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

Year 2026 2025 2024 Total
Questions 8 15 8 31
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A motor operating on 100~V draws a current of 1 A. If the efficiency of the motor is 91.6%, then the loss of power in units of cal/s is :

Solution & Explanation

Related Formula

The electrical input power Pᵢₙ delivered to the motor is:

Pᵢₙ = V · I

The fraction of power lost is determined by the efficiency η:

Plost = Pᵢₙ - Pout = Pᵢₙ (1 - η)

To convert watts (J/s) to calories per second:

Plost (cal/s) = Plost (W)4.2
Core Logic

Given parameters:

  • Potential
Step 1: Calculate Input Power ($P_{\text{in}})
Pᵢₙ = 100 × 1 = 100~W
Step 2: Calculate Power Loss in Watts
Plost = 100 × (1 - 0.916) = 100 × 0.084 = 8.4~W
Step 3: Convert Power Loss to cal/s
Plost = (8.4)/(4.2) = 2~cal/s
Pattern Recognition

Efficiency calculations often feature clean conversion rates. Recognizing that

Pattern Recognition

Efficiency calculations often feature clean conversion rates. Recognizing that $8.4is exactly twice4.2$ (the conversion factor for Joules to calories) simplifies the final step.

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_2025_03_april_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%)