1
JEE Advanced 2013 Paper 2 Offline
+3
-1

A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers' usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with power factor unity. All the currents and voltages mentioned are rms values.

In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1 : 10. If the power of the consumers has to be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is

A
200 : 1
B
150 : 1
C
100 : 1
D
50 : 1
2
JEE Advanced 2013 Paper 2 Offline
+3
-1

A point charge Q is moving in a circular orbit of radius R in the xy-plane with an angular velocity $$\omega$$. This can be considered as equivalent to a loop carrying a steady current $${{Q\omega } \over {2\pi }}$$. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant $$\gamma$$.

The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is
A
$${{BR} \over 4}$$
B
$${{BR} \over 2}$$
C
BR
D
2BR
3
JEE Advanced 2013 Paper 2 Offline
+3
-1

A point charge Q is moving in a circular orbit of radius R in the xy-plane with an angular velocity $$\omega$$. This can be considered as equivalent to a loop carrying a steady current $${{Q\omega } \over {2\pi }}$$. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant $$\gamma$$.

The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is

A
$$- \gamma BQ{R^2}$$
B
$$- \gamma {{BQ{R^2}} \over 2}$$
C
$$\gamma {{BQ{R^2}} \over 2}$$
D
$$\gamma BQ{R^2}$$
4
JEE Advanced 2013 Paper 2 Offline
+3
-1

The mass of a nucleus $$_Z^AX$$ is less than the sum of the masses of (A-Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m1 and m2 only if (m1 + m2) < M. Also two light nuclei of masses m3 and m4 can undergo complete fusion and form a heavy nucleus of mass M' only if (m3 + m4) > M'. The masses of some neutral atoms are given in the table below :

$$_1^1H$$ 1.007825 u $$_1^2H$$ 2.014102 u
$$_3^6Li$$ 6.015123 u $$_3^7Li$$ 7.016004 u
$$_{64}^{152}Gd$$ 151.919803 u $$_{82}^{206}Pb$$ 205.974455 u
$$_1^3H$$ 3.016050 u $$_2^4He$$ 4.002603 u
$$_{30}^{70}Zn$$ 69.925325 u $$_{34}^{82}Se$$ 81.916709 u
$$_{83}^{209}Bi$$ 208.980388 u $$_{84}^{210}Po$$ 209.982876 u

(1 u = 932 MeV/c2)

The correct statement is

A
the nucleus $$_3^6Li$$ can emit an alpha particle.
B
the nucleus $$_{84}^{210}Po$$ can emit a proton.
C
deuteron and alpha particle can undergo complete fusion.
D
the nuclei $$_{30}^{70}Zn$$ and $$_{34}^{82}Se$$ can undergo complete fusion.
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