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TS EAMCET 2020 Agriculture and Medical Question Paper with Answer Key PDFs (September 29 - Forenoon Session)
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Collegedunia Team

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TS EAMCET 2020 Agriculture and Medical Question paper with answer key pdf conducted on September 29 in Forenoon Session 9 AM to 12 PM is available for download. The exam was successfully organized by Jawaharlal Nehru Technological University, Hyderabad (JNTUH). The question paper comprised a total of 160 questions divided among 4 sections.

TS EAMCET 2020 Agriculture and Medical Question Paper with Answer Key PDFs Forenoon Session

TS EAMCET 2020 Agriculture and Medical Question Paper PDF TS EAMCET 2020 Agriculture and Medical Answer Key PDF
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TS EAMCET Questions

1.

If i=√-1 then 

\[Arg\left[ \frac{(1+i)^{2025}}{1+i^{2022}} \right] =\]

    • \(\frac{-π}{4}\)

    • \(\frac{π}{4}\)

    • \(\frac{3π}{4}\)

    • \(\frac{-3π}{4}\)

    2.
    The energy of second orbit of hydrogen atom is -5.45x10-19 J. What is the energy of first orbit of Li2+ ion (in J)

      • -1.962x10-18

      • -1.962x10-17

      • -3.924x10-17

      • -3.924x10-18

      3.

      The number of diagonals of a polygon is 35. If A, B are two distinct vertices of this polygon, then the number of all those triangles formed by joining three vertices of the polygon having AB as one of its sides is:

        • 1

        • 8

        • 10

        • 12

        4.
        If the ratio of densities of two substances is 5:6 and the ratio of their specific heat capacities is 3:5, then the ratio of heat energies required per unit volume so that the two substances can have same temperature rise is: 

          • 1:1

          • 1:4

          • 1:2

          • 1:3

          5.

          lim n→∞ \(\frac{1}{n^3} \) 

          \[\sum_{k=1}^{n} k^{2} = \]

           

            • x

            • \(\frac{x}{2}\)

            • \(\frac{x}{3}\)

            • \(\frac{x}{4}\)

            6.

            5 persons entered a lift cabin in the cellar of a 7-floor building apart from cellar. If each of the independently and with equal probability can leave the cabin at any floor out of the 7 floors beginning with the first, then the probability of all the 5 persons leaving the cabin at different floors is

              • \(\frac{360}{2401}\)

              • \(\frac{5}{54}\)

              • \(\frac{51}{71}\)

              • \(\frac{5}{18}\)

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