Free Objective Test 02 Practice Test - 11th and 12th 

Question 1

You are given 48 cells each of emf 2 V and internal resistance 1 ohm. How will you connect them so that the current through an external resistance of 3 ohm is the maximum?

A. 8 cells in series 6 such groups in parallel
B. 12 cells in series 4 such groups in parallel
C. 16 cells in series 3 such groups in parallel
D. 24 cells in series 2 such groups in parallel

SOLUTION

Solution : B

Let m cells be connected in series and n such groups are connected in parallel
If the emf of each cell is E and internal resistance r then the total emf of m in series is mE and  the total internal resistance is mr. When n such groups are in parallel the effective internal resistance is mr/n. Then the current through an external resistance R is 
l=mER+mrn=mnEnR+mr=mnE(nRmr)2+2mnRr
Now i will be maximum if the denominator is the minimum i.e. if  nR=mr
Using  R= 3 ohm and r=1 ohm we have 3n = m
But  mn = 48
therefore m×m3=48
which gives m= 12
thus n=4

Question 2

In the electrical circuit shown in figure the current through the galvanometer will be zero if the value of resistance x is 

A. 4 ohm
B. 8 ohm
C. 16 ohm
D. 24 ohm

SOLUTION

Solution : D

The given circuit is Wheatstone's bridge.The current through the galvanometer will be  zero if the bridge is balanced 
PQ+RS where P = 2+3 =5 ohm, Q = 10 ohm and R=4 ohm
The value of S is given by 
510=4S
or S= 8  ohm
thus the effective resistance of the parallel combination of 12 ohm and x ohm must be 8 ohm
therefore 
112+1x=18
which gives x = 24 ohm

Question 3

Six resistors each of resistance R are connected as shown in figure. What is the effective resistance between points A and B? 

A. R3
B. R  
C. 3R 
D. 6R

SOLUTION

Solution : A

Resistance R,(R+R)= 2R, (R+R)=2R and R are in parallel 
Hence effective resistance is given by
1Reff=1R+12R+12R+1R
which gives Reff=R3

Question 4

A constant voltage is applied between the two ends of a metallic wire. Some heat is developed in it. The heat developed is doubled if

A. both the length and the radius of the wire are halved
B. both the length and the radius of the wire are doubled
C. the radius of the wire is halved
D. the length of the wire is doubled

SOLUTION

Solution : B

Q=V2R
But R=plπ r2
Therefore Q=(π V2p)r21
Q is doubled if both l and r are doubled

Question 5

Eight cells marked 1 to 8 each of emf 5 V and internal resistance 0.2 ohm are connected as shown in figure. What is the reading of the ideal voltmeter V?

A. 40 V  
B. 20 V
C. 5 V   
D. 0

SOLUTION

Solution : D

The emfs of cells connected in reverse polarity cancel each other. Hence cells marked 2, 3 and 4 together cancel the effect of cells marked 5, 6 and 7 and the circuit reduces to that shown in figure. Now cells 1 and 8 are in reverse polarity. Hence the voltmeter reading = 5 – 5 = 0 V

Question 6

A piece of copper and another of germanium are cooled from room temperature to 40 K. The resistance of

A. each of them decreases 
B. each of them increases
C. copper increases and of germanium decreases  
D. copper decreases and of germanium increases

SOLUTION

Solution : D

copper is a conductor whereas Germanium is a semi conductor

Question 7

A steady current flows in a metallic conductor of non uniform cross-section. The quantity that remains constant along the length of the conductor is

A. current, electric field and drift speed   
B. drift speed only
C. current and drift speed  
D. current only

SOLUTION

Solution : D

The drift speed depends on A the cross sectional area of the conductor but the current is independent of A

Question 8

In the circuit shown in figure the current through

           

A. the current in 3 ohm resistor is 0.5 A
B. the current in 3 ohm resistor is 0.25 A
C. the current in 4 ohm resistor is 0.5 A 
D. the current in 4 ohm resistor is 0.25 A

