JEE Advanced 2013 Paper 2 Offline | JEE Advanced Year Wise Previous Years Questions

JEE Advanced 2013 Paper 2 Offline

MCQ (More than One Correct Answer)

-2

The function $$f(x) = 2\left| x \right| + \left| {x + 2} \right| - \left| {\left| {x + 2} \right| - 2\left| x \right|} \right|$$ has a local minimum or a local maximum at x =

$$-$$2

$${{ - 2} \over 3}$$

$${{ 2} \over 3}$$

JEE Advanced 2013 Paper 2 Offline

MCQ (More than One Correct Answer)

-1

Two non-conducting spheres of radii $${R_1}$$ and $${R_2}$$ and carrying uniform volume charge densities $$ + \rho $$ and $$ - \rho ,$$ respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region

JEE Advanced 2013 Paper 2 Offline Physics - Electrostatics Question 45 English

JEE Advanced 2013 Paper 2 Offline Physics - Electrostatics Question 45 English

The electrostatic field is zero

The electrostatic potential is constant

The electrostatic field is constant in magnitude

The electrostatic field has same direction

JEE Advanced 2013 Paper 2 Offline

MCQ (More than One Correct Answer)

-0.75

Two bodies, each of mass M, are kept fixed with a separation $$2L$$. A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The correct statement(s) is (are)

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is $$4\sqrt {{{GM} \over L}} $$

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is $$2\sqrt {{{GM} \over L}} $$

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is $$\sqrt {{{2GM} \over L}} $$

The energy of the mass m remains constant.

JEE Advanced 2013 Paper 2 Offline

MCQ (More than One Correct Answer)

-0.75

A particle of mass m is attached to one end of a mass-less spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t = 0 with an initial velocity u₀. When the speed of the particle is 0.5 u₀. It collides elastically with a rigid wall. After this collision,

the speed of the particle when it returns to its equilibrium position is u₀.

the time at which the particle passes through the equilibrium position for the first time is $$t = \pi \sqrt {{m \over k}} $$

the time at which the maximum compression of the spring occurs is $$t = {{4\pi } \over 3}\sqrt {{m \over k}} $$

the time at which the particle passes through the equilibrium position for the second time is $$t = {{5\pi } \over 3}\sqrt {{m \over k}} $$

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