Content Curator
TS EAMCET 2020 Engineering Question paper with answer key pdf conducted on September 14 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 2020 Engineering Question Paper with Answer Key PDFs Afternoon Session
| TS EAMCET 2020 Engineering Question Paper PDF | TS EAMCET 2020 Engineering Answer Key PDF |
|---|---|
| Download PDF | Download PDF |
TS EAMCET Previous Year Question Papers
Similar Exam Question Papers:
TS EAMCET Questions
1. If i=√-1 then
\[Arg\left[ \frac{(1+i)^{2025}}{1+i^{2022}} \right] =\]
If i=√-1 then
\[Arg\left[ \frac{(1+i)^{2025}}{1+i^{2022}} \right] =\]\(\frac{-π}{4}\)
\(\frac{π}{4}\)
\(\frac{3π}{4}\)
\(\frac{-3π}{4}\)
2. The number of significant figures in the measurement of a length 0.079000 m is:
The number of significant figures in the measurement of a length 0.079000 m is:
7
2
5
4
3. 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
4. If the line x cos α + y sin α = 2√3 is tangent to the ellipse \(\frac{x^2}{16} + \frac{y^2}{8} = 1\) and α is an acute angle then α =
If the line x cos α + y sin α = 2√3 is tangent to the ellipse \(\frac{x^2}{16} + \frac{y^2}{8} = 1\) and α is an acute angle then α =
\(\frac{π}{6}\)
\(\frac{π}{4}\)
\(\frac{π}{3}\)
\(\frac{π}{2}\)
5. 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
6. The locus of z such that \(\frac{|z-i|}{|z+i|}\)= 2, where z = x+iy. is
The locus of z such that \(\frac{|z-i|}{|z+i|}\)= 2, where z = x+iy. is
3x2 + 3y2 +10y + 3
3x2 - 3y2 - 10y - 3 = 0
3x2 + 3y2 + 10y + 3 = 0
x2 + y2 - 5y + 3 = 0
.png?h=160&w=320&mode=stretch)
.jpeg?h=160&w=320&mode=stretch)
.png?h=160&w=320&mode=stretch)
.png?h=160&w=320&mode=stretch)
Comments