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JEE ADVANCED 2017 PAPER 2 OFFLINE
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54
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180 mins
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IIT-JEE Advance - Previous Year Papers
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2017 · Physics · Mechanics · Units And Measurements
JEE ADVANCED 2017 PAPER 2 OFFLINE
A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is \(\delta T = 0.01\) seconds and he measures the depth of the well to be \(L=20\) meters. Take the acceleration due to gravity \(g = 10m{s^{ - 2}}\) and the velocity of sound is \(300\) \(m{s^{ - 1}}\). Then the fractional error in the measurement, \(\delta L/L,\) is closest to
2017 · Chemistry · Physical Chemistry · Thermodynamics
JEE ADVANCED 2017 PAPER 2 OFFLINE
For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant \(K\) in terms of change in entropy is described by
2017 · Mathematics · Calculus · Functions
JEE ADVANCED 2017 PAPER 2 OFFLINE
Let S = {1, 2, 3, .........., 9}. For k = 1, 2, .........., 5, let Nk be the number of subsets of S, each containing five elements out of which exactly k are odd. Then N1 + N2 + N3 + N4 + N5 =
2017 · Physics · Mechanics · Motion
JEE ADVANCED 2017 PAPER 2 OFFLINE
Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density \(\rho\) remains uniform throughout the volume. The rate of fractional change in density \(\left( {{1 \over \rho } {{d\rho } \over {dt}}} \right)\) is constant. The velocity \(v\) of any point on the surface of the expanding sphere is proportional to
2017 · Chemistry · Physical Chemistry · Thermodynamics
JEE ADVANCED 2017 PAPER 2 OFFLINE
The standard state Gibbs free energies of formation of \(C\)(graphite) and \(C\)(diamond) at \(T=298\) \(K\) are
\({\Delta _f}{G^0}\) [\(C\)(graphite)] \(= 0kJmo{l^{ - 1}}\)
\({\Delta _f}{G^0}\) [\(C\)(diamond)] \(= 2.9kJmo{l^{ - 1}}\)
The standard state means that the pressure should be \(1\) bar, and substance should be pure at a given temperature. The conversion of graphite [\(C\)(graphite)] to diamond [\(C\)(diamond)] reduces its volume by \(2 \times {10^{ - 6}}\,{m^3}\,mo{l^{ - 1}}\) If \(C\)(graphite) is converted to \(C\)(diamond) isothermally at \(T=298\) \(K,\) the pressure at which \(C\)(graphite) is in equilibrium with \(C\)(diamond), is
[Useful information : \(1\) \(J=1\) \(kg\,{m^2}{s^{ - 2}};1\,Pa = 1\,kg\,{m^{ - 1}}{s^{ - 2}};\) \(1\) bar \(= {10^5}\) \(Pa\)]
\({\Delta _f}{G^0}\) [\(C\)(graphite)] \(= 0kJmo{l^{ - 1}}\)
\({\Delta _f}{G^0}\) [\(C\)(diamond)] \(= 2.9kJmo{l^{ - 1}}\)
The standard state means that the pressure should be \(1\) bar, and substance should be pure at a given temperature. The conversion of graphite [\(C\)(graphite)] to diamond [\(C\)(diamond)] reduces its volume by \(2 \times {10^{ - 6}}\,{m^3}\,mo{l^{ - 1}}\) If \(C\)(graphite) is converted to \(C\)(diamond) isothermally at \(T=298\) \(K,\) the pressure at which \(C\)(graphite) is in equilibrium with \(C\)(diamond), is
[Useful information : \(1\) \(J=1\) \(kg\,{m^2}{s^{ - 2}};1\,Pa = 1\,kg\,{m^{ - 1}}{s^{ - 2}};\) \(1\) bar \(= {10^5}\) \(Pa\)]
2017 · Mathematics · Trigonometry · Trigonometric Functions And Equations
JEE ADVANCED 2017 PAPER 2 OFFLINE
Let \(\alpha\) and \(\beta\) be non zero real numbers such that \(2(\cos \beta - \cos \alpha ) + \cos \alpha \cos \beta = 1\). Then which of the following is/are true?