Physics Archives

191. An incandescent electric bulb converts 20% of its power consumption in

An incandescent electric bulb converts 20% of its power consumption into light, and the remaining power is dissipated as heat. The bulb’s filament has a resistance of 200 Ω and 2 A current flows through it. If the bulb remains ON for 10 h and the rate of electricity charge is ₹5/unit, then which among the following is the correct amount for the money spent on producing light ?

[amp_mcq option1=”₹5″ option2=”₹6″ option3=”₹7″ option4=”₹8″ correct=”option4″]

This question was previously asked in

UPSC NDA-2 – 2024

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The power consumed by the incandescent bulb is calculated using the given resistance and current. The total energy consumed over 10 hours is then calculated. Since only 20% of the power (and thus energy) is converted into light, the energy spent on producing light is 20% of the total energy consumed. The cost is calculated based on the rate per unit (kWh) for this amount of energy.

Power consumed (P) = I²R = (2 A)² * 200 Ω = 4 * 200 = 800 W = 0.8 kW.
Total energy consumed (E_total) = Power * Time = 0.8 kW * 10 h = 8 kWh.
Total cost of electricity = E_total * Rate = 8 kWh * ₹5/kWh = ₹40.
Energy converted into light (E_light) = 20% of E_total = 0.20 * 8 kWh = 1.6 kWh.
Cost spent on producing light = E_light * Rate = 1.6 kWh * ₹5/kWh = ₹8.

Incandescent bulbs are inefficient in converting electrical energy into visible light; a large portion of the energy is dissipated as heat, which is why the question specifies that only 20% is converted to light. The remaining 80% (1.6 kWh * 4 = 6.4 kWh) is dissipated as heat, costing ₹32 (₹40 – ₹8).

192. A pumpkin weighs 7.5 N. On submerging it completely in water, ¾ L of w

A pumpkin weighs 7.5 N. On submerging it completely in water, ¾ L of water gets displaced. The acceleration due to gravity at the place where the pumpkin was weighed is 10 m/s². Which one of the following is the correct value of the density of the pumpkin ?

[amp_mcq option1=”10 kg/m³” option2=”100 kg/m³” option3=”1000 kg/m³” option4=”10000 kg/m³” correct=”option3″]

This question was previously asked in

UPSC NDA-2 – 2024

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The weight of the pumpkin is given as 7.5 N. Using the formula Weight = mass × gravity, the mass of the pumpkin can be calculated. When the pumpkin is completely submerged, the volume of water displaced is equal to the volume of the pumpkin. Given the volume of displaced water, we can calculate the volume of the pumpkin. Density is then calculated as mass divided by volume.

Mass of pumpkin (m) = Weight / gravity = 7.5 N / 10 m/s² = 0.75 kg.
Volume of water displaced (V_displaced) = 3/4 L = 0.75 L.
Converting litres to cubic meters: 1 L = 0.001 m³. So, V_displaced = 0.75 × 0.001 m³ = 0.00075 m³.
When completely submerged, Volume of pumpkin (V_pumpkin) = V_displaced = 0.00075 m³.
Density of pumpkin (ρ) = Mass / Volume = 0.75 kg / 0.00075 m³.
ρ = 0.75 / (7.5 × 10⁻⁴) = (7.5 × 10⁻¹) / (7.5 × 10⁻⁴) = 10³ kg/m³ = 1000 kg/m³.

The density of water is approximately 1000 kg/m³. A pumpkin with a density equal to or slightly less than water would float or remain suspended when fully submerged. The fact that it displaces ¾ L of water upon complete submersion means its volume is ¾ L. The calculation of density confirms it is very close to the density of water, which is reasonable for a pumpkin.

193. A vehicle starts moving along a straight line path from rest. In first

A vehicle starts moving along a straight line path from rest. In first `t` seconds it moves with an acceleration of 2 m/s² and then in next 10 seconds it moves with an acceleration of 5 m/s². The total distance travelled by the vehicle is 550 m. The value of time `t` is

[amp_mcq option1=”10 s” option2=”13 s” option3=”20 s” option4=”25 s” correct=”option1″]

This question was previously asked in

UPSC NDA-2 – 2024

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The motion consists of two phases of constant acceleration. We need to find the time ‘t’ in the first phase such that the total distance covered in both phases is 550 m. By calculating the distance in each phase and setting the total distance equal to 550 m, we arrive at a quadratic equation in ‘t’. Solving this equation gives t = 10 s (the positive solution).

