Explanation​:
The magnetic field between the magnets is from N to S, therefore towards the left.

The magnetic field of the current carrying conductor is towards the left at A, towards the right at B (right-hand rule)

At B and C, the two fields will be perpendicular to each other.

At D they will be opposite to each other and therefore partially cancel out.

Since they are in the same direction at A, the resultant will be greatest at A.

Explanation​:

Because the electric field lines point toward the charge $q$, this charge must be negative, meaning that statement I is false.

The relatively greater number of field lines around charge $q$ means that this charge magnitude is greater than the charge +Q, indicating that statement II is also untrue.

Electric field line density is proportional to electric field strength. Because the field line density near $q$ is greater than that near $+Q$, statement III is correct.

Therefore the correct answer is option C.

Explanation​:
The electric field strength direction is the same as the direction of the electric force.

The electric field will be directed out of the rod in all directions, as it contains uniformly distributed positive charges.

Getting the vector sum for the electric field, due to symmetry the total electric field strength will be directed out of the rod with an angle away from the midpoint of the rod as shown

Also we can think of B as the only correct solution, because it’s a positively charged rod, so the direction must be out of the rod.

Explanation​:
Summing the four electric field vectors from the individual point charges, the fields from the charge at the top left and the bottom right will cancel out. The fields from the bottom left and top right will add together to give a net field in the direction of the charge in the top right.

Since object moves at a constant velocity, vertical forces are balanced and Fe=mg, where Fe=Eq.

So, $B=0$ and E=mg/q​.

Explanation​:
Centripetal acceleration of an orbiting satellite is the gravitational acceleration of the planet, which is given by

g=GM/r2

Since the gravitational acceleration is inversely proportional to the square of the orbital radius, the correct graphical representation is given in C.

Explanation​:

Statement I is incorrect. This is due to Newton’s Law of Gravitation. The astronaut will experience gravitational force from the Earth that acts as the centripetal force.

Since the centripetal force is the gravitational force, then the centripetal acceleration is equal to the gravitational field strength at that point. Therefore both the astronaut and the space station experience the same acceleration. Then statements II and III are correct.

Explanation​:
The $20nC$ charge produces a leftward electric field in region 1. Therefore, the electric field generated by the $X$ charge should be rightward to cancel the electric field of the 20nC charge. This is achieved by a negative charge at X. The magnitude of the electric field decreases by distance. All locations in region 1 are closer to point charge $X$ than they are to the $20nC$ charge. Therefore $X$ should be smaller than the charge of $20nC$. The answer is D.

Explanation​:
Recall that magnetic force acts perpendicular to current flow according to the Fleming Left hand rule. That immediately eliminates option A and D. Then using the right hand grip rule, we can determine that the magnetic field experienced by wire $Y$ points into the page. The force direction and magnetic field direction allows current direction to be found with Fleming’s left-hand rule.

Explanation​:
The direction of the current is perpendicular to the magnetic field in both regions and the magnitude of the field is unchanged. So the magnitude of the force is $F$.

Using Fleming's left hand rule, direction of the force is

Explanation​:
Choice A is incorrect as it equates the gravitational field strength with the gravitational force and these two are two different physical quantities.

Choice B is the correct answer, since the gravitational field strength equals to force per unit mass; g=F/m.

Choice C is incorrect, because the gravitational field strength is inversely proportional to the square of the distance between the center of mass of the planet and that point; g=GM/r2

Choice D is incorrect, because the gravitational field strength depends on the mass of the planet; g=GM/r2​.

Explanation​:
The strong nuclear force is the strongest of the four fundamental forces, then comes electromagnetic, then weak nuclear force and finally gravitation.

Explanation​:
Conventional current will flow anticlockwise due to the orientation of the cell. To determine the force experienced by a current carrying conductor in a magnetic field, use (any appropriate hand-rule convention or) Flemming's left hand rule which states that if we arrange our thumb, forefinger and middle finger of the left-hand perpendicular to each other, then the thumb points towards the direction of the force experienced by the conductor, the forefinger points towards the direction of the magnetic field and the middle finger points towards the direction of the electric current.

Explanation​:
The alpha, beta and gamma radiation have different penetrating powers, where alpha is the one with the least penetrating power and won’t pass through the paper sheet, while gamma and beta radiations will pass through the paper sheet.

When entering the magnetic field, gamma radiation will continue in its path without deflection because they have no charge. On the other hand, beta particles have negative charges, so they will be affected by a magnetic force that causes deflection in their path.

To determine the direction of the magnetic force, the right hand rule is used keeping in mind that the velocity direction in the right hand rule is that of a positive charge. So we have the direction of velocity is to the left, the magnetic field is into page, thus the magnetic force will be directed downwards.

*Any other hand-rule convention can also be used.

Explanation​:

The electric field at X due to -Q points toward -Q. The field at X due to +Q points away from +Q, with the same magnitude at X as the field from -Q. Adding the electric field vectors gives a vector pointing to the left.