
The position of a particle at time t is given by the relation \[x\left( t \right) = \left( {\dfrac{{{v_0}}}{\alpha }} \right)\left( {1 - {c^{ - \alpha t}}} \right)\], where \[{v_0}\] is a constant and \[\alpha > 0\]. Find the dimensions of ${v_0}$ and $\alpha $
A)\[{M^0}L{T^{ - 1}}\,{\text{and}}\,{T^{ - 1}}\]
B)\[{M^0}L{T^1}\,{\text{and}}\,{T^{ - 1}}\]
C)\[{M^0}L{T^1}\,{\text{and}}\,L{T^{ - 2}}\]
D)\[{M^0}L{T^1}\,{\text{and}}\,T\]
Answer
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Hint: In this solution, we will use the rules of dimensional formula analysis to determine the necessary dimensions. Any term in an exponential must be dimensionless and to equate two terms, they must have the same dimensions.
Complete step by step answer:
We’ve been given that the position of a particle is given by \[x\left( t \right) = \left( {\dfrac{{{v_0}}}{\alpha }} \right)\left( {1 - {c^{ - at}}} \right)\] and we want to find the dimensional formula of \[{v_0}\] and \[\alpha \].
Now, we know that according to the rules of dimensional formula, the term in the exponential must be dimensionless. This implies that the term $ - \alpha t$ must be dimensionless. So, we can write
$[\alpha t] = [\alpha ]{T^1} = {M^0}{L^0}{T^0}$
Dividing both sides in the above equation by ${T^1}$, we get
$[\alpha ] = {M^0}{L^0}{T^{ - 1}}$
Now, to find the dimensions of \[{v_0}\], we can use the rule that the two terms that are being equated must have the same dimensional formula. Now we know the dimensional formula of velocity as
$[v] = {M^0}{L^1}{T^{ - 1}}$
Now the term on the right side, we know that the term inside the bracket must be dimensionless since there is subtraction with a constant. So the dimensions of the right side will be the ratio of \[{v_0}\] and \[\alpha \].
So, we can write
$[x] = \dfrac{{[{v_0}]}}{{[\alpha ]}}$
Hence the dimensions of $\alpha $ will be
${M^0}{L^1}{T^0} = \dfrac{{[{v_0}]}}{{{T^{-1}}}}$
Taking the inverse on both sides, the dimensions of $\alpha $ will be
$[\alpha ] = {M^0}{L^1}{T^{ - 1}}$
Hence the correct choice is option (A).
Note: In such questions, we must know how to apply the rules of dimensional formula analysis to figure out the different dimensional formula. We should also know the dimensional formula of basic quantities of kinematics such as velocity, acceleration, distance, etc.
Complete step by step answer:
We’ve been given that the position of a particle is given by \[x\left( t \right) = \left( {\dfrac{{{v_0}}}{\alpha }} \right)\left( {1 - {c^{ - at}}} \right)\] and we want to find the dimensional formula of \[{v_0}\] and \[\alpha \].
Now, we know that according to the rules of dimensional formula, the term in the exponential must be dimensionless. This implies that the term $ - \alpha t$ must be dimensionless. So, we can write
$[\alpha t] = [\alpha ]{T^1} = {M^0}{L^0}{T^0}$
Dividing both sides in the above equation by ${T^1}$, we get
$[\alpha ] = {M^0}{L^0}{T^{ - 1}}$
Now, to find the dimensions of \[{v_0}\], we can use the rule that the two terms that are being equated must have the same dimensional formula. Now we know the dimensional formula of velocity as
$[v] = {M^0}{L^1}{T^{ - 1}}$
Now the term on the right side, we know that the term inside the bracket must be dimensionless since there is subtraction with a constant. So the dimensions of the right side will be the ratio of \[{v_0}\] and \[\alpha \].
So, we can write
$[x] = \dfrac{{[{v_0}]}}{{[\alpha ]}}$
Hence the dimensions of $\alpha $ will be
${M^0}{L^1}{T^0} = \dfrac{{[{v_0}]}}{{{T^{-1}}}}$
Taking the inverse on both sides, the dimensions of $\alpha $ will be
$[\alpha ] = {M^0}{L^1}{T^{ - 1}}$
Hence the correct choice is option (A).
Note: In such questions, we must know how to apply the rules of dimensional formula analysis to figure out the different dimensional formula. We should also know the dimensional formula of basic quantities of kinematics such as velocity, acceleration, distance, etc.
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