JEE Main · Physics ↓ Falling

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

Year 2026 2025 2024 Total
Questions 8 15 8 31
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The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10 m. If it dissipates 10% of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is: [Use g = 10 m s⁻²]

Solution & Explanation

Related Formula

Potential energy at the horizontal release point relative to the lowest point:

Uᵢ = mg

If 10% of energy is dissipated, the remaining 90% is converted entirely to kinetic energy at the lowest point:

Ef = 0.90 × Uᵢ = (1)/(2)mv²
Core Logic

Let the horizontal position be our reference for potential energy relative to the lowest point.

  • Initial potential energy: Uᵢ = mg
  • Energy remaining after 10% loss: 0.90(mg )

    • Equating this to kinetic energy at the bottom:

(9)/(10) mg = (1)/(2) mv²
Step 1: Calculate the Velocity

We can cancel m from both sides:

(9)/(10) g = (1)/(2) v²

Substitute = 10 m and g = 10 m s⁻²:

(9)/(10) (10)(10) = (1)/(2) v² 90 = (1)/(2) v² v² = 180 v = √(180) = √(36 × 5) = 6√(5) m s⁻¹

Pendulum trajectory diagram for Q37
Pendulum trajectory diagram for Q37

Pattern Recognition

Whenever there is a fractional energy loss x, use the formula: v = √(2(1-x)g ). Substituting x = 0.1 gives v = √(1.8g ) directly.

Chapter Mix

Class 11 Physics: Work, Energy and Power

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

More Work, Energy and Power Questions — jee_main_2024_29_january_evening

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Rankbit System
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%)