Physics Archives

81. Let an electric current of 1.5 A flow through an incandescent lamp in

Let an electric current of 1.5 A flow through an incandescent lamp in a circuit. What is the amount of charge that flows through it in 10 ms?

[amp_mcq option1=”0.015 C” option2=”0.15 C” option3=”1.5 C” option4=”15 C” correct=”option1″]

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Electric current (I) is defined as the rate of flow of electric charge (Q) per unit time (t). The relationship is given by Q = I * t.

Given the current (I) = 1.5 A and the time (t) = 10 ms. First, convert the time into seconds: 10 ms = 10 * 10^-3 seconds = 0.01 seconds. Now, calculate the charge: Q = 1.5 A * 0.01 s = 0.015 Coulombs (C).

The unit of electric current is the Ampere (A), and the unit of electric charge is the Coulomb (C). One Ampere is equal to the flow of one Coulomb of charge per second (1 A = 1 C/s). Milliseconds (ms) are a common unit of time, where 1 second = 1000 milliseconds.

82. Which one among the following is not a luminous object?

Which one among the following is not a luminous object?

[amp_mcq option1=”Sun” option2=”Proxima Centauri” option3=”Jupiter” option4=”Alpha Centauri” correct=”option3″]

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UPSC CAPF – 2021

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Luminous objects are those that produce and emit their own light. Non-luminous objects do not produce light but are visible because they reflect light from a luminous source.

The Sun, Proxima Centauri, and Alpha Centauri are all stars. Stars are massive, luminous celestial bodies that emit light through nuclear fusion. Jupiter is a planet, which is a large celestial body orbiting a star. Planets are non-luminous; they are visible because they reflect the light from their star (in Jupiter’s case, the Sun).

Examples of luminous objects include stars, lamps, and candles. Examples of non-luminous objects include planets, moons, tables, and chairs.

83. The following figure shows the image of a clock in a plane mirror : [I

The following figure shows the image of a clock in a plane mirror :
[Image of a clock face with hands]
Which one of the following is the correct time?

[amp_mcq option1=”2:35″ option2=”3:45″ option3=”9:15″ option4=”9:25″ correct=”option4″]

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UPSC CAPF – 2021

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A plane mirror produces a laterally inverted image. To find the actual time from a mirror image of a clock, one can imagine flipping the clock face horizontally or subtract the mirror time from 12:00.

Observing the image provided, the minute hand points exactly at the ‘7’, indicating 35 minutes. The hour hand is positioned between ‘2’ and ‘3’, closer to ‘2’. This position of the hour hand, coupled with the minute hand at 7 (35 minutes past the hour), suggests the time shown in the mirror image is 2:35.

If the mirror image shows 2:35, the actual time can be calculated as 12:00 – 2:35. 12:00 is equivalent to 11 hours and 60 minutes. So, (11 hours – 2 hours) and (60 minutes – 35 minutes) gives 9 hours and 25 minutes. Thus, the correct time is 9:25. This matches option D.

84. When a white light beam is made to fall on a hollow prism filled with

When a white light beam is made to fall on a hollow prism filled with water, it breaks into seven constituent colours. Which of the following colours suffers the maximum angle of deviation?

[amp_mcq option1=”Red” option2=”Violet” option3=”Green” option4=”All suffer by equal amount of deviation” correct=”option2″]

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UPSC CAPF – 2021

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When a beam of white light passes through a prism, it is dispersed into its seven constituent colours due to the difference in the refractive index of the prism material for different wavelengths of light. The extent to which a colour is deviated depends on the refractive index for that colour; a higher refractive index leads to a greater deviation.

The refractive index of a material is highest for shorter wavelengths (like violet light) and lowest for longer wavelengths (like red light). Consequently, violet light undergoes the maximum deviation, while red light undergoes the minimum deviation.

This phenomenon of splitting white light into its constituent colours is called dispersion. The band of seven colours produced is called the spectrum. The acronym VIBGYOR represents the order of colours in the spectrum from maximum deviation (Violet) to minimum deviation (Red). A hollow prism filled with water still functions as a prism made of water, exhibiting the same dispersion property.

85. Which one of the following statements with regard to a short-sighted p

Which one of the following statements with regard to a short-sighted person is correct?

