
Which of the following pairs have the same dimensions?
a) Torque and work
b) Angular momentum and work
c) Energy and Young’s modulus
d) Light year and wavelength.
A. a and b are correct
B. b and c are correct
C. c and d are correct
D. d and a correct
Answer
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Hint:To check whether which pair has the same dimensions. We will write their SI units. After that, we will write the SI units in the dimensional formula. If the dimensional formula for the pairs are the same then they have the same dimensions. We will, one by one, check the dimensions of the four pairs.
Complete step by step answer:
Given:
We will check for the dimensions of the quantities given in different options one by one.
In option a), We have torque and work. We know that torque is the product of force and the distance. The SI unit of force is Newton and that of distance is metre.
Hence SI unit of torque is Newton metre, which can be written as ${\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}} \cdot {{\rm{s}}^{ - 2}}$. In the dimensional formula, it can be written as $M{L^2}{T^{ - 2}}$.
Now for work the SI unit is Joule which is expressed as Newton metre, which can be written as ${\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}} \cdot {{\rm{s}}^{ - 2}}$. In the dimensional formula it can be written as $M{L^2}{T^{ - 2}}$. Therefore, we can say that the dimensions of pair a) are the same.
In option b) we have angular momentum and work.
As we have written in option a) for work, the SI unit is Joule, which is expressed as Newton metre which can be written as ${\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}} \cdot {{\rm{s}}^{ - 2}}$. In the dimensional formula, it can be written as $M{L^2}{T^{ - 2}}$
Now for angular momentum, the SI unit is kilogram metres square per second that is ${\rm{kg}} \cdot {{\rm{m}}^2} \cdot {{\rm{s}}^{ - 1}}$. In the dimensional formula it can be written as $M{L^2}{T^{ - 1}}$. Therefore, the pair b) does not have the same dimensions.
In option c) we have energy and young’s modulus,
The SI unit of energy is the same as work. Hence it can be expressed as Newton metre which can be written as ${\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}} \cdot {{\rm{s}}^{ - 2}}$. In the dimensional formula, it can be written as $M{L^2}{T^{ - 2}}$.
Now for young’s modulus, the SI unit is Newton per square metre that is ${\rm{kg}} \cdot {{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^{ - 2}}$ which can be written as $M{L^{ - 1}}{T^{ - 2}}$ in the dimensional formula. We can say that the dimensions for this pair are not the same.
In option d) we have light year and wavelength. We know that light years measures as distance; hence its SI unit will be metre, which can be expressed as $M{L^0}{T^0}$.
Wavelength is also measured in metre; hence it can be expressed as $M{L^0}{T^0}$.
Thus, pair d) has the same dimensions.
Hence, D is the correct answer.
Note:We always write the dimensional formula in the form ${M^a}{L^b}{T^c}$ where the powers $a$, $b$, and $c$ are the dimension of a physical quantity. We always write the dimensional formula in terms of fundamental units. The fundamental units are the units that do not depend on any other quantity.
Complete step by step answer:
Given:
We will check for the dimensions of the quantities given in different options one by one.
In option a), We have torque and work. We know that torque is the product of force and the distance. The SI unit of force is Newton and that of distance is metre.
Hence SI unit of torque is Newton metre, which can be written as ${\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}} \cdot {{\rm{s}}^{ - 2}}$. In the dimensional formula, it can be written as $M{L^2}{T^{ - 2}}$.
Now for work the SI unit is Joule which is expressed as Newton metre, which can be written as ${\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}} \cdot {{\rm{s}}^{ - 2}}$. In the dimensional formula it can be written as $M{L^2}{T^{ - 2}}$. Therefore, we can say that the dimensions of pair a) are the same.
In option b) we have angular momentum and work.
As we have written in option a) for work, the SI unit is Joule, which is expressed as Newton metre which can be written as ${\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}} \cdot {{\rm{s}}^{ - 2}}$. In the dimensional formula, it can be written as $M{L^2}{T^{ - 2}}$
Now for angular momentum, the SI unit is kilogram metres square per second that is ${\rm{kg}} \cdot {{\rm{m}}^2} \cdot {{\rm{s}}^{ - 1}}$. In the dimensional formula it can be written as $M{L^2}{T^{ - 1}}$. Therefore, the pair b) does not have the same dimensions.
In option c) we have energy and young’s modulus,
The SI unit of energy is the same as work. Hence it can be expressed as Newton metre which can be written as ${\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}} \cdot {{\rm{s}}^{ - 2}}$. In the dimensional formula, it can be written as $M{L^2}{T^{ - 2}}$.
Now for young’s modulus, the SI unit is Newton per square metre that is ${\rm{kg}} \cdot {{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^{ - 2}}$ which can be written as $M{L^{ - 1}}{T^{ - 2}}$ in the dimensional formula. We can say that the dimensions for this pair are not the same.
In option d) we have light year and wavelength. We know that light years measures as distance; hence its SI unit will be metre, which can be expressed as $M{L^0}{T^0}$.
Wavelength is also measured in metre; hence it can be expressed as $M{L^0}{T^0}$.
Thus, pair d) has the same dimensions.
Hence, D is the correct answer.
Note:We always write the dimensional formula in the form ${M^a}{L^b}{T^c}$ where the powers $a$, $b$, and $c$ are the dimension of a physical quantity. We always write the dimensional formula in terms of fundamental units. The fundamental units are the units that do not depend on any other quantity.
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