Given below are two statements :Statement (I) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The formula for heat energy is $\Delta Q = mS \Delta T$,where $S$ is the specific heat capacity.Therefore,$S = \frac{\Delta Q}{m \Delta T}$.The di

Vedclass · Accepted Answer

Statement $(I)$ is correct but statement $(II)$ is incorrect.

Vedclass · Answer

(C) The formula for heat energy is $\Delta Q = mS \Delta T$,where $S$ is the specific heat capacity.Therefore,$S = \frac{\Delta Q}{m \Delta T}$.The dimensions are $[S] = \frac{[ML^2 T^{-2}]}{[M][K]} = [L^2 T^{-2} K^{-1}]$.Thus,Statement $(I)$ is correct.The ideal gas equation is $PV = nRT$,where $R$ is the universal gas constant.Therefore,$R = \frac{PV}{nT}$.The dimensions are $[R] = \frac{[ML^{-1} T^{-2}][L^3]}{[mol][K]} = [ML^2 T^{-2} mol^{-1} K^{-1}]$.Comparing this with the given statement,Statement $(II)$ is incorrect.

Given below are two statements :
Statement $(I)$ : Dimensions of specific heat are $\left[L^2 \,T^{-2} \,K^{-1}\right]$.
Statement $(II)$ : Dimensions of gas constant are $\left[ML^2 \,T^{-1} \,K^{-1}\right]$.

Similar Questions

The dimensional formula of potential difference is ........

The dimensional formula $M{L^2}{T^{ - 3}}$ represents:

Dimensions of potential energy are

Write the dimensional formula for impulse of a force.

What is the dimensional formula of electrical resistance?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Given below are two statements :Statement $(I)$ : Dimensions of specific heat are $\left[L^2 \,T^{-2} \,K^{-1}\right]$.Statement $(II)$ : Dimensions of gas constant are $\left[ML^2 \,T^{-1} \,K^{-1}\right]$.