All questions of Units and Measurements for NEET Exam

Distance associated with Parse

Unit Conversions

Unit conversion is the process of converting one unit of measurement to another. It is important to ensure that the correct unit conversion is used to avoid errors in calculations and measurements.

Given the following unit conversions, the wrong unit conversion is:

a) 1 angstrom = 10-10 m
b) 1 fermi = 10-15 m
c) 1 light year = 9.46 x 1015 m
d) 1 astronomical unit = 1.496 x 10-11 m

Explanation

- Angstrom (Å) to meter (m): 1 Å = 10-10 m. This is a correct conversion.
- Fermi (fm) to meter (m): 1 fm = 10-15 m. This is a correct conversion.
- Light year (ly) to meter (m): 1 ly = 9.46 x 1015 m. This is a correct conversion.
- Astronomical unit (AU) to meter (m): 1 AU = 1.496 x 10-11 m. This is an incorrect conversion. The correct conversion is 1 AU = 1.496 x 108 km.

Conclusion

The wrong unit conversion among the given options is option 'D' which states that 1 astronomical unit is equal to 1.496 x 10-11 m. The correct conversion is 1 astronomical unit is equal to 1.496 x 108 km.

Explanation:
To determine which combination can never be a meaningful quantity, we need to consider the dimensions of the quantities involved.

The dimensions of a physical quantity represent its physical nature and are usually expressed in terms of fundamental dimensions such as length, mass, time, etc.

Let's analyze each combination and determine their dimensions:

a) (P - Q)/R:
The dimensions of (P - Q) will be the same as the dimensions of P and Q since subtraction is only valid between quantities with the same dimensions. Therefore, the dimensions of (P - Q)/R will be the same as the dimensions of P and Q divided by the dimensions of R.

b) PQ - R:
The dimensions of PQ will be the product of the dimensions of P and Q. The dimensions of R will be the same as the dimensions of P and Q since subtraction is only valid between quantities with the same dimensions. Therefore, the dimensions of PQ - R will be the same as the dimensions of P and Q.

c) PQ/R:
The dimensions of PQ will be the product of the dimensions of P and Q. The dimensions of R will be the same as the dimensions of P and Q since division is only valid between quantities with the same dimensions. Therefore, the dimensions of PQ/R will be the same as the dimensions of P and Q divided by the dimensions of R.

d) (PR - Q^2)/R:
The dimensions of PR will be the product of the dimensions of P and R. The dimensions of Q^2 will be the square of the dimensions of Q. The dimensions of R will be the same as the dimensions of P and Q since subtraction and division are only valid between quantities with the same dimensions. Therefore, the dimensions of (PR - Q^2)/R will be the same as the dimensions of P and Q.

From the above analysis, we can conclude that the combination (P - Q)/R can never be a meaningful quantity because subtraction is only valid between quantities with the same dimensions. Therefore, the correct answer is option 'A'.

Explanation:

To determine the dimension of the variable "b" in the given equation v = at + bt^2, we need to analyze the dimensions of both sides of the equation.

The dimension of velocity (v) is given by [L][T]^-1, where [L] represents length and [T] represents time.

Dimension of the left-hand side (LHS):
The dimension of velocity (v) is [L][T]^-1.

Dimension of the right-hand side (RHS):
The equation v = at + bt^2 consists of two terms: at and bt^2.

1. Dimension of the first term (at):
The dimension of acceleration (a) is [L][T]^-2.
The dimension of time (t) is [T].

Therefore, the dimension of the first term (at) is [L].

2. Dimension of the second term (bt^2):
The dimension of time (t) is [T].

Therefore, the dimension of the second term (bt^2) is [L][T]^2.

Combining the dimensions:
Since both terms on the RHS have different dimensions, we cannot directly add them. However, for addition to be possible, the dimensions of both terms must be the same.

Comparing the dimensions of the first term (at) and the second term (bt^2), we can see that they have the same dimension of [L].

Therefore, the dimension of b must be such that the dimension of bt^2 is [L]. To cancel out the [T]^2 term, the dimension of b must be [L][T]^-2.

Hence, the correct dimension of b is [LT]^-2, which can be rearranged to [LT]^-3.

Answer:
The correct answer is option D, [LT]^-3.