SOLUTION

Solution : D

The equivalent resistance between points A and B to the right of AB is 4 ohm. Therefore total resistance  =3+4+2= 9 ohm. Current I =99=1 A. This current is equally divided in the 8 ohm resistor between A and B and the remainder 8 ohm resistor. Hence current in  AC =0.5 A. This current is equally divided between the 8 ohm resistor in CD and the circuit to the right of CD. Therefore current in the 4 ohm resistor  = 0.25  A

Question 9

In the given circuit with steady current the potential drop across the capacitor must be 

A. V
B. V2
C. V3
D. 2V3

SOLUTION

Solution : C



In the steady state no current flows in the branch containing the capacitor .Thus the current say l flows in the  branches containing R and 2R. Applying Kirchoff's second rule to the loop abcdefa
2V-I(2R)-IR-V=0
l=V3R
Potential drop across capacitor  =2V-V-l(2R)=Vv3R× 2R
=V2v3=v3
 

Question 10

The effective resistance between points P and Q of the electrical circuit shown in figure is 

A. 2Rr(R+r)
B. 8R(R+r)(3R+r) 
C. 2r + 4R
D. 5R2+2r

SOLUTION

Solution : A



The branches ABPQ and PQCD are a balanced Wheatstone's bridge. Therefore resistances between E and E and between F and G do not contribute and the circuit simplies to the one show in figure. The effective resistance Re between P and Q is given by
1Re=14R+12r+14R
which gives Re=2Rr(R+r)

Question 11

 A  100 W bulb B1 and two 60 W bulbs B2 and B3 are connected to a 250 V source as shown in figure. W1, W2and W3are the output powers of the bulbs B1, B2 and B3 respectively then 

A. W1> W2 = W3
B. W1> W2> W3
C. W1< W2 = W3
D. W1< W2< W3

SOLUTION

Solution : D

The resistances of bulbs B1,B2 and B3 respectively are 
R1v2w1=(250)2100=625 ohm
R2=(250)260=1042 ohm =R3
Voltage across B3 is V3= 250 V
Voltage across B1 is V1=VR1(R1+R2)=250× 625(625+1042)= 93.7 V
Voltage across B2 is V2=25093.7=156.3 V
Power output W1=V2R1=(93.7)2625=14 W
W2=(156.3)21042 = 23 w
W3=(250)21042 = 60 w
Hence W1< W2< W3

Question 12

In the meter bridge experiment shown in figure, the balance length AC corresponding to null deflection of the galvanometer is x. What would be the balance length if the radius of the wire AB is doubled? 

A. x2
B. x  
C. 2x
D. 4x

SOLUTION

Solution : B

The condition for no delflection of the galvanometer is 
R1R2=RACRCB
where RAC and RCB are the resistances of the bridge wire of length AC and CB respectively. If the radius of the wire AB is doubled the ratio
RACRCB will remain unchanged.Hence the balance  length will remain  the same 

Question 13

Three resistances of equal value are connected in four different combinations as shown in figure. Arrange them in increasing order of power dissipation 

A. III < II < IV < I    
B. II< III <IV < I
C. I< IV < III < II
D. I < III < II < IV

SOLUTION

Solution : B

Let R be the value of each resistance. The resistances of combinations I, II, III and IV are  3R, 2R3, R3 and 3R2 respectively. Now power dissipation is inversely proportional to resistance

Question 14

Which of the circuits shown in figure can be used to verify Ohm’s law? 

A. a
B. b
C. c
D. d

SOLUTION

Solution : A

The voltmeter must be connected in parallel with the resistor and the ammeter must be connected in series with the resistor

Question 15

The balancing length for a cell is 560 cm in a potentiometer experiment. When an external resistance of 10 ohm is connected in parallel to the cell the balancing length changes by 60 cm. The internal resistance of the cell in ohms is

A. 1.6 
B. 1.4
C. 1.2
D. 0.12

SOLUTION

Solution : C

rR=l1l2l2
here R=10  ohm
l1 = 560 cm
l2 =560-60=500 cm
r=R × (l1l2l2)
=10 × (560500500)
=1.2 ohm