
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
585k+ 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.
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