# IIT JEE Physics Test Paper 2

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## IIT JEE Physics Test Paper 2

IIT JEE Physics Test Paper 2

Part – I (Physics)
Section-(A) One option correct
Each questions has 4 choices (A), (B), (C) and (D) out of which only one is correct. 4 marks each question and 1 marks negative.

1. The electric field inside a sphere which carries a volume charge density proportional to the distance from the origin rho = alpha * r( alpha is a constant) is:
(A)alpha r^3/ 4E0 (B)alpha r^2/ 4E0 (C)alpha r^2/3E0 (D)None of these

2.A particle of charge -q & mass m moves in a circle of radiur r around an infinitely long line charge of linear charge density +lamda. Then time period of revolution of charge will be:  where k=1/4E0

3. Figure shows three circular arcs, each of radius R and total charge as indicated. The net electric potential at the centre of curvature is: (A)Q/2piE0R   (B)Q/4piE0R   (C) 2Q/piE0R     (D)   Q/piE0R

4. A point charge Q is placed at a distance d from the centre of an uncharged conducting sphere of radius R. The potential of the sphere (d > R): 5. Two point dipoles of dipole moments  pk and p/2 k  are located at (0, 0, 0) & (1 m, 0, 2m) respectively. The resultant electric field due to the two dipoles at the point (1m, 0, 0) is: 06.. Electrical potential ‘v’ in space as a function of co-ordinates is given by, v=1/x + 1/y + 1/z Then the electric field intensity at (1, 1, 1) is given by: 7. In the figure shown, initially the spring of negligible mass in is underformed state and the block has zero velocity E is a uniform electri field, then: (K = spring constant) (i) The maximum speed of the block will be QE/sqrt(mK)
(ii) The maximum speed of the block will be 2QE/sqrt(mK)
(iii) The maximum compression of the spring will be QE/K
(iv) The maximum compression of the spring will be 2QE/K

(A) only (i) and (iii) are correct (B) only (i) and (iv) are correct
(C) only (ii) and (iii) are correct (D) only (ii) and (iv) are correct

8.In the figure initial status of capacitance and their connection is shown. Which of the following is incorrect about this circuit: (A) Final charge on each capacitor will be zero

(B) Final total electrical energy of the capacitors will be zero

(C) Total charge flown from A to D is 30µC

(D) Total charge flown from A to D is –30µC

9.A parallel plate capacitor of capacitance C (without dielectrics) is filled by dielectric slabs as shown in the figure. Then the new capacitance of the capacitor is: (A) 3.9C (B) 4C (C) 2.4C (D) 3C

10.In the figure a capacitor of capacitance 2µF is connected to a cell of emf 20 volt. the plates of the capacitor are drawn apart slowly to double the dis-tance between them. The work done by the external agent on the plates is: (A) –200 µJ (B) 200 µJ
(C) 400 µJ (D) –400 µJ

Section-(B) More than one option

Each questions has 4 choices (A), (B), (C) and (D) out of which one and more correct. 4 marks each question and 1 marks negative.

12. Two concentric rings of radii R1 = sqrt(6)m and R2 = 4m are placed in y-z plane with their centres at origin. They have uniform charge –q and +Q = 2 sqrt(2)q on the inner and outer rings respectively. Consider the electrostatic potential to be zero at infinity. Then (A) The electric potential is zero at origin
(B) The electric field intensity is zero at r = 2 m
(C) A positive charged particle disturbed from origin along the x-axis restore back to origin.
(D) Where potential is maximum on the x-axis, field intensity is zero.

13. A parallel plate capacitor of capacitance ‘C’ has charges on its plates
initially as shown in the figure. Now at t = 0, the switch ‘S’ is closed.
Select the correct alternative(s) for this circuit diagram. (A) In steady state the charges on the outer surfaces of plates ‘A’ and ‘B’ will be
same in magnitude and sign.
(B) In steady state the charges on the outer surfaces of plates ‘A’ and ‘B’ will be same in magnitude and
opposite in sign.
(C) In steady state the charges on the inner surface of the plates ‘A’ and ‘B’ will be same in magnitude
and opposite in sign.
(D) The work done by the cell by the time steady state is reached is 5E^2C/2

14. The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a
dielectric of dielectric constant K is inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor.
(A) the energy stored in the capacitor will become K-times
(B) the electric field inside the capacitor will decrease K-times
(C) the force of attraction between the plates will become K²-times
(D) the charge on the capacitor will become K-times.

15. A parallel plate capacitor of capacitance 10μF is connected to a cell of emf 10 Volt and fully charged. Now
a dielectric slab ( k = 3) of thickness equal to the gap between the plates, is very slowly inserted to
completely fill in the gap, keeping the cell connected. During the filling process:
(A) the increase in charge on the capacitor is 200 μC
(B) the heat produced is zero.
(C) energy supplied by the cell = increase in stored potential energy + work done on the person who is
filling the dielectric slab.
(D) energy supplied by the cell = increase in stored potential energy + work done on the person who is
filling the dielectric slab + heat produced.

Section-(C)
Comprehension
Each questions has 4 choices (A), (B), (C) and (D) out of which one and more correct. 4 marks each
question and 1 marks negative.

Comprehension-01 (Questions : 16 to 17)
A metal ball (Neutral) with radius r is concentric with hollow metal sphere
of radius ‘R’, having charge ‘Q’ as shown in figure, Now ball is connected
with a very long wire to earth. Then: 16. Potential difference between sphere and metal ball, after grounding is: 17. After grounding:
(A) net electric field between sphere and ball is zero
(B) electric field between ball and sphere is zero due to ball only
(C) electric field between sphere and ball due to ball is non zero
(D) electric field between sphere and ball is non-zero due to sphere

Comprehension-02 (Questions : 18 to 20)
Two positive point charges A and B have charge +q and +2q; mass m and 2m respectively as shown. Both
the charges are released from rest when they are at a distance l0 apart. Neglect gravity and also assume 18. The speed of charge A at the instant separation between both charges is 2l0 is :

(A)sqrt(q^2/12pi E0 ml)
(B)sqrt(q^2/6pi E0 ml)
(C)sqrt(q^2/4pi E0 ml)
(D)sqrt(q^2/3pi E0 ml) Recent Posts
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