Where E is the energy, f is the frequency of oscillator and ‘h’ is Planck’s constant. To find the dimensions of h, therefore, we have to find the dimensions f E and f. Now, joul is the unit of energy, the dimensions of energy can be calculated as
Dimensions of energy = [J] =[N m] = [kg ms-2 m] = [kg m2 s-2]
⇒ Dimensions of energy = [M][L2][T-2] = [ML2T-2] ….. (A)
Similarly, dimensions of frequency, f = [T-1] ….. (B)
Now to find the dimensions of Planck’s constant, h, we divide the dimensions of energy on the dimensions of frequency.
Dimensions of ‘h’ = [ML2T-2]/[T-1] = [ML2T-1]
There is a mistake in numerical 6 part à in the last step
Yes, thanks Faizan. T is raised to power -1. Now it is corrected. Thanks once again.
Very very thankful 💞
Muhammad Abid
Your are welcome.
thanks
thankssssssssssssssssssssssssss
Pingback:Numerical Problem 7, Measurement … msa – msa
Pingback:Numerical Problem 5, Measurement … msa – msa
Pingback:Numerical Problems on Measurement … msa – msa