DPP#25 to 31 Electrostatics 7 Capacitance 15.06.2013 HOMEWORK
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DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
VACATIONS HOMEWORK REVISION
Electrostatics & Capacitance CONTENTS DPP#25 to 30
Single Answer Type
Page 2–25
DPP#31
Multiple Answer Type
Page 26–32
Page #1
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
Physics For IIT–JEE by Shiv R. Goel (B.Tech., IIT–Delhi) DPP#25: 17/06/2013 Monday 1.
Time Allotted 1 Hour
Single Answer Correct
In the figure shown the plates of a parallel plate capacitor have unequal charges. Its capacitance is ‘C’ P is a point outside the capacitor and close to the plate of charge –Q. The distance between the plates is ‘d’ select incorrect alternative 2Q –Q
P (A) A point charge at point ‘P’ will experience electric force due to capacitor (B) The potencial difference between the plates will be 3Q/2C 9Q 2 (C) The energy stored in the electric field in the region between the plates is 8C 2 Q (D) The force on one plate due to the other plate is 2π∈0 d 2 2.
Three capacitors A,B and C each of capacitance 1µF are connected as shown. The charge on capacitor C is – 6V
A
B
C
18 V
3.
(A) 4 µC (B) 6 µC (C) 8 µC (D) 2 µC Separation between the plates of a parallel plate capacitor is 5 mm. This capacitor, having air as the dielectric medium between the plates, is charged to a potential difference 25 V using a battery. The battery is then disconnected and a dielectric slab of thickness 3 mm and dielectric constant K = 10 is placed between the plates, as shown. Potential difference between the plates after the dielectric slab has been introduced is -
3mm 5mm (A) 18.5 V
4.
(B) 13.5 V
(C) 11.5 V
(D) 6.5 V
Current versus time and voltage versus time graphs of a circuit element are shown in figure. V(Volt) I(A)
1.0 amp
4.0 Volt 4.0 sec
t(s)
4.0 sec
t(s)
The type of the circuit element is : (A) capacitance of 2 F
(B) resistance of 2Ω
(C) capacitance of 1 F
(D) a voltage source of e.m.f 1 V
Page #2
DPP No. #25 TO 31/HOMEWORK/17/05/2013
5.
Target IIT-JEE 2014
A circuit element is placed in a closed box. At time t = 0, a constant current generator supplying a current of I amp is connected across the box. Potential difference across the box varies according to graph shown in the figure. The element in the box is : V (volts) 8
2
6.
t/s 3 (A) a resistance of 2 ohm (B) a battery of e.m.f 6 V (C) an inductance of 2 H (D) a capacitance of 0.5 F A,B,C,D,E,F are conducting plates each of area A and any two consecutive plates separated by a distance d. The net energy stored in the system after the switch S is closed is :
A B
(A) 7.
3ε 0 A 2 V 2d
(B)
C D
5ε 0 A 2 V 12d
E F
(C)
ε0A 2 V 2d
(D)
ε0A 2 V d
Three identical metal plates of area 'A' are at distance d1 & d2 from each other. Metal plate A is uncharged, while plate B & C have respective charges +q & – q. If metal plates A &C are connected by switch K through a consumer of unknown resistance, what energy does the consumer give out to its surrounding ? Assume d1 = d2 = d A B
+q
C
–q
K 2
(A) 8.
q d 4ε 0 A
2
(B)
q d ε0A
(C)
q 2d 2ε 0 A
(D)
2q 2d ε0A
Initially K1 is closed, now if K2 is also closed, find the heat dissipated in the resistances of connecting wires C
C
C K2
C V
K1
1 2 1 1 CV2 (B) CV2 (C) CV2 (D) CV2 2 3 3 4 Capacity of an isolated sphere is increased n times when it is enclosed by an earthed concentric sphere. The ratio of their radii is –
(A) 9.
Page #3
DPP No. #25 TO 31/HOMEWORK/17/05/2013
(A) 10.
n2 n −1
(B)
n n −1
(C)
2n n +1
Target IIT-JEE 2014
(D)
2n + 1 n +1
A capacitor is composed of three parallel conducting plates. All three plates are of same area A. The first pair of plates are kept a distance d1 apart and the space between them is filled with a medium of a dielectric ε1. The corresponding data for the second pair are d2 & ε2 respectively. What is the surface charge density on the middle plate ? d1 d2 ε1
ε2
V0
⎡ε ε ⎤ (A) ε 0 V ⎢ 1 + 2 ⎥ ⎣ d1 d 2 ⎦
11.
⎡ε ε ⎤ (B) − ε 0 V ⎢ 1 + 2 ⎥ ⎣ d1 d 2 ⎦
⎡ε ε ⎤ (C) 2ε 0 V ⎢ 1 + 2 ⎥ ⎣ d1 d 2 ⎦
⎡ε ε ⎤ (D) − 2ε 0 V ⎢ 1 + 2 ⎥ ⎣ d1 d 2 ⎦
A thin conducting plate is inserted in half way between the plates of a parallel plates capacitor of capacitance C. Conducting plate
What does the value of capacitance, if both the plate of capacitor is shortened ? (A) C 12.
(B) 2C
(C) 3C
(D) 4C
Following figure represents a parallel plate capacitor with a grid halfway with the voltages shown. An electron starting from plate P1 will : P2 +100V G – 60V P1 0V (A) not be able to reach the plate P2
(B) reach P2 with 100eV kinetic energy
(C) reach P2 with 40 eV kinetic energy
(D) reach P2 with 140 eV kinetic energy
13.
The distance between the plates of a circular parallel plate condenser of diameter 40 mm, in order to make its capacity equal to that of a metallic sphere of radius 1m, will be – (A) 0.1 mm (B) 10 mm (C) 0.01 mm (D) 1 mm
14.
Five identical plates each of area A are joined as shown in the figure the distance between the plates is d. The plates are connected to a potential difference of V volts. The charges on plates 1 and 4 will be –
1
(A)
ε 0 AV 2ε 0 AV , d d
(B)
– V +
2 3 4 5
−ε 0 AV 2ε 0 AV , d d
(C)
ε 0 AV −2ε 0 AV , d d
(D)
−ε 0 AV , d
−2ε 0 AV d
Page #4
DPP No. #25 TO 31/HOMEWORK/17/05/2013
15.
Target IIT-JEE 2014
Four equal capacitors , each with a capacitance (C) lare connected to a battery of E.M.F 10 volts as shown in the adjoining figure. The mid point of the capacitor system is connected to earth. Then the potentials of B and D are respectively – A
C
B
D Earth
(A) + 10 volts, zero volts
(B) + 5 volts, – 5 volts
(C) – 5 volts, + 5volts
(D) zero volts, 10 volts
ANSWER KEY 1.
placed between the plates, the thickness of air between
[D] 2Q Q 3Q E= + ⇒E= 2Aε 0 2Aε 0 2Aε 0 3Q 3 Q ⇒ Ed = =V E= 2C 2 Cd (ii) F = EQ/2 ⎛ 2Q ⎞ (– Q) – Q 2 ⎟⎟ × = F = ⎜⎜ 1 Aε 0 ⎝ 2Aε 0 ⎠
the plates will be 5 – 3 = 2 mm. Electric field strength in air will have the same value (5000 N/C) but inside the dielectric, it will be = 500 N/C so potential difference = Eair dair + Emed dmed
Q2 F= Aε 0 1 (iii) Energy = ε0 E2 Ad 2
= 5000 × (2 × 10–3) + 500 × (3 × 10–3) 4.
2
=
2.
1 ⎛ 3Q ⎞ 9 Q2 ε0 ⎜ ⎟ Ad = 2 ⎝ 2cd ⎠ 8 C
[A] Equivalent circuit B A
5. 6.
C 12V
3.
5000 5000 = K 10
[C] The capacitor is charged by a battery of 25 V. Let the 7. magnitude of surface charge density on each plate be σ. Before inserting the dielectric slab, electric field strength between the plates, σ V E= = d ε0 σ 25 = = 5000 N/C or E= ε0 5 × 10 –3 The capacitor is disconnected from the battery but
= 11.5 V [C] In case of a capacitor q = CV dq ⎛ dV ⎞ ∴ i= =C ⎜ ⎟ dt ⎝ dt ⎠ dV 4.0 = V/s = 1.0 V/s dt 4.0 Therefore, if C = 1 F then i = 1×1 = 1A (constant) [D] [C] ε A Ceff = 0 since effective capacitance between plates d A and E is zero 1 ε A ∴U= CV2 = 0 V2 2 2d [A]
+q
Energy loss =
charge on it will not change so that σ has the same value. When a dielectric slab of thickness 3mm is 8.
–q
=
1 C2 1 C q2 ×V2 = × 2 C+C 2 2 C2
q 2d q2 = 4C 4ε 0 A
[C]
Page #5
DPP No. #25 TO 31/HOMEWORK/17/05/2013
When K1 is closed
Given that C2 = nC1 C
+Q –Q V
Q = CV 1 CV2 2 When K2 is also closed
energy Ui =
10.
or
R 2 R1 = nR1 R 2 − R1
or
R 2 / R1 =n R 2 / R1 − 1
or
R2 n = R1 n −1
[A]
Equivalent circuit
2
1
Target IIT-JEE 2014
5
6
C1V0
4
3
ε1 11' 33'
V
9..
1 1'
Equivalent circuit C C C 4 1 2 5 C 3 6 2C 5C Ceq = C + = 3 3 1 5C 2 5 Energy Uf = × V = CV2 2 3 6 Charge supplied by battery after closing K2 5 2 = CV – CV = CV 3 3 Energy supplied by battery = Uf – Ui + ∆H 2 5 1 CV2 = CV2 – CV2 + ∆H 3 6 2 1 ∴ ∆H = CV2 3 [B]
⎡ε ε ⎤ σ = ε0V ⎢ 1 + 2 ⎥ ⎣ d1 d 2 ⎦
11.
[D] 1 1′ 2 2′ 3 3′
and
⎛ R R ⎞ C2 = 4πε0 ⎜⎜ 1 2 ⎟⎟ ⎝ R 2 − R1 ⎠
d/2
Cl 1 1′ 3 3′
C2
C1 = 4πε0R1
V0
⎡ε ε A ε ε A⎤ Total charge on 2 & 2' plate = ⎢ 1 0 + 2 0 ⎥ V d2 ⎦ ⎣ d1
R2
C1
3 3'
Middle plate
R1 R1
2
2, 2'
ε2 C2V0
2'
=2× 12. 13.
[A] [A]
14. 15.
[C] [B]
Cl
2 2′
(∈0 A) × 2 ∈ A =4. 0 = 4C d d
Page #6
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
Physics For IIT–JEE by Shiv R. Goel (B.Tech., IIT–Delhi) DPP#26: 18/06/2013 Tuesday 1.
Time Allotted 1 Hour
Single Answer Correct
What is the capacitance of the capacitor of square plates of area A, Shown in figure –
K3
d d
K1
B
(A) 2.
∈0 A K1K 2 4d K 1 + K 2
(B)
∈0 A K1 (K1 + K 2 ) d 3K1 + K 2
(C)
∈0 A K 1 (K 1 + 3K 2 ) 4d K1 + K 2
(D)
∈0 A 4d(K1 + 3K 2 )
A dielectric slab of area A and thickness d is inserted between the plates of a capacitor of area 2A and distance between plates d with a constant speed v as shown in the fig. The capacitor is connected to a battery of emf E. The current in the circuit varies with time as
E
v
I
I
I
I
(B)
(A)
t
(C)
t
(D)
t
3.
t Find the capacitance of a system of three parallel plates each of area A separated by distance d1 and d2. The space
between them is filled with dielectrics of relative dielectric constants ε1 and ε2. The dielectric constant of free space is ε0 –
4.
(A)
ε1ε 2 ε 0 A ε1d 2 + ε 2 d1
(B)
ε1ε 2 ε 0 A ε1d1 + ε 2 d 2
(C)
ε1ε 2 A ε 0 (ε1 + ε 2 )d1d 2
(D)
A ε1ε 2 ε 0 (ε1d1 + ε 2 d 2 )
For making a parallel plate capacitor , two plates of copper, a sheet of mica (thickness = 0.1 mm, K = 5.4), a sheet of glass (thickness = 0.2 mm, K = 7) and a slab of paraffin (thickness = 1.0 cm, K = 2) are available. To obtain the largest capacitance, which sheet should you place between the copper plates ? (A) Mica
5.
(B) Glass
(C) Paraffin
(D) None of these
The capacity of a parallel plate condenser is C0. If a dielectric of relative permittivity εr and thickness equal to one fourth the plate separation is placed between the plates, then its capacity becomes C. Then value of
(A)
5ε r 4ε r + 1
(B)
4ε r 3ε r + 1
(C)
3ε r 2ε r + 1
C will beC0
(D)
2ε r εr +1
Page #7
DPP No. #25 TO 31/HOMEWORK/17/05/2013
6.
Target IIT-JEE 2014
Two metal plates form a parallel plate condenser. The distance between the plates is d. A metal plate of thickness d/2 and of the same area is inserted completely between the plates. The ratio of capacitances in the two cases (later to initial) is (A) 2 : 1
7.
(B) 1 : 2
(C) 1 : 1
(D) 1 : 4
A parallel plate capacitor is completely filled by an insulating material 3 mm thick and of relative permittivity 4. The distance between the plates is increased to allow the insertion of a second sheet 5 mm thick and of relative permittivity ∈r. If the capacitance of the capacitor so formed is one-half of the original capacitance, then the value of ∈r (A)
8.
20 3
(B)
16 3
(C)
10 3
(D) None of these
A positive charge q is given to each plate of a parallel plate air capacitor having area of each plate A and separation between them, d (A) Since both the plates are identically charged, therefore, capacitance becomes equal to zero (B) Energy stored in the space between the capacitor plates is equal to
q 2A ε0A
(C) No charge appears on inner surface of the plates (D) Potential difference between the plates is equal to 9.
2qd ε0A
Two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series, i.e., the positive plate of one is connected to the negative of the other(A) The charge on the plates connected together are destroyed (B) The charge on free plates are enhanced (C) The energy stored in the system increases (D) The P.D. between the free plates is 2V
10.
Two spherical conductors A1 and A2 of radii r1 and r2 are placed concentrically in air. The two are connected by a copper wire as shown in fig. Then the equivalent capacitance of the system isA2 A1 r1 r2
(A) 11.
4π ∈0 k r1.r2 r2 − r1
(B) 4π∈0 (r1 + r2)
(C) 4π∈0 r2
(D) 4π∈0 r1
Three concentric thin spherical shells are of radii a , b and c [a < b < c]. The first and third are connected by a fine wire through a small hole in the second, and the second is connected to earth through a small hole in the third, the capacity of the capacitor so formed is ⎡ ab c2 ⎤ (A) 4π∈0 ⎢ + ⎥ ⎢⎣ b − a c − b ⎥⎦
⎡ ab c2 ⎤ + (B) 4π∈0 ⎢ ⎥ ⎢⎣ b + a a + b ⎥⎦
⎡b −a c − b⎤ (C) 4π∈0 ⎢ + ⎥ c2 ⎦ ⎣ ab
⎡b + a c + b⎤ (D) 4π∈0 ⎢ + ⎥ c2 ⎦ ⎣ ab
Page #8
DPP No. #25 TO 31/HOMEWORK/17/05/2013
12.
Target IIT-JEE 2014
A parallel plate capacitor has a dielectric slab in it. The slab just fills the space inside the capacitor. The capacitor is charged by a battery and then battery is disconnected. Now the slab is started to pull out slowly at t = 0. If at time t, capacitance of the capacitor is C, then which of the following graphs is correct ? C
(B)
(A) O
13.
C
C
(D)
(C)
O
t
C
O
t
O
t
t
If the current, charging a capacitor, is kept constant then the potential difference V across the capacitor varies with time t as – V
V
(B)
(A) O
14.
V
(C)
(D) O
O
V
O
A capacitor of capacitance 160 µF is charged to a potential difference of 200V and then connected across a discharge tube which conducts until the potential deference across it has failed to 100 V. The energy dissipated in the tube is (A) 6.4 J
15.
(B) 4.8 J
(C) 3.2 J
(D) 2.4 J
Two capacitors each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be – (A) 2C and 2V
(B) C/2 and V/2
(C) 2C and V/2
(D) C/2 and 2V
ANSWER KEY 1.
[C]
2.
[B]
3.
[A]
4.
[A]
5.
[A]
6.
[A]
7.
[A]
8.
[C]
9.
[D]
10.
[C]
11.
[A]
12.
[D]
13.
[C]
14.
[D]
15.
[D]
Page #9
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
Physics For IIT–JEE by Shiv R. Goel (B.Tech., IIT–Delhi) DPP#27: 19/06/2013 Wednesday
Time Allotted 1 Hour
Single Answer Correct
1.
Two metal plates having charges Q, – Q face each other at some separation and are dipped into an oil tank. If the oil is pumped out, the electric field between the plates will (A) increase (B) decrease (C) remain the same (D) become zero
2.
A parallel plate capacitor has plates of unequal area. The larger plate is connected to the positive terminal of the battery and the smaller plate to its negative terminal. Let Q+ and Q – be the charges appearing on the positive and negative plates respectively (B) Q+ = Q – (C) Q+ < Q – (A) Q+ > Q – (D) The information is not sufficient to decide the relation between Q+ and Q –
3.
A parallel plate capacitor without any dielectric has capacitance C0. A dielectric slab is made up to two dielectric slabs of dielectric constants K and 2K and is of same dimensions as that of capacitor plates and both the parts are of equal dimensions arranged serially as shown. If this dielectric slab is introduced (dielectric K enters first) in between the plates at constant speed, then variation of capacitance with time will be best represented byA V
A
A
K
2K
L
L
L
C
(A) C0 4.
5.
C
C
(B) C0
(C) C0
C
(D) C0
t t t t The metal plate on the left in the fig. carries a charge +q. The metal plate on the right has a charge of –2q. What charge will flow through S when it is closed if the central plate is initially neutral ? q –2q + – + +
– –
+
–
+ +
– –
(A) zero (B) – q (C) + q (D) + 2q You have a parallel plate capacitor, a spherical capacitor and cylindrical capacitor. Each capacitor is charged by and then removed from the same battery. Consider the following situations (i) separation between the plates of parallel plate capacitor is reduced (ii) radius of the outer spherical shell of the spherical capacitor is increased (iii) radius of the outer cylinder of cylindrical capacitor is increased Which of the following is correct ? (A) In each of these situations (i), (ii) and (iii), charge on the given capacitor remains the same and potential difference across it also remains the same (B) In each of these situations (i), (ii) and (iii), charge on the given capacitor remains the same but potential difference, in situations (i) and (iii), decreases, and in situation (ii), increases
Page #10
DPP No. #25 TO 31/HOMEWORK/17/05/2013
6.
Target IIT-JEE 2014
(C) In each of these situations (i), (ii) and (iii), charge on the given capacitor remains the same but potential difference, in situations (i), decreases, and in situations (ii) and (iii), increases (D) Charge on the capacitor in each situation changes. It increases in all these situations but potential difference remains the same A closed body, whose surface F is made of metal foil, has an electrical capacitance C with respect to an uniformly distant point. The foil is now dented in such a way that the new surface F* is entirely inside or an the original surface as shown in the figure. Then – F
F*
7.
(A) Capacitance of F* > capacitance of F (B) Capacitance of F* < capacitance of F * (D) Nothing can be concluded from given (C) Capacitance of F = capacitance of F Two capacitor C1 & C2, charged with q1 & q2 are connected in series with an uncharged capacitor C, as shown in figure. As the switch S is closedC
C1
+ q1 –
q2 + C2 –
S
(B) C gets charged only when q1C2 > q2 C1 (D) C gets charged when q1C2 ≠ q2 C1
(A) C gets charged in any condition (C) C gets charged only when q1C2 < q2C1 8.
An electron with a kinetic energy of 100 eV enters the space between the plates of plane capacitor made of two dense metal grids at an angle of 30° with the plates of capacitor and leaves this space at an angle of 45° with the plates. What is the potential difference of the capacitor – 30°
45°
9.
(A) 100 V (B) 50 V (C) 150 V In the given fig. the work done by battery after the switch S is closed –
(D) 200 V
10 µF
4V
10 µF
S
(A) 100 µJ 10.
(B) – 100 µJ
(C) 80 µJ
(D) – 80 µJ
In the condenser show in the circuit is charged to 5V and left in the circuit, in 12 s the charge on the condenser will become –
Page #11
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
6Ω 2F
(A) 11.
10 C e
(B)
e C 10
10
(C)
e2
(D)
e2 C 10
Two capacitors are joined in series as shown in figure. The area of each plate is A. The equivalent of the combination is – d b
a
d
(A)
ε0A d1 − d 2
(B)
ε0A a−b
⎛1 1⎞ (C) ε 0 A⎜ − ⎟ ⎝a b⎠
⎛ 1 1 (D) ε 0 A⎜⎜ − ⎝ d1 d 2
⎞ ⎟⎟ ⎠
12.
The two spherical shells are at large separation one of them has radius 10 cm and has 1.25 µC charge. The other is of 20 cm radius and has 0.75 µC charge. If they are connected by a conducting wire of negligible capacity, the charge on the shells are2 4 4 2 (C) µC, µC (D) 0.25 µC, 0.25 µC (A) 1 µC, 1µC (B) µC, µC 3 3 3 3
13.
The net capacitance between A and B is C
C
A C
C
C
B C
2C 2C (C) (D) None of these 5 3 In the following circuit composed of identical capacitor, across which terminals would you connect a battery in order for all capacitors to charge up –
(A) 6 C 14.
C
(B)
A
B
(A) AB 15.
(B) AC
D
C
(C) BD
(D) None of these
Five identical plates of equal area A are placed parallel to and at equal distance d from each other as shown in figure. The effective capacity of the system between the terminals A and B isB
A
(A)
3 ∈o A 5 d
(B)
5 ∈o A 4 d
(C)
5 ∈o A 3 d
(D)
4 ∈o A 5 d
Page #12
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
ANSWER KEY 1. 2. 3. 4. 5.
[A] [B] [B] [C] [C] Each capacitor is charged and then removed from the battery. Changing the plate separation in a parallel plate capacitor or the radius of any shell or cylinder in the spherical or the cylindrical capacitors will not change the charge on any capacitor. Each capacitor, in the disconnected state, is an isolated system. ε A In situation (i), since C = 0 , capacity will increase d
as d is reduced. Therefore, V =
Q decreases, Q being C
the same. ab a = 4πε0 b−a 1− a / b 9. a and b are the radii of the inner and the outer spherical shells, respectively. As b is increased, capacity will reduce and In situation (ii), C = 4πε0
Q increases, Q being the same. C L In situation (iii), C = 2πε0 ln ( b / a )
V=
6.
L is the length of cylinders; a and b are the radii of inner and outer cylinders, respectively. As b is increased, capacity will decrease and Q increases, Q being the same. V= C 10. [B] 2 Q Ui = 2C i 2
Q 2C f As surface was deformed in such a way that charge on original surface are coming closer or moving perpendicular to electric force acting on them total energy of foil get increase. Uf > Ui Q2 Q2 > 11. 2C f 2C i Ci > Cf [D] Charge in the circuit flows only when potential difference across C1 is either greater or less than that across C2 q q ie. 1 ≠ 2 ∴ q1 C2 ≠ q2C1 C1 C2 [B]
Uf =
7.
8.
v1
v2 Horizontal velocity is constant v1 cos 30° = v2 cos 45° 1 3 = v2 × v1 2 2
3v12 v2 = 2 4 2 3 1 1 × mv12 = mv 22 2 2 2 3 KEf = × KEi 2 = 1.5 × 100 = 150 eV ∆KE (150 − 100)eV P.d. = = = 50 V e e
[C]
Before the switch S is closed Qi = Ceq V = 5 × 4 = 20 µC When the switch S is closed Qf = CV = 10 × 4 = 40 µC W(cell) = ∆QE = (Qf – Qi) × E = 20 × 4 = 80 µJ [A]
During discharge ⎛ −t ⎞ qc = CE⎜ e RC ⎟ ⎜ ⎟ ⎝ ⎠ ⎛ − 12 ⎞ = 2 × 5 ⎜ e 12 ⎟ ⎜ ⎟ ⎝ ⎠ 10 = e [B] When two capacitors are in series
x1 ⇒
x2
x1 x x + x2 1 + 2 = 1 = C eq ∈0 A ∈0 A ∈0 A
∴ Ceq =
∈0 A x1 + x 2
Page #13
DPP No. #25 TO 31/HOMEWORK/17/05/2013
=
12.
∈0 A Sum of separations between plates
Potential across C and D is same. Capacitance between C and D is uncharged.
Now in given arrangement Capacitors are in series & sum of separations = a – b ∈ A ∴ Ceq = 0 a−b [A]
Battery is connected across AC i A
B
Total charge = 1.25 µC + 0.75 µC = 2 µC q1' : q2' = R1 : R2 = 1 : 2 1 ×2 = 3 2 q2' = ×2 = 3
∴ q1 ' =
13.
D
i
C
Now B and D will remain uncharged.
2 ⎤ µC ⎥ 3 ⎥ 4 ⎥ µC 3 ⎥⎦
Battery is connected across BD A
[B] Use Junction method
i
x x
x C
X
A and C are at same potential ∴ Capacitance between A and C is uncharged. [C]
X
Y
15. A
B
X Y 1 1 1 1 = + + C eq C 2C C
14.
D
i – 2x
X B
i
x
x
B
X
A
Target IIT-JEE 2014
5 4
2C ∴ Ceq = 5 [D]
B
3 2
A
1 1
2
A
≡
C
B
D
A
3
5
2
3
4
B
4
Battery is connected between A and B ∈0 A d
The capacity of each capacitor is, C0 = x
x
iA
x
x
From fig. it is clear that Ceq =
x
B
5 C0 = 3
5 ∈o A 3 d
D
x i C
Page #14
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
Physics For IIT–JEE by Shiv R. Goel (B.Tech., IIT–Delhi) DPP#28: 20/06/2013 Thursday 1.
