\[\Delta Q = m S \Delta T\]
\[s = \frac{\Delta Q}{m \Delta T}\]
\[[s] = \frac{\left[ M L^2 T^{-2} \right]}{M \cdot K}\]
\[[s] = \left[ L^2 T^{-2} K^{-1} \right]\]
Statement-(I) is correct.
From \( PV = nRT \), we have:
\[R = \frac{PV}{nT}\]
Substitute dimensions:
\[[R] = \frac{\left[ M L^{-1} T^{-2} L^3 \right]}{\left[ \text{mol} \right] \cdot \left[ K \right]}\]
Simplify:
\[[R] = \left[ M L^2 T^{-2} \text{mol}^{-1} K^{-1} \right]\]
Statement-(II) is incorrect.
Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter $ D $ of a tube. The measured value of $ D $ is:
If \[ f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx, \quad f(0) = -6 \], then f(1) is equal to:
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is: