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Units And Measurement

Question

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If force F, velocity v and momentum p are taken as fundamental quantities, then what is the dimensional formula of time?

Solution

Let time, T = K F^av^bp^c
So, [M^oL^oT¹] = [MLT ^-2]^a [ LT ^-1]^b [ M¹L¹T^-1]^c
= M^a+c L^a+b+cT^-2a-b-c
That is,
a+c = 0
a+b+c = 0
-2a-b-c = 1
On solving the above three equations, we get
a = -1 ; b= 0 and c= 1
Therefore, the dimensional formula of time is [F^-1v⁰p¹]

Some More Questions From Units and Measurement Chapter

What is the need of measurement?

Define physical quantity.

Give four examples of physical quantities.

Define unit.

What are the two main required characteristics of a unit?

Can we use same units for different physical quantities?

Can we use different units to measure same physical quantity?

To measure a physical quantity X, the unit U is repeatedly used for n times. What is the magnitude of physical quantity?

Does the numerical factor in the quantitative measurement depend on the system of unit used?

Is one unit sufficient to measure a physical quantity?

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