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Gravitation question

2022 · 28 Jun · Shift 2 · Q54
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  5. /2022 · 28 Jun · Shift 2 · Q54

Gravitation question

2022 · 28 Jun · Shift 2 · Q54

JEE MainPhysicsGravitationMCQ+4 / −1
Water falls from a 40 m high dam at the rate of 9 ×\times× 104 kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydroelectric energy number of 100 W lamps, that can be lit, is : (Take g = 10 ms −-− 2)
  1. A
    25
  2. B
    50
  3. C
    100
  4. D
    18
View written solutionFree

Correct answer: B

  1. Given data
  • Height of dam: h=40 mh = 40\,\text{m}h=40m
  • Mass of water falling per hour: 9×104 kg/hour9 \times 10^4\,\text{kg/hour}9×104kg/hour
  • Acceleration due to gravity: g=10 m s−2g = 10\,\text{m s}^{-2}g=10m s−2
  • Efficiency of conversion: 50%=0.550\% = 0.550%=0.5
  • Power of each lamp: 100 W100\,\text{W}100W
  1. Gravitational potential energy of water falling in 1 hour

Potential energy released in 1 hour:

E=mghE = mghE=mgh E=(9×104)(10)(40)E = (9 \times 10^4)(10)(40)E=(9×104)(10)(40) E=3.6×107 JE = 3.6 \times 10^7\,\text{J}E=3.6×107J
  1. Electrical energy obtained in 1 hour

Only 50%50\%50% is converted into electrical energy:

Eelec=0.5×3.6×107E_{\text{elec}} = 0.5 \times 3.6 \times 10^7Eelec​=0.5×3.6×107 Eelec=1.8×107 JE_{\text{elec}} = 1.8 \times 10^7\,\text{J}Eelec​=1.8×107J
  1. Convert this energy into power

Since this energy is produced in 111 hour = 360036003600 s,

P=Et=1.8×1073600P = \frac{E}{t} = \frac{1.8 \times 10^7}{3600}P=tE​=36001.8×107​ P=5×103 W=5000 WP = 5 \times 10^3\,\text{W} = 5000\,\text{W}P=5×103W=5000W
  1. Number of 100 W lamps
N=5000100=50N = \frac{5000}{100} = 50N=1005000​=50
  1. Check options
  • A: 25 ❌
  • B: 50 ✅
  • C: 100 ❌
  • D: 18 ❌

Therefore, the correct answer is Option B.

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