Measurement/Unit

21. The formula for conversion between Fahrenheit and Celsius is °F=X+(1.8

The formula for conversion between Fahrenheit and Celsius is
°F=X+(1.8×°C)
What is factor X ?

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32
22
98
42
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The correct option is A) 32. The formula for converting Celsius to Fahrenheit is °F = (1.8 × °C) + 32.
– The standard formula for temperature conversion from Celsius (°C) to Fahrenheit (°F) is °F = (°C × 9/5) + 32.
– Since 9/5 is equal to 1.8, the formula can be written as °F = (1.8 × °C) + 32.
– The given formula is °F = X + (1.8 × °C).
– Comparing the standard formula with the given formula, we can see that X must be 32.
The Fahrenheit and Celsius scales have different reference points and different increments per degree. The freezing point of water is 0°C or 32°F. The boiling point of water is 100°C or 212°F. The factor 1.8 (or 9/5) accounts for the different degree sizes, and the factor 32 accounts for the different zero points. The inverse formula for converting Fahrenheit to Celsius is °C = (°F – 32) / 1.8 or °C = (°F – 32) × 5/9.

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22. 1 dyne (a unit of force in CGS system) equals to

1 dyne (a unit of force in CGS system) equals to

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$10^3$ g cm/s²
$10^{-3}$ g cm/s²
$10^5$ kg m/s²
$10^{-5}$ kg m/s²
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1 dyne, a unit of force in the CGS system, is equal to $10^{-5}$ kg m/s² in the SI system.
– The CGS unit of force is the dyne, defined as 1 g ⋅ cm/s².
– The SI unit of force is the Newton (N), defined as 1 kg ⋅ m/s².
– To convert dyne to Newton, we convert the units:
– 1 gram (g) = $10^{-3}$ kilogram (kg)
– 1 centimeter (cm) = $10^{-2}$ meter (m)
– So, 1 dyne = 1 g ⋅ cm/s² = ($10^{-3}$ kg) ⋅ ($10^{-2}$ m) / s² = $10^{-3} \times 10^{-2}$ kg ⋅ m/s² = $10^{-5}$ kg ⋅ m/s².
Therefore, 1 dyne = $10^{-5}$ N. Conversely, 1 Newton = $10^5$ dyne. Understanding unit conversions between different systems like CGS and SI is fundamental in physics.

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23. Light year is a unit of measurement of

Light year is a unit of measurement of

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very large distances
time interval in years
amount of light received on earth in a year
mass of atoms
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Light year is a unit of measurement of very large distances.
A light-year is defined as the distance that light travels in vacuum in one Julian year (365.25 days). It is a unit of length used to express astronomical distances.
The speed of light in vacuum is approximately 3 x 10⁸ meters per second. One light-year is approximately equal to 9.461 x 10¹² kilometers or 5.879 x 10¹² miles. This unit is convenient for measuring the vast distances between stars and galaxies.

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24. Light year is a unit for measurement of

Light year is a unit for measurement of

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age of universe
very small time intervals
very high temperature
very large distance
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The correct answer is D.
A light-year is defined as the distance that light travels in vacuum in one Julian year (365.25 days). Since the speed of light is constant and very high, a light-year represents a very large distance. It is commonly used in astronomy to measure the distances between celestial objects, such as stars and galaxies. It is not a unit of time, temperature, or a measure related to the age of the universe (although the size of the observable universe is often expressed in light-years, its age is in years).
One light-year is approximately 9.461 trillion kilometers (5.879 trillion miles). Other units for measuring astronomical distances include the Astronomical Unit (AU) and the Parsec.

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25. Which one of the following is the value of 1 kWh of energy converted i

Which one of the following is the value of 1 kWh of energy converted into joules ?

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1.8 × 10⁶ J
3.6 × 10⁶ J
6.0 × 10⁶ J
7.2 × 10⁶ J
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1 kWh is a unit of energy commonly used for electrical energy consumption. It stands for kilowatt-hour. To convert it to joules (the SI unit of energy), we need to convert the power unit (kW) to watts (W) and the time unit (hour) to seconds (s).
1 kilowatt (kW) = 1000 Watts (W)
1 hour (h) = 60 minutes * 60 seconds = 3600 seconds (s)
Energy (E) = Power (P) × Time (t)
1 kWh = 1 kW × 1 h = (1000 W) × (3600 s) = 3,600,000 Ws
Since 1 Watt-second (Ws) is equal to 1 Joule (J),
1 kWh = 3,600,000 J = 3.6 × 10⁶ J.
– kWh is a unit of energy, not power.
– 1 Watt = 1 Joule per second (1 W = 1 J/s).
– 1 kilowatt = 1000 Watts.
– 1 hour = 3600 seconds.
Joules are the standard SI unit for energy. kWh is a practical unit for measuring large amounts of energy, like electricity consumption in households or industries. Power is the rate at which energy is consumed or transferred; 1 Watt is 1 Joule per second.

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26. Which one of the following physical quantity has the same unit as that

Which one of the following physical quantity has the same unit as that of pressure ?

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Angular momentum
Stress
Strain
Work
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Stress has the same unit as that of pressure.
Pressure is defined as force per unit area (P = F/A). Its unit is Newton per square meter (N/m²), also known as Pascal (Pa). Stress is also defined as force per unit area, representing the internal force per unit area within a material resisting deformation (Stress = F/A). Therefore, stress also has the unit of N/m² or Pascal (Pa).
Angular momentum is the product of moment of inertia and angular velocity; its unit is kg⋅m²/s or J⋅s. Strain is a dimensionless quantity, defined as the ratio of change in dimension to the original dimension (e.g., change in length/original length). Work is defined as force times distance; its unit is Newton-meter (N⋅m) or Joule (J). Out of the given options, only stress shares the same unit (N/m² or Pa) with pressure.

