1
JEE Main 2020 (Online) 6th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
For a plane electromagnetic wave, the magnetic field at a point x and time t is

$$\overrightarrow B \left( {x,t} \right)$$ = $$\left[ {1.2 \times {{10}^{ - 7}}\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]$$ T

The instantaneous electric field $$\overrightarrow E $$ corresponding to $$\overrightarrow B $$ is :
(speed of light c = 3 × 108 ms–1)
A
$$\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {1 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat i} \right]$$ $${V \over m}$$
B
$$\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]{V \over m}$$
C
$$\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {1 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}$$
D
$$\overrightarrow E \left( {x,t} \right) = \left[ { - 36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}$$
2
JEE Main 2020 (Online) 6th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Given the masses of various atomic particles
mp = 1.0072 u, mn = 1.0087 u, me = 0.000548 u,
$${m_{\overline v }}$$ = 0, md = 2.0141 u, where p $$ \equiv $$ proton, n $$ \equiv $$ neutron,
e $$ \equiv $$ electron, $$\overline v $$ $$ \equiv $$ antineutrino and d $$ \equiv $$ deuteron. Which of the following process is allowed by momentum and energy conservation?
A
n + n $$ \to $$ deuterium atom
(electron bound to the nucleus)
B
n + p $$ \to $$ d + $$\gamma $$
C
p $$ \to $$ n + e+ + $$\overline v $$
D
e+ + e- $$ \to $$ $$\gamma $$
3
JEE Main 2020 (Online) 6th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In the figure shown, the current in the 10 V battery is close to : JEE Main 2020 (Online) 6th September Evening Slot Physics - Current Electricity Question 197 English
A
0.36 A from negative to positive terminal
B
0.42 A from positive to negative terminal
C
0.71 A from positive to negative terminal
D
0.21 A from positive to negative terminal
4
JEE Main 2020 (Online) 6th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings 5.50 mm, 5.55 mm, 5.45 mm, 5.65 mm. The average of these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The average diameter
of the pencil should therefore be recorded as :
A
(5.54 $$ \pm $$ 0.07) mm
B
(5.5375 $$ \pm $$ 0.0740) mm
C
(5.5375 $$ \pm $$ 0.0739) mm
D
(5.538 $$ \pm $$ 0.074) mm
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