# Free Objective Test 01 Practice Test - 11th and 12th

### Question 1

Seven resistors each of resistance 5 ohm are connected as shown in figure. The equivalent resistance between points A and B is

#### SOLUTION

Solution :B

The circuit can be redrawn as shown in figure. Let a bettery of emf E and of negligible resistance be connected as shown. The current in various branches are shown.Applying Kirchoff's second law to loops ACDA,CBDC and ADBGFA we have

5i2+5i3−10i1=0

10(i2−i3)−5(i1+i3)−i3=0

10i1+5(i1+i3)−E=0

equation (i) and (ii) give i2=3i12 and i3=i12. If R is the effective resistance between A and B

then R=Ei1+i2=2E5i1

from equation (iii) we have

E=15i1+5i3=35i12

R=2E5i1=2∗35i15i1 ∗2=7 Ohm

### Question 2

cell supplies a current of 0.9 A through a 2 ohm resistor and current of 0.3 A through a 7 ohm resistor. What is the internal resistance of the cell?

#### SOLUTION

Solution :A

If E the emf of the cell and r its internal resistance then

E2+r=0.9 and

E7+r=0.3

dividing the two equations we get

7+r2+r=93

r=0.5 ohm

### Question 3

The deflection in a moving cell galvanometer falls from 50 to 10 divisions when a shunt of 12 ohm is connected across it. The resistance of the galvanometer coil is

#### SOLUTION

Solution :C

If k is the current sensitivity of the galvanometer the current in the galvanometer is

lg=50 K

when a shunt S is connected across it the current through the galavnometer becomes

l′g=lgSG+S=10 K

where G is the resistance of the galvanometer dividing the equations we get

G+SS=5

which gives G=4× S=4× 12=48 ohm

### Question 4

An electric kettle has two coils. When one coil is connected to the AC mains the water in the kettle boils in 10 minutes. When the other coil is used the same quantity of water takes 15 minutes to boil. How long will it take for the same quantity of water to boil if the two coils are connected in parallel?

#### SOLUTION

Solution :A

Let H be the amount of the heat energy needed to boil the given quantity of water. If R1 and R2 are the resistances of coils and V is the applied voltage then

H=V2t1R1=V2t2R2

R2R1=t2t1=1510=32

when the coils are connected in parallel the combined resistance is given by

1R=1R1+1R2=R1+R2R1R2

R=R1R2R1+R2

If the water takes t minutes to boil then

v2tR=V2t1R1

t=t1×RR1=t1× R1R2R1+R2×1R1=t1×1(R1R2+1)=10 × 1(23+1)=6 min

### Question 5

An electric bell has a rating of 500 W 100 V. It is used in a circuit having a 200 V supply. What resistance must be connected in series with the bulb so that it delivers 500 W?

#### SOLUTION

Solution :B

The current flowing in the bulb of 500 W operating at 100 V is

I=500100=5 A

resistance of bulb =1005=20 ohm say R1 to deliver 500 W the current in the bulb must remain 5 A when it is operated with 200 V supply the resistance R to be connected in series for this purpose is given by

200R1+R2=5

R1+R2=40

R1+20=40

R=20 ohm

### Question 6

The current voltage (I – V) graphs for a given metallic wire at two different temperatures T1 and T2 are shown in figure. If follows from the graphs that

#### SOLUTION

Solution :B

It is clear from figure that at a given voltage V0 the current I1 in the wire at temperature T1 is greater than the current I2 in the wire at temperature T2, Therefore the resistance of the wire at temperature T1 is less than that of temperature T2. This can happen if T1 is less than T2 because the resistance of wire increases with increase in temperature

### Question 7

A wire of resistance 5 ohm is drawn out so that its new length is three times its original length. What is the resistance of the new wire ?

#### SOLUTION

Solution :D

We know that Volume=Massdensity

Let m be the mass of the given wire, d its density and ρ its resistivity.

These quantities remain unchanged when the wire is drawn out.

Let l1 be the length of the smaller wire and r1 its corresponding radius of cross section. If l2 and r2 represent the length and radius of cross section of the elongated wire,

⟹m=π r21l2 d for smaller wire and for elongated wire, m=π r22l2 d.

Since mass and density are constant

⟹r21l1=r22l2

If R1 is the resistance of the smaller wire and R2 is the resistance of the elongated wire, we have,

R1=ρl1π r21

R2=ρl2π r22

Now, R2R1=l2l1× r21r22

It is given that, l2=3l1.

Therefore r22=r213

Substituting the values,

R2R1=9

Since, R1=5 Ω

R2=45 Ω.

