1
JEE Main 2021 (Online) 22th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
An electron of mass me and a proton of mass mp are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is
A
$${{{m_e}} \over {{m_p}}}$$
B
1
C
$${{{m_p}} \over {{m_e}}}$$
D
$$\sqrt {{{{m_p}} \over {{m_e}}}} $$
2
JEE Main 2021 (Online) 22th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
What will be the average value of energy for a monoatomic gas in thermal equilibrium at temperature T?
A
$${3 \over 2}{k_B}T$$
B
$${k_B}T$$
C
$${2 \over 3}{k_B}T$$
D
$${1 \over 2}{k_B}T$$
3
JEE Main 2021 (Online) 22th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
T0 is the time period of a simple pendulum at a place. if the length of the pendulum is reduced to $${1 \over {16}}$$ times of its initial value, the modified time period is :
A
4 T0
B
$${1 \over {4}}$$ T0
C
T0
D
8$$\pi$$ T0
4
JEE Main 2021 (Online) 22th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
The motion of a mass on a spring, with spring constant K is as shown in figure.

JEE Main 2021 (Online) 22th July Evening Shift Physics - Waves Question 64 English
The equation of motion is given by
x(t) = A sin$$\omega$$t + B cos$$\omega$$t with $$\omega$$ = $$\sqrt {{K \over m}} $$

Suppose that at time t = 0, the position of mass is x(0) and velocity v(0), then its displacement can also be represented as x(t) = C cos($$\omega$$t $$-$$ $$\phi$$), where C and $$\phi$$ are :
A
$$C = \sqrt {{{2v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{x(0)\omega } \over {2v(0)}}} \right)$$
B
$$C = \sqrt {{{v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{x(0)\omega } \over {v(0)}}} \right)$$
C
$$C = \sqrt {{{v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{v(0)} \over {x(0)\omega }}} \right)$$
D
$$C = \sqrt {{{2v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{v(0)} \over {x(0)\omega }}} \right)$$
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