\[\Delta Q = m S \Delta T\]
\[s = \frac{\Delta Q}{m \Delta T}\]
\[[s] = \frac{\left[ M L^2 T^{-2} \right]}{M \cdot K}\]
\[[s] = \left[ L^2 T^{-2} K^{-1} \right]\]
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.
Match the LIST-I with LIST-II
LIST-I | LIST-II | ||
---|---|---|---|
A. | Boltzmann constant | I. | \( \text{ML}^2\text{T}^{-1} \) |
B. | Coefficient of viscosity | II. | \( \text{MLT}^{-3}\text{K}^{-1} \) |
C. | Planck's constant | III. | \( \text{ML}^2\text{T}^{-2}\text{K}^{-1} \) |
D. | Thermal conductivity | IV. | \( \text{ML}^{-1}\text{T}^{-1} \) |
Choose the correct answer from the options given below :
Choose the correct set of reagents for the following conversion:
A bead of mass \( m \) slides without friction on the wall of a vertical circular hoop of radius \( R \) as shown in figure. The bead moves under the combined action of gravity and a massless spring \( k \) attached to the bottom of the hoop. The equilibrium length of the spring is \( R \). If the bead is released from the top of the hoop with (negligible) zero initial speed, the velocity of the bead, when the length of spring becomes \( R \), would be (spring constant is \( k \), \( g \) is acceleration due to gravity):