### Question 8

What is the equivalent resistance of the network shown in figure? The numbers indicate resistance in ohms

#### SOLUTION

Solution :B

The network consists of four loops connected in series. Each loop consists of two 1 ohm resistors in series connected in parallel with two ohm resistors in series. Thus the effective resistance R of each loop is given by 1R = 12+14 giving R = 43 ohm. Therefore the equivalent resistance of the network = resistance of four loops connected in series each having an effective resistance of 43 ohm = 4× 43 = 163 ohm

### Question 9

The current I in the circuit shown in figure is

#### SOLUTION

Solution :C

The equivalent resistance is given by

1R=160+130=360

R = 20 ohm. Therefore I=220 =110A

### Question 10

Which of the following statements is correct about the circuit shown in figure where 1 ohm and 0.5 ohm are the internal resistances of the 6 V and 12 V batteries respectively?

The potential at point P is 6 V

The potential at point Q is +0.5 V

If a voltmeter is connected across the 6 V battery it will read 7 V

If a voltmeter is connected across the 6 V battery it will read 5 V

#### SOLUTION

Solution :C

Total resistance = 4 + 1 + 0.5 + 0.5 = 6 ohm

Net voltage in the circuit is 6 V

Current l=66=1 A in the anticlockwise direction

VPR=1× 4=4 V

Since R is connected to earth VR=0 hence Vp=4 V

VSQ=0.5× 1 =0.5 V

S is at a higher potential than Q

VQ=−0.5 V

Current is being forced into the 6 V battery in the opposite direction.Hence V6 =E+lr=6+1 X

1 = 7 V

### Question 11

In the potentiometer circuit shown in figure the internal resistance of the 6 V battery is 1 ohm and the length of the wire AB is 100 cm. When AD = 60 cm the galvanometer shows no deflection. The emf of cell C is (the resistance of wire AB is 2 ohm)

0.7 V

0.8 V

0.9 V

1 V

#### SOLUTION

Solution :C

Current in the circuit due to 6 V battery is

l=61+5+2=34A

Now emf of cell C =potential difference across AD

=34× 3 × 60100=0.9 V

### Question 12

The voltage across a bulb is decreased by 2%. Assuming that the resistance of the filament remains unchanged the power of the bulb will

#### SOLUTION

Solution :C

Power P=v2R

Therefore △ P=2 V△ VR

Hence △ PP=2△ VV

Now △ VV=−2%

Therefore △ PP=−2× 2=−4% i.e. the power decreases by 4%

### Question 13

Figure shows a network of resistances connected to a 2 V battery. If the internal resistance of the battery is negligible current I in the circuit is

#### SOLUTION

Solution :B

Because PQ=RS the circuit is balanced Wheatstone's bride.Therefore potential at B = potential at D. Hence no current flows through the 5 ohm resistor. It is therefore not effective The equivalent resistance between A and C is the resistance of a parallel combination of P + R = 2 + 4 = 6 ohm and Q + S = 4 + 8 = 12 ohm which is given by

1R+16+112

R = 4 ohm

Therefore current l is

l=24=0.5 A

### Question 14

A battery of emf E and internal resistance r is connected to a resistor of resistance r1 and Q joules of heat is produced in a certain time t. When the same battery is connected to another resistor of resistance r2 the same quantity of heat is produced in the same time t the value of r is

#### SOLUTION

Solution :D

In the first case the current in the circuit is

l1=Er1+r

Q1=l2 r1 t=(Er1+r)2 ×r1× t

In the second case

Q2=(Er1+r)2 ×r2× t

equating above equations we get

r1(r1+r)2=r2(r2+r)2

r1(r2+r)2=r2(r1+r)2

r1(r22+2 r r2+r2)=r2(r21+2r r1+r2)

r2(r1−r2)=r1r2(r1−r2)

r=√r1r2

### Question 15

A uniform wire of resistance 4 ohm is bent into the form of a circle of radius r. A specimen of the same wire is connected along the diameter of the circle. What is the equivalent resistance across the ends of this wire?

#### SOLUTION

Solution :A

Circumference of the circle = 2 π r

Therefore the resistance per unit length of the wire = R2 π r

where r= 4 ohm is the resistance of wire

Now the length of the specimen connected along the diameter = 2r

Therefore the resistance of this specimen is

R1=R2π r× 2 r =Rπ

also the resistance of each semicircle is

R2=R2

equivalent resistance R' across the specimen is given by

1R′=2R+2R+πR=4+πR

R′=R4+π=44+π ohm