > V_p. The lighter particle (electron) achieves a much higher speed than the heavier particle (proton) for the same kinetic energy.
The mass of an electron is approximately 9.11 x 10^-31 kg, and the mass of a proton is approximately 1.67 x 10^-27 kg. The proton mass is about 1836 times the electron mass. Thus, the electron’s speed will be about sqrt(1836) β‰ˆ 43 times the proton’s speed.

8. If the wavelengths corresponding to ultraviolet, visible and infrared

If the wavelengths corresponding to ultraviolet, visible and infrared radiations are given as Ξ»UV, Ξ»VIS and Ξ»IR respectively, then which one of the following gives the correct relationship among these wavelengths?

”λ<sub>UV</sub>
”λ<sub>UV</sub>
”λ<sub>UV</sub>
”λ<sub>UV</sub>
This question was previously asked in
UPSC CDS-1 – 2019
Download PDFAttempt Online
The correct option is D.
The electromagnetic spectrum arranges different types of radiation by wavelength and frequency. Wavelength increases from gamma rays to radio waves. The order of the given radiations by increasing wavelength is: Ultraviolet (UV), Visible, Infrared (IR). Therefore, the correct relationship is Ξ»UV < Ξ»VIS < Ξ»IR.
The approximate wavelength ranges are: UV (10 nm to 400 nm), Visible (400 nm to 700 nm), and IR (700 nm to 1 mm). This confirms the order of wavelengths.

9. Two identical solid pieces, one of gold and other of silver, when imme

Two identical solid pieces, one of gold and other of silver, when immersed completely in water exhibit equal weights. When weighed in air (given that density of gold is greater than that of silver)

the gold piece will weigh more
the silver piece will weigh more
both silver and gold pieces weigh equal
weighing will depend on their masses
This question was previously asked in
UPSC CDS-1 – 2019
Download PDFAttempt Online
The correct option is B.
When immersed in water, the apparent weight of an object is its true weight in air minus the buoyant force. The buoyant force is equal to the weight of the water displaced, which is (volume of the object) * (density of water) * g. The two pieces have equal apparent weight in water: W_air,gold – B_gold = W_air,silver – B_silver. Since both are solid and immersed, B_gold = V_gold * density_water * g and B_silver = V_silver * density_water * g. Given that the density of gold is greater than the density of silver, and they displace the same weight of water to have equal apparent weight difference from their air weight, the object with lower density (silver) must have a larger volume to displace the amount of water needed to satisfy the equality in apparent weight.
W_air = density * Volume * g.
W_air,gold – V_gold * density_water * g = W_air,silver – V_silver * density_water * g
V_gold * (density_gold – density_water) = V_silver * (density_silver – density_water)
Since density_gold > density_silver, it follows that (density_gold – density_water) > (density_silver – density_water). For the equation to hold, V_gold must be less than V_silver.
Now comparing weight in air: W_air,gold = density_gold * V_gold * g and W_air,silver = density_silver * V_silver * g.
We know V_gold < V_silver. To determine which one weighs more in air, consider the relation derived from the apparent weight equation: W_air,gold - W_air,silver = (V_gold - V_silver) * density_water * g. Since V_gold < V_silver, (V_gold - V_silver) is negative. Therefore, W_air,gold - W_air,silver is negative, meaning W_air,gold < W_air,silver. The silver piece weighs more in air.
This problem highlights the effect of buoyancy, which is dependent on the volume of the object. Although gold is denser, the condition of equal apparent weight in water necessitates that the less dense silver piece has a larger volume, which compensates for its lower density when determining its weight in air.

10. Which of the following planets of our solar system has least mass?

Which of the following planets of our solar system has least mass?

Neptune
Jupiter
Mars
Mercury
This question was previously asked in
UPSC CDS-1 – 2019
Download PDFAttempt Online
The correct option is D.
Comparing the masses of the given planets:
– Neptune: approx. 1.02 x 10^26 kg
– Jupiter: approx. 1.90 x 10^27 kg (most massive in the solar system)
– Mars: approx. 6.42 x 10^23 kg
– Mercury: approx. 3.30 x 10^23 kg
Comparing these values, Mercury has the smallest mass. In fact, Mercury is the smallest planet in the solar system both in terms of mass and radius.
The order of planets by mass (from least to most massive) is: Mercury, Mars, Venus, Earth, Uranus, Neptune, Saturn, Jupiter.
Page 1Page 2Page 3Page 4Page 5
Test 1Test 2Test 3Test 4Test 5Test 6Test 7Test 8Test 9Test 10Test 11Test 12Test 13Test 14Test 15Test 16Test 17Test 18Test 19Test 20Test 21Test 22
Exit mobile version