GATE 2010 Mechanical Engineering (ME) Question Paper with Answer Key PDFs (February 14- Forenoon Session)

GATE 2010 Mechanical Engineering (ME) Question paper with answer key pdfs are available for download. GATE 2010 ME Paper was conducted by the Indian Institute of Technology (IIT) Guwahati in the Forenoon Session on February 14. GATE 2010 IN had a moderate difficulty level. There were a total of 65 questions with 1 mark and 2 mark MCQs from General Aptitude and Mechanical Engineering.

GATE 2010 Mechanical Engineering (ME) Question Paper with Answer Key Pdfs Offline (February 14 - Forenoon Session)

GATE 2010 Mechanical Engineering (ME) Question Paper GATE 2010 Mechanical Engineering (ME) Answer Key
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GATE 2010 Mechanical Engineering (ME) Paper Analysis

  • There were 65 multiple-Choice questions in the test. Questions 1 to 25 carry 1mark and 26 to 55 carries 2 marks and Question 56 to 65 belong to General Aptitude with 56 to 60 being 1 mark and 61 to 65 comprising 2mark. The technical component was graded as 'easy to moderate,' while the general aptitude section was evaluated as 'easy.'
  • GATE 2010 IN has 65 MCQ questions, according to IIT Roorkee. Please see the table below for information on the MCQs that were listed based on the marks earned.
Question Types Question Frequency Carried Marks
MCQs with 1 mark 30 30
MCQs with 2 marks 35 70
Total 65 100

GATE Previous Year Question Papers:

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GATE Questions

  • 1.
    Question Text

      • A
      • B

    • 2.

      The area of a rhombus whose vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order, is:

        • 12 sq.units

        • 24 sq.units


      • 3.

        In p-type semiconductor density of mobile holes exceeds that of conduction electrons. Hence, minority carriers in p -type semiconductor are conduction (free) electrons.


         

          • a

          • b


        • 4.

          Write a balanced equation for the reaction of molecular nitrogen (N2) and oxygen (O2) to form dinitrogen pentoxide. \((f(x)=x^2)\)

            • N2 + O2 → N2O5 (unbalanced equation)
            • 2N2 + 5O2 → 2N2O5
            • option 3

          • 5.
            For the reaction, $H_{2} + I_{2} {\rightleftharpoons} 2HI, K= 47.6.$ If the initial number of moles of each reactant and product is 1 mole then at equilibrium

              • $\left[I_{2}\right]=\left[H_{2}\right], \left[I_{2}\right] > \left[HI\right]$
              • $({\frac{x^3}{9}})$
              • \(\left[I_{2}\right]>\left[H_{2}\right], \left[I_{2}\right] = \left[HI\right]\)

              • $\omega\propto\,n^{\frac{1}{3}}$

            • 6.

              test

                • test

                • test

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