Understanding the effect of Earth's rotation on gravity.
Due to the rotation of the Earth, the effective acceleration due to gravity is given by: \[ g_{\text{eff}} = g - \omega^2 R \cos^2 \theta \] Where \( \theta \) is the angle made with the equator.
At the poles, where \( \theta = 90^\circ \), the change in gravity is zero because \( \cos(90^\circ) = 0 \). This shows no effect on the poles. For the equator, where \( \theta = 0^\circ \), the change in gravity is maximum: \[ g_{\text{eff}} = g - \omega^2 R \] Thus, the change in gravity is maximum at the equator and zero at the poles.
This contradicts Statement II, but Statement I is correct.
The acceleration due to gravity at a height of 6400 km from the surface of the earth is \(2.5 \, \text{ms}^{-2}\). The acceleration due to gravity at a height of 12800 km from the surface of the earth is (Radius of the earth = 6400 km)
Let A = \(\begin{bmatrix} \log_5 128 & \log_4 5 \log_5 8 & \log_4 25 \end{bmatrix}\) \). If \(A_{ij}\) is the cofactor of \( a_{ij} \), \( C_{ij} = \sum_{k=1}^2 a_{ik} A_{jk} \), and \( C = [C_{ij}] \), then \( 8|C| \) is equal to:
A molecule with the formula $ \text{A} \text{X}_2 \text{Y}_2 $ has all it's elements from p-block. Element A is rarest, monotomic, non-radioactive from its group and has the lowest ionization energy value among X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is:
A transition metal (M) among Mn, Cr, Co, and Fe has the highest standard electrode potential $ M^{n}/M^{n+1} $. It forms a metal complex of the type $[M \text{CN}]^{n+}$. The number of electrons present in the $ e $-orbital of the complex is ... ...
Consider the following electrochemical cell at standard condition. $$ \text{Au(s) | QH}_2\text{ | QH}_X(0.01 M) \, \text{| Ag(1M) | Ag(s) } \, E_{\text{cell}} = +0.4V $$ The couple QH/Q represents quinhydrone electrode, the half cell reaction is given below: $$ \text{QH}_2 \rightarrow \text{Q} + 2e^- + 2H^+ \, E^\circ_{\text{QH}/\text{Q}} = +0.7V $$
0.1 mol of the following given antiviral compound (P) will weigh .........x $ 10^{-1} $ g.