Connected Bodies, Strings & Pulley Systems — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Connected Bodies, Strings & Pulley Systems MCQs with step-by-step solutions (30 questions). Part of Newton's Laws of Motion & Friction. Practise online on Prepizo — no login needed.
▶ Practise Connected Bodies, Strings & Pulley Systems online (free)
Questions with solutions
Q1 — Connected Bodies, Strings & Pulley Systems · medium · theory
For an ideal (massless, inextensible) string, the tension:
A. Can push as well as pull
B. Is zero everywhere
C. Varies along its length
D. Is the same at every point and the string can only pull ✓ Correct
Solution: A massless inextensible string has a uniform tension throughout and exerts only pulling (never pushing) forces along its length.
Q2 — Connected Bodies, Strings & Pulley Systems · medium · theory
An ideal (massless, frictionless) pulley:
A. Doubles the tension
B. Merely changes the direction of the string tension, keeping its magnitude the same on both sides ✓ Correct
C. Halves the tension
D. Adds friction to the string
Solution: A massless frictionless pulley changes only the direction of the tension; the magnitude is equal on both sides of the string.
Q3 — Connected Bodies, Strings & Pulley Systems · hard · theory
In a string that has appreciable mass (a massful rope), the tension:
A. Is the same everywhere
B. Acts perpendicular to the string
C. Varies from point to point along the string ✓ Correct
D. Is zero at the point of support
Solution: For a heavy rope, tension at a section must support the weight of the rope below/behind it, so it varies along the length.
Q4 — Connected Bodies, Strings & Pulley Systems · medium · theory
Two blocks of masses m₁ and m₂ in contact on a smooth floor are pushed by a horizontal force F. Their common acceleration is:
A. $\dfrac{F}{m_1 + m_2}$ ✓ Correct
B. $\dfrac{F(m_1+m_2)}{m_1 m_2}$
C. $\dfrac{F}{m_1}$
D. $\dfrac{F}{m_2}$
Solution: Treating the two blocks as one system, a = F/(m₁ + m₂).
Q5 — Connected Bodies, Strings & Pulley Systems · hard · theory
When force F pushes block m₁ which is in contact with block m₂ (smooth floor), the contact (normal) force between them equals:
A. Zero
B. The weight of the second block
C. The full applied force F
D. The force needed to accelerate the second block: m₂·a ✓ Correct
Solution: The contact force is what accelerates the block that is NOT pushed directly: N = m₂·a = m₂F/(m₁+m₂). Pitfall: the contact force depends on which block is pushed.
Q6 — Connected Bodies, Strings & Pulley Systems · medium · theory
In an Atwood machine with masses m₁ > m₂ over an ideal pulley, the acceleration of the system is:
A. $\dfrac{(m_1 - m_2)g}{m_1 + m_2}$ ✓ Correct
B. $(m_1 - m_2)g$
C. $\dfrac{(m_1 + m_2)g}{m_1 - m_2}$
D. $\dfrac{m_1 g}{m_2}$
Solution: Adding the equations of motion of the two masses gives a = (m₁ − m₂)g/(m₁ + m₂).
Q7 — Connected Bodies, Strings & Pulley Systems · hard · theory
The tension in the string of an ideal Atwood machine (masses m₁ and m₂) is:
A. $\dfrac{(m_1 - m_2)g}{2}$
B. $\dfrac{m_1 m_2 g}{m_1 + m_2}$
C. $\dfrac{2 m_1 m_2 g}{m_1 + m_2}$ ✓ Correct
D. $(m_1 + m_2)g$
Solution: Substituting a into either mass's equation gives T = 2m₁m₂g/(m₁ + m₂) (the harmonic-mean-like result).
Q8 — Connected Bodies, Strings & Pulley Systems · medium · theory
Two blocks m₁ and m₂ connected by a light string on a smooth floor are pulled by force F applied to m₁. The tension in the connecting string is:
A. $\dfrac{F}{2}$
B. $\dfrac{m_1 F}{m_1 + m_2}$
C. F
D. $\dfrac{m_2 F}{m_1 + m_2}$ ✓ Correct
Solution: The tension only has to accelerate the far block m₂: T = m₂·a = m₂F/(m₁+m₂).
Q9 — Connected Bodies, Strings & Pulley Systems · hard · theory
A uniform rope hangs from a ceiling. The tension in the rope:
A. Is greatest at the top and least at the bottom ✓ Correct
B. Is zero at the top
C. Is uniform throughout
D. Is greatest at the bottom
Solution: At any section the tension supports the weight of the rope below it, so it increases from bottom (≈0) to top (= full weight).
