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LAKSHYA JEE AIR (2025)
Electrostatics DPP-01
1. Three charges each of +1 μC are placed at the 5. Two small balls, each having equal positive charge
corners of an equilateral triangle. If the force Q are suspended by two insulating strings of equal
between any two charges be F, then the net force on length L from a hook fixed to a stand. If the whole
either charge will be
set-up is transferred to a satellite in orbit around the
(A) 2F (B) F 3 earth, the tension in equilibrium in each string is
(C) 2F (D) 3F equal to:
2. Charge 𝑞2 of mass 𝑚 revolves around a stationary
charge 𝑞1 in a circular orbit of radius 𝑟. The orbital
periodic time of 𝑞2 would be L L
1/ 2
4 2 mr 3
(A) qm mq
kq1q2
kQ
kq q
1/ 2 (A) zero (B)
(B) 21 2 2 L2
4 mr
kQ 2 kQ 2
1 (C) (D)
4 2 mr 4 2 2L2 4L2
(C)
kq1q 2
1 6. In two cases, two identical conducting spheres are
4 2 mr 2 2
(D) given equal charges, in one case of the same type
kq1q 2
whereas in another case of opposite type. The
distance between the spheres is not large comparing
3. The distance between two-point charges is
with the diameter. Let F1 and F2 be the magnitude
increased by 10%. The force of interaction between
of the force of interaction between the spheres, as
them
(A) Increased by 10% (B) decreased by 10% shown, then
(C) decreased by 17% (D) decreased by 21%
4. Two nucleons are at a separation of one Fermi.
Protons have a charge of +1.6 × 10–19 C. The net
nuclear force between them is F1, if both are
(A) F1 > F2
neutrons, F2 if both are protons and F3 if one is
(B) F1 = F2
proton and the other is neutron. Then:
(C) F1 < F2
(A) F1 = F2 > F3
(B) F1 = F2 = F3 (D) information is not sufficient to draw the
(C) F1 < F2 < F3 conclusion
(D) F1 > F2 > F3
LAKSHYA JEE AIR (2025)
Electrostatics DPP-01
1. Three charges each of +1 μC are placed at the 5. Two small balls, each having equal positive charge
corners of an equilateral triangle. If the force Q are suspended by two insulating strings of equal
between any two charges be F, then the net force on length L from a hook fixed to a stand. If the whole
either charge will be
set-up is transferred to a satellite in orbit around the
(A) 2F (B) F 3 earth, the tension in equilibrium in each string is
(C) 2F (D) 3F equal to:
2. Charge 𝑞2 of mass 𝑚 revolves around a stationary
charge 𝑞1 in a circular orbit of radius 𝑟. The orbital
periodic time of 𝑞2 would be L L
1/ 2
4 2 mr 3
(A) qm mq
kq1q2
kQ
kq q
1/ 2 (A) zero (B)
(B) 21 2 2 L2
4 mr
kQ 2 kQ 2
1 (C) (D)
4 2 mr 4 2 2L2 4L2
(C)
kq1q 2
1 6. In two cases, two identical conducting spheres are
4 2 mr 2 2
(D) given equal charges, in one case of the same type
kq1q 2
whereas in another case of opposite type. The
distance between the spheres is not large comparing
3. The distance between two-point charges is
with the diameter. Let F1 and F2 be the magnitude
increased by 10%. The force of interaction between
of the force of interaction between the spheres, as
them
(A) Increased by 10% (B) decreased by 10% shown, then
(C) decreased by 17% (D) decreased by 21%
4. Two nucleons are at a separation of one Fermi.
Protons have a charge of +1.6 × 10–19 C. The net
nuclear force between them is F1, if both are
(A) F1 > F2
neutrons, F2 if both are protons and F3 if one is
(B) F1 = F2
proton and the other is neutron. Then:
(C) F1 < F2
(A) F1 = F2 > F3
(B) F1 = F2 = F3 (D) information is not sufficient to draw the
(C) F1 < F2 < F3 conclusion
(D) F1 > F2 > F3
