
Dimensions of impulse are
\[
A.{\text{ }}\left[ {ML{T^{ - 1}}} \right] \\
B.{\text{ }}\left[ {ML{T^2}} \right] \\
C.{\text{ }}\left[ {M{T^{ - 2}}} \right] \\
D.{\text{ }}\left[ {M{L^{ - 1}}{T^{ - 3}}} \right] \\
\]
Answer
590.7k+ views
Hint: In order to deal with this question first we will define the term impulse, further we will write its equation in the simplified form or some standard basic form in order to put the dimension of basic quantity and then according to the formula we will simplify and will evaluate its dimensions.
Complete step by step answer:
Formula used- $\left( I \right) = {\text{Force}} \times {\text{Time}}$
${\text{Force}} = {\text{Mass}} \times {\text{Acceleration}}$
Impulse is the change of momentum of an object as a force acts on it for the period of time.
Impulse $\left( I \right) = {\text{Force}} \times {\text{Time}}.......{\text{(1)}}$
Since, as we know the formula of force is:
${\text{Force}} = {\text{Mass}} \times {\text{Acceleration}}.......{\text{(2)}}$
So from equation (1) and equation (2), we have impulse as:
$
\because \left( I \right) = {\text{Force}} \times {\text{Time}} \\
\Rightarrow I = {\text{Mass}} \times {\text{Acceleration}} \times {\text{Time}}........{\text{(3)}} \\
$
We know the dimensional formula of these basic quantities
The dimensional formula of mass is $\left[ M \right]$ .
The dimensional formula of acceleration is $\left[ {L{T^{ - 2}}} \right]$ .
The dimensional formula of time is $\left[ T \right]$ .
On substituting these dimensional formulas in equation (3), we get:
\[
\because I = {\text{Mass}} \times {\text{Acceleration}} \times {\text{Time}} \\
\Rightarrow \left[ I \right] = \left[ M \right] \times \left[ {L{T^{ - 2}}} \right] \times \left[ T \right] \\
\Rightarrow \left[ I \right] = \left[ {{M^1}{L^1}{T^{ - 1}}} \right] \\
\]
Hence, the impulse is dimensionally represented as \[\left[ {{M^1}{L^1}{T^{ - 1}}} \right]\]
So, the correct option is A.
Additional information:
In classical mechanics, impulse (symbolized by J or Imp) is the integral of a force on which it operates, F, over the time interval, t. Because force is a quantity of vectors impulse is also a quantity of vectors. Impulse applied to an object results in an equal change of the vector of its angular velocity, also in the same direction. The SI unit of impulse is the newton second, and the kilogram meter per second is the dimensionally equivalent unit of momentum.
Note- The measure of a physical substance can be represented as a sum of the basic physical dimensions such as length, space, and time, each elevated to a logical power. A physical quantity's dimension is more fundamental than some unit of scale used to express the quantity of that physical amount. The dimensional equations have three uses: to test a physical equation's correctness. To derive the relation between a physical phenomenon involving different physical quantities. Move from one control configuration to another.
Complete step by step answer:
Formula used- $\left( I \right) = {\text{Force}} \times {\text{Time}}$
${\text{Force}} = {\text{Mass}} \times {\text{Acceleration}}$
Impulse is the change of momentum of an object as a force acts on it for the period of time.
Impulse $\left( I \right) = {\text{Force}} \times {\text{Time}}.......{\text{(1)}}$
Since, as we know the formula of force is:
${\text{Force}} = {\text{Mass}} \times {\text{Acceleration}}.......{\text{(2)}}$
So from equation (1) and equation (2), we have impulse as:
$
\because \left( I \right) = {\text{Force}} \times {\text{Time}} \\
\Rightarrow I = {\text{Mass}} \times {\text{Acceleration}} \times {\text{Time}}........{\text{(3)}} \\
$
We know the dimensional formula of these basic quantities
The dimensional formula of mass is $\left[ M \right]$ .
The dimensional formula of acceleration is $\left[ {L{T^{ - 2}}} \right]$ .
The dimensional formula of time is $\left[ T \right]$ .
On substituting these dimensional formulas in equation (3), we get:
\[
\because I = {\text{Mass}} \times {\text{Acceleration}} \times {\text{Time}} \\
\Rightarrow \left[ I \right] = \left[ M \right] \times \left[ {L{T^{ - 2}}} \right] \times \left[ T \right] \\
\Rightarrow \left[ I \right] = \left[ {{M^1}{L^1}{T^{ - 1}}} \right] \\
\]
Hence, the impulse is dimensionally represented as \[\left[ {{M^1}{L^1}{T^{ - 1}}} \right]\]
So, the correct option is A.
Additional information:
In classical mechanics, impulse (symbolized by J or Imp) is the integral of a force on which it operates, F, over the time interval, t. Because force is a quantity of vectors impulse is also a quantity of vectors. Impulse applied to an object results in an equal change of the vector of its angular velocity, also in the same direction. The SI unit of impulse is the newton second, and the kilogram meter per second is the dimensionally equivalent unit of momentum.
Note- The measure of a physical substance can be represented as a sum of the basic physical dimensions such as length, space, and time, each elevated to a logical power. A physical quantity's dimension is more fundamental than some unit of scale used to express the quantity of that physical amount. The dimensional equations have three uses: to test a physical equation's correctness. To derive the relation between a physical phenomenon involving different physical quantities. Move from one control configuration to another.
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