Modulus of Elasticity (Young's Modulus) = \( \frac{\text{Stress}}{\text{Strain}} \)
Strain is dimensionless, so the dimension of Modulus of Elasticity = Dimension of Stress.
Stress = \( \frac{\text{Force}}{\text{Area}} = \frac{[M^1 L^1 T^{-2}]}{[L^2]} = [M^1 L^{-1} T^{-2}] \)
Torque = \( \text{Force} \times \text{Distance} \)
\( [M^1 L^1 T^{-2}] \times [L] = [M^1 L^2 T^{-2}] \)
The measured quantity is Modulus of Elasticity per unit Torque.
\[
\text{Dimension} = \frac{[M^1 L^{-1} T^{-2}]}{[M^1 L^2 T^{-2}]} = [M^{1 - 1} L^{-1 - 2} T^{-2 + 2}]
\]
\[
= [M^{0} L^{-3} T^{0}] = [L^{-3}]
\]
Comparing the dimensions: \[ [M^0 L^{-3} T^0] = [M^a L^b T^c] \] \[ a = 0, \, b = -3, \, c = 0 \]
Since we need to compare it with \( b = 3 \), we take the magnitude of the value \( b = -3 \).
Therefore, the value of \( c \) is:
\[
\boldsymbol{0}
\]
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: