1
JEE Main 2020 (Online) 2nd September Morning Slot
+4
-1
In a reactor, 2 kg of 92U235 fuel is fully used up in 30 days. The energy released per fission is 200 MeV. Given that the Avogadro number, N = 6.023 $$\times$$ 1026 per kilo mole and 1 eV = 1.6 × 10–19 J. The power output of the reactor is close to
A
125 MW
B
60 MW
C
54 MW
D
35 MW
2
JEE Main 2020 (Online) 8th January Evening Slot
+4
-1
An electron (mass m) with initial velocity $$\overrightarrow v = {v_0}\widehat i + {v_0}\widehat j$$ is in an electric field $$\overrightarrow E = - {E_0}\widehat k$$. If $$\lambda _0$$ is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by :
A
$${{{\lambda _0} } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}$$
B
$${{{\lambda _0}\sqrt 2 } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}$$
C
$${{{\lambda _0} } \over {\sqrt {1 + {{{e^2}{E^2}{t^2}} \over {2{m^2}v_0^2}}} }}$$
D
$${{{\lambda _0}} \over {\sqrt {2 + {{{e^2}{E^2}{t^2}} \over {{m^2}v_0^2}}} }}$$
3
JEE Main 2020 (Online) 8th January Morning Slot
+4
-1
When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV end de-Broglie wavelength $$\lambda _A$$. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA – 1.5) eV. If the de-Broglie wavelength of these photoelectrons $$\lambda _B$$ = 2$$\lambda _A$$, then the work function of metal B is :
A
1.5eV
B
4eV
C
2eV
D
3eV
4
JEE Main 2020 (Online) 8th January Morning Slot
+4
-1
The graph which depicts the results of Rutherford gold foil experiment with $$\alpha$$-particales is :
$$\theta$$ : Scattering angle
Y : Number of scattered $$\alpha$$-particles detected (Plots are schematic and not to scale)
A
B
C
D
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