[amp_mcq option1=”A short-sighted person sees the near objects blurred but distant objects clearly and a convex lens is used to correct this defect.” option2=”A short-sighted person sees the near objects clearly but distant objects blurred and a concave lens is used to correct this defect.” option3=”A short-sighted person sees the near objects blurred but distant objects clearly and a concave lens is used to correct this defect.” option4=”A short-sighted person sees the near objects clearly but distant objects blurred and a convex lens is used to correct this defect.” correct=”option2″]

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UPSC CAPF – 2021

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Short-sightedness is medically known as Myopia.
A person with myopia has difficulty seeing distant objects clearly, while they can see near objects clearly. This occurs because the eye converges distant light rays too strongly, causing the image to form in front of the retina instead of on the retina.
To correct myopia, a diverging lens (concave lens) is used. A concave lens spreads out the light rays slightly before they enter the eye, reducing the overall convergence and allowing the image of distant objects to be focused correctly on the retina.
Let’s examine the options:
A) States near objects are blurred and distant objects are clear, which is the opposite of myopia (this describes hyperopia or long-sightedness).
B) States near objects are clear and distant objects are blurred, which correctly describes myopia, and also correctly states that a concave lens is used for correction.
C) States near objects are blurred and distant objects are clear (hyperopia), and suggests a concave lens (corrects myopia). Incorrect combination.
D) States near objects are clear and distant objects are blurred (myopia), but suggests a convex lens (corrects hyperopia). Incorrect lens.

– Understand the definition and symptoms of short-sightedness (Myopia).
– Know the type of lens (converging/convex or diverging/concave) used to correct Myopia.
– Myopia: Near objects clear, distant objects blurred. Image forms in front of retina. Corrected by concave lens.
– Hyperopia: Near objects blurred, distant objects clear. Image forms behind retina. Corrected by convex lens.

The power of a lens used to correct myopia is negative, which corresponds to a concave lens. The power needed depends on the severity of the myopia. Myopia can be caused by an eyeball that is too long or a cornea/lens that is too curved.

86. An object is placed 10 cm in front of a lens. The image formed is real

An object is placed 10 cm in front of a lens. The image formed is real, inverted and of same size as the object. What is the focal length and nature of the lens?

[amp_mcq option1=”5 cm, converging” option2=”10 cm, diverging” option3=”20 cm, converging” option4=”20 cm, diverging” correct=”option1″]

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UPSC CAPF – 2021

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An object is placed 10 cm in front of a lens. So, object distance u = 10 cm.
The image formed is real, inverted, and of the same size as the object.
A real and inverted image is formed by a converging lens (convex lens). Diverging lenses (concave lenses) always produce virtual, erect, and diminished images.
For a lens, the magnification (m) is given by the ratio of image distance (v) to object distance (u), with a sign convention: m = v/u for erect images and m = -v/u for inverted images.
Since the image is inverted, m = -v/u.
The image is of the same size as the object, so the magnitude of magnification |m| = 1.
Thus, m = -1.
-v/u = -1 => v = u.
Given u = 10 cm, so v = 10 cm.
Using the lens formula (thin lens equation): 1/f = 1/v – 1/u.
Applying the sign convention: object distance u is typically taken as negative when placed in front of the lens, so u = -10 cm. The image is real and formed on the opposite side of the lens from the object, so image distance v is positive, v = +10 cm.
1/f = 1/(+10 cm) – 1/(-10 cm)
1/f = 1/10 + 1/10
1/f = 2/10
1/f = 1/5
f = 5 cm.
The focal length is positive (f > 0), which confirms that the lens is a converging lens (convex lens).
Alternatively, for a converging lens, a real image of the same size as the object is formed only when the object is placed at a distance of 2f from the lens, and the image is also formed at 2f on the other side. So, u = 2f.
Given u = 10 cm, 10 cm = 2f => f = 5 cm.
The nature of the lens is converging.

– Identify the nature of the lens based on the image characteristics (real, inverted implies converging).
– Use the magnification information (same size implies |m|=1) and image type (inverted implies m=-1) to relate object and image distances (v=u).
– Apply the lens formula (1/f = 1/v – 1/u) with appropriate sign conventions or use the special case rule for an image of the same size formed by a converging lens (object at 2f).
– Determine the sign of the focal length to confirm the nature of the lens (positive f for converging).

For a converging lens (convex lens):
– Object at infinity: real, inverted, point image at F.
– Object beyond 2F: real, inverted, diminished image between F and 2F.
– Object at 2F: real, inverted, same size image at 2F.
– Object between F and 2F: real, inverted, magnified image beyond 2F.
– Object at F: real, inverted, image at infinity.
– Object between the lens and F: virtual, erect, magnified image on the same side as the object.
The condition “real, inverted, and of same size” is unique to the object being placed at 2F for a converging lens.