2.
Time Allotted 1 Hour
Single Answer Correct
A capacitor consists of two parallel metal plate of area A separated by a distance d. A dielectric slab of area A, thickness b and dielectric constant k is placed inside the capacitor. If CK is the capacitance of capacitor with dielectric, how much K and b be restricted so that CK = 2C, where C is capacitance without dielectric – d 4b 2b d & 0 2b − d d−b+
2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.
d
d 2
[B] [C] [A] [C] [A] [A] [D] [B] [C] [D] [A] [C] [A] [A]
Page #17
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
Physics For IIT–JEE by Shiv R. Goel (B.Tech., IIT–Delhi) DPP#29: 21/06/2013 Friday
Time Allotted 1 Hour
Single Answer Correct
1.
Two capacitors of capacities C1 and C2 are charged to voltages V1 and V2 respectively. There will be no exchange of energy in connecting them in parallel . IfC C (A) C1 = C2 (B) C1V1 = C2V2 (C) V1 = V2 (D) 1 = 2 V1 V2
2.
Two spheres of radii R1 and R2 have equal charge are joint together with a copper wire. If the potential on each sphere 1 )– after they are separated to each other is V, then initial charge on any sphere was (k = 4π ∈0 (A)
3.
V (R1+ R2) k
(B)
V (R1 + R2) 2k
(C)
V (R1 + R2) k
(D)
V ( R 1R 2 ) k R1 + R 2
Calculate the reading of voltmeter between X and Y then (Vx – Vy ) is equal to 1F 2F A
X V
3F
Y
1F
B
6F
20V
(A) 10 V 4.
(B) 13.33V
(C) 3.33 V
(D) 10.33 V
In the figure shown, the capacity of the condenser C is 2µF. The current in 2Ω resistor is 2Ω
2µF
3Ω
4Ω
+ – 6V
(A) 9 A 5.
2.8Ω
(B) 0.9 A
(C)
1 A 9
(D)
1 A 0.9
In the circuit shown here C1 = 6µF, C2 = 3µF and battery B = 20V. The Switch S1 is first closed. It is then opened and afterwards S2 is closed. What is the charge finally on C2 C2 3µF C1
6µF
S2 S1
B=20V
(A) 120µC 6.
(B) 80µC
(C) 40µC
(D) 20µC
Three identical capacitors are given a charge Q each and they are then allowed to discharge through resistance R1, R2 and R3. Their charges, as a function of time shown in the graph below. The smallest of the three resistance is -
Page #18
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
Q
R3 R2
R1
t
(A) R3 7.
(B) R2
(D) Cannot be predicted
The work done in placing a charge of 8 × 10–18 coulomb on a condenser of capacity 100 micro-farad is – (A) 3.1 × 10–26 J
8.
(C) R1
(B) 4 × 10–10 J
(C) 32 × 10–32 J
(D) 16 × 10–32 J
A fully charged capacitor has a capacitance 'C'. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity 's' and mass 'm'. If the temperature of the block is raised by '∆T', the potential difference 'V' across the capacitance is – (A)
9.
2m C∆T s
(B)
m C∆T s
(C)
ms∆T C
(D)
2ms∆T C
A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is ‘d’. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k1 = 3 and thickness
(A) 45 pF 10.
d 2d while the other one has dielectric constant k2 = 6 and thickness . Capacitance of the capacitor is now – 3 3
(B) 40.5 pF
(C) 20.25 pF
(D) 1.8 pF
A series combination of 0.1 MΩ resistor and a 10 µF capacitor is connected across a 1.5V source of negligible resistance. The time required for the capacitor to get charged up to 0.75 V approximately (in seconds) is (A) ∞
11.
(B) loge2
(C) log102
(D) zero
A capacitor of capacity C is charged in RC circuit. The variation of log I Vs time is shown in the figure by dotted line when net resistance of circuit is X. When resistance changes to 2X the variation is now shown by Log I S R Q P
(A) P
(B) Q
t (C) R
(D) S
12.
A capacitor of capacitance 4µF is charged through a resistor 2.5 MΩ connected in series to a battery of emf 12 volt having negligible internal resistance . Then time in which potential drop across capacitor is 3 times the potential drop across the resistor (A) 13.86 sec (B) 6.93 sec (C) 27.72 sec (D) 3. 46 sec
13.
Time constant for the given circuits are –
Page #19
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
2µF
2µF
1Ω 2µF
4µF
1Ω
4µF
2Ω
1Ω
2Ω
4µF
2Ω 12V
14.
12V 12 V 8 8 8 8 (B) 18 µs, 4 µs, µs (C) 4µs, µs, 18 µs (D) µs, 18µs, 4µs (A) 18 µs, µs, 4µs 9 9 9 9 A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is 6µF 3µF X
S 6Ω
3Ω Y 9V
(B) 54 µC
(A) 0 15.
(C) 27 µC
Calculate the equivalent capacitance between a & b A/3
(D) 81 µC
a A/3 A/3
d/2 K1=3 K2=6 K3=9 d/2
K=2
b
(A) 3
∈0 A d
(B) 2
∈0 A d
(C)
∈0 A d
(D)
∈0 A 2d
Page #20
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
ANSWER KEY C3 =
∈0 A × K3 = 6C0 d 3× 2
C4 =
∈0 A × K4 = 4C0 d 2
1.
[C]
2.
[B]
3.
[C]
4.
[B]
5.
[C]
6.
[C]
7.
[C]
8.
[D]
9.
[B]
10.
[B]
11.
[B]
12.
[A]
C1,C2 & C3 are in parallel and C4 is in series with
13.
[A]
C1,C2 & C3
14.
[C]
15.
[A]
a A/3 A/3
A/3 d/2
C1
d/2
C2
C3
C4
b
∴ C1 || C2 || C3 → 2C0 + 4 C0 + 6C0 = 12C0
C1 =
∈0 A × K1 = 2C0 d 3× 2
C2 =
∈0 A × K2 = 4C0 d 3× 2
Ceq
=
12 C 0 × 4 C 0 16 C 0
=
3C0
=
3∈0 A d
Page #21
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
Physics For IIT–JEE by Shiv R. Goel (B.Tech., IIT–Delhi) DPP#30: 22/06/2013 Saturday 1.
