COMEDK UGET 2020 Question Paper(Available): Download Solution PDF with Answer Key PDF

COMEDK UGET Questions

• 1.

  The difference in energy levels of an electron at two excited levels is 13.75 eV. If it makes a transition from the higher energy level to the lower energy level then what will be the wavelength of the emitted radiation? 
  Given:
  $ h = 6.6 \times 10^{-34} \, \text{m}^2 \, \text{kg} \, \text{s}^{-1} $, $ c = 3 \times 10^8 \, \text{ms}^{-1} $, $ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} $

  • 900 nm
  • \( 90^\circ \, \text{A} \)
  • 9000 nm
  • 900  $\text{A}$

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• 2.

  A particle executes a simple harmonic motion of amplitude A. The distance from the mean position at which its kinetic energy is equal to its potential energy is:

  • 0.91 A
  • 0.71 A
  • 0.81 A
  • 0.51 A
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• 3.

  For a reaction $ 5X + Y \to 3Z $, the rate of formation of Z is $ 2.4 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1} $. Calculate the average rate of disappearance of X.
  

  • \( 4.8 \times 10^{-7} \, \text{mol L}^{-1} \text{s}^{-1} \)
  • \( 13.33 \times 10^{-6} \, \text{mol L}^{-1} \text{s}^{-1} \)
  • \( 4.0 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1} \)
  • \( 12.0 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1} \)
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• 4.

  At 700 K, the Equilibrium constant value for the formation of HI from $ H_2 $ and $ I_2 $ is 49.0. 0.7 mole of HI(g) is present at equilibrium. What will be the concentrations of $ H_2 $ and $ I_2 $ gases if we initially started with HI(g) and allowed the reaction to reach equilibrium at the same temperature?

  • 0.1195
  • 0.3442
  • 0.4692
  • 0.521
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• 5.

  An inorganic compound W undergoes the following reactions: $ W + \text{Na}_2\text{CO}_3 \xrightarrow{\text{O}_2 / \text{heat}} X + H^+ \xrightarrow{} Y(s) $ $ Y(aq) + \text{KCl} (aq) \xrightarrow{} Z(s) $ Z appears in the form of orange crystals and is used as an oxidising agent in acid medium. Identify the compound W.

  • \(\text{K}_2\text{Cr}_2\text{O}_4\)
  • \(\text{FeCr}_2\text{O}_4\)
  • \(\text{Cu}(\text{Cr}_2\text{O}_4)_2\)
  • \(\text{Na}_2\text{Cr}_2\text{O}_7\)
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• 6.

  In a single slit diffraction experiment, for slit width $ \alpha $, the width of the central maxima is $ \beta $. If we double the slit width then the corresponding width of the central maxima will be:
  

  • \( 4\beta \)
  • \( \beta \)
  • \( \frac{\beta}{2} \)
  • \( 2\beta \)
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