Physics Archives

241. A current of 0·6 A is drawn by an electric bulb for 10 minutes. Which

A current of 0·6 A is drawn by an electric bulb for 10 minutes. Which one of the following is the amount of electric charge that flows through the circuit?

[amp_mcq option1=”6 C” option2=”0·6 C” option3=”360 C” option4=”36 C” correct=”option3″]

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The amount of electric charge that flows through the circuit is 360 C.

Electric charge (Q) is defined as the product of electric current (I) and time (t): Q = I × t.
The current given is I = 0.6 A.
The time given is t = 10 minutes. This needs to be converted to seconds: 10 minutes × 60 seconds/minute = 600 seconds.

Now, calculate the charge: Q = 0.6 A × 600 s = 360 Ampere-seconds. The unit of electric charge is the Coulomb (C), where 1 Coulomb = 1 Ampere × 1 second. Therefore, Q = 360 C.

242. The amplitude of sound waves is measured in the units of

The amplitude of sound waves is measured in the units of

[amp_mcq option1=”pressure” option2=”distance” option3=”time” option4=”speed” correct=”option1″]

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UPSC NDA-2 – 2022

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The amplitude of sound waves is measured in the units of pressure.

Sound waves in a medium like air are longitudinal waves involving compressions and rarefactions. The amplitude of a sound wave is the maximum variation in pressure from the equilibrium pressure (atmospheric pressure) or the maximum displacement of particles from their mean position.

While amplitude can be expressed as displacement amplitude (measured in meters, a unit of distance), it is more commonly and directly related to the intensity or loudness of the sound when expressed as pressure amplitude. Pressure is measured in Pascals (Pa), which are units of pressure. The options provided include both pressure and distance, but pressure is a more direct measure of the force per unit area variation that constitutes the wave, and its amplitude correlates directly with loudness.

243. When the pitch of sound increases, which one of the following

When the pitch of sound increases, which one of the following increases?

[amp_mcq option1=”Intensity” option2=”Loudness” option3=”Wavelength” option4=”Frequency” correct=”option4″]

This question was previously asked in

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The correct answer is Frequency.

The pitch of a sound is determined by its frequency. A higher frequency corresponds to a higher pitch, and a lower frequency corresponds to a lower pitch.

Intensity is related to the amplitude of the sound wave and is perceived as loudness. Wavelength is inversely proportional to frequency for a constant speed of sound. Loudness is the subjective perception of sound intensity.

244. An electric bulb is connected to 220 V generator. The current drawn is

An electric bulb is connected to 220 V generator. The current drawn is 600 mA. What is the power of the bulb?

[amp_mcq option1=”132 W” option2=”13.2 W” option3=”1320 W” option4=”13200 W” correct=”option1″]

This question was previously asked in

UPSC NDA-2 – 2022

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The power (P) of an electrical device connected to a circuit is given by the formula:
P = V × I
Where V is the voltage across the device and I is the current flowing through it.
Given:
Voltage (V) = 220 V
Current (I) = 600 mA
First, convert the current from milliamperes (mA) to amperes (A):
1 A = 1000 mA
So, 600 mA = 600 / 1000 A = 0.6 A.
Now, calculate the power:
P = 220 V × 0.6 A
P = 220 × (6/10) W
P = 22 × 6 W
P = 132 W.

Power is calculated as the product of voltage and current (P=VI). Ensure units are in Volts and Amperes to get Power in Watts.

This calculation is a fundamental application of Ohm’s law and the power formula in basic electricity. The resistance of the bulb could also be calculated using Ohm’s law (V=IR), but it is not needed to find the power. The power rating (in Watts) indicates the rate at which the bulb consumes electrical energy.

245. Which one of the following wavelengths corresponds to the wavelength o

Which one of the following wavelengths corresponds to the wavelength of X-rays?

