JEE Main Question Paper 2003 (Available): Check Previous Year Question Paper with Solution PDF

JEE Main 2003 Questions

• 1.

  The electric field of an electromagnetic wave in free space is \[ \vec{E} = 57 \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] \left( 4\hat{i} - 3\hat{j} \right) \, \text{N/C}. \] The associated magnetic field in Tesla is:

  • \( \vec{B} = \frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (5\hat{k}) \)
    
  • \( \vec{B} = \frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (\hat{k}) \)
    
  • \( \vec{B} = - \frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (5\hat{k}) \)
    
  • \( \vec{B} = - \frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t - 5 \times 10^{-3} (3x + 4y)\right] (\hat{k}) \)
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• 2.

  A transparent block A having refractive index $ \mu_2 = 1.25 $ is surrounded by another medium of refractive index $ \mu_1 = 1.0 $ as shown in figure. A light ray is incident on the flat face of the block with incident angle $ \theta $ as shown in figure. What is the maximum value of $ \theta $ for which light suffers total internal reflection at the top surface of the block ?
  

  • \( \tan^{-1}(4/3) \)
  • \( \tan^{-1}(3/4) \)
  • \( \sin^{-1}(3/4) \)
  • \( \cos^{-1}(3/4) \)
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• 3.

  A particle is subjected to simple harmonic motions as: $ x_1 = \sqrt{7} \sin 5t \, \text{cm} $ $ x_2 = 2 \sqrt{7} \sin \left( 5t + \frac{\pi}{3} \right) \, \text{cm} $ where $ x $ is displacement and $ t $ is time in seconds. The maximum acceleration of the particle is $ x \times 10^{-2} \, \text{m/s}^2 $. The value of $ x $ is:

  • 175
  • 25 \(\sqrt{7}\)
  • \( 5 \sqrt{7} \)
  • 125
    
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• 4.

  Let A be the point of intersection of the lines $$ L_1 : \frac{x - 7}{1} = \frac{y - 5}{0} = \frac{z - 3}{-1} \quad \text{and} \quad L_2 : \frac{x - 1}{3} = \frac{y + 3}{4} = \frac{z + 7}{5}$$ Let B and C be the points on the lines $ L_1 $ and $ L_2 $, respectively, such that $ AB - AC = \sqrt{15} $. Then the square of the area of the triangle ABC is:

  • 54
  • 63
  • 57
  • 60
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• 5.

  The molar mass of the water insoluble product formed from the fusion of chromite ore \(FeCr_2\text{O}_4\) with \(Na_2\text{CO}_3\) in presence of \(O_2\) is ....... g mol\(^{-1}\):

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