Content Curator | Updated On - May 12, 2024
TS EAMCET 2024 Question Paper May 7 Shift 2 with Answer Key PDF is available here for download. JNTU, Hyderabad on behalf of TSCHE conducted TS EAMCET on May 7 from 3 PM to 6 PM. TS EAMCET 2024 Question Paper consists of 160 questions carrying 1 mark each. TS EAMCET 2024 Question Paper May 7 Shift 2 PDF for BiPC includes four subjects, Physics, Chemistry and Biology with Botany & Zoology. Each subject includes 40 questions.
TS EAMCET 2024 Question Paper with Answer Key May 7 Shift 2 PDF
| TS EAMCET 2024 Question Paper PDF | TS EAMCET 2024 Answer Key PDF |
|---|---|
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Also Check:
TS EAMCET Previous Year Question Papers
| TS EAMCET 2023 Question Paper | TS EAMCET 2022 Question Paper |
| TS EAMCET 2021 Question Paper | TS EAMCET 2020 Question Paper |
| TS EAMCET 2019 Question Paper | TS EAMCET 2018 Question Paper |
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TS EAMCET Questions
1. If \(\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1}+ \frac{Bx+C}{2x^2+3}\), then A - B + C=
If \(\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1}+ \frac{Bx+C}{2x^2+3}\), then A - B + C=
2
1
3
6
2. If i=√-1 then
\[Arg\left[ \frac{(1+i)^{2025}}{1+i^{2022}} \right] =\]
If i=√-1 then
\[Arg\left[ \frac{(1+i)^{2025}}{1+i^{2022}} \right] =\]\(\frac{-π}{4}\)
\(\frac{π}{4}\)
\(\frac{3π}{4}\)
\(\frac{-3π}{4}\)
3. 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
4. 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)
5. 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 =
\(\frac{π}{6}\)
\(\frac{π}{4}\)
\(\frac{π}{3}\)
\(\frac{π}{2}\)
6. If order and degree of the differential equation corresponding to the family of curves y2 = 4a(x+a)(a is parameter) are m and n respectively, then m+n2 =
If order and degree of the differential equation corresponding to the family of curves y2 = 4a(x+a)(a is parameter) are m and n respectively, then m+n2 =
3
4
5
2
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