1
JEE Main 2018 (Online) 16th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A heating element has a resistance of 100 $$\Omega $$ at room temperature. When it is connected to a supply of 220 V, a steady current of 2 A passes in it and temperature is 500oC more than room temperature. what is the temperature coefficient of resistance of the heating element ?
A
0.5 $$ \times $$ 10$$-$$4 oC$$-$$1
B
5 $$ \times $$ 10$$-$$4 oC$$-$$1
C
1 $$ \times $$ 10$$-$$4 oC$$-$$1
D
2 $$ \times $$ 10$$-$$4 oC$$-$$1
2
JEE Main 2018 (Online) 16th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Both the nucleus and the atom of some element arein their respective first excited states. They get de-excted by emitting photons of wavelengths $$\lambda $$N, $$\lambda $$A respectively. The ratio $${{{}^\lambda N} \over {{}^\lambda A}}$$is closest to :
A
10$$-$$6
B
10
C
10$$-$$10
D
10$$-$$1
3
JEE Main 2018 (Online) 16th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The de-Broglie wavelength ($$\lambda $$B) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state ($$\lambda $$G) by :
A
$$\lambda $$B = 2$$\lambda $$G
B
$$\lambda $$B = 3$$\lambda $$G
C
$$\lambda $$B = $$\lambda $$G/2
D
$$\lambda $$B = $$\lambda $$G/3
4
JEE Main 2018 (Online) 16th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Unpolarized light of intensity I is incident on a system of two polarizers, A followed by B. The intensity of emergent light is I/2. If a third polarizer C is placed between A and B, the intensity of emergent light is reduced to I/3. The angle between the polarizers A and C is $$\theta $$. Then :
A
cos$$\theta $$ = $${\left( {{2 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
B
cos$$\theta $$ = $${\left( {{2 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 4$}}}}$$
C
cos$$\theta $$ = $${\left( {{1 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
D
cos$$\theta $$ = $${\left( {{1 \over 3}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 4$}}}}$$
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