• 4. The SI units are based on six base units.

1. Meter (m) for length
2. Kilogram (kg) for mass
3. Second (s) for time
4. Ampere (A) for electric current
5. Kelvin (K) for thermodynamic temperature
6. Mole (mol) for amount of substance
7. Candela (cd) for luminous intensity

• 1. Direct and indirect methods are used for the measurement of physical quantities. (True - We can measure directly or use indirect methods like displacement for volume.)
• 2. Scientific notation and the prefixes are used to simplify numerical computation. (True - They help express large/small numbers conveniently.)
• 3. A dimensionally correct equation need not be a correct equation. (True - Matching units doesn't guarantee a factual equation.)

Explanation:

Calorie, kilowatt hour, joule:
- Calorie, kilowatt hour, and joule are all units of energy measurement.
- Calorie is a unit of energy commonly used in nutrition to measure the energy content of food.
- Kilowatt hour is a unit of energy used to measure electricity consumption.
- Joule is the SI unit of energy.

Watt:
- Watt, on the other hand, is a unit of power, not energy.
- Power is the rate at which energy is transferred or converted.
- Watt is commonly used to measure the power of electrical devices or the rate at which energy is consumed or produced.

Conclusion:
- While calorie, kilowatt hour, and joule are all units of energy, watt is a unit of power. Therefore, the odd one out in this list is 'watt'.

Time is the physical quantity that has the same unit in all the three systems of units, i.e., CGS (Centimeter-Gram-Second) system, FPS (Foot-Pound-Second) system, and SI (Systeme International) system.

CGS System:
In the CGS system, the unit of time is second (s).

FPS System:
In the FPS system, the unit of time is second (s).

SI System:
In the SI system, the unit of time is second (s).

Explanation:
Mass and length have different units in different systems of units. For example, the unit of mass in CGS system is gram (g), in FPS system it is pound (lb), and in SI system it is kilogram (kg). Similarly, the unit of length in CGS system is centimeter (cm), in FPS system it is foot (ft), and in SI system it is meter (m). Therefore, time is the only physical quantity that has the same unit in all the three systems of units.

Explanation:

Dimensions of ω/k:
The dimensions of a quantity are the physical units in which it is measured. In this case, we need to determine the dimensions of the ratio ω/k in the given wave equation y = A sin(ωt - kx).

Relation with Velocity:
The velocity of a wave is given by the ratio of the angular frequency (ω) to the wave number (k) in the form v = ω/k. Therefore, the dimensions of ω/k are the same as the dimensions of velocity.

Conclusion:
Hence, the correct answer is option 'A' - velocity. The dimensions of ω/k are the same as those of velocity.

Understanding Dimensionless Quantities
Dimensionless quantities are unique in the realm of physics and mathematics, as they do not possess any physical dimensions. Here is an explanation of why option 'C' is the correct answer:
Definition of Dimensionless Quantity
- A dimensionless quantity is a number without any associated unit of measurement.
- It is often used to describe ratios or coefficients that are independent of the system of units.
Examples of Dimensionless Quantities
- Reynolds Number: This is a ratio used in fluid mechanics to predict flow patterns and is calculated using velocity, length, and viscosity but results in a dimensionless number.
- Strain: In mechanics, strain is defined as the ratio of change in length to the original length and is also dimensionless.
Possible Units for Dimensionless Quantities
- Although dimensionless quantities do not have units, they can be expressed in terms of ratios of the same units.
- For example, when calculating a ratio of two lengths (like height to height), the units cancel out, resulting in a dimensionless number.
Conclusion
- Hence, while dimensionless quantities themselves do not carry units, they may be derived from quantities that do have units.
- Therefore, the correct choice is option 'C' – a dimensionless quantity may have a unit, depending on how it is derived or expressed in relation to other quantities.
This understanding is crucial in fields like physics and engineering, where dimensionless numbers play a significant role in analysis and comparisons.

Explanation:

To determine which of the given relations is dimensionally incorrect, we need to check if the units on both sides of the equation are consistent.

Relation a:
1 u = 931.5 MeV/c²

The unit on the left side is atomic mass unit (u), and the unit on the right side is MeV/c² (megaelectronvolt per square of the speed of light). These units are consistent because MeV/c² is commonly used to express the mass-energy equivalence in particle physics. Therefore, relation a is dimensionally correct.

