
Given that \[Y = a\sin \omega x + bt + c{t^2}\cos \omega x\]. The unit of abc is same as that of
A. y
B. \[\left( {\dfrac{y}{t}} \right)\]
C. \[{\left( {\dfrac{y}{t}} \right)^2}\]
D. \[{\left( {\dfrac{y}{t}} \right)^3}\]
Answer
558.9k+ views
Hint:We will be comparing the units of variables present on the left hand side and right hand side because from the basics of physics and mathematics we know that two quantities can be added and subtracted only when they have the same units.
Complete step by step answer:
Given : \[Y = \underbrace {a\sin \omega x}_{1{\rm{st}}} + \underbrace {bt}_{2{\rm{nd}}} + \underbrace {c{t^2}\cos \omega x}_{3{\rm{rd}}}\]
From the basic chapter of physics that is units and dimensions we know that two units can be added or subtracted from each other when their units are the same. In other words, it can be said that two quantities can be added or subtracted only when their units are the same.It is also known to us that the trigonometric functions do not have any units, that is they are dimensionless.On comparing the first term of right hand side and left hand side, we get:
Unit of a = unit of Y
On comparing the second term of right hand side and left hand side, we get:
Unit of \[bt\]= unit of Y
Unit of b = unit of \[\dfrac{Y}{t}\]
On comparing the first term of right hand side and left hand side, we get:
Unit of \[c{t^2}\]= unit of Y
Unit of c = unit of \[\dfrac{Y}{{{t^2}}}\]
We have to calculate the unit of abc.
\[{\rm{Unit of abc = }}\left( {{\rm{unit of a}}} \right) \times \left( {{\rm{unit of b}}} \right) \times \left( {{\rm{unit of c}}} \right){\rm{ }}\]......(1)
Substitute unit of Y for unit of a, unit of \[\dfrac{Y}{{{t^2}}}\] for unit of c and unit of \[\dfrac{Y}{t}\] for unit of b in equation (1).
\[
{\rm{Unit of }}abc{\rm{ }} = {\rm{ }}\left( {{\rm{unit of }}Y} \right) \times \left( {{\rm{unit of }}\dfrac{Y}{t}} \right) \times \left( {{\rm{unit of }}\dfrac{Y}{{{t^2}}}} \right){\rm{ }}\\
\therefore\rm{Unit of }}abc{\rm{ }} = {\rm{ unit of }}{\left( {\dfrac{Y}{t}} \right)^3}\]
Therefore, the unit of abc is equal to the unit of \[{\left( {\dfrac{Y}{t}} \right)^3}\] and option (D) is correct.
Note:Do not calculate the unit of trigonometric terms as they are dimensionless. This type of problems can also be solved using the comparison of the dimensions of the given quantities as per the dimensioning rules.
Complete step by step answer:
Given : \[Y = \underbrace {a\sin \omega x}_{1{\rm{st}}} + \underbrace {bt}_{2{\rm{nd}}} + \underbrace {c{t^2}\cos \omega x}_{3{\rm{rd}}}\]
From the basic chapter of physics that is units and dimensions we know that two units can be added or subtracted from each other when their units are the same. In other words, it can be said that two quantities can be added or subtracted only when their units are the same.It is also known to us that the trigonometric functions do not have any units, that is they are dimensionless.On comparing the first term of right hand side and left hand side, we get:
Unit of a = unit of Y
On comparing the second term of right hand side and left hand side, we get:
Unit of \[bt\]= unit of Y
Unit of b = unit of \[\dfrac{Y}{t}\]
On comparing the first term of right hand side and left hand side, we get:
Unit of \[c{t^2}\]= unit of Y
Unit of c = unit of \[\dfrac{Y}{{{t^2}}}\]
We have to calculate the unit of abc.
\[{\rm{Unit of abc = }}\left( {{\rm{unit of a}}} \right) \times \left( {{\rm{unit of b}}} \right) \times \left( {{\rm{unit of c}}} \right){\rm{ }}\]......(1)
Substitute unit of Y for unit of a, unit of \[\dfrac{Y}{{{t^2}}}\] for unit of c and unit of \[\dfrac{Y}{t}\] for unit of b in equation (1).
\[
{\rm{Unit of }}abc{\rm{ }} = {\rm{ }}\left( {{\rm{unit of }}Y} \right) \times \left( {{\rm{unit of }}\dfrac{Y}{t}} \right) \times \left( {{\rm{unit of }}\dfrac{Y}{{{t^2}}}} \right){\rm{ }}\\
\therefore\rm{Unit of }}abc{\rm{ }} = {\rm{ unit of }}{\left( {\dfrac{Y}{t}} \right)^3}\]
Therefore, the unit of abc is equal to the unit of \[{\left( {\dfrac{Y}{t}} \right)^3}\] and option (D) is correct.
Note:Do not calculate the unit of trigonometric terms as they are dimensionless. This type of problems can also be solved using the comparison of the dimensions of the given quantities as per the dimensioning rules.
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