1
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let ƒ : R $$ \to $$ R be such that for all x $$ \in $$ R
(21+x + 21–x), ƒ(x) and (3x + 3–x) are in A.P.,
then the minimum value of ƒ(x) is
A
2
B
0
C
3
D
4
2
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If a, b and c are the greatest value of 19Cp, 20Cq and 21Cr respectively, then :
A
$${a \over {11}} = {b \over {22}} = {c \over {21}}$$
B
$${a \over {10}} = {b \over {22}} = {c \over {21}}$$
C
$${a \over {10}} = {b \over {11}} = {c \over {42}}$$
D
$${a \over {11}} = {b \over {22}} = {c \over {42}}$$
3
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In finding the electric field using Gauss Law the formula $$\left| {\overrightarrow E } \right| = {{{q_{enc}}} \over {{\varepsilon _0}\left| A \right|}}$$ is applicable. In the formula $${{\varepsilon _0}}$$ is permittivity of free space, A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface. The equation can be used in which of the following situation?
A
Only when $$\left| {\overrightarrow E } \right|$$ = constant on the surface.
B
For any choice of Gaussian surface.
C
Only when the Gaussian surface is an equipotential surface.
D
Only when the Gaussian surface is an equipotential surface and $$\left| {\overrightarrow E } \right|$$ is constant on the surface.
4
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
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