Time Allotted 1 Hour
Single Answer Correct
The ratio of potential difference between 1 µF and 5 µF capacitors – 1 µF 2 µF
2 µF 3 µF 5 µF
10 V
2.
3.
(A) 1 : 2 (B) 3 : 1 (C) 1 : 5 (D) 10 : 1 Two parallel plate capacitors of capacitance C and 3C are connected in parallel and charged to a potential difference V with the help of a battery. The battery is disconnected and the space between these two capacitors are filled with dielectric of constant K and K/2 respectively. The voltage across the combination will be 4V 6V 8V 3V (A) (B) (C) (D) 3K 3K 5K 5K In the given system a capacitor of plate area A is charged upto charge q. The mass of each plate is m2. The lower plate is rigidly fixed. Find the value of m1 so that the system is in equilibrium –
q++++ ––––
(A) m2 + 4.
q2 ε 0 Ag
(B) m2
m1
(C)
q2 + m2 2Aε 0 g
(D) None of these
Find the amount of charge that will flow through the switch when it is closed – a 6µF 3µF S V=0
+200V 3µF
6µF b
5.
(A) 300 µC (B) 400 µC (C) 500 µC (D) 100 µC A slab of copper of thickness 'b' is inserted in between the plates of parallel plate capacitor. The separation of the plates is d d. If b = , then the ratio of the capacities of capacitor after and before inserting the slab is 2 (A)
6.
7.
2 :1
(B) 2 : 1
(C) 1 : 1
(D) 2 :
2
An electron is first placed at A and then at B between the two plates of a parallel plate capacitor charged to a potential difference of V volt as shown. The electrostatic potential energy of electron at A is + • – + B– – + – + • A – + – + – + (A) more than at B (B) less than at B Find the charge on each capacitor during steady state -
(C) equal to that at B
(D) nothing can be said
Page #22
DPP No. #25 TO 31/HOMEWORK/17/05/2013 3Ω
3µF
Target IIT-JEE 2014
3µF
3Ω 3Ω
R = 4Ω 6Ω 6Ω 6Ω
3µF
10 V
(A) 3µC 8.
1Ω
(B) 6µC
(C) 9µC
(D) 12µC
In the given circuit switch K is open. The charge on the capacitor is 'C' in steady state is q1. Now key is closed and steady state charge on C is q2. The ratio of charges q1/q2 is R K C
E
(A) 3/2
9.
3µF
2R
(B) 3/2
(C) 1
(D) 1/2
If each plate has area A and separation between successive plates is d from equivalent capacitance between A and B is A
B
(A) 10.
Aε 0 d
(B)
Aε 0 2d
(C)
4Aε 0 d
(D)
3Aε 0 d
In the circuit shown in figure switch S is at position 1 for long time. The total heat generated in resistors 2r0 and r0 when switch is shifted from position 1 to position 2. r0 C0 S 1
2
2r0 E0
(A) 11.
C 0 E 02 2
(B)
C 0 E 02
(C)
3
C 0 E 02 3r0
(D) None
In the given circuit diagram. Find the heat generated on closing the switch S (Initially the capacitor of capacitance C is uncharged) C 2C S V
3 CV 2 (B) CV2 2 If each capacitor has capacitance is C then equivalent capacitance between A & B is -
(A) 12.
3C
(C)
1 CV2 2
(D) 2 CV2
Page #23
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
B
A
(A)
4C 3
(B)
8C 3
(C) 12 C
(D)
5C 12
ANSWSER KEY 1.
[C] 1 µF 5 µF
T = Fe + m2g ….(1) (for upper plate) T = m1g ….(2) (for m1)
2 µF & 2 µF are reduced to 1 capacitor of 1 µF and 3
∴ m1g =
q2 + m2g 2Aε 0
⇒ m1 =
q2 + m2 2Aε 0g
µF & 1 µF is reduced to 4 µF. 10 × 1 = =2 5
(potential difference across 4 µF capacitor is calculated by voltage division Rule) ∴ V5µF = 10 V
∴ 2.
V5µF
q1 = CV
+ + + + + +
+ + + + + +
3µF 6µF
V=0
3C
q2 = 3CV
Hence charge on each capacitor is 400 µC. When switch's closed
K + – + – + –
+ + + +
6µF + +
+ – K/2 + – + –
+200V 3µF + +
q1 + q 2 = CV + 3CV K C1 + C 2 KC +
3.
6µF + +
q = Ceq × V (for both branches) q = 2 × 200 = 400
C
4CV 8V = = 5KC 5K 2 [C]
When switch is open
+ + + +
[C]
Common potential =
[A]
+200V 3µF + +
2 1 = = 10 5
V
4.
+ + + +
(potential difference across 5 µF is same as Battery) V1µF
m1g
V4µF
10 V
∴ V4µF
T
T Fe + + + + m2g ––––
4 µF
2
+ + + +
+ + + + + +
+ + + + + +
3µF 6µF V = 0
⎯Equivalent ⎯ ⎯ ⎯→ 6µF
3µF
3C 3µF 100V
4 6µF 100V 200 V
q6 = 600 µC ⇒ q3 = 300 µC Hence final arrangement Initial was
Page #24
DPP No. #25 TO 31/HOMEWORK/17/05/2013
6µF+ +
a
+ + + +
600µC
3µF
+ + + + + +
+ + + + + +
300µC
400µC
+ + + + + +
Target IIT-JEE 2014
Before closing switch Ui =
400µC
3C × 2C 1 × × V2 3C + 2C 2
= + + + + + +
+ + + +
+ + + + + +
b ++ 3µF, 300µC 6µF, 600µC
400µF
+ + + + + +
400µC
After closing switch Uf =
Final charge on both plates Initial charge on both is 300 plates is zero 5.
CV2 = Uf – Ui + ∆H
6.
[B]
7.
[D]
CV2 =
∈0 A (d – t ) +
t ∈r
∆H = 12.
1 CV2 + ∆H 2 1 CV2 2 5
4Ω
A
3 10
[B]
6
3µF
3µF
4 B 12
11 B
9
7 3µF
∴ 8.
A
1Ω
10 V = = 2A; ∴VA – VB = IR = 2 × 4 = 8 V R eq. 5
I=
P.D across each capacitor is 4 V hence q = 4 × 3 = 12 µC [A] Initially when switch is open R
C
E
2 8
3µF 10 V
3 1 CV2 + CV2 5 2
Energy supply by battery = CV × V = CV2
Hence charge flow is 300µC. [B] Cpartial medium =
3 CV2 5
1 A
(1 to 7)
8
11
10
9 B
q1 = CE
Finally when switch is closed I
R
A C
E
2R B
E ; ∴ VA – VB = I 2R = E/3R × 2R = 2E/3 3R C2E ∴ Charge on capacitor = CV ; q2 = 3 q1 CE ∴ = =3:2 q2 2 / 3CE [A] It is a Wheatstone bridge 2C C I=
9. 10. 11.
