Statement-wise analysis: A. Electrostatic field lines form closed loops. False. Electrostatic field lines always originate from positive charges and terminate on negative charges. They never form closed loops. Closed loops are characteristic of magnetic field lines.
B. The electric field lines point radially outward when charge is greater than zero. True. For a positive point charge, electric field lines emerge radially outward, indicating the direction of force on a positive test charge.
C. The Gauss’s Law is valid only for inverse-square force. True. Gauss’s law strictly holds when the force follows an inverse-square dependence on distance, as is the case for electrostatic (Coulomb) force.
D. The work done in moving a charged particle in a static electric field around a closed path is zero. True. Electrostatic fields are conservative. Hence, the work done over any closed loop is zero.
E. The motion of a particle under Coulomb’s force must take place in a plane. True. Coulomb force is a central force. Motion under any central force is always confined to a plane.
Step 2: Collect the true statements Correct statements are: \[ \boxed{B,\ C,\ D,\ E} \] Final Answer: \[ \boxed{\text{(C) B, C, D, E Only}} \]
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

A small block of mass \(m\) slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration \(a_0\). The angle between the inclined plane and ground is \(\theta\) and its base length is \(L\). Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is _______. 
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)