Q10 — Connected Bodies, Strings & Pulley Systems · medium · theory
The tension in a string attached to a block always acts:
A. Perpendicular to the string
B. Towards the block (a push)
C. Along the string, directed away from the block (a pull) ✓ Correct
D. Vertically downward always
Solution: A string can only pull, so tension acts along the string, directed away from the body it is attached to.
Q11 — Connected Bodies, Strings & Pulley Systems · hard · theory
A block on a smooth incline is connected over a pulley to a hanging block. The system accelerates in the direction of the hanging block only if:
A. The hanging block is lighter
B. The two masses are equal
C. The weight of the hanging block exceeds the incline component m₁g sinθ ✓ Correct
D. The incline is frictionless only
Solution: The net driving force is m₂g − m₁g sinθ; the hanging mass descends when m₂g > m₁g sinθ.
Q12 — Connected Bodies, Strings & Pulley Systems · medium · theory
Two equal masses hang from the two ends of a string over an ideal pulley. The system is:
A. In equilibrium, with tension = mg on each side ✓ Correct
B. Accelerating downward on both sides
C. Moving with tension 2mg
D. Having zero tension
Solution: Equal masses balance, so a = 0 and the tension equals the weight mg on each side.
Q13 — Connected Bodies, Strings & Pulley Systems · medium · numerical
Blocks of 2 kg and 3 kg are placed in contact on a smooth floor and a horizontal force of 10 N pushes the 2 kg block. Their common acceleration is:
A. 5 m/s²
B. 3.3 m/s²
C. 10 m/s²
D. 2 m/s² ✓ Correct
Solution: a = F/(m₁ + m₂) = 10/(2 + 3) = 2 m/s².
Q14 — Connected Bodies, Strings & Pulley Systems · medium · numerical
For a 2 kg and 3 kg block in contact (smooth floor), with 10 N pushing the 2 kg block, the contact force between them is:
A. 6 N ✓ Correct
B. 5 N
C. 10 N
D. 4 N
Solution: a = 2 m/s². Contact force = m(far block)·a = 3 × 2 = 6 N.
Q15 — Connected Bodies, Strings & Pulley Systems · hard · numerical
The same 2 kg and 3 kg blocks are in contact on a smooth floor, but now the 10 N force pushes the 3 kg block. The contact force between them is:
A. 10 N
B. 6 N
C. 5 N
D. 4 N ✓ Correct
Solution: a = 2 m/s² still, but now the contact force accelerates the 2 kg block: N = 2 × 2 = 4 N. Pitfall: the contact force changed even though a did not.
Q16 — Connected Bodies, Strings & Pulley Systems · medium · numerical
In an Atwood machine with masses 3 kg and 2 kg (g = 10 m/s²), the acceleration of the system is:
A. 10 m/s²
B. 1 m/s²
C. 2 m/s² ✓ Correct
D. 5 m/s²
Solution: a = (m₁ − m₂)g/(m₁ + m₂) = (3 − 2)(10)/(3 + 2) = 10/5 = 2 m/s².
Q17 — Connected Bodies, Strings & Pulley Systems · medium · numerical
For the Atwood machine with 3 kg and 2 kg masses (g = 10 m/s²), the tension in the string is:
A. 24 N ✓ Correct
B. 20 N
C. 30 N
D. 12 N
Solution: T = 2m₁m₂g/(m₁ + m₂) = 2 × 3 × 2 × 10/5 = 120/5 = 24 N.
Q18 — Connected Bodies, Strings & Pulley Systems · hard · numerical
A 4 kg and a 6 kg block are in contact on a smooth floor. A horizontal force of 20 N pushes the 4 kg block. The contact force between the blocks is:
A. 20 N
B. 6 N
C. 12 N ✓ Correct
D. 8 N
Solution: a = 20/10 = 2 m/s². Contact force = 6 × 2 = 12 N.
Q19 — Connected Bodies, Strings & Pulley Systems · medium · numerical
Three blocks of 1 kg, 2 kg, 3 kg are joined by light strings (in that order) and pulled along a smooth floor by 12 N applied to the 1 kg block. The tension in the string between the 2 kg and 3 kg blocks is:
A. 4 N
B. 10 N
C. 12 N
D. 6 N ✓ Correct
Solution: a = 12/6 = 2 m/s². That string pulls only the 3 kg block: T = 3 × 2 = 6 N.
Q20 — Connected Bodies, Strings & Pulley Systems · hard · numerical
A 2 kg block on a smooth incline of 30° is connected over an ideal pulley to a 3 kg hanging block (g = 10 m/s²). The acceleration of the system is:
A. 2 m/s²
B. 4 m/s² ✓ Correct
C. 6 m/s²
D. 10 m/s²
Solution: a = (m₂g − m₁g sinθ)/(m₁ + m₂) = (30 − 2×10×0.5)/5 = (30 − 10)/5 = 4 m/s².
