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?
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(√nR−√mr)2+2√mnRr
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
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?
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
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?
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
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
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
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
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)=V−v3R× 2R
=V−2v3=v3
Question 10
The effective resistance between points P and Q of the electrical circuit shown in figure is
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
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=250−93.7=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?
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
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?
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
SOLUTION
Solution : C
rR=l1−l2l2
here R=10 ohm
l1 = 560 cm
l2 =560-60=500 cm
r=R × (l1−l2l2)
=10 × (560−500500)
=1.2 ohm