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Work, Energy and Power appeared 31 times across 3 years — 3.6% of Physics. This question is from Vertical Circular Motion.

Year 2026 2025 2024 Total
Questions 8 15 8 31
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In case of vertical circular motion of a particle by a thread of length r if the tension in the thread is zero at an angle 30° shown in figure, the velocity at the bottom point (A) of the circular path is (g = gravitational acceleration)
Vertical Circular Motion diagram for Q30 - JEE Main 2026 Evening
A particle on a circular track of radius r with tension becoming zero at 30 degrees to the horizontal.

Solution & Explanation

Related Formula
T + mg θ = (mV²)/(r)

Einitial = Efinal (Mechanical Energy Conservation)

Core Logic

Vertical Circular Motion diagram for Q30 - JEE Main 2026 Evening
A particle on a circular track of radius r with tension becoming zero at 30 degrees to the horizontal.

At the point where the string makes an angle of 60° with the upward vertical (which corresponds to 30° above the horizontal diameter):

T + mg 60° = (mV²)/( )

Given that T = 0:

V² = (g )/(2)

Here, V is the speed at the topmost point.

Step 1: Energy Conservation

Applying Mechanical Energy Conservation (M.E.C.) between the bottom point and the point of zero tension:

(1)/(2) m u² = mg( + 60°) + (1)/(2) m V²
Step 2: Solving for u
(1)/(2) m u² = mg ( + ( )/(2)) + (1)/(2) m ((g )/(2)) u² = 3g + (g )/(2) u = √((7g )/(2))

(Note: The problem uses r as the length, so substituting r for yields √((7)/(2) gr)).

Pattern Recognition

When tension is zero at an angle θ above the horizontal, the required bottom velocity is derived by finding the critical velocity (mv²)/(r) = mg θ and then equating energy.

Chapter Mix

Class 11 Physics: Work, Energy and Power Class 11 Physics: Laws of Motion

Reference Study Guides

Work Energy And Power

More Work, Energy and Power Previous-Year Questions — Page 7

Q49 jee_main_2024_31_jan_morning Conservation Of Momentum
An artillery piece of mass M₁ fires a shell of mass M₂ horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is :
  • A. M₁ / (M₁ + M₂)
  • B. (M₂)/(M₁)
  • C. M₂ / (M₁ + M₂)
  • D. (M₁)/(M₂)

Solution

Related Formula
KE = (p²)/(2m)
Core Logic

By conservation of linear momentum (since no external horizontal force acts on the system):

0 = M₁ v₁ + M₂ v₂ | p₁| = | p₂| = p

Both the artillery and the shell acquire the exact same magnitude of momentum during firing.

Step 2: Kinetic Energy Ratio

The kinetic energy is related to momentum by KE = (p²)/(2m). Since p is identical for both bodies:

KE ∝ (1)/(m)

Therefore, the ratio of kinetic energy of the artillery (M₁) to the shell (M₂) is:

KE₁KE₂ = ((p²)/(2M₁))/((p²)/(2M₂)) = (M₂)/(M₁)
Chapter Mix

Class 11 Physics: Work, Energy And Power

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)