Relation b:
1 u = 931.5 MeV

The unit on the left side is atomic mass unit (u), and the unit on the right side is MeV (megaelectronvolt). These units are not consistent because atomic mass unit represents mass, while MeV represents energy. Therefore, relation b is dimensionally incorrect.

Relation c:
1 u = 1.67 x 10⁻²⁷ kg

The unit on the left side is atomic mass unit (u), and the unit on the right side is kg (kilogram). These units are consistent because atomic mass unit is defined as 1/12th the mass of a carbon-12 atom, and the kilogram is the SI unit of mass. Therefore, relation c is dimensionally correct.

Conclusion:
Among the given relations, relation b (1 u = 931.5 MeV) is dimensionally incorrect because the units on both sides of the equation are not consistent.

Understanding Relative Density
Relative density (also known as specific gravity) is a dimensionless quantity that compares the density of a substance to the density of water at a specific temperature (usually 4°C). For lead, the relative density is given as 11.3.
Calculating Density in SI Units
To find the density of lead in SI units (Kg/m³), we use the following relationship:
- The density of water is approximately 1000 Kg/m³.
- The formula for relative density is:
Relative Density = Density of Substance / Density of Water
By rearranging this formula, we can calculate the density of lead:
- Density of Lead = Relative Density × Density of Water
- Density of Lead = 11.3 × 1000 Kg/m³ = 11,300 Kg/m³
Interpreting the Answer
The calculated density of lead is 11,300 Kg/m³. In scientific notation, this can be expressed as:
- 11,300 Kg/m³ = 1.13 × 10^4 Kg/m³
Thus, the correct answer is option 'C' (1.13 × 10^4 Kg/m³).
Options Overview
- a) 1.13 × 10³ Kg/m³ → Incorrect
- b) 1.13 × 10² Kg/m³ → Incorrect
- c) 1.13 × 10⁴ Kg/m³ → Correct
- d) 11.3 Kg/m³ → Incorrect
Conclusion
Hence, the density of lead in SI units is indeed 1.13 × 10^4 Kg/m³, confirming that option 'C' is the right choice. Understanding these conversions is essential for physics and chemistry, especially in examinations like NEET.

The angular diameter of the Sun is observed to be 1920 arcseconds.

Mass Spectrograph for Measuring the Mass of Atoms and Molecules

Mass spectrograph is a device used for measuring the mass of atoms and molecules. It is an important tool in analytical chemistry and is used in a wide range of applications, including chemical analysis, biomedical research, and environmental monitoring.

Principle of Mass Spectrograph

The principle of mass spectrograph is based on the separation of ions according to their mass and charge. The sample is ionized by bombarding it with high-energy electrons, causing the atoms or molecules to lose one or more electrons and become positively charged ions. These ions are then accelerated by an electric field and passed through a magnetic field. The magnetic field causes the ions to follow a curved path, with the degree of curvature depending on the mass-to-charge ratio of the ion. The ions are detected by a detector, which generates a signal that is proportional to the number of ions of a particular mass-to-charge ratio.

Components of Mass Spectrograph

- Ion source: The ion source is used to ionize the sample and produce a beam of ions. There are several types of ion sources, including electron impact ionization, chemical ionization, and electrospray ionization.

- Accelerator: The accelerator is used to accelerate the ions to a high velocity. This is typically done using an electric field.

- Analyzer: The analyzer is used to separate the ions according to their mass-to-charge ratio. There are several types of analyzers, including magnetic sector analyzers, time-of-flight analyzers, and quadrupole analyzers.

- Detector: The detector is used to detect the ions that have passed through the analyzer. There are several types of detectors, including Faraday cups, electron multipliers, and photomultiplier tubes.

Applications of Mass Spectrograph

Mass spectrograph is used in a wide range of applications, including:

- Chemical analysis: Mass spectrograph is used to identify the components of a sample and determine their relative abundances.

- Biomedical research: Mass spectrograph is used to study the structure and function of proteins, peptides, and other biomolecules.

- Environmental monitoring: Mass spectrograph is used to monitor the levels of pollutants in the air, water, and soil.

