In the expression $R{{V}_{ext}}=ma$, R stands for
(A). Resistance
(B). Rate of fuel consumption
(C). Rate of change of velocity
(D). Rate of change of mass
Answer
669.6k+ views
Hint: Try to understand the concepts of dimensional analysis. Express the derived physical quantities in terms of fundamental quantities. After finding the dimension of each quantity try to equate them to make the equation correct. Then you will find the dimension of the unknown quantity.
Complete step by step answer:
If we want to find an unknown quantity, we can check it from the dimensional formula of the equation. All equations should be always dimensionally correct.
To check if the equation is dimensionally correct, we should check the dimension of each quantity involved in the equation.
In the equation $R{{V}_{ext}}=ma$, the known quantities are ${{V}_{ext}}$, m and a, where ${{V}_{ext}}$ is the velocity, m is the mass and a is the acceleration of the object. The only unknown quantity is R.
Now, Dimension of ${{V}_{ext}}$ is = $\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]$
Dimension of m is = $\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]$
Dimension of acceleration is = $\left[ {{M}^{0}}{{L}^{1}}{{T}^{-2}} \right]$
Now, to check if the unknown quantity,
$\begin{align}
& \left[ R \right]\times \left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]=\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]\otimes \left[ {{M}^{0}}{{L}^{1}}{{T}^{-2}} \right] \\
& \left[ R \right]\times \left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]=\left[ {{M}^{1}}{{L}^{1}}{{T}^{-2}} \right] \\
& \left[ R \right]=\dfrac{\left[ {{M}^{1}}{{L}^{1}}{{T}^{-2}} \right]}{\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]} \\
& \left[ R \right]=\left[ {{M}^{1}}{{L}^{0}}{{T}^{-1}} \right] \\
\end{align}$
Hence, the unit of R should be mass per unit time (or Kg per second)
Hence, R stands for rate of change of mass
The correct option is (D)
Additional information: With the help of the dimensional formula we can either check the correctness of an equation or we can find some unknown quantity in the equation.
Note: Never try to remember the dimensional formula of a physical quantity. Always express them in terms of the fundamental quantities. Then you can find the dimensions of a physical quantity easily.
Complete step by step answer:
If we want to find an unknown quantity, we can check it from the dimensional formula of the equation. All equations should be always dimensionally correct.
To check if the equation is dimensionally correct, we should check the dimension of each quantity involved in the equation.
In the equation $R{{V}_{ext}}=ma$, the known quantities are ${{V}_{ext}}$, m and a, where ${{V}_{ext}}$ is the velocity, m is the mass and a is the acceleration of the object. The only unknown quantity is R.
Now, Dimension of ${{V}_{ext}}$ is = $\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]$
Dimension of m is = $\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]$
Dimension of acceleration is = $\left[ {{M}^{0}}{{L}^{1}}{{T}^{-2}} \right]$
Now, to check if the unknown quantity,
$\begin{align}
& \left[ R \right]\times \left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]=\left[ {{M}^{1}}{{L}^{0}}{{T}^{0}} \right]\otimes \left[ {{M}^{0}}{{L}^{1}}{{T}^{-2}} \right] \\
& \left[ R \right]\times \left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]=\left[ {{M}^{1}}{{L}^{1}}{{T}^{-2}} \right] \\
& \left[ R \right]=\dfrac{\left[ {{M}^{1}}{{L}^{1}}{{T}^{-2}} \right]}{\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]} \\
& \left[ R \right]=\left[ {{M}^{1}}{{L}^{0}}{{T}^{-1}} \right] \\
\end{align}$
Hence, the unit of R should be mass per unit time (or Kg per second)
Hence, R stands for rate of change of mass
The correct option is (D)
Additional information: With the help of the dimensional formula we can either check the correctness of an equation or we can find some unknown quantity in the equation.
Note: Never try to remember the dimensional formula of a physical quantity. Always express them in terms of the fundamental quantities. Then you can find the dimensions of a physical quantity easily.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