2C
C
2C
C
2C
C
2C
C
Ceq. = 4 ×
[C] S V
2C × C 4 × 2C 8C = = 2C + C 3 3
3C
Page #25
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
Physics For IIT–JEE by Shiv R. Goel (B.Tech., IIT–Delhi) DPP#31: 24/06/2013 Monday 1.
Time Allotted 1 Hour
Multiple Answer Correct
Three capacitors each having capacitance C = 2µF are connected with a battery of emf 30 V as shown in figure. When the switch S is closed : S C
C
C 30 V (A) the amount of charge flown through the battery is 20 µC
(B) the heat generated in the circuit is 0.6 mJ (C) the energy supplied by the battery is 0.6 mJ (D) the amount of charge flown through the switch S is 60µC 2.
In the circuit shown in figure. 1Ω
2Ω
C1
3Ω
C2
2Ω
1Ω
3Ω
120V
3.
C1 = C2 = 2µF. Then charge stored in : (A) capacitor C1 is zero (B) capacitor C2 is zero (C) both capacitors is zero (D) capacitor C1 is 40 µC Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a battery. Now (A) The facing surfaces of the capacitor have equal and opposite charges (B) The two plates of the capacitor have equal and opposite charges (C) The battery supplies equal and opposite charges to the two plates (D) The outer surfaces of the plates have equal charges
4.
A charge Q is imparted to two identical capacitors in parallel. Separation of the plates in each capacitor is d0. Suddenly, the first plate of the first capacitor and the second plate of the second capacitor starts moving to the left with speed v, then – Q(d 0 − vt ) Q(d 0 + vt ) (A) Charge on the two capacitor as a function of time are , 2d 0 2d 0 (B) Charge on the two capacitors as a function of time are
Qd 0 Qd 0 , 2(d 0 − vt ) 2(d 0 + vt )
(C) Current in the circuit will increase as time passes (D) Current in the circuit will be constant 5.
Two plates of a parallel plate capacitors carry charges q and –q and are separated by a distance x from each other. The capacitor is connected to a constant voltage source V0. The distance between the plates is changed to x + dx. Then in steady state : V0 +q
–q x
(A) Change in electrostatic energy stored in the capacitor is
− Udx x
Page #26
DPP No. #25 TO 31/HOMEWORK/17/05/2013
− Ux dx
(B) Change in electrostatic energy stored in the capacitor is
6.
7.
(C) Attraction force between the plates is 1/2 qE (D) Attraction force between the plates is qE (where E is electric field between the plates) A parallel plate condenser is charged by a battery. The battery is removed and a thick glass slab is placed between the plates. Now(A) the capacity of the condenser is increased (B) the electrical energy stored in the condenser is decreased (C) the potential across the plates is decreased (D) the electric field between the plates is decreased Identical dielectric slabs are inserted into two identical capacitors A and B. These capacitors and battery are connected as shown in figure. Now the slab of capacitor B is pulled out with battery remaining connected-
B
A a
8.
Target IIT-JEE 2014
b
(A) during the process charge flows from a to b (B) final charge on capacitor B will be less than on capacitor A (C) during the process, work is done by external force F which appears as heat in the circuit (D) during the process, battery receive energy The two plates x and y of parallel plate capacitor of capacitance C are given a charge of amount Q each. x is now joined to the Q positive terminal and y to the negative terminal of a cell of emf e = . Then – C (A) a charge of amount Q will flow from the positive terminal to the negative terminal of the cell through the capacitor (B) the total charge on the x will be 2Q (C) the total charge on the plate y will be zero (D) the cell will supply Ce2 amount of energy
9.
The plates of a parallel plate capacitor are not exactly parallel then (A) surface charge density is therefore higher at the closer end (B) the surface charged density will not be uniform (C) each plate will have the same potential at each point (D) the electric field is smallest where the plates are closest
10.
The figure shows three infinite thin non-conducting charged plates perpendicular to the plane of the paper with charge per unit area +σ, +2σ and –3σ. As we move from plate 1 to plate 2 magnitude of potential change occured is V12, that from plate 2 to 3 is V23 and that from plate 1 to 3 is V13. Then1 + σ +
2
3
+ 2σ +
A
– −3σ B
–
+
+
–
Plate +
+
–
2m 4m
2m 6m
(A) Ratio of net electric field at point A to that at point B is 1/3 (B) 9V12 = 2 V23 (C) 2V13 = 7 V12 (D) 3V13 = 5V23
Page #27
DPP No. #25 TO 31/HOMEWORK/17/05/2013
11.
A uniformly charged ring charge +q has radius R. A uniformly charged solid sphere (over its total volume) has charge Q is placed on the axis of ring at a distance x =
3 R from centre of ring as shown. Radius of sphere is also R, then –
q +++ + + +R + + + +O + + + + + + +
(A) Electric force on ring is (B) Electric force on ring is
Q + + + R+ + + + ++ + + + + + ++ + + + ++ + + O ++ + + + + + + + + ++ + + ++ + + + ++
x = 3R
3kQq 8R 2 kQq 2R 2
(C) The increase in tension in ring due to sphere is (D) The increase in tension in ring due to sphere is
12.
Target IIT-JEE 2014
kQq 16πR 2 kQq 8πR 2
Four sphere each of different radius named 'a', 'b', 'c', 'd' are given, each are solid insulating uniformly charged and carrying equal total charge. Variation of electric field with distance r from the centre is given. Then E b
c a d
r
(A) Radius of sphere a > Radius of sphere b (B) Radius of sphere a < Radius of sphere b (C)Volume charged density of c > volume charged density of b (D) Volume charged density of c < volume charged density of b
13.
We have an infinite non-conducting sheet of negligible thickness carrying a uniform surface charge density – σ and next to it, an infinite parallel slab of thickness D with uniform volume charge density +ρ. All charges are fixed. –σ
D
+ρ
(A) Magnitude of electric field at a distance h above the negatively charged sheet is
ρD − σ 2ε 0
(B) Magnitude of electric field inside the slab at a distance h below the negatively charged sheet (h < D) is σ + ρ(D − 2h ) 2ε 0
Page #28
DPP No. #25 TO 31/HOMEWORK/17/05/2013
(C) Magnitude of electric field at a distance h below the bottom of the slab is–
(D) Magnitude of electric field at a distance h below the bottom of the slab is 14.
15.