[amp_mcq option1=”500 nm” option2=”5000 nm” option3=”100 nm” option4=”1 nm” correct=”option4″]

This question was previously asked in

UPSC NDA-2 – 2022

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Electromagnetic radiation is classified by wavelength or frequency. The electromagnetic spectrum includes (in order of decreasing wavelength): Radio waves, Microwaves, Infrared radiation, Visible light, Ultraviolet radiation, X-rays, and Gamma rays.
Visible light wavelengths are typically in the range of 400 nm to 700 nm.
Ultraviolet (UV) radiation has wavelengths shorter than visible light, usually in the range of 10 nm to 400 nm.
X-rays have even shorter wavelengths, generally ranging from about 0.01 nm to 10 nm.
Gamma rays have the shortest wavelengths, typically less than 0.1 nm.

Comparing the given options to the typical range of X-ray wavelengths (0.01 nm – 10 nm):
A) 500 nm falls in the visible light spectrum.
B) 5000 nm (5 µm) falls in the infrared spectrum.
C) 100 nm falls in the ultraviolet spectrum.
D) 1 nm falls within the typical range of X-ray wavelengths.

X-rays are high-energy photons commonly used in medical imaging (radiography, CT scans) and material analysis (X-ray crystallography, fluorescence). They are produced when high-speed electrons strike a metal target or by electron transitions in atoms.

246. A pressure cooker cooks food faster by

A pressure cooker cooks food faster by

[amp_mcq option1=”increasing the boiling point of water” option2=”decreasing the boiling point of water” option3=”increasing the melting point of water” option4=”decreasing the melting point of water” correct=”option1″]

This question was previously asked in

UPSC NDA-2 – 2022

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A pressure cooker works by sealing the cooking pot tightly, which prevents steam from escaping. As water heats up, it turns into steam, increasing the pressure inside the cooker. According to the relationship between pressure and boiling point, increasing the pressure on a liquid increases its boiling point.

By increasing the pressure inside, the boiling point of water is raised from 100°C (at standard atmospheric pressure) to a higher temperature, typically around 120-125°C. Food cooks faster at these higher temperatures.

Conversely, at high altitudes where atmospheric pressure is lower, water boils at temperatures below 100°C, which means food takes longer to cook compared to cooking at sea level. Pressure cookers counteract this effect or speed up cooking at any altitude by artificially increasing the pressure and thus the boiling point.

247. An object is made of two equal parts by volume; one part has density $

An object is made of two equal parts by volume; one part has density $\rho_0$ and the other part has density $2\rho_0$. What is the average density of the object?

[amp_mcq option1=”$3\rho_0$” option2=”$\frac{3}{2}\rho_0$” option3=”$\rho_0$” option4=”$\frac{1}{2}\rho_0$” correct=”option2″]

This question was previously asked in

UPSC NDA-2 – 2022

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Let V be the total volume of the object. The object is made of two equal parts by volume, so the volume of each part is $V_1 = V_2 = V/2$.
Let $\rho_1$ be the density of the first part and $\rho_2$ be the density of the second part.
Given: $\rho_1 = \rho_0$ and $\rho_2 = 2\rho_0$.
The mass of the first part is $m_1 = \rho_1 \times V_1 = \rho_0 \times (V/2)$.
The mass of the second part is $m_2 = \rho_2 \times V_2 = 2\rho_0 \times (V/2) = \rho_0 V$.
The total mass of the object is $M = m_1 + m_2 = \rho_0 (V/2) + \rho_0 V = \rho_0 V (\frac{1}{2} + 1) = \rho_0 V (\frac{3}{2})$.
The total volume of the object is $V_{\text{total}} = V_1 + V_2 = V/2 + V/2 = V$.
The average density of the object is $\rho_{\text{avg}} = \frac{M}{V_{\text{total}}} = \frac{\rho_0 V (3/2)}{V} = \frac{3}{2}\rho_0$.

When calculating average density for parts of equal volume, the average density is the simple arithmetic mean of the densities. However, in this case, the masses are different. The calculation involves finding the total mass and dividing by the total volume.

If the parts were of equal mass instead of equal volume, the calculation would be different, involving the reciprocal of the average of reciprocals (harmonic mean of densities).

248. Which one of the following statements regarding a current-carrying sol

Which one of the following statements regarding a current-carrying solenoid is not correct?