For the first phase (0 to t seconds): Initial velocity (u₁) = 0, acceleration (a₁) = 2 m/s². Distance s₁ = u₁t + (1/2)a₁t² = 0*t + (1/2)*2*t² = t². Velocity at time t (v₁) = u₁ + a₁t = 0 + 2t = 2t m/s.
For the second phase (t to t+10 seconds): Initial velocity (u₂) = v₁ = 2t m/s, acceleration (a₂) = 5 m/s², time (t₂) = 10 s. Distance s₂ = u₂t₂ + (1/2)a₂t₂² = (2t)*10 + (1/2)*5*(10)² = 20t + 250 m.
Total distance = s₁ + s₂ = t² + 20t + 250. Given total distance is 550 m, t² + 20t + 250 = 550, which simplifies to t² + 20t – 300 = 0.
Factoring the quadratic: (t + 30)(t – 10) = 0. Possible solutions are t = -30 or t = 10. Since time cannot be negative, t = 10 s.

We can verify the answer: If t=10 s, s₁ = 10² = 100 m. Velocity after 10s is v₁ = 2*10 = 20 m/s. In the next 10s, starting at 20 m/s with acceleration 5 m/s², distance s₂ = 20*10 + (1/2)*5*10² = 200 + 250 = 450 m. Total distance = s₁ + s₂ = 100 + 450 = 550 m, which matches the given information.

194. Which one among the following is correct for a person suffering from m

Which one among the following is correct for a person suffering from myopia ?

[amp_mcq option1=”The person can see near objects clearly” option2=”The person can see distant objects clearly” option3=”The person cannot distinguish colours” option4=”The person can neither see near objects nor distant objects clearly” correct=”option1″]

This question was previously asked in

UPSC NDA-2 – 2024

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Myopia, or nearsightedness, is a common refractive error where light focuses in front of the retina instead of on it. This results in distant objects appearing blurred, while near objects can be seen clearly.

A person with myopia has difficulty seeing objects that are far away. The eye might be longer than normal, or the cornea might be too curved, causing light to converge too strongly.

Myopia can be corrected using diverging lenses (concave lenses), which cause the light rays to diverge slightly before entering the eye, thereby shifting the focal point back onto the retina. Hyperopia (farsightedness) is the opposite condition, where near objects are blurred. Astigmatism causes blurred vision at all distances due to an irregularly shaped cornea or lens.

195. A point object is placed at the centre of curvature of a spherical con

A point object is placed at the centre of curvature of a spherical concave mirror. Which one among the following would be the correct location of image formed ?

[amp_mcq option1=”At infinity” option2=”At the centre of curvature” option3=”At the focal point” option4=”Between the focal point and the centre of curvature” correct=”option2″]

This question was previously asked in

UPSC NDA-2 – 2024

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For a spherical concave mirror, when an object is placed at the centre of curvature (C), the image formed is also located at the centre of curvature (C).

When the object is at the centre of curvature of a concave mirror, the image formed is real, inverted, and has the same size as the object. Ray diagrams show that rays from the object passing through the focal point become parallel after reflection, and rays hitting the mirror perpendicularly pass through the centre of curvature after reflection. These reflected rays intersect at the centre of curvature.

The relationship between object distance (u), image distance (v), and focal length (f) for a spherical mirror is given by the mirror formula: 1/f = 1/u + 1/v. For a concave mirror, f is positive. The centre of curvature (C) is located at a distance 2f from the pole of the mirror. If u = 2f (object at C), then 1/f = 1/(2f) + 1/v. Solving for 1/v gives 1/v = 1/f – 1/(2f) = 2/(2f) – 1/(2f) = 1/(2f). Thus, v = 2f, meaning the image is also formed at the centre of curvature.

196. For an electric circuit given below, the correct combination of voltag

For an electric circuit given below, the correct combination of voltage (V) and current (I) is

[amp_mcq option1=”V = 900 V; I = 18 A” option2=”V = 300 V; I = 5.5 A” option3=”V = 600 V; I = 1 A” option4=”V = 300 V; I = 2 A” correct=”option4″]

This question was previously asked in

UPSC NDA-2 – 2024

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Assuming the circuit diagram (not visible) represents a setup where a voltage of 300 V results in a current of 2 A, then option D is the correct combination. This implies the equivalent resistance of the circuit connected to the voltage source is 150 Ω (since V = IR, R = V/I = 300V / 2A = 150 Ω).

