JEE (Advanced) 2012 Paper-2
Most materials have therefractive index, n > 1. So, when a light ray from air enters a naturally occurring material, then by Snells' law, $$\frac{\sin \theta_1}{\sin \theta_2} = \frac{n_2}{n_1}$$, it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray.According to electromagnetism, the
refractive index of themedium is given by the relation, $$n = \left(\frac{c}{v}\right) = \pm \sqrt{ε_r\mu_r}$$ where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, $$ε_r$$ and $$\mu_r$$ are negative, one one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are calledmeta-materials.
They exhibit significantly different optical behavior, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normalmaterials, the frequency of light remains unchanged upon refraction even inmeta-materials.
The general motion of a rigid body can be considered to be a combination of (i) a motion --- centre of mass about an axis, and (ii) its motion about an instantanneous axis passing through center of mass. These axes need not be stationary. Consider, for example, a thin uniform welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. Where disc-stick system is rotated about the origin ona horizontal frictionless plane with angular sp--- $$\omega$$, the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass the disc about the z-axis, and (ii) a rotation of the disc through an instantaneous vertical axis pass through its centre of mass (as is seen from the changed orientation of points P and Q). Both the motions have the same angular speed $$\omega$$ in the case.
Now consider two similar systems as shown in the figure: case (a) the disc with its face ver--- and parallel to x-z plane; Case (b) the disc with its face making an angle of $$45^\circ$$ with x-y plane its horizontal diameter parallel to x-axis. In both the cases, the disc is weleded at point P, and systems are rotated with constant angular speed $$\omega$$ about the z-axis.
Which of the following statement regarding the angular speed about the istantaneous axis (passing through the centre of mass) is correct?
Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct?
For the following questions answer them individually
Two solid cylinders P andQ of same mass and same radius start rolling down a fixed inclined plane form the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement (s) is (are) correct?
A current carrying infinitely long wire is kept along the diameter of a circular wire loop,without touching it. The correct statement (s) is (are) :
In the given circuit, the AC source has $$ \omega = 100$$ rad/s. considering the inductor and capacitor to be ideal, the correct choice (s) is(are)
Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shown in the figure. Given that $$K = \frac{1}{4 \pi ε_0}\frac{q}{L^2}$$, which of the following statement (s) is (are) correct?
Two spherical planets P andQ have the same unfirom density $$\rho$$, masses $$M_P$$ and $$M_Q$$, an surface areas A and 4A, respectively. A spherical planet R also has unfirom density $$\rho$$ and its mass is $$(M_P + M_Q)$$. The escape velocities from the planets P, Q and R, are $$V_P, V_Q$$ and $$V$$ respectively. Then
The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed $$\omega$$ and (ii) an inner disc of radius 2R rotating anti-clockwisewith angular speed $$\frac{\omega}{2}$$. The ring and disc are separated b frictionaless ball bearings. The system is in the x-z plane. The
point P on the inner disc is at distance R from the origin, where OP makes an angle of $$30^\circ$$ with the horizontal. Then with respect to the horizontal surface,
