Question

What is the dimension of the quantity\[l\sqrt {\dfrac{l}{g}} \], l being the length and g is the acceleration due to gravity.

Answer 1

Hint:To answer this question the first step would be writing down the formula for gravity in terms of velocity and time, later we will write down the formula for velocity in terms of time and distance. We will observe that we get all in terms of length and time and then we will find the dimensional formula for this.

Complete step-by-step solution:
Let us start by looking at what’s given:
It is given that:
Quantity=\[l \times \sqrt {\dfrac{l}{g}} .....(1)\]
Let us look at acceleration due to gravity: The acceleration of a body in free fall under the influence of earth's gravity, expressed as the rate of increase in velocity per unit of time, with a standard value of 980.665 centimeters per second.
Acceleration due to gravity= \[g = \dfrac{{velocity}}{{time}}\]
Now let's substitute the value of velocity in this equation for acceleration due to gravity. We know that The primary indicator of an object's position as well as its speed is its velocity. It is the distance travelled by an object in a given amount of time. The displacement of an item in unit time is defined as velocity.
We know that, \[velocity = \dfrac{d}{{time}}\]
Where d- distance and t-time.
Hence substituting this in acceleration due to gravity we get:
\[g = \dfrac{d}{{time \times time}}\]
Or, \[g = \dfrac{d}{{tim{e^2}}}\]
We already know that length is a fundamental unit. Hence its formula is only \[L\]
Let us substitute all these in equation 1:
We get
\[l \times \sqrt {\dfrac{l}{g}} .....(1)\]
\[l \times \sqrt {\dfrac{{l\,}}{{\dfrac{d}{{tim{e^2}}}}}} \]
Or, \[l \times \sqrt {\dfrac{{l\, \times tim{e^2}}}{d}} \]
Again time is a fundamental quantity and is represented by T and length is L. We see that all quantities are in the terms of length and time, hence we will now find the dimensions:
\[\left[ L \right] \times \sqrt {\dfrac{{\left[ L \right]\, \times {{\left[ T \right]}^2}}}{{\left[ L \right]}}} \]
Or, \[\left[ L \right] \times \sqrt {{{\left[ T \right]}^2}} \]
Or, \[\left[ L \right] \times \left[ T \right]\]
Or, \[{L^1}{T^1}\]
Hence the dimensional formula for \[l\sqrt {\dfrac{l}{g}} \] will be \[{L^1}{T^1}\].

 Note:Another way to answer this question is a very simple one, first we know that formula for time-period of pendulum is \[2\pi \sqrt {\dfrac{l}{g}} \].. while considering dimensional formula we do not consider \[2\pi \]. Hence the dimensional formula for \[\sqrt {\dfrac{l}{g}} \] is T. now in the given question it is asked \[l\sqrt {\dfrac{l}{g}} \], hence we can write the dimensional formula as \[{L^1} \times {T^1}\].