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Units And Measurement

Question
CBSEENPH11016604

If velocity of light in air (3x108m/s), acceleration due to gravity (9.8m/s2) and density of mercury (13600kg/m3) be chosen as fundamental units, then find the unit of mass, length and time.

Solution

Given that,
V e l o c i t y space o f space l i g h t space equals space open square brackets L to the power of 1 T to the power of negative 1 end exponent close square brackets equals 3 cross times 10 to the power of 8 space m divided by s space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
A c c e l e r a t i o n space d u e space t o space g r a v i t y equals space open square brackets L to the power of 1 T to the power of negative 2 end exponent close square brackets space equals space 9.8 m divided by s squared space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
D e n s i t y space o f space m e r c u r y comma space open square brackets M to the power of 1 L to the power of negative 3 end exponent close square brackets equals 13600 space k g divided by m cubed space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis
Now, dividing equation (1) by (2),
         {T} = 3.061 x 107s                  ....(4)
From(1) and (4) 
             L = 3 x 108 x T
              
                = 9.183 x 1015 m              ....(5)
Therefore, from equation (3) and (5), we have
             M = 13600 x L3
                = 13600 x (9.183 X 1015)3
            M = 1.053 x 1052 kg