87. Which electromagnetic radiation(s) is/are used to cook food?

Which electromagnetic radiation(s) is/are used to cook food?

[amp_mcq option1=”Infrared radiation only” option2=”Microwave radiation only” option3=”Infrared and microwave radiations” option4=”All electromagnetic radiations” correct=”option3″]

This question was previously asked in

UPSC CAPF – 2021

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Electromagnetic radiation is used in various forms for cooking food. Common examples include:
– Microwave ovens use microwave radiation to heat food. Microwaves are absorbed by water, fats, and sugars, causing them to vibrate and produce heat.
– Conventional ovens and grills use infrared radiation (thermal radiation) emitted by heating elements or flames to heat the food surface, and then heat is transferred into the food by conduction and convection.
Therefore, both infrared and microwave radiations are used in cooking. While visible light and other forms of EM radiation also carry energy and can potentially heat things, infrared and microwave are the primary forms utilized in common cooking appliances.

– Understand the different forms of electromagnetic radiation.
– Identify which forms of EM radiation are used in common cooking methods and appliances.
– Microwave ovens use microwaves.
– Ovens and grills use infrared radiation.

The electromagnetic spectrum includes radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays, in order of increasing frequency and energy. Cooking methods primarily utilize the energy transfer mechanisms of infrared radiation (heating surfaces) and microwave radiation (volumetric heating of polar molecules).

88. Which one of the following statements on photoelectric effect is NOT c

Which one of the following statements on photoelectric effect is NOT correct ?

[amp_mcq option1=”Albert Einstein received the Nobel Prize in Physics for explaining photoelectric effect” option2=”For each metal, there is a threshold frequency v₀ below which this effect is not observed” option3=”At a frequency v > v₀, the kinetic energy of ejected electrons does not change on increasing the frequency of incident light” option4=”The number of electrons ejected is proportional to the intensity of incident light” correct=”option3″]

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UPSC CAPF – 2020

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The statement “At a frequency v > v₀, the kinetic energy of ejected electrons does not change on increasing the frequency of incident light” is NOT correct.

According to Einstein’s explanation of the photoelectric effect, the maximum kinetic energy (KE_max) of the ejected electrons is given by KE_max = h(v – v₀), where h is Planck’s constant, v is the frequency of incident light, and v₀ is the threshold frequency. This equation shows that for v > v₀, the maximum kinetic energy is directly proportional to the frequency of the incident light; increasing v leads to increased KE_max.

Statement A is correct; Einstein received the Nobel Prize in Physics in 1921 for his explanation of the photoelectric effect. Statement B is correct; a minimum frequency (threshold frequency) is required for electron emission. Statement D is correct; the number of emitted electrons is proportional to the intensity of the incident light (for v > v₀), as higher intensity means more photons striking the surface.

89. Latent heat corresponds to the change in heat at constant

Latent heat corresponds to the change in heat at constant

[amp_mcq option1=”temperature only” option2=”volume only” option3=”pressure only” option4=”temperature, volume and pressure” correct=”option1″]

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Latent heat corresponds to the change in heat required for a substance to undergo a phase transition (like melting, boiling, or condensation) at a constant temperature.

During a phase change at constant pressure, the temperature remains constant while heat energy is absorbed (for melting/boiling/sublimation) or released (for freezing/condensation/deposition). This energy is used to change the state (break or form intermolecular bonds) rather than increase the kinetic energy of the molecules, which would result in a temperature change.

For typical phase transitions under standard conditions, pressure is also constant. However, the defining characteristic of latent heat is that it is the energy involved in a phase change *without* a change in temperature. Volume typically changes during a phase transition.

90. At triple point the substance co-exists in 1. Liquid phase 2. Solid

At triple point the substance co-exists in

1. Liquid phase
2. Solid phase
3. Vapour phase

Select the correct answer using the code given below :

[amp_mcq option1=”1 only” option2=”1 and 2 only” option3=”2 and 3 only” option4=”1, 2 and 3″ correct=”option4″]

This question was previously asked in

UPSC CAPF – 2020

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At the triple point of a substance, the solid, liquid, and vapour (gaseous) phases of that substance coexist in thermodynamic equilibrium.

The triple point is a unique specific temperature and pressure for a substance, distinct from the melting point (solid-liquid equilibrium) or boiling point (liquid-vapour equilibrium) which vary with pressure.

For water, the triple point is at 0.01°C (273.16 K) and 611.657 pascals (about 0.006 atm). This specific point is used to define the Kelvin temperature scale.