Q21 — Connected Bodies, Strings & Pulley Systems · hard · numerical
For the previous system (2 kg on 30° smooth incline, 3 kg hanging, a = 4 m/s²), the string tension is:
A. 18 N ✓ Correct
B. 24 N
C. 30 N
D. 12 N
Solution: For the hanging block: m₂g − T = m₂a ⇒ T = 3(10 − 4) = 18 N. (Check on incline: T − m₁g sinθ = 18 − 10 = 8 = 2×4 ✓.)
Q22 — Connected Bodies, Strings & Pulley Systems · medium · numerical
Two 5 kg masses hang from the ends of a string over an ideal pulley (g = 10 m/s²). The tension in the string is:
A. 100 N
B. 25 N
C. 0 N
D. 50 N ✓ Correct
Solution: Equal masses ⇒ a = 0 ⇒ T = mg = 5 × 10 = 50 N.
Q23 — Connected Bodies, Strings & Pulley Systems · hard · numerical
A uniform rope of mass 2 kg lies on a smooth floor and is pulled by a horizontal force of 12 N at one end. The tension at the midpoint of the rope is:
A. 9 N
B. 12 N
C. 6 N ✓ Correct
D. 3 N
Solution: a = F/m = 12/2 = 6 m/s². The tension at the midpoint accelerates the half of the rope beyond it (1 kg): T = 1 × 6 = 6 N.
Q24 — Connected Bodies, Strings & Pulley Systems · medium · numerical
An 8 kg block on a smooth floor is pulled by a massless rope with a force of 16 N. The tension in the rope is:
A. 128 N
B. 2 N
C. 16 N ✓ Correct
D. 8 N
Solution: A massless rope transmits the applied force undiminished, so the tension equals 16 N throughout.
Q25 — Connected Bodies, Strings & Pulley Systems · hard · numerical
Three blocks of 1 kg, 2 kg and 3 kg hang vertically from a ceiling, joined by strings (1 kg at top). The tension in the topmost string (g = 10 m/s²) is:
A. 30 N
B. 60 N ✓ Correct
C. 20 N
D. 10 N
Solution: The top string supports all three blocks: T = (1 + 2 + 3)g = 6 × 10 = 60 N.
Q26 — Connected Bodies, Strings & Pulley Systems · hard · numerical
A monkey of mass 10 kg climbs up a rope with an upward acceleration of 2 m/s² (g = 10 m/s²). The tension in the rope is:
A. 20 N
B. 80 N
C. 100 N
D. 120 N ✓ Correct
Solution: T − mg = ma ⇒ T = m(g + a) = 10(10 + 2) = 120 N. Pitfall: the rope must exceed the weight to accelerate the monkey upward.
Q27 — Connected Bodies, Strings & Pulley Systems · medium · numerical
A 5 kg block is pulled vertically upward by a string with an acceleration of 2 m/s² (g = 10 m/s²). The tension in the string is:
A. 40 N
B. 10 N
C. 50 N
D. 60 N ✓ Correct
Solution: T = m(g + a) = 5(10 + 2) = 60 N.
Q28 — Connected Bodies, Strings & Pulley Systems · hard · numerical
A 4 kg block on a smooth table is connected over a pulley at the edge to a 6 kg hanging block (g = 10 m/s²). The acceleration of the system is:
A. 2 m/s²
B. 10 m/s²
C. 4 m/s²
D. 6 m/s² ✓ Correct
Solution: a = m₂g/(m₁ + m₂) = 60/10 = 6 m/s².
Q29 — Connected Bodies, Strings & Pulley Systems · hard · numerical
For a 3 kg block on a smooth table connected over a pulley to a 2 kg hanging block (g = 10 m/s²), the string tension is:
A. 20 N
B. 6 N
C. 15 N
D. 12 N ✓ Correct
Solution: a = 2×10/5 = 4 m/s². Tension = m(table)·a = 3 × 4 = 12 N.
Q30 — Connected Bodies, Strings & Pulley Systems · hard · numerical
A 10 kg block is joined below a 20 kg block by a string; the pair is pulled straight up by a force of 450 N applied to the top (20 kg) block (g = 10 m/s²). The tension in the connecting string is:
A. 450 N
B. 150 N ✓ Correct
C. 100 N
D. 300 N
Solution: a = (F − (m₁+m₂)g)/(m₁+m₂) = (450 − 300)/30 = 5 m/s². The connecting string carries the lower 10 kg block: T = m(g + a) = 10(10 + 5) = 150 N.