Solveeit Logo
Login

Question

Question: A force is given by \(F = at + b{t^2}\), where \(t\) is time, the dimensions of \(a\) and \(b\) are ...

A force is given by F=at+bt2F = at + b{t^2}, where tt is time, the dimensions of aa and bb are respectively-
(A) [MLT−4]and[MLT−1]\left[ {ML{T^{ - 4}}} \right]and\left[ {ML{T^{ - 1}}} \right]
(B) [MLT1]and[MLT0]\left[ {ML{T^1}} \right]and\left[ {ML{T^0}} \right]
(C) [MLT−3]and[MLT−4]\left[ {ML{T^{ - 3}}} \right]and\left[ {ML{T^{ - 4}}} \right]
(D) [MLT−3]and[MLT0]\left[ {ML{T^{ - 3}}} \right]and\left[ {ML{T^0}} \right]

Explanation

Solution

Hint
Here, we have to find the dimensions of aa and bb, for this the given equation is F=at+bt2F = at + b{t^2}.
From this equation, the dimension of at is equal to the dimension of force and also the dimension of bt2bt^2% has also the dimension of force. As the dimension of force is MLT−2{ML{T^{ - 2}}}, therefore on equating the dimensions we will get the required dimensions.

Complete step by step solution
The given equation is F=at+bt2F = at + b{t^2}………………. (1)
Where, aa and bb are constants.
We have to find the dimension of aa and bb. for this we will equate the dimension on both sides of the equation.
As the dimension of force is [MLT−2]\left[ {ML{T^{ - 2}}} \right] ………………… (2)
From the equation (1) and (2), we get
⇒[at]=[a][T]=[F]\Rightarrow \left[ {at} \right] = \left[ a \right]\left[ T \right] = \left[ F \right]
⇒[a][T]=[MLT−2]\Rightarrow \left[ a \right]\left[ T \right] = \left[ {ML{T^{ - 2}}} \right]
⇒[a]=[MLT−2][T]\Rightarrow \left[ a \right] = \dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ T \right]}}
⇒[a]=[MLT−3]\Rightarrow \left[ a \right] = \left[ {ML{T^{ - 3}}} \right] …………………. (3)
Similarly, we can find the dimensionality of b, for this again from equation (1) and (2), we get
⇒[bt2]=[F]\Rightarrow \left[ {b{t^2}} \right] = \left[ F \right]
⇒[b][T2]=[MLT−2]\Rightarrow \left[ b \right]\left[ {{T^2}} \right] = \left[ {ML{T^{ - 2}}} \right]
⇒[b]=[MLT−2][T2]\Rightarrow \left[ b \right] = \dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{T^2}} \right]}}
⇒[b]=[MLT−4]\Rightarrow \left[ b \right] = \left[ {ML{T^{ - 4}}} \right] …………………. (4)
Hence, from equation (3) and (4) the dimensionality of a and b are [MLT−3]and[MLT−4]\left[ {ML{T^{ - 3}}} \right]and\left[ {ML{T^{ - 4}}} \right] respectively.
Thus, option (C) is correct.

Note
For these types of the questions, we must remember that the dimensionality on both sides of the equation are equal. For these types of the equation we should remember the formulas and from there we can directly find out the dimensions. And it must be also notice that the dimensions are always written in the square brackets.