1
JEE Advanced 2015 Paper 2 Offline
Numerical
+4
-0
If $$\alpha = \int\limits_0^1 {\left( {{e^{9x + 3{{\tan }^{ - 1}}x}}} \right)\left( {{{12 + 9{x^2}} \over {1 + {x^2}}}} \right)} dx$$ where $${\tan ^{ - 1}}x$$ takes only principal values, then the value of $$\left( {{{\log }_e}\left| {1 + \alpha } \right| - {{3\pi } \over 4}} \right)$$ is
Your input ____
2
JEE Advanced 2015 Paper 2 Offline
Numerical
+3
-1
Let m and n be two positive integers greater than 1. If $$$\mathop {\lim }\limits_{\alpha \to 0} \left( {{{{e^{\cos \left( {{\alpha ^n}} \right)}} - e} \over {{\alpha ^m}}}} \right) = - \left( {{e \over 2}} \right)$$$ then the value of $${m \over n}$$ is _________.
Your input ____
3
JEE Advanced 2015 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-2
Consider a uniform spherical charge distribution of radius $${R_1}$$ centred at the origin $$O.$$ In this distribution, a spherical cavity of radius $${R_2},$$ centred at $$P$$ with distance $$OP=a$$ $$ = {R_1} - {R_2}$$ (see figure) is made. If the electric field inside the cavity at position $$\overrightarrow r $$ is $$\overrightarrow E \overrightarrow {\left( r \right)} ,$$ then the correct statement(s) is (are)

JEE Advanced 2015 Paper 2 Offline Physics - Electrostatics Question 47 English
A
$$\overrightarrow E $$ is uniform, its magnitude is independent of $${R_2}$$ but its direction depends on $$\overrightarrow r .$$
B
$$\overrightarrow E $$ is uniform, its magnitude depends on $${R_2}$$ and its direction depends on $$\overrightarrow r .$$
C
$$\overrightarrow E $$ is uniform, its magnitude is independent of a but its direction depends on $$\overrightarrow a $$
D
$$\overrightarrow E $$ is uniform and both its magnitude and direction depend on $$\overrightarrow a $$
4
JEE Advanced 2015 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-2
A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If P(r) is the pressure at r (r < R), then the correct option(s) is(are)
A
$$P(r = 0) = 0$$
B
$${{P(r = 3R/4)} \over {P(r = 2R/3)}} = {{63} \over {80}}$$
C
$${{P(r = 3R/5)} \over {P(r = 2R/5)}} = {{16} \over {21}}$$
D
$${{P(r = R/2)} \over {P(r = R/3)}} = {{20} \over {27}}$$
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