Answer

Verified

21k+ views

- Hint: As we know that all physical quantities can be expressed in terms of seven fundamental base quantities such as mass, length, time, temperature, electric current, luminous intensity and amount of substance so use this concept to get the dimension of all these given quantities.

As we know that the formula of energy is given as

$E = {i^2}Rt$ joule.

Now we know that the dimension of current I = [A]

Dimension of time t = [T]

Dimension of resistance R = [$M{L^2}{T^{ - 3}}{A^{ - 2}}$]

So the dimension of energy is E = [$M{L^2}{T^{ - 2}}$]

Now as we know that the unit of the volume is ${m^3}$ or $c{m^3}$ both dimensions are equal which is given as [L].

So the dimension of the volume is V = [${L^3}$]

So the dimension of energy per unit volume is = $\dfrac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {{L^3}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$.

Now we know that the dimension of work done is the same as energy (as work done in a particular time is the same as energy consumed in the same time interval).

So the dimension of work done = [$M{L^2}{T^{ - 2}}$]

Now as we know pressure is the ratio of force and area.

$ \Rightarrow p = \dfrac{F}{A} = \dfrac{{Ma}}{{{m^2}}}$, Where M = mass, a = acceleration (m/s2) and A = area (m2).

So the dimension of pressure is given as,

$ \Rightarrow p = \dfrac{F}{A} = \dfrac{{Ma}}{{{m^2}}} = \dfrac{{\left[ M \right]\left[ {L{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$,

Now as we know dimension of stress is same as pressure definition is not same but dimension is same.

So the dimension of stress is $\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$.

Now as we know that young’s modulus is the ratio of stress to strain and we all know strain is dimensionless so the dimension of Young’s modulus is also same as stress $\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$

So as we see that the dimension of energy per unit volume is the same as stress, pressure and Young’s modulus.

So this is the required answer.

Note - We can use symbols instead of the names of the base quantities and they are represented in square brackets.

[M] for mass, [L] for length, [T] for time, [K] for temperature, [A] for current, [cd] for luminous intensity and [mol] for the amount of substance.

__Complete step by step answer:__-As we know that the formula of energy is given as

$E = {i^2}Rt$ joule.

Now we know that the dimension of current I = [A]

Dimension of time t = [T]

Dimension of resistance R = [$M{L^2}{T^{ - 3}}{A^{ - 2}}$]

So the dimension of energy is E = [$M{L^2}{T^{ - 2}}$]

Now as we know that the unit of the volume is ${m^3}$ or $c{m^3}$ both dimensions are equal which is given as [L].

So the dimension of the volume is V = [${L^3}$]

So the dimension of energy per unit volume is = $\dfrac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {{L^3}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$.

Now we know that the dimension of work done is the same as energy (as work done in a particular time is the same as energy consumed in the same time interval).

So the dimension of work done = [$M{L^2}{T^{ - 2}}$]

Now as we know pressure is the ratio of force and area.

$ \Rightarrow p = \dfrac{F}{A} = \dfrac{{Ma}}{{{m^2}}}$, Where M = mass, a = acceleration (m/s2) and A = area (m2).

So the dimension of pressure is given as,

$ \Rightarrow p = \dfrac{F}{A} = \dfrac{{Ma}}{{{m^2}}} = \dfrac{{\left[ M \right]\left[ {L{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$,

Now as we know dimension of stress is same as pressure definition is not same but dimension is same.

So the dimension of stress is $\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$.

Now as we know that young’s modulus is the ratio of stress to strain and we all know strain is dimensionless so the dimension of Young’s modulus is also same as stress $\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$

So as we see that the dimension of energy per unit volume is the same as stress, pressure and Young’s modulus.

So this is the required answer.

__Hence option (B), (C) and (D) are the correct answer.__Note - We can use symbols instead of the names of the base quantities and they are represented in square brackets.

[M] for mass, [L] for length, [T] for time, [K] for temperature, [A] for current, [cd] for luminous intensity and [mol] for the amount of substance.

Recently Updated Pages

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main

Find the points of intersection of the tangents at class 11 maths JEE_Main

For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main

The path difference between two waves for constructive class 11 physics JEE_MAIN

What is the difference between solvation and hydra class 11 chemistry JEE_Main

Other Pages

Assertion Acidic character of group 16 hydrides increases class 11 chemistry JEE_Main

Write short notes on the following carbylamine rea class 12 chemistry JEE_Main

What is the pH of 001 M solution of HCl a 1 b 10 c class 11 chemistry JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main

If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main