BITSAT 2023 21 May Question Paper: Download Question Paper with Answers PDF

BITSAT Questions

• 1.

  Let the foot of perpendicular from a point \( P(1,2,-1) \) to the straight line \( L: \frac{x}{1} = \frac{y}{0} = \frac{z}{-1} \) be \( N \). Let a line be drawn from \( P \) parallel to the plane \( x + y + 2z = 0 \) which meets \( L \) at point \( Q \). If \( \alpha \) is the acute angle between the lines PN and PQ, then \( \cos \alpha \) is equal to
  

  • \( \frac{1}{\sqrt{5}} \)
    
  • \( \frac{\sqrt{3}}{2} \)
    
  • \( \frac{1}{\sqrt{3}} \)
    
  • \( \frac{1}{2\sqrt{3}} \)
  View Solution

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• 2.

  If \( \vec{a} = 2\hat{i} + \hat{j} + 2\hat{k} \), then the value of \( |\hat{i} \times (\vec{a} \times \hat{i})|^2 + |\hat{j} \times (\vec{a} \times \hat{j})|^2 + |\hat{k} \times (\vec{a} \times \hat{k})|^2 \) is equal to
  

  • 17
    
  • 18
    
  • 19
    
  • 20
  View Solution

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• 3.

  Two lighter nuclei combine to form a comparatively heavier nucleus by the relation \[ \frac{2}{1}\text{X} + \frac{2}{1}\text{X} = \frac{4}{2}\text{Y} \] The binding energies per nucleon of \( \frac{2}{1}\text{X} \) and \( \frac{4}{2}\text{Y} \) are 1.1 MeV and 7.6 MeV respectively. The energy released in this process is
  

  • 26 MeV
    
  • 56 MeV
    
  • 78 MeV
    
  • 108 MeV
  View Solution

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• 4.

  The value of \( \int e^{\tan \theta} (\sec \theta - \sin \theta) \, d\theta \) is
  

  • \( e^{\tan \theta} \sec \theta + c \)
    
  • \( e^{\tan \theta} \sin \theta + c \)
    
  • \( e^{\tan \theta} (\sec \theta + \sin \theta) + c \)
    
  • \( e^{\tan \theta} \cos \theta + c \)
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• 5.

  A first-order reaction is 25% complete in 30 minutes. How much time will it take for the reaction to be 75% complete?
  

  • 90 min
    
  • 60 min
    
  • 120 min
    
  • 150 min
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• 6.

  What is the IUPAC name of the compound: CH\(_3\) – CH\(_2\) – CH(CH\(_3\)) – CH\(_2\) – CH\(_3\)
  

  • 3-Methylpentane
    
  • 2-Methylpentane
    
  • 3-Methylbutane
    
  • 2-Methylbutane
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