1
IIT-JEE 2012 Paper 2 Offline
+3
-1

The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed $$\omega$$, the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as it seen from the changed orientation of points P and Q). Both these motions have the same angular speed $$\omega$$ in this case. Now consider two similar systems as shown in the figure.

Case (a) : The disc with its face vertical and parallel to x-z axis;

Case (b) : The disc with its face making an angle of 45$$^\circ$$ with xy-plane and its horizontal diameter parallel to x-axis.

In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed $$\omega$$ about the z-axis. Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct?

A
It is $$\sqrt2$$$$\omega$$ for both cases.
B
It is $$\omega$$ for case (a); and $$\omega$$ / $$\sqrt2$$ for case (b).
C
It is $$\omega$$ for case (a); and $$\sqrt2$$$$\omega$$ for case (b).
D
It is $$\omega$$ for both cases.
2
IIT-JEE 2012 Paper 2 Offline
+3
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The $$\beta$$-decay process, discovered in around 1900, is basically the decay of a neutron (n). In the laboratory, a proton (p) and an electron (e$$-$$) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has continuous spectrum. Considering a three-body decay process, that is, n $$\to$$ p + e$$-$$ + $${\overline v _e}$$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ($${\overline v _e}$$) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 $$\times$$ 106 eV. The kinetic energy carried by the proton is only the recoil energy.

What is the maximum energy of the anti-neutrino?

A
Zero.
B
Much less than 0.8 $$\times$$ 106 eV.
C
Nearly 0.8 $$\times$$ 106 eV.
D
much larger than 0.8 $$\times$$ 106 eV.
3
IIT-JEE 2012 Paper 2 Offline
+3
-1

The $$\beta$$-decay process, discovered in around 1900, is basically the decay of a neutron (n). In the laboratory, a proton (p) and an electron (e$$-$$) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has continuous spectrum. Considering a three-body decay process, that is, n $$\to$$ p + e$$-$$ + $${\overline v _e}$$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ($${\overline v _e}$$) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 $$\times$$ 106 eV. The kinetic energy carried by the proton is only the recoil energy.

If the anti-neutrino had a mass of 3 eV/c2 (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron?

A
0 $$\le$$ K $$\le$$ 0.8 $$\times$$ 106 eV
B
3.0 eV $$\le$$ K $$\le$$ 0.8 $$\times$$ 106 eV
C
3.0 eV $$\le$$ K < 0.8 $$\times$$ 106 eV
D
0 $$\le$$ K < 0.8 $$\times$$ 106 eV
4
IIT-JEE 2012 Paper 2 Offline
+3
-1

Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring material, then by Snell's law, $${{\sin {\theta _1}} \over {\sin {\theta _2}}} = {{{n_2}} \over {{n_1}}}$$, 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 the medium is given by the relation $$n = \left( {{c \over v}} \right) = \pm \,\sqrt {{\varepsilon _r}{\mu _r}}$$, where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, $$\varepsilon$$r and $$\mu$$r, are the relative permittivity and permeability of the medium, respectively.

In normal materials, both $$\varepsilon$$r and $$\mu$$r , are positive, implying positive n for the medium. When both $$\varepsilon$$r and $$\mu$$r are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behaviour, without violating and physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.

For light incident from air on a meta-material, the appropriate ray diagram is

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