[amp_mcq option1=”The magnetic field inside the solenoid is uniform.” option2=”The current-carrying solenoid behaves like a bar magnet.” option3=”The magnetic field inside the solenoid increases with increase in current.” option4=”If a soft iron bar is inserted inside the solenoid, the magnetic field remains the same.” correct=”option4″]

This question was previously asked in

UPSC NDA-2 – 2022

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Let’s analyze each statement about a current-carrying solenoid:
A) The magnetic field inside a *long* solenoid is approximately uniform and directed along the axis of the solenoid, except near the ends. This statement is correct for an ideal or long solenoid.
B) A current-carrying solenoid creates a magnetic field pattern similar to that of a bar magnet, with magnetic poles at its ends. This statement is correct.
C) The magnitude of the magnetic field inside a solenoid is given by $B = \mu n I$, where $\mu$ is the permeability of the core material, $n$ is the number of turns per unit length, and $I$ is the current. The field is directly proportional to the current ($I$). So, increasing the current increases the magnetic field. This statement is correct.
D) If a soft iron bar (a ferromagnetic material with high permeability) is inserted inside the solenoid, the magnetic field inside increases significantly. This happens because the soft iron gets strongly magnetized in the direction of the solenoid’s field, and its own magnetic field adds to the field produced by the current. The permeability of soft iron is much greater than the permeability of air or vacuum ($\mu >> \mu_0$). The magnetic field does *not* remain the same; it increases. This statement is incorrect.

Ferromagnetic materials like soft iron dramatically increase the magnetic field strength when placed inside a solenoid or coil because they become strongly magnetized, effectively increasing the magnetic permeability of the core.

This property is used to make electromagnets, where a soft iron core is inserted into a solenoid to produce a very strong magnetic field when current flows.

249. The commercial unit of electrical energy is kilowatt-hour (kWh), which

The commercial unit of electrical energy is kilowatt-hour (kWh), which is equal to

[amp_mcq option1=”$3.6 \times 10^6$ J” option2=”$3.6 \times 10^3$ J” option3=”$10^3$ J” option4=”1 J” correct=”option1″]

This question was previously asked in

UPSC NDA-2 – 2022

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The commercial unit of electrical energy is kilowatt-hour (kWh). Energy consumed is calculated as Power multiplied by Time.
1 kilowatt (kW) = 1000 Watts (W)
1 hour (h) = 3600 seconds (s)
Energy (Joules) = Power (Watts) × Time (seconds)
1 kWh = 1 kW × 1 h = 1000 W × 3600 s = 3,600,000 Joules (J)
In scientific notation, this is $3.6 \times 10^6$ J.

1 kWh is the energy used by a 1 kW device operating for 1 hour. The conversion factor from kilowatt-hour to Joules is obtained by converting kilowatts to watts and hours to seconds.

The Joule is the SI unit of energy, but for practical purposes, especially for calculating electricity consumption in households and industries, kilowatt-hour is commonly used. It represents a larger unit of energy. 1 unit of electricity on a meter typically corresponds to 1 kWh.

250. A negative work is done when an applied force F and the corresponding

A negative work is done when an applied force F and the corresponding displacement S are

[amp_mcq option1=”perpendicular to each other.” option2=”parallel to each other.” option3=”anti-parallel to each other.” option4=”equal in magnitude.” correct=”option3″]

This question was previously asked in

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A negative work is done when an applied force F and the corresponding displacement S are anti-parallel to each other.

The work done (W) by a constant force (F) when an object undergoes a displacement (S) is given by the dot product of the force and displacement vectors: $W = \vec{F} \cdot \vec{S} = |\vec{F}| |\vec{S}| \cos\theta$, where $\theta$ is the angle between the force and displacement vectors. Work done is positive when the force and displacement are in the same general direction ($0^\circ \le \theta < 90^\circ$), zero when they are perpendicular ($\theta = 90^\circ$), and negative when they are in opposite directions ($90^\circ < \theta \le 180^\circ$). When the force and displacement are anti-parallel ($\theta = 180^\circ$), $\cos\theta = -1$, and the work done is maximum negative: $W = -FS$.

Examples of negative work include the work done by friction when an object moves (friction acts opposite to motion), or the work done by gravity when an object is lifted upwards (gravity acts downwards while displacement is upwards).