A block is pulled along a horizontal frictionless surface by a rope. The tension in the rope will be the same at all points on it:
A. If and only if the rope is not accelerated
B. If and only if the rope is massless
C. If either the rope is not accelerated or it is massless
D. Always

Hint: Use Newton’s second law of motion. Apply Newton’s law of motion to the block only in horizontal direction.

Formula used:
Newton’s second law of motion is
\[ \Rightarrow {F_{net}} = ma\] …… (1)
Here, \[{F_{net}}\] is the net force on the object, \[m\] is the mass of the object and \[a\] is the acceleration of the object.

Complete step by step answer:
The net force acting on the block in the horizontal direction is the force of tension \[T\] in the rope in a direction in which the block is being pulled.
Rewrite equation (1) for the net force on the block in the horizontal direction.
\[ \Rightarrow T = \left( {{m_b} + {m_r}} \right)a\] …… (2)
Here, \[{m_b}\] is the mass of the block, \[{m_r}\] is the mass of rope and \[a\] is the acceleration of the block.
Consider the mass of the rope is zero.
Substitute \[0\,{\text{kg}}\] for \[{m_r}\] in equation (2).
\[ \Rightarrow T = \left( {{m_b} + 0\,{\text{kg}}} \right)a\]
\[ \Rightarrow T = {m_b}a\]
Hence, if the mass of the rope is zero then the tension is constant at all points on the rope.
Consider the block is not accelerated.
Substitute \[0\,{\text{m/}}{{\text{s}}^2}\] for \[a\] in equation (2).
\[ \Rightarrow T = \left( {{m_b} + {m_r}} \right)\left( {0\,{\text{m/}}{{\text{s}}^2}} \right)\]
\[ \Rightarrow T = 0\,{\text{N}}\]
Hence, if the block is not accelerated, the tension on the rope is zero at all the points.
Therefore, the tension is the same at all the points if the block is not accelerated or the rope is massless.

Hence, the correct option is C.

Note: Newton’s law is not applied in the vertical direction as the weight of the block in downward direction is balanced by the normal force of the surface on the block in vertical direction.