\[ \text{Graph 1: } \]
\[ \text{Graph 2: } \]
\[ \text{Graph 3: } \]
\[ \text{Graph 4: } \]
The kinetic energy \( K \) of a particle is given by: \[ K = \frac{1}{2} m v^2, \] where \( v \) is the velocity of the particle. The velocity is the derivative of the displacement \( x(t) \) with respect to time: \[ v(t) = \frac{d}{dt} \left( x_0 \sin^2 \left( \frac{\pi t}{T} \right) \right) = 2x_0 \sin \left( \frac{\pi t}{T} \right) \cos \left( \frac{\pi t}{T} \right) \frac{\pi}{T}. \] Thus, the velocity is proportional to \( \sin \left( \frac{\pi t}{T} \right) \), and the kinetic energy is proportional to the square of the velocity, which results in a graph where the kinetic energy increases as the particle moves from the origin to its maximum displacement and decreases symmetrically thereafter.
Final Answer: Graph 1.
To obtain the given truth table, the following logic gate should be placed at G:
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)
For the reaction, \[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \] Attainment of equilibrium is predicted correctly by:
Match List - I with List - II:
List - I:
(A) \([ \text{MnBr}_4]^{2-}\)
(B) \([ \text{FeF}_6]^{3-}\)
(C) \([ \text{Co(C}_2\text{O}_4)_3]^{3-}\)
(D) \([ \text{Ni(CO)}_4]\)
List - II:
(I) d²sp³ diamagnetic
(II) sp²d² paramagnetic
(III) sp³ diamagnetic
(IV) sp³ paramagnetic