Target IIT-JEE 2014
ρD − σ 4ε 0 ρD − σ 2ε 0
Three charged particles are in equilibrium under their electrostatic forces only – (A) The particles must be collinear
(B) All the the charges cannot have the same magnitude
(C) All the charges cannot have the same sign.
(D) The equilibrium is unstable
Two infinitely large plane sheets are separated by a distance l and carry a uniform charge of surface density +σ and – σ. The planes have coaxial holes of radius R (l >> R). Find the potential f and field E at a point on the axis of the holes at a distance x from the midpoint O between the holes – – – –– –– – – – – –– – – –– – –
O
+ + + ++ ++ + + + + ++ + + ++ + +
P
x l
(A) φ =
16.
σ 2ε 0 x ( R + x ) 2
2 1/ 2
(B) E =
σlR 2 2ε 0 (R + x ) 2
2 3/ 2
(C) φ =
σ 2ε 0 x
(D) E = −
2
σ 2ε 0 R 5
If at distance r from a positively charged particle, the electric field strength, potential and energy density are E, V and U respectively, which of the following graphs are correct? (B)
(A) V O
V2
(C)
(D)
V2
E E
O
17.
E
O
U
U
O
Two large conducting plates having surface charge densities + σ & – σ are fixed d distance apart. A small test charge q of mass m is attached to two identical springs as shown in the adjacent figure. The charge q is now released from rest with springs in natural length. Then q will [neglect gravity] +σ
–σ k m k q
(A) perform SHM with angular frequency
2k m
(C) not perform SHM but will have a periodic motion
18.
(B) perform SHM with amplitude
σq 2k ∈0
(D) remain stationary
Consider three infinite line charge kept in X–Y Plane parallel to X-axis, wire with density +λ is along X-axis and other two wires with density –λ passes through (0, a) and (0, – a) respectively. Now choose the correct alternative(s)
Page #29
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
Y – – – – – – – –λ (0, a) ++++++++λ X O (0, –a) – – – – – – – –λ Z
(A) E at point (0, 0, a) is zero (B) work done in displacing +q charge from (0, 0. a) to (0, 0, 2a) is
(C) work done in displacing +q charge from (0, 0, a) to (0, 0, 3a)
(D) Electric field at (0, 0, 2a) is E = –
19.
λ ⎛5⎞ log ⎜ ⎟ 2πε 0 ⎝4⎠
λ ⎛5⎞ log ⎜ ⎟ 2πε 0 ⎝3⎠
3λ kˆ 20πε 0 a
Five point charges, each of charge +q coulomb are placed on five vertices of a regular hexagon of side h as shown in figure. Then – +q +q D E O
F
C
–q +q
A
B
+q
+q
(A) the forces on (–q) at O due to charges +q at A and D are balanced (B) the forces on (–q) due to charges at D and E are balanced (C) the resultant force on –q at O is
1 q2 along OE 4πε 0 h 2
(D) the resultant force on –q at O is
1 q2 along OC 4πε 0 h 2
Page #30
DPP No. #25 TO 31/HOMEWORK/17/05/2013
Target IIT-JEE 2014
ANSWER KEY 1.
[A,C,D]
The charges stored in different capacitors before and
–Q KC +Q A
after closing the switch S are 20µC + –
40µC + –
60µC + –
+ – 20µC
a Q=
30 V before
30 V after
+Q KC –Q B
E b
KCE 2
on removal of dielectric from B
x
The amount of charge flown through the battery is q =
–Q+x KC Q–xA
20 µC ∴
Energy supplied by the battery is U = qV = (20 × 10–6) (30) J
Q−x Q−x + =E KC C
switch S is
⎛ K ⎞ Q–x= ⎜ ⎟CE ⎝ K +1⎠
1 1⎛4 ⎞ Cnet V2 = ⎜ × 10 −6 ⎟ (30)2 2 2⎝3 ⎠
from x is coming = positive
= 0.6 mJ
∴
and after closing the switch
Charge will flow from a to b, (A) is right
answer.
1 1 CnetV2 = (2 × 10–6) (30)2 2 2
∴
= 0.9 mJ
2.
Final charge on B = Final charge on A, B is
wrong answer
∴ Heat generated H = U – (Uf – Ui) = 0.3 mJ and
∴
charge flown through the switch is 60 µC.
= E × x. (D) is right answer.
[B,D]
8.
Potential difference across C2 = 0 V ∴ Charge stored on C1 = 40µC
and Charge stored on C2 = 0 µC 40V 2Ω
2Ω 40V
1Ω 20V 120V
3.
[A,C,D]
4.
[A,D]
5.
[A,C]
6.
[A,B,C,D]
7.
[A,D]
Q+x Q
Q–x Q
Q +x
–x Q
x
60V 3Ω C2
C1
During the process energy received by battery
[A,B,C,D]
Potential difference across C1 = 20 V
20V 1Ω
C
Applying kirchoff
Energy stored in all the capacitors before closing the
Uf =
–Q+x
a x b E x
U = 0.6 mJ
Ui =
Q–x
3Ω 60V
Q/C
x
x=
Q ×C=Q C
Total charge on plate x = Q + x = Q + Q = 2Q Total charge on plate y = Q – x =Q–Q=0 Cell will supply =
Q2 Q ×x= C C
Page #31
DPP No. #25 TO 31/HOMEWORK/17/05/2013
= 9.
Target IIT-JEE 2014
At a distance h below the top surface of slab
C 2e 2 = Ce2 C
Eslab =
[A,B,C]
ρ(D − 2h ) 2ε 0
E = Esheet + Eslab =
=
σ ρ(D − 2h ) + 2ε 0 2ε 0
σ + ρ(D − 2h ) 2ε 0
As distance between plate decreases C is higher on
At a distance h below the bottom surface of the slab =
that point.
−σ ρD ρD − σ + = 2ε 0 2ε 0 2ε 0
∴
More charge will accumulate.
∴
a is correct
∴
E =
14.
15. V when distance between plate is less, E is d 16.
higher.
17.
[A,B,C,D] [A,B] [B,C] [B]
10. [A,B]
Electric field between plates is uniform hence
11.
[A, C]
electrostatic force due to this field is constant. Angular
12.
[B,C]
frequency does not change by constant force.
13.
[A,B,D]
Equilibrium position will be at the point where 2kX0
At a distance h above the sheet
=q×
E = Esheet + Eslab −σ ρD ρD − σ = + = 2ε 0 2ε 0 2ε 0
σ σq X0 = ∈0 2k ∈0
X0 is also the amplitude of oscillation. 18.
[A,B,D]
19.
[A,D]
Page #32
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