Physics

171. Which one of the following pairs of rays is *not* electromagnetic in n

Which one of the following pairs of rays is *not* electromagnetic in nature ?

[amp_mcq option1=”X-rays and cathode rays” option2=”Gamma rays and X-rays” option3=”Alpha rays and beta rays” option4=”Beta rays and gamma rays” correct=”option3″]

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UPSC CAPF – 2010
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Alpha rays and beta rays are *not* electromagnetic in nature.
– Electromagnetic (EM) radiation consists of waves of oscillating electric and magnetic fields that travel at the speed of light. Examples include radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays.
– X-rays and gamma rays are high-energy forms of electromagnetic radiation.
– Alpha rays are streams of alpha particles, which are helium nuclei (2 protons, 2 neutrons). They are massive particles with a positive charge.
– Beta rays are streams of beta particles, which are high-energy electrons (β⁻ decay) or positrons (β⁺ decay). They are particles with a negative or positive charge.
– Cathode rays are streams of electrons, typically observed in vacuum tubes. They are also particles.
Alpha, beta, and cathode rays are beams of charged particles, subject to electric and magnetic fields in a way that EM waves are not (EM waves are only affected by changes in the medium they travel through or by strong gravitational fields in extreme cases).

172. Spectacles used for viewing 3-Dimensional films have :

Spectacles used for viewing 3-Dimensional films have :

[amp_mcq option1=”convex lens.” option2=”polaroids.” option3=”concave lens.” option4=”bifocal lens.” correct=”option2″]

This question was previously asked in
UPSC CAPF – 2010
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Spectacles used for viewing 3-Dimensional films often use polaroids (polarizing filters).
– One common method for displaying 3D films uses polarized light. Two images are projected simultaneously, each polarized differently (e.g., linearly polarized at 90 degrees to each other, or circularly polarized in opposite directions).
– The 3D glasses contain polarizing filters (polaroids) that match the polarization of the projected images. The filter over the left eye allows only the image intended for the left eye to pass through, and the filter over the right eye allows only the image intended for the right eye.
– The brain then combines these two slightly different images to create the perception of depth.
Older 3D systems used colored filters (anaglyph glasses, typically red and cyan). Modern active 3D systems use glasses with electronic shutters (LCD lenses) that rapidly open and close in sync with the display showing alternating images for the left and right eyes. The question likely refers to passive polarization glasses commonly used in cinemas or with some 3D TVs.

173. During free fall of an object :

During free fall of an object :

[amp_mcq option1=”its potential energy increases and its kinetic energy decreases.” option2=”its potential energy decreases and its kinetic energy increases.” option3=”both its potential energy and kinetic energy increase.” option4=”both its potential energy and kinetic energy decrease.” correct=”option2″]

This question was previously asked in
UPSC CAPF – 2010
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During free fall of an object, its potential energy decreases and its kinetic energy increases.
– Free fall is defined as motion under the influence of gravity only.
– Potential energy (PE) due to gravity is given by mgh, where m is mass, g is acceleration due to gravity, and h is height. As an object falls, its height (h) decreases, so its potential energy decreases.
– Kinetic energy (KE) is given by 0.5mv², where m is mass and v is velocity. As an object falls under gravity, its speed (v) increases (due to acceleration ‘g’), so its kinetic energy increases.
– In ideal free fall (neglecting air resistance), the total mechanical energy (PE + KE) is conserved. The decrease in potential energy is equal to the increase in kinetic energy.
If air resistance is considered, some of the potential energy is converted into heat and sound energy due to air friction, and the increase in kinetic energy is less than the decrease in potential energy. However, the question refers to ‘free fall’ which typically implies ideal conditions unless specified otherwise.

174. The Earth travels on its orbit at a speed of approximately 4400 km per

The Earth travels on its orbit at a speed of approximately 4400 km per hour. Why do we not feel this high speed ?

[amp_mcq option1=”We are too small compared to the size of the Earth” option2=”Our relative speed with respect to the Earth along the Earth’s orbit is zero” option3=”The gravity of the Earth constantly pulls us towards the Earth’s centre” option4=”The solar system as a whole is also moving” correct=”option2″]

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UPSC CAPF – 2010
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We do not feel the high speed of Earth’s orbit because our relative speed with respect to the Earth along the Earth’s orbit is zero.
– We are moving along with the Earth at the same speed and in the same direction as it orbits the sun.
– In physics, we typically feel acceleration (changes in speed or direction), not constant velocity. Since our velocity relative to the Earth’s surface (for the orbital motion) is zero and remains zero, we don’t feel the orbital speed.
– This is related to the concept of inertial frames of reference. We are in the same inertial frame as the Earth’s surface for this motion.
While the Earth is constantly accelerating towards the sun due to gravity (causing its curved path), this acceleration is relatively small (about 0.006 m/s²) and is the same for everything on Earth, so we don’t feel it as a distinct force pulling us away from the vertical. Other motions, like Earth’s rotation, do have slight effects (e.g., centrifugal force), but the primary reason for not feeling the vast orbital speed is being co-moving with the Earth.

