Question

Which of the following has the same dimensional formula for both the quantities?
i) Kinetic energy and torque
ii) Resistance and inductance
iii) Young’s modulus and pressure
A. (i) only
B. (ii) only
C. (i) and (iii) only
D. All the three

Answer 1

Hint: As we know the dimensional formulas of mass, length and time are \[\left[ M \right]\], \[\left[ L \right]\] and \[\left[ T \right]\] so we will calculate the dimensional formula for all and if result of any of them is same then we will check the options according to it.

Complete step by step answer
 As we know the dimensional formula for mass, length and time so to calculate the dimensional formula for above given terms like for kinetic energy we are knowing the formula for this that is given as, \[K.E = \dfrac{{m{v^2}}}{2}\], and the dimensional formula for \[m\] is \[\left[ M \right]\] and for \[v\] it is given as \[\left[ L \right]{\left[ T \right]^{ - 1}}\], it is calculated from velocity.

So after substituting the values in formula above we get \[K.E = \left[ M \right] \times {\left( {\left[ L \right]{{\left[ T \right]}^{ - 1}}} \right)^2}\],and during calculating we can neglect constants in formula if any so after calculating we get\[K.E = \left[ M \right]{\left[ L \right]^2}{\left[ T \right]^{ - 2}}\]

And now for torque we know the formula for torque is \[T = F \times R\]and substituting the dimensional formula in the formula we get \[T = \left[ M \right]\left[ L \right]{\left[ T \right]^{ - 2}} \times \left[ L \right]\]

\[T = \left[ M \right]{\left[ L \right]^2}{\left[ T \right]^{ - 2}}\], as we can see dimensional formula for kinetic energy and torque are same so now to check for other options as we did in this now for second option as for resistance we know resistance is product of voltage per unit current and voltage is product of electric field and distance and electric field is force per unit charge and charge is product of current and time, so we get the formula as
\[R = \dfrac{{\left( {\dfrac{F}{{I \times t}} \times l} \right)}}{I}\]
And after substituting the dimensions we get dimensional formula for \[R\] as
\[R = \left[ M \right]{\left[ L \right]^2}{\left[ T \right]^{ - 3}}{\left[ A \right]^{ - 2}}\],where \[\left[ A \right]\] is current.
Whereas for inductance we are knowing that inductance is voltage per unit (current per unit time) that is
\[L = \dfrac{V}{{\dfrac{I}{t}}}\], so after substituting the values of dimensional formulas we get the result as
\[L = \left[ M \right]{\left[ L \right]^2}{\left[ T \right]^{ - 2}}{\left[ A \right]^{ - 2}}\], so these are not equal therefore second option is wrong, now do repeat the same steps for third option and we will get result of young’s modulus as \[Y = \left[ M \right]{\left[ L \right]^{ - 1}}{\left[ T \right]^{ - 2}}\] and for pressure are same as young’s modulus so third is right.

therefore correct answer is C. option.

Note: Alternative method to do this question is that if we are knowing the S.I. units of all these terms then it will become easy as then we have to only substitute the values in these terms and calculate the result.