1
JEE Main 2020 (Online) 3rd September Evening Slot
+4
-1
Concentric metallic hollow spheres of radii R and 4R hold charges Q1 and Q2 respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference V(R) – V(4R) is :
A
$${{3{Q_2}} \over {4\pi {\varepsilon _0}R}}$$
B
$${{3{Q_1}} \over {4\pi {\varepsilon _0}R}}$$
C
$${{3{Q_1}} \over {16\pi {\varepsilon _0}R}}$$
D
$${{{Q_2}} \over {4\pi {\varepsilon _0}R}}$$
2
JEE Main 2020 (Online) 3rd September Morning Slot
+4
-1
Two isolated conducting spheres S1 and S2 of radius $${2 \over 3}R$$ and $${1 \over 3}R$$ have 12 $$\mu$$C and –3 $$\mu$$C charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on S1 and S2 are respectively :
A
4.5 $$\mu$$C on both
B
+4.5 $$\mu$$C and –4.5 $$\mu$$C
C
6 $$\mu$$C and 3 $$\mu$$C
D
3 $$\mu$$C and 6 $$\mu$$C
3
JEE Main 2020 (Online) 2nd September Evening Slot
+4
-1
A charge Q is distributed over two concentric conducting thin spherical shells radii r and R (R > r). If the surface charge densities on the two shells are equal, the electric potential at the common centre is :
A
$${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + r} \right)} \over {\left( {{R^2} + {r^2}} \right)}}$$Q
B
$${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + r} \right)} \over {2\left( {{R^2} + {r^2}} \right)}}Q$$
C
$${1 \over {4\pi {\varepsilon _0}}}{{\left( {R + 2r} \right)Q} \over {2\left( {{R^2} + {r^2}} \right)}}$$
D
$${1 \over {4\pi {\varepsilon _0}}}{{\left( {2R + r} \right)} \over {\left( {{R^2} + {r^2}} \right)}}Q$$
4
JEE Main 2020 (Online) 2nd September Morning Slot
+4
-1
A charged particle (mass m and charge q)
moves along X-axis with velocity V0. When it
passes through the origin it enters a region having uniform electric field
$$\overrightarrow E = - E\widehat j$$ which extends upto x = d.
Equation of path of electron in the region x > d is
A
y = $${{qEd} \over {mV_0^2}}\left( {x - d} \right)$$
B
y = $${{qEd} \over {mV_0^2}}\left( {{d \over 2} - x} \right)$$
C
y = $${{qEd} \over {mV_0^2}}x$$
D
y = $${{qE{d^2}} \over {mV_0^2}}x$$
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