175. The Stethoscope used by a medical practitioner is based on the phenome

The Stethoscope used by a medical practitioner is based on the phenomenon of :

[amp_mcq option1=”multiple reflection of sound waves.” option2=”scattering of sound waves.” option3=”refraction of sound waves.” option4=”none of the above.” correct=”option1″]

This question was previously asked in
UPSC CAPF – 2010
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The stethoscope works based on the phenomenon of multiple reflection of sound waves.
– Sound waves generated inside the body (like heartbeats or breathing sounds) are collected by the diaphragm (or bell) of the stethoscope.
– These sound waves travel through the hollow tube. As they travel, they undergo multiple reflections off the inner walls of the tube.
– These repeated reflections guide the sound waves towards the earpieces, concentrating the sound and making it audible to the medical practitioner.
While sound can scatter or refract, the primary mechanism by which sound is transmitted and focused through the stethoscope tube to the listener’s ears is multiple internal reflection.

176. The following figure shows the displacement time (x-t) graph of a body

The following figure shows the displacement time (x-t) graph of a body in motion. The ratio of the speed in first second and that in next two seconds is :

[amp_mcq option1=”1 : 2″ option2=”1 : 3″ option3=”3 : 1″ option4=”2 : 1″ correct=”option4″]

This question was previously asked in
UPSC CAPF – 2009
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The correct option is D) 2 : 1.
The speed of the body is given by the magnitude of the slope of the displacement-time (x-t) graph.
In the first second (from t=0 to t=1), the displacement changes from 0 to 3. The velocity is `(3 – 0) / (1 – 0) = 3/1 = 3` units/second. The speed in the first second (v1) is `|3| = 3`.
In the next two seconds (from t=1 to t=3), the displacement changes from 3 to 0. The velocity is `(0 – 3) / (3 – 1) = -3/2` units/second. The speed in the next two seconds (v2) is `|-3/2| = 3/2`.
The ratio of the speed in the first second and that in the next two seconds is `v1 : v2 = 3 : (3/2)`. To express this ratio in integers, we can multiply both parts by 2: `(3 * 2) : (3/2 * 2) = 6 : 3`. Dividing by 3, we get the simplified ratio `2 : 1`.
The displacement-time graph provides information about the position of the body over time. Velocity is the rate of change of displacement, and speed is the magnitude of velocity. A positive slope indicates movement in one direction, while a negative slope indicates movement in the opposite direction. In this graph, the body moves from x=0 to x=3 in the first second and then back from x=3 to x=0 in the subsequent two seconds.

177. A person moves along a circular path by a distance equal to half the c

A person moves along a circular path by a distance equal to half the circumference in a given time. The ratio of his average speed to his average velocity is :

[amp_mcq option1=”0.5″ option2=”0.5π” option3=”0.75π” option4=”1.0″ correct=”option2″]

This question was previously asked in
UPSC CAPF – 2009
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The correct option is B.
Let the circular path have radius R. The circumference is $C = 2\pi R$.
The distance covered by the person is half the circumference, $d = \frac{1}{2} C = \pi R$.
Let the time taken be $t$.
Average speed is defined as the total distance traveled divided by the total time taken.
Average speed = $\frac{d}{t} = \frac{\pi R}{t}$.

The person moves along a circular path by a distance equal to half the circumference. This means the person starts at one point on the circle and ends at the diametrically opposite point.
Let the starting point be A and the ending point be B, where AB is a diameter of the circle.
The displacement is the shortest straight-line distance from the initial position to the final position. In this case, the displacement is the length of the diameter.
Displacement = $2R$.

Average velocity is defined as the total displacement divided by the total time taken.
Average velocity = $\frac{\text{Displacement}}{t} = \frac{2R}{t}$.

The ratio of average speed to average velocity is:
Ratio = $\frac{\text{Average speed}}{\text{Average velocity}} = \frac{\pi R / t}{2R / t} = \frac{\pi R}{t} \times \frac{t}{2R} = \frac{\pi}{2}$.
The value $\frac{\pi}{2}$ is equivalent to $0.5\pi$.

This question highlights the difference between speed (scalar, based on distance) and velocity (vector, based on displacement). Distance is the path length, while displacement is the change in position vector. For motion along a curved path, the distance is generally greater than the magnitude of the displacement. For a half circle, the distance is $\pi R$ and the displacement magnitude is $2R$.