Conclusion

Mass spectrograph is an important tool for measuring the mass of atoms and molecules. It is based on the separation of ions according to their mass and charge and is used in a wide range of applications, including chemical analysis, biomedical research, and environmental monitoring.

Significant figures refer to the number of digits in a number that have meaning or contribute to the accuracy of the number. Here, we need to determine the number of significant figures in the numbers 4.8000 x 104 and 48000.50.

4.8000 x 104:
- There are five digits after the decimal point, but trailing zeros after a decimal point are not significant.
- The number 4.8000 has five significant figures because all the zeros between the non-zero digits are significant.
- The exponent 104 indicates that the decimal point is shifted 4 places to the right, so there are 4 additional significant figures.
- Therefore, the total number of significant figures is 5 + 4 = 9.

48000.50:
- There are six digits before the decimal point and two digits after the decimal point.
- All the digits are non-zero, so they are all significant.
- Therefore, the total number of significant figures is 6 + 2 = 8.

Therefore, the correct answer is option B - 5 and 7.

Precision in Measuring Length

Measuring length is one of the fundamental aspects of physics and engineering. The accuracy and precision of any measurement depend on the instrument used to take the measurement. Precision refers to the degree of exactness with which a measurement is made. The most precise instrument for measuring length is the one that has the smallest least count.

Definition of Least Count

Least count refers to the smallest measurement that an instrument can make. For example, if the least count of a measuring instrument is 0.1 cm, then the instrument can measure lengths that are multiples of 0.1 cm. The smaller the least count, the more precise the instrument.

Comparison of Instruments

a) Metre Rod of Least Count 0.1 cm

A metre rod is a simple instrument used to measure length. It is a straight rod that is one metre in length. The least count of a metre rod is usually 0.1 cm. This means that the rod can measure lengths that are multiples of 0.1 cm. However, the metre rod is not a very precise instrument for measuring length.

b) Vernier Callipers of Least Count 0.01 cm

A vernier caliper is a more precise instrument for measuring length. It consists of two jaws, an upper and a lower, that can be adjusted to fit the object being measured. The jaws are connected to a scale that can be read to the nearest 0.1 mm. The least count of a vernier caliper is 0.01 cm, which means that it can measure lengths that are multiples of 0.01 cm.

c) Screw Gauge of Least Count 0.001 cm

A screw gauge is the most precise instrument for measuring length. It consists of a U-shaped frame with a screw attached to one end. The screw has a scale that can be read to the nearest 0.01 mm. The least count of a screw gauge is 0.001 cm, which means that it can measure lengths that are multiples of 0.001 cm.

Conclusion

In conclusion, the screw gauge is the most precise instrument for measuring length because it has the smallest least count. The smaller the least count, the more precise the instrument. While a metre rod and vernier calipers can also be used to measure length, they are not as precise as a screw gauge.

SI unit of pressure gradient:

The correct answer is option 'D': N m-3.

Explanation:

Pressure gradient:
The pressure gradient is a measure of the change in pressure over a given distance. It represents how quickly the pressure is changing in a particular direction.

SI unit of pressure:
The SI unit of pressure is the Pascal (Pa), which is defined as one Newton per square meter (N m-2).

Derivation of SI unit of pressure gradient:
To derive the SI unit of pressure gradient, we need to consider the definition of pressure gradient.

Pressure gradient = Change in pressure / Distance

The change in pressure is measured in Pascal (Pa), and the distance is measured in meters (m). Therefore, the SI unit of pressure gradient can be determined by dividing the unit of pressure by the unit of distance.

SI unit of pressure gradient = Pascal / meter

Since Pascal is equivalent to one Newton per square meter (N m-2), the SI unit of pressure gradient can be further simplified as:

SI unit of pressure gradient = (N m-2) / meter

Simplifying further:

SI unit of pressure gradient = N m-2 m-1

Using the laws of exponents:

SI unit of pressure gradient = N m-3

Therefore, the SI unit of pressure gradient is N m-3.

Summary:

The SI unit of pressure gradient is N m-3, which means Newton per cubic meter. It represents the change in pressure per unit volume and is derived by dividing the unit of pressure (Pascal) by the unit of distance (meter).