1
JEE Advanced 2015 Paper 1 Offline
MCQ (More than One Correct Answer)
+4
-2

Let $$f(x) = \sin \left( {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right)$$ for all $$x \in R$$ and g(x) = $${{\pi \over 2}\sin x}$$ for all x$$\in$$R. Let $$(f \circ g)(x)$$ denote f(g(x)) and $$(g \circ f)(x)$$ denote g(f(x)). Then which of the following is/are true?

A
Range of f is $$\left[ { - {1 \over 2},{1 \over 2}} \right]$$.
B
Range of f $$\circ$$ g is $$\left[ { - {1 \over 2},{1 \over 2}} \right]$$.
C
$$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {g(x)}} = {\pi \over 6}$$.
D
There is an x$$\in$$R such that (g $$\circ$$ f)(x) = 1.
2
JEE Advanced 2015 Paper 1 Offline
MCQ (More than One Correct Answer)
+4
-2
Planck's constant h, speed of light c and gravitational constant G are used to form a unit of length L and a unit of mass M. Then the correct option(s) is(are)
A
$$M \propto \sqrt c$$
B
$$M \propto \sqrt G$$
C
$$L \propto \sqrt h$$
D
$$L \propto \sqrt G$$
3
JEE Advanced 2015 Paper 1 Offline
MCQ (More than One Correct Answer)
+4
-2
Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
A
If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.
B
If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.
C
If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.
D
If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.
4
JEE Advanced 2015 Paper 1 Offline
Numerical
+4
-0
A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet’s gravity is $${\left( {{1 \over 4}} \right)^{th}}$$ of its value at the surface of the planet. If the escape velocity from the planet is $${v_{esc}} = v\sqrt N$$, then the value of N is (ignore energy loss due to atmosphere)