178. The explanation of, why we get thrown back with a jerk when the statio

The explanation of, why we get thrown back with a jerk when the stationary bus we are sitting in starts moving forward is given by :

[amp_mcq option1=”Zeroth law of gravity” option2=”Newton’s first law” option3=”Newton’s second law” option4=”Newton’s third law” correct=”option2″]

This question was previously asked in
UPSC CAPF – 2009
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The explanation of why you get thrown back with a jerk when a stationary bus starts moving forward is given by Newton’s first law.
Newton’s first law of motion, also known as the Law of Inertia, states that an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. When the bus is stationary, you are also at rest relative to the ground. When the bus suddenly accelerates forward, your body, due to inertia, tends to maintain its state of rest. This relative motion causes you to feel a backward jerk with respect to the accelerating bus.
Newton’s second law relates force, mass, and acceleration (F=ma). Newton’s third law states that for every action, there is an equal and opposite reaction. The concept of inertia directly explains the resistance to changes in motion, which is fundamental to the first law. The “Zeroth law of gravity” is not a standard physics term; gravity is described by Newton’s law of universal gravitation or Einstein’s theory of general relativity.

179. Which one of the following lenses would you prefer to use while readin

Which one of the following lenses would you prefer to use while reading very small letters printed on a label ?

[amp_mcq option1=”Convex lens of large focal length” option2=”Concave lens of large focal length” option3=”Convex lens of small focal length” option4=”Concave lens of small focal length” correct=”option3″]

This question was previously asked in
UPSC CAPF – 2009
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A Convex lens of small focal length is preferred for reading very small letters printed on a label.
To read small letters, a magnifying glass is used. A magnifying glass is a convex lens. A convex lens forms a magnified, virtual, and erect image when the object is placed within its focal length. The magnification power of a simple magnifying glass is given by M = 1 + (D/f), where D is the least distance of distinct vision (approx. 25 cm) and f is the focal length of the lens. To achieve higher magnification, the focal length (f) must be smaller.
A concave lens is a diverging lens and produces reduced or same-size images (virtual and erect or real and inverted), not magnified images for viewing small objects directly. A convex lens of large focal length will produce less magnification compared to one with a small focal length.

180. The figure given below shows the direction of the two forces P and Q a

The figure given below shows the direction of the two forces P and Q acting on a skydiver :

The figure shows two forces acting on a skydiver: Force P downwards and Force Q upwards. In the context of a skydiver falling through the air, the dominant downward force (P) is gravity (weight = mg), and the dominant upward force (Q) is air resistance (drag).
Let’s analyze the options:
A) Force P is caused by the gravity and force Q is caused by the friction. This statement correctly identifies the source of the forces. P is gravity, and Q is air resistance, which is a form of fluid friction. This statement is correct.
B) When the force P is bigger than the force Q, the speed of the skydiver remains the same. If P > Q, there is a net downward force (P – Q). According to Newton’s Second Law (F_net = ma), this net force causes acceleration in the downward direction, meaning the speed will increase, not remain the same. This statement is incorrect.
C) After the parachute opens, force P remains the same while force Q increases. Force P is gravity (mg), which depends on the skydiver’s mass and the acceleration due to gravity. Opening a parachute does not significantly change the skydiver’s mass or gravity. Thus, P remains essentially the same. Air resistance (Q) depends on the skydiver’s speed, the density of the air, and the skydiver’s shape and size (drag coefficient and area). Opening a parachute dramatically increases the surface area and drag coefficient, causing the air resistance force Q to increase significantly at the same speed. This statement is correct.
D) After the parachute opens, force P decreases while force Q increases. As explained above, P (gravity) remains essentially the same. Q increases dramatically. This statement is incorrect because it says P decreases.

Both A and C are factually correct statements about the forces. However, MCQs typically have a single best answer. Option C describes a crucial dynamic event in skydiving (parachute deployment) and its direct impact on the forces and resulting motion, which is a common physics concept tested. Option A is a static identification of the forces. Given the options and the nature of physics questions regarding skydiving, Option C is likely considered the intended answer as it addresses a key change in the system’s dynamics.

The forces acting on a skydiver are primarily gravity (downwards) and air resistance (upwards). Gravity is constant (for a constant mass). Air resistance depends on speed, shape, and size. Opening a parachute significantly increases air resistance.
When falling, a skydiver reaches a terminal velocity when the air resistance force (Q) equals the gravitational force (P), resulting in zero net force and zero acceleration (constant speed). Opening a parachute increases Q dramatically, making Q temporarily much larger than P, causing rapid deceleration until a new, much lower terminal velocity is reached.
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