– Ohm’s Law (V = IR) relates the voltage (V) across a component or circuit, the current (I) flowing through it, and its resistance (R).
– For a complete circuit with a voltage source and an equivalent resistance R_eq, the total voltage V supplied by the source is related to the total current I flowing out of the source by V = I * R_eq.
– The question provides possible pairs of total voltage (V) and total current (I) for an electric circuit. We can calculate the implied equivalent resistance (R_eq = V/I) for each pair:
– A) V = 900 V, I = 18 A => R_eq = 900/18 = 50 Ω
– B) V = 300 V, I = 5.5 A => R_eq = 300/5.5 ≈ 54.55 Ω
– C) V = 600 V, I = 1 A => R_eq = 600/1 = 600 Ω
– D) V = 300 V, I = 2 A => R_eq = 300/2 = 150 Ω
– The circuit diagram would define the value of R_eq based on the configuration and values of its components (resistors, etc.). Without the diagram, we must assume that one of these calculated R_eq values corresponds to the actual circuit’s equivalent resistance. Assuming option D is correct, the circuit must have an equivalent resistance of 150 Ω when connected to a 300 V source or when a 2 A current flows through it resulting in a 300 V drop.

To definitively solve this problem, the circuit diagram showing the voltage source(s) and resistor network is required. The calculation of the equivalent resistance from the diagram would then confirm which (V, I) pair from the options is consistent with Ohm’s Law and the circuit’s resistance. Standard resistor values and combinations often yield resistances like 50, 150, 600 ohms.

197. Which one of the following sketches correctly describes a lever of sec

Which one of the following sketches correctly describes a lever of second class ?

[amp_mcq option1=”Sketch (a) showing Fulcrum at one end, Load in the middle, Effort at the other end.” option2=”Sketch (b) showing Fulcrum at one end, Effort in the middle, Load at the other end.” option3=”Sketch (c) showing Fulcrum in the middle, Load at one end, Effort at the other end.” option4=”Sketch (d) showing Fulcrum at one end, Load in the middle, Effort at the other end.” correct=”option1″]

This question was previously asked in

UPSC NDA-2 – 2024

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Assuming Sketch (a) shows the configuration with the Fulcrum at one end, the Load in the middle, and the Effort at the other end, it correctly describes a lever of the second class.

– Levers are simple machines consisting of a rigid bar that pivots around a fixed point called the fulcrum. Forces are applied to the lever: the effort (the force applied by the user) and the load (the force exerted by the object being moved or worked on).
– Levers are classified into three classes based on the relative positions of the fulcrum (F), the load (L), and the effort (E).
– Class 1 Lever: F is between E and L (E – F – L or L – F – E). Examples: seesaw, crowbar.
– Class 2 Lever: L is between F and E (F – L – E). Examples: wheelbarrow, nutcracker, bottle opener.
– Class 3 Lever: E is between F and L (F – E – L). Examples: tweezers, fishing rod, forearm lifting a weight.
– The question asks for a second-class lever. In a second-class lever, the Load is always located between the Fulcrum and the Effort.
– Option A’s description (“Fulcrum at one end, Load in the middle, Effort at the other end”) matches the configuration of a Class 2 lever (F – L – E).
– Option B describes a Class 3 lever (F – E – L).
– Option C describes a Class 1 lever (E – F – L or L – F – E, specifically F in the middle).
– Option D’s description is identical to Option A’s. Without seeing the sketches, we rely on the description provided in the options. Assuming Sketch (a) corresponds to the description in A and is a visual representation of F-L-E, then A is correct.

Second-class levers always provide a mechanical advantage greater than 1 (MA > 1), meaning the effort required is less than the load force, because the effort arm (distance from fulcrum to effort) is always longer than the load arm (distance from fulcrum to load).

198. Which one among the following diagrams may correctly represent the mot

Which one among the following diagrams may correctly represent the motion of a skydiver during a jump ?

[amp_mcq option1=”Diagram (a)” option2=”Diagram (b)” option3=”Diagram (c)” option4=”Diagram (d)” correct=”option2″]

This question was previously asked in

UPSC NDA-2 – 2024

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Assuming Diagram (b) represents a velocity-time graph showing an initial increase in velocity with decreasing acceleration, followed by reaching a constant terminal velocity, it correctly represents the motion of a skydiver during a jump.

– When a skydiver jumps, their initial velocity is zero.
– The primary force acting initially is gravity, causing downward acceleration approximately equal to g (acceleration due to gravity). The skydiver’s velocity increases rapidly.
– As the skydiver’s velocity increases, air resistance (drag force) becomes significant. Drag force opposes the motion and increases with speed (often proportional to v or v²).
– The net downward force is the difference between gravity and drag (F_net = mg – F_drag). The acceleration is F_net/m.
– As speed increases, drag increases, so the net force decreases. This means the acceleration decreases over time.
– The velocity continues to increase, but the rate of increase slows down.
– Eventually, the drag force becomes equal in magnitude to the force of gravity. At this point, the net force is zero, and the acceleration is zero. The skydiver then falls at a constant velocity called terminal velocity.
– A velocity-time graph for this motion would start at v=0, show the velocity increasing with a slope (acceleration) that decreases over time, and finally level off at the terminal velocity. Diagram (b) is the standard representation of this type of motion.

