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The time dependence of a physical quantity ?
P = P0exp(_at2)
where a is a constant and t is time
The constant a
- a)Will be dimensionless
- b)Will have dimensions of T_2
- c)Will have dimensions as that of P
- d)Will have dimensions equal to the dimension of P multiplied by T_2
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The time dependence of a physical quantity ?P = P0exp(_at2)where a is ...
Given:
The time dependence of physical quantity P is found to be of the form
p=poexp(-α t2)
αt2 is dimensionless
Therefore:
α=1/t2
=1/[T2]
=[T-2] or[T^-2]
Conclusion :
constant α will [T-2] or[T^-2]
Most Upvoted Answer
The time dependence of a physical quantity ?P = P0exp(_at2)where a is ...
Dimensional Analysis of the given quantity:
?P = P0exp(_at2)
Here, P0 is the initial value of P.
Let's determine the dimensions of the various quantities involved in the expression:
Dimensions of P0 = [P]
Dimensions of t = [T]
Dimensions of ?P = [P]
Dimensions of a = [?P/T2]
Since the exponential function is dimensionless, the dimensions of the right-hand side of the equation will be the same as the dimensions of P0.
Therefore, [?P] = [P] = M^L^T^-2
Now, substituting the dimensions of ?P and a in the given equation:
[P] = [P0][exp(-[?P/T2][T^2])]
[P] = [P0][exp(-[?P][T^-1])]
[P] = [P0][exp(-[M^L^T^-2][T^-1])]
[P] = [P0][exp(-L^2/T)]
From the above equation, it is clear that the dimensions of the constant 'a' are the same as the dimensions of T^-2, i.e., T^2.
Therefore, option B is the correct answer.
?P = P0exp(_at2)
Here, P0 is the initial value of P.
Let's determine the dimensions of the various quantities involved in the expression:
Dimensions of P0 = [P]
Dimensions of t = [T]
Dimensions of ?P = [P]
Dimensions of a = [?P/T2]
Since the exponential function is dimensionless, the dimensions of the right-hand side of the equation will be the same as the dimensions of P0.
Therefore, [?P] = [P] = M^L^T^-2
Now, substituting the dimensions of ?P and a in the given equation:
[P] = [P0][exp(-[?P/T2][T^2])]
[P] = [P0][exp(-[?P][T^-1])]
[P] = [P0][exp(-[M^L^T^-2][T^-1])]
[P] = [P0][exp(-L^2/T)]
From the above equation, it is clear that the dimensions of the constant 'a' are the same as the dimensions of T^-2, i.e., T^2.
Therefore, option B is the correct answer.
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The time dependence of a physical quantity ?P = P0exp(_at2)where a is a constant and t is timeThe constant aa)Will be dimensionlessb)Will have dimensions of T_2c)Will have dimensions as that of Pd)Will have dimensions equal to the dimension of P multiplied by T_2Correct answer is option 'B'. Can you explain this answer?
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