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}}] $



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