1
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A small spherical droplet of density d is floating exactly half immersed in a liquid of density $$\rho $$ and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet) :
A
$$r = \sqrt {{T \over {\left( {d - \rho } \right)g}}} $$
B
$$r = \sqrt {{{2T} \over {3\left( {d + \rho } \right)g}}} $$
C
$$r = \sqrt {{T \over {\left( {d + \rho } \right)g}}} $$
D
$$r = \sqrt {{{3T} \over {\left( {2d - \rho } \right)g}}} $$
2
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In LC circuit the inductance L = 40 mH and
capacitance C = 100 $$\mu $$F. If a voltage
V(t) = 10sin(314t) is applied to the circuit, the
current in the circuit is given as :
A
0.52 cos 314 t
B
5.2 cos 314 t
C
0.52 sin 314 t
D
10 cos 314 t
3
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A plane electromagnetic wave is propagating along the direction $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$ , with its polarization along the direction $$\widehat k$$ . The correct form of the magnetic field of the wave would be (here B0 is an appropriate constant) :
A
$${B_0}{{\widehat i - \widehat j} \over {\sqrt 2 }}\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
B
$${B_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
C
$${B_0}{{\widehat j - \widehat i} \over {\sqrt 2 }}\cos \left( {\omega t + k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
D
$${B_0}\widehat k\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
4
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A rod of length L has non-uniform linear mass
density given by $$\rho $$(x) = $$a + b{\left( {{x \over L}} \right)^2}$$ , where a
and b are constants and 0 $$ \le $$ x $$ \le $$ L. The value
of x for the centre of mass of the rod is at :
A
$${3 \over 2}\left( {{{a + b} \over {2a + b}}} \right)L$$
B
$${4 \over 3}\left( {{{a + b} \over {2a + 3b}}} \right)L$$
C
$${3 \over 4}\left( {{{2a + b} \over {3a + b}}} \right)L$$
D
$${3 \over 2}\left( {{{2a + b} \over {3a + b}}} \right)L$$
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