BITSAT 2026 May 26 Shift 2 Question Paper: Download Answer Key with Solutions PDF

BITSAT Questions

• 1.

  Two capillary tubes of radii \(r_1\) and \(r_2\) \((r_1 > r_2)\) are dipped vertically in the same liquid. The rise of liquid in the tubes \(h_1\) and \(h_2\) satisfies:

  • \(h_1 > h_2\)
    
  • \(h_1 < h_2\)
    
  • \(h_1 = h_2\)
    
  • \(h_1 r_2 = h_2 r_1\)
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• 2.

  The value of enthalpy change (\(\Delta H\)) for the reaction \(\text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)\), at \(27^\circ\text{C}\) is \(-1366.5 \text{ kJ mol}^{-1}\). The value of internal energy change for the above reaction at this temperature will be

  • \( -1371.5 \text{ kJ mol}^{-1} \)
    
  • \( -1369.0 \text{ kJ mol}^{-1} \)
    
  • \( -1364.0 \text{ kJ mol}^{-1} \)
    
  • \( -1361.5 \text{ kJ mol}^{-1} \)
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• 3.

  1.1 mole of A mixed with 2.2 moles of B and the mixture is kept in a 1 L flask and the equilibrium, \(\text{A} + 2\text{B} \rightleftharpoons 2\text{C} + \text{D}\) is reached. If at equilibrium 0.2 mole of C is formed then the value of \(\text{K}_c\) will be

  • \( 0.1 \)
    
  • \( 0.01 \)
    
  • \( 0.001 \)
    
  • \( 1 \)
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• 4.

  What is \([\text{NH}_4^+]\) in a solution that is \(0.02\text{M } \text{NH}_3\) and \(0.01\text{ M KOH}\)? \([K_b(\text{NH}_3) = 1.8 \times 10^{-5}]\)

  • \( 3.6 \times 10^{-5}\text{ M} \)
    
  • \( 1.8 \times 10^{-5}\text{ M} \)
    
  • \( 0.9 \times 10^{-5}\text{ M} \)
    
  • \( 7.2 \times 10^{-5}\text{ M} \)
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• 5.

  If \( |z_1| = 2 \), \( |z_2| = 3 \), \( |z_3| = 4 \) and \( |2z_1 + 3z_2 + 4z_3| = 4 \), then absolute value of \( 8z_2 z_3 + 27z_3 z_1 + 64z_1 z_2 \) equals
  

  • 24
    
  • 48
    
  • 72
    
  • 96
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• 6.

  The magnetic field at the center of a circular current-carrying loop of radius $R$ is $B_0$. The magnetic field at an axial point at a distance $x = R$ from the center of the loop is:

  • \( B_0/2 \)
    
  • \( B_0/2\sqrt{2} \)
    
  • \( B_0/4 \)
    
  • \( B_0\sqrt{2} \)
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