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27. The S.I. unit of acceleration is

The S.I. unit of acceleration is

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ms⁻¹
ms⁻²
cms⁻²
kms⁻²
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The correct answer is B) ms⁻².
Acceleration is the rate of change of velocity with respect to time. The S.I. unit for velocity is meters per second (m/s or ms⁻¹) and the S.I. unit for time is seconds (s).
The unit of acceleration is derived by dividing the unit of velocity by the unit of time: (m/s) / s = m/s² = ms⁻². Options A, C, and D represent units of velocity (ms⁻¹), acceleration in CGS units (cms⁻²), and acceleration in a non-standard or larger unit (kms⁻²), respectively.

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28. Match List I with List II and select the correct answer using the code

Match List I with List II and select the correct answer using the code given below the Lists :

List I
(Physical quantity)
List II
(Unit)

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A. Distance1. Mole
B. Amount of material2. Coulomb
C. Amount of electrical charge3. Light year
D. Energy4. Watt hour
3 1 2 4
3 2 1 4
4 2 1 3
4 1 2 3
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Matching the physical quantities with their appropriate units:
A. Distance – 3. Light year (A light-year is the distance light travels in one year, used for astronomical distances).
B. Amount of material – 1. Mole (The mole is the SI unit for amount of substance).
C. Amount of electrical charge – 2. Coulomb (The Coulomb is the SI unit for electric charge).
D. Energy – 4. Watt hour (A watt-hour is a unit of energy, calculated as power (Watts) multiplied by time (hours). It is commonly used for measuring electrical energy).
This results in the sequence 3 1 2 4, which matches option A.
Matching common physical quantities with their standard or relevant units.
Other common units for distance include meters, kilometers, miles, etc. Other common units for energy include Joules (the SI unit), calories, kilowatt-hours (1 kWh = 1000 Wh). The units provided in List II are all valid units for the corresponding quantities, although some are not SI units (like light-year or watt-hour).

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29. Which one of the following expressions has dimensions of energy (here

Which one of the following expressions has dimensions of energy (here V is the voltage across a resistor of resistance R and I is the current through the resistor, and t is the time)?

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V<sup>2</sup> / I t
V<sup>2</sup> / R t
I<sup>2</sup> / R t
I<sup>2</sup> / V t
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Energy is defined as the capacity to do work. In electrical circuits, energy dissipated or transferred is often related to power and time. Power (P) is the rate of energy transfer, so Energy (E) = Power (P) × time (t). For a resistor, power can be expressed as P = V²/R or P = I²R or P = VI.
Option A: V²/I t – The dimensions of V/I are R (resistance), so V²/I t is dimensionally (V/I) * (V/t) * t = R * (V/t) * t. This doesn’t directly yield energy dimensions.
Option B: V²/R t – V²/R represents power dissipated in a resistor. Multiplying power by time (t) gives energy. Thus, (V²/R) × t has the dimensions of energy.
Option C: I²/R t – I²R represents power dissipated in a resistor. I²/R is dimensionally Current² / Resistance, which is not power. I²R * t would be energy.
Option D: I²/V t – I²/V is dimensionally Current² / Voltage. This does not represent power.
– Energy = Power × Time.
– Power in a resistor = V²/R = I²R = VI.
– Dimensions of Energy = [Force × Distance] = [MLT⁻² × L] = [ML²T⁻²].
Checking dimensions formally:
V has dimensions [ML²I⁻¹T⁻³]. R has dimensions [ML²I⁻²T⁻³]. I has dimensions [I]. t has dimensions [T].
Option B: (V²/R) * t = ([ML²I⁻¹T⁻³]² / [ML²I⁻²T⁻³]) * [T] = ([M²L⁴I⁻²T⁻⁶] / [ML²I⁻²T⁻³]) * [T] = [ML²T⁻³] * [T] = [ML²T⁻²], which are the dimensions of energy.
Option C: (I²/R) * t would have been energy. I²/R * t as written in option C is [I²] / ([ML²I⁻²T⁻³] * [T]) = [I²] / [ML²I⁻²T⁻²] = [M⁻¹L⁻²I⁴T²].

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30. Which one of the following quantities is dimensionless?

Which one of the following quantities is dimensionless?

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Specific weight
Specific heat
Specific density
Specific gravity
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A dimensionless quantity is a quantity without any physical units or dimensions.
Specific weight is weight per unit volume (Force/Volume), with units like N/m³ or lb/ft³. It has dimensions [M L⁻² T⁻²].
Specific heat (or specific heat capacity) is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree. It has units like J/(kg·K) or cal/(g·°C), and dimensions [L² T⁻² θ⁻¹].
Specific density is often used interchangeably with relative density or specific gravity, which is the ratio of the density of a substance to the density of a reference substance (like water). In this context, it is dimensionless. However, sometimes ‘specific density’ might be used ambiguously.
Specific gravity is precisely defined as the ratio of the density of a substance to the density of a reference substance. As it is a ratio of two quantities with the same dimensions (density), it is dimensionless.
Given the options, Specific Gravity (D) is unambiguously a dimensionless quantity. Specific density (C) would also be dimensionless if it means relative density, but specific gravity is the more commonly used term for this dimensionless ratio. If specific density were interpreted as density per unit mass (1/volume), it would be dimensional. Therefore, Specific Gravity is the correct answer.
Specific gravity is defined as the ratio of the density of a substance to the density of a standard substance, making it a dimensionless quantity.
Dimensionless quantities are useful as they can be used in ratios and provide relative measures without depending on the system of units. Examples include Reynolds number, Mach number, friction coefficient, and angles (in radians).

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