Content Curator | Updated On - May 24, 2024
JEE Advanced 2009 Paper 2 Question paper with answer key pdf conducted on April 12, 2009 is available for download. The exam was successfully organized by Indian Institute of Technology, Guwahati. In terms of difficulty level, JEE Advanced was of Moderate level. The question paper comprised a total of 57 questions divided among Three sections.
JEE Advanced 2009 Paper 2 Question Paper with Answer Key PDF in English
JEE Advanced 2009 Paper 2 Question Paper | JEE Advanced 2009 Paper 2 Answer Key |
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Similar Engineering Exam Question Papers
JEE Advanced Questions
1. The option(s) with correct sequence of reagents for the conversion of P to Q is(are)
2. In the following reaction sequence, the major product Q is
\(\text{L-Glucose}\xrightarrow[ii)Cr_2,775K,10-20 \text{atm}]{i)HI, \Delta}P\xrightarrow[UV]{Cl_2(excess)}Q\)
\(\text{L-Glucose}\xrightarrow[ii)Cr_2,775K,10-20 \text{atm}]{i)HI, \Delta}P\xrightarrow[UV]{Cl_2(excess)}Q\)
3. A closed vessel contains 10 g of an ideal gas X at 300 K, which exerts 2 atm pressure. At the same temperature, 80 g of another ideal gas Y is added to it and the pressure becomes 6 atm. The ratio of root mean square velocities of X and Y at 300 K is
- 2√2:√3
- 2√2:1
- 1:2
- 2:1
4. Let f(x) be a continuously differentiable function on the interval (0, ∞) such that f(1) = 2 and
\(\lim\limits_{t→x}\frac{t^{10}f(x)-x^{10}f(t)}{t^9-x^9}=1\)
for each x > 0. Then, for all x > 0, f(x) is equal to
\(\lim\limits_{t→x}\frac{t^{10}f(x)-x^{10}f(t)}{t^9-x^9}=1\)
for each x > 0. Then, for all x > 0, f(x) is equal to
- \(\frac{31}{11x}-\frac{9}{11}x^{10}\)
- \(\frac{9}{11x}+\frac{13}{11}x^{10}\)
- \(\frac{-9}{11x}+\frac{31}{11}x^{10}\)
- \(\frac{13}{11x}+\frac{9}{11}x^{10}\)
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