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TS EAMCET 2021 Engineering Question paper with answer key pdf conducted on August 4 in Afternoon Session 3 PM to 6 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 3 sections.
TS EAMCET 2021 Engineering Question Paper with Answer Key PDFs Afternoon Session
| TS EAMCET 2021 Engineering Question Paper PDF | TS EAMCET 2021 Engineering Answer Key PDF |
|---|---|
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TS EAMCET Questions
1. The orthocenter of the triangle whose sides are given by x + y + 10 = 0, x - y - 2 = 0 and 2x + y - 7 = 0 is
The orthocenter of the triangle whose sides are given by x + y + 10 = 0, x - y - 2 = 0 and 2x + y - 7 = 0 is
(-4, -3)
(-4, -6)
(4,6)
(3,6)
2. The number of electrons with (n+1) values equal to 3,4 and 5 in an element with atomic number (z) 24 are respectively (n = principal quantum number and l = azimuthal quantum number)
The number of electrons with (n+1) values equal to 3,4 and 5 in an element with atomic number (z) 24 are respectively (n = principal quantum number and l = azimuthal quantum number)
7,8,5
6,8,6
8,7,5
8,8,5
3. If i=√-1 then
\[Arg\left[ \frac{(1+i)^{2025}}{1+i^{2022}} \right] =\]
If i=√-1 then
\[Arg\left[ \frac{(1+i)^{2025}}{1+i^{2022}} \right] =\]\(\frac{-π}{4}\)
\(\frac{π}{4}\)
\(\frac{3π}{4}\)
\(\frac{-3π}{4}\)
4. 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:
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
5. lim n→∞ \(\frac{1}{n^3} \)
\[\sum_{k=1}^{n} k^{2} = \]
lim n→∞ \(\frac{1}{n^3} \)
\[\sum_{k=1}^{n} k^{2} = \]
x
\(\frac{x}{2}\)
\(\frac{x}{3}\)
\(\frac{x}{4}\)
6. 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
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