The relation between force F and density d is $F = \dfrac{x}{{\sqrt d }}$. The dimensions of x are
$
{\text{A}}{\text{. }}\left[ {{L^{ - 1/2}}{M^{3/2}}{T^{ - 2}}} \right] \\
{\text{B}}{\text{. }}\left[ {{L^{ - 1/2}}{M^{1/2}}{T^{ - 2}}} \right] \\
{\text{C}}{\text{. }}\left[ {{L^{ - 1}}{M^{3/2}}{T^{ - 2}}} \right] \\
{\text{D}}{\text{. }}\left[ {{L^{ - 1}}{M^{1/2}}{T^{ - 2}}} \right] \\
$
Answer
653.7k+ views
Hint: We are given the expression connecting the three quantities. We know the dimensional formula for force and density. Then solving for x, we can obtain the required dimensions.
Complete step-by-step solution:
We are given a relation between force F and density d as the following expression:
$F = \dfrac{x}{{\sqrt d }}$
We need to find out the dimensions of x by using the dimensional formulas for force F and density d.
$x = F\sqrt d $
The dimensional formula of force is given as $\left[ {ML{T^{ - 2}}} \right]$ while the dimensional formula for density is given as $\left[ {M{L^{ - 3}}{T^0}} \right]$.
Now we can use the given expression and insert the dimensions of force and density there to obtain dimensions of x.
Dimension of x are:
$
x = \left[ {ML{T^{ - 2}}} \right]{\left[ {M{L^{ - 3}}{T^0}} \right]^{\dfrac{1}{2}}} \\
= \left[ {ML{T^{ - 2}}} \right]\left[ {{M^{\dfrac{1}{2}}}{L^{ - \dfrac{3}{2}}}{T^0}} \right] \\
= \left[ {{M^{\dfrac{3}{2}}}{L^{\dfrac{{ - 1}}{2}}}{T^{ - 2}}} \right] \\
$
These are the required dimensions. Hence, the correct answer is option A.
Additional information:
Dimensional formula: A dimensional formula of a physical quantity is an expression, which describes the dependence of that quantity on the fundamental quantities.
All physical quantities can be expressed in terms of certain fundamental quantities. The following table contains the fundamental quantities and their units and dimensional notation respectively.
Note: 1. Mass, length, and time are most commonly encountered fundamental quantities so they must be specified in all dimensional formulas. The square bracket notation is used only for dimensional formulas.
2. In case if the student does not remember the dimensions of force and density, then my making use of basic formulas connecting them with fundamental quantities given in the table, the dimensions of force, and density can be obtained.
Complete step-by-step solution:
We are given a relation between force F and density d as the following expression:
$F = \dfrac{x}{{\sqrt d }}$
We need to find out the dimensions of x by using the dimensional formulas for force F and density d.
$x = F\sqrt d $
The dimensional formula of force is given as $\left[ {ML{T^{ - 2}}} \right]$ while the dimensional formula for density is given as $\left[ {M{L^{ - 3}}{T^0}} \right]$.
Now we can use the given expression and insert the dimensions of force and density there to obtain dimensions of x.
Dimension of x are:
$
x = \left[ {ML{T^{ - 2}}} \right]{\left[ {M{L^{ - 3}}{T^0}} \right]^{\dfrac{1}{2}}} \\
= \left[ {ML{T^{ - 2}}} \right]\left[ {{M^{\dfrac{1}{2}}}{L^{ - \dfrac{3}{2}}}{T^0}} \right] \\
= \left[ {{M^{\dfrac{3}{2}}}{L^{\dfrac{{ - 1}}{2}}}{T^{ - 2}}} \right] \\
$
These are the required dimensions. Hence, the correct answer is option A.
Additional information:
Dimensional formula: A dimensional formula of a physical quantity is an expression, which describes the dependence of that quantity on the fundamental quantities.
All physical quantities can be expressed in terms of certain fundamental quantities. The following table contains the fundamental quantities and their units and dimensional notation respectively.
| No. | Quantities | unit | Dimensional formula |
| 1. | Length | metre (m) | $\left[ {{M^0}{L^1}{T^0}} \right]$ |
| 2. | Mass | kilogram (g) | $\left[ {{M^1}{L^0}{T^0}} \right]$ |
| 3. | Time | second (s) | $\left[ {{M^0}{L^0}{T^1}} \right]$ |
| 4. | Electric current | ampere (A) | $\left[ {{M^0}{L^0}{T^0}{A^1}} \right]$ |
| 5. | Temperature | kelvin [K] | $\left[ {{M^0}{L^0}{T^0}{K^1}} \right]$ |
| 6. | Amount of substance | mole [mol] | $\left[ {{M^0}{L^0}{T^0}mo{l^1}} \right]$ |
| 7. | Luminous intensity | candela [cd] | $\left[ {{M^0}{L^0}{T^0}C{d^1}} \right]$ |
Note: 1. Mass, length, and time are most commonly encountered fundamental quantities so they must be specified in all dimensional formulas. The square bracket notation is used only for dimensional formulas.
2. In case if the student does not remember the dimensions of force and density, then my making use of basic formulas connecting them with fundamental quantities given in the table, the dimensions of force, and density can be obtained.
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
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

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

What do you mean by retardation What is its SI uni 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

Difference between physical and chemical change class 11 chemistry CBSE