Opening a parachute significantly increases the drag force, causing the skydiver to rapidly decelerate to a much lower terminal velocity, allowing for a safe landing. The graph described here typically represents the motion before the parachute is opened.

199. In which one among the following situations, the bulb would glow the m

In which one among the following situations, the bulb would glow the most ? (Consider all batteries are the same)

[amp_mcq option1=”Diagram (a)” option2=”Diagram (b)” option3=”Diagram (c)” option4=”Diagram (d)” correct=”option2″]

This question was previously asked in

UPSC NDA-2 – 2024

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Assuming Diagram (b) shows two identical batteries connected in series to a single bulb, this configuration would result in the highest voltage across the bulb, making it glow the brightest.

– The brightness of a bulb (assumed to be resistive) is proportional to the power dissipated, P. Power can be calculated as P = V²/R or P = I²R, where V is the voltage across the bulb, I is the current through the bulb, and R is its resistance. For a given bulb (fixed R), maximum brightness corresponds to maximum voltage or maximum current.
– Assuming each battery provides a voltage V:
– Diagram (a) (one battery, one bulb): Voltage across bulb = V.
– Diagram (b) (two batteries in series, one bulb): The voltages of batteries in series add up. Total voltage = 2V. This voltage is across the bulb.
– Diagram (c) (two batteries in parallel, one bulb): Batteries in parallel maintain the same voltage as a single battery (assuming identical ideal batteries). Total voltage = V. This voltage is across the bulb. (Parallel connection increases total current capacity, but not voltage).
– Diagram (d) (one battery, two bulbs in series): The voltage V from the battery is divided between the two bulbs. Assuming identical bulbs, voltage across each bulb = V/2.
– Comparing the voltage across the bulb in each case: (a) V, (b) 2V, (c) V, (d) V/2.
– Since the power dissipated (and thus brightness) is proportional to V², the bulb will glow brightest when the voltage across it is highest, which is in case (b) with two batteries in series (2V).

Connecting batteries in series increases the total voltage supplied, while connecting identical batteries in parallel increases the total charge capacity and allows the supply of a larger current for a longer duration, without increasing the voltage (ideally). Connecting bulbs in series increases the total resistance of the circuit, reducing the current and the voltage across each individual bulb (and thus brightness, assuming they are identical).

200. Lightning is due to

Lightning is due to

[amp_mcq option1=”The flow of charges between different parts of the cloud” option2=”The short-circuiting of charges between the upper and lower surfaces of the cloud” option3=”The collection of positively charged particles on the base and collection of negatively charged particles at the top of the cloud” option4=”The induction of positive charge on the ground below the negative charge at the base of the cloud” correct=”option2″]

This question was previously asked in

UPSC NDA-2 – 2024

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The correct answer is B. Lightning is fundamentally a rapid, large-scale electrical discharge, often described as a “short-circuiting” of charges, which frequently occurs between different charged regions within a thundercloud, typically between the upper and lower parts.

– Lightning is an electrical discharge phenomena occurring in the atmosphere.
– It is caused by the build-up of electric charge differences within cumulonimbus clouds (thunderclouds) or between the cloud and the ground.
– Charge separation within a cloud typically results in positive charges accumulating at the top and negative charges accumulating at the bottom, although complex charge distributions exist.
– When the potential difference between these regions (or between the cloud and ground) exceeds the dielectric strength of the air, an electrical breakdown occurs, creating a conductive channel.
– The lightning strike is the rapid flow of electric charge (electrons and ions) through this channel. This discharge can happen within the cloud (intra-cloud lightning), between clouds (inter-cloud lightning), or between the cloud and the ground (cloud-to-ground lightning).
– Option B describes the rapid discharge (“short-circuiting”) between upper and lower charged regions within a cloud, which is a common type of lightning (intra-cloud lightning). While option A is also a flow of charges, B is more specific about the location and uses the term “short-circuiting” which captures the rapid discharge aspect. Option C describes the cause (charge separation) but gets the common polarity wrong (usually positive at top, negative at bottom). Option D describes ground charge induction, a condition for ground strikes, not the general cause of lightning.

Cloud-to-ground lightning is the most dangerous type. The rapid heating of the air along the lightning channel causes it to expand explosively, creating the sound waves we perceive as thunder.