Physics And Measurement Question 232
Question: Dimensions of specific heat are
Options:
A) $ [ML^{2}{{T}^{-2}}K] $
B) $ [ML^{2}{{T}^{-2}}{{K}^{-1}}] $
C) $ [ML^{2}T^{2}{{K}^{-1}}] $
D) $ [L^{2}{{T}^{-2}}{{K}^{-1}}] $
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Answer:
Correct Answer: D
Solution:
[d] $ s=\frac{Q}{m\theta }=\frac{ML^{2}{{T}^{-2}}}{MK}=[L^{2}{{T}^{-2}}{{K}^{-1}}] $