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

Question 1

A battery of internal resistance $4 Ω$ is connected to the network of resistances as shown. In order to give the maximum power to the network, the value of R (in $Ω$ ) should be

\frac{4}{9}

\frac{8}{9}

C. 2

D. 18

SOLUTION

Solution : C

The equivalent circuit becomes a balanced wheatstone bridge

For maximum power transfer, external resistance should be equal to internal resistance of source
$\Rightarrow \frac{(R + 2 R) (2 R + 4 R)}{(R + 2 R) + (2 R + 4 R)} = 4$ i.e. $\frac{3 R \times 6 R}{3 R + 6 R} = 4$ or $R = 2 Ω$

Question 2

The scale of a galvanometer of resistance $100 Ω$ contains 25 divisions. It gives a deflection of one division on passing a current of $4 \times 10^{- 4} A$ . The resistance in ohms to be added to it, so that it may become a voltmeter of range 2.5 volt is

A. 100

B. 150

C. 250

D. 300

SOLUTION

Solution : B

Current sensitivity of galvanometer $= 4 \times 10^{- 4}$ Amp/div.
So full scale deflection current $(i_{g}) = C u r r e n t s e n s i t i v i t y \times T o t a l n u m b e r o f d i v i s i o n = 4 \times 10^{- 4} \times 25 = 10^{- 2} A$
To convert galvanometer in to voltmeter, resistance to be put in series is
$R = \frac{V}{i_{g}} - G = \frac{2.5}{10^{- 2}} - 100 = 150 Ω$

Question 3

When the two identical cells are connected either in series or in parallel across a 4 ohm resistor they send the same current through it. The internal resistance of the cell is

1.2 Ω

2 Ω

4 Ω

4.8 Ω

SOLUTION

Solution : C

In series the current through the resistor R is $\frac{2 E}{R + 2 r}$
In parallel connection, current through the resistor R is $\frac{E}{R + \frac{r}{2}}$
$I = \frac{2 E}{R + 2 r} = \frac{E}{R + \frac{r}{2}} 2 R + r = R + 2 r R = r r = 4 Ω$

Question 4

The potential difference across the terminals of a battery is 50V when 6A of current is drawn and 60V when 1A is drawn. The emf and internal resistance of the battery are

A. 62V, 2

Ω

B. 63V, 1

Ω

C. 61V, 1

Ω

D. 64V, 2

Ω

SOLUTION

Solution : A

V= E - Ir
50 = E - 6r $\to$ (1)
60 = E - 1r $\to$ (2)
(2) - (1) 10 = 5r; r = 2 $Ω$
E = 62V

Question 5

24 cells of each emf 4V are connected in series. Among them if 5 cells are connected wrongly, the effective emf of the combination is

A. 18V

B. 56V

C. 24V

D. 4V

SOLUTION

Solution : B

$E^{'} = (N - 2 n) E = (24 - 2 \times 5) E = 14 E = 14 \times 4 E^{'} = 56 V$

Question 6

36 cells each of e.m.f. 2 volt and internal resistance 0.5 ohm are connected in mixed grouping so as to deliver maximum current to a bulb of resistance 2 ohm. How are they connected. What is the maximum current through the bulb (n=no. of bulbs in series and, m = no. of bulbs in parallel)

A. n=12 ; m=3; 6 A

B. n = 9; m=4; 5A

C. n=5; m=7’ 6A

D. n=7; m=4; 8A

SOLUTION

Solution : A

To deliver maximum current, the external resistance should be equal to internal resistance.
mn = 36; Rm = r n
$\frac{m}{n} = \frac{1}{4} = \frac{2}{8} = \frac{3}{12} \frac{m}{n} = \frac{r}{R} = \frac{0.5}{2} = \frac{1}{4} l_{m a x} = \frac{m n E}{2 n r} = \frac{3 \times 12 \times 2}{2 \times 12 \times 0.5} 6 A \to (1)$

Question 7

In the given circuit, if the galvanometer shows zero reading, then
x= ___ ohms

A. 20

B. 30

C. 50

D. 100

SOLUTION

Solution : A

$V_{x} = 2 V V_{100} = 10 V I = \frac{V_{100}}{R} = \frac{10}{100} = 0.1 A V_{x} = I X 2 = 0.1 X X = 20 Ω$

Question 8

When resistance of 30 $Ω$ is connected to a battery, the current in the circuit is 0.4 A. if this resistance is replaced by 10 $Ω$ resistance, the current is 0.8 A. Then e.m.f. and the internal resistance of thebattery are

10 V, 16 Ω

3 V, 20 Ω

16 V, 10 Ω

24 V, 32 Ω

SOLUTION

Solution : C

$E = I_{1} [R_{1} + r] = I_{2} [R_{2} + r] 0.4 [30 + r] = 0.8 [10 + r] 30 + r = 20 + 2 r r = 10 E = 0.4 [30 + 10] E = 16 V$

Question 9

The P.D between A and B will be zero when the resistance R is

A. 4.5 ohm

B. 12 ohm

C. 9 ohm

D. 6 ohm

SOLUTION

Solution : A

From the circuit
$\frac{15}{4} = \frac{R + 3}{2} 15 = 2 R + 6 R = \frac{9}{2} = 4.5 Ω$

Question 10

A cell is connected in the secondary circuit of a potentiometer and a balance point is obtained at 84 cm When the cell is shunted by a $5 Ω$ resistor, the balance point changes to 70 cm. If the $5 Ω$ resistor is replaced with $3 Ω$ resistor then the balance point will be at

A. 53 cm

B. 63 cm

C. 42 cm

D. 112 cm

SOLUTION

Solution : B

$r = R [\frac{l_{1} - l_{2}}{l_{2}}] = R^{1} [\frac{l_{1} - l_{2}^{1}}{l_{2}^{1}}] 5 [\frac{84 - 70}{70}] = 3 [\frac{84 - l_{2}^{1}}{l_{2}^{1}}] \frac{l_{2}^{1} \times 14}{14 \times 3} = 84 - l_{2}^{1} l_{2}^{1} + \frac{l_{2}^{1}}{3} = 84 4 l_{2}^{1} = 84 \times 3 l_{2}^{1} = 63 c m$

Question 11

Two resistances X and Y are in the left and right gaps of a meter bridge. The balance point is 40 cm from left.Two resistances of $10 Ω$ each are connected in series with X and Y separately. The balance point is 45 cm. The value of X and Y are

12 Ω, 8 Ω

4 Ω, 6 Ω

8 Ω, 12 Ω

12 Ω, 16 Ω

SOLUTION

Solution : C

$\frac{x}{y} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3} \to (1) \frac{x + 10}{y + 10} = \frac{45}{55} = \frac{9}{11} \frac{\frac{2}{3} y + 10}{y + 10} = \frac{9}{11} \frac{2}{3} \times 11 y + 110 = 9 y + 90 20 = 9 y - \frac{22}{3} y 20 \times 3 = 5 y y = 12 Ω x = \frac{2}{3} \times 12 = 8 Ω$

Question 12

In a metre bridge experiment, null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y?

A. 50 cm

B. 80 cm

C. 40 cm

D. 70 cm

SOLUTION

Solution : A

$\frac{X}{Y} = \frac{20}{100 - l} = \frac{20}{80} = \frac{1}{4} Y = 4 X \frac{4 X}{Y} = \frac{l^{'}}{100 - l^{'}}; l^{'} = 50 c m$

Question 13

In an experiment for calibration of voltmeter, a standard cell of emf 1.5V is balanced at 375cm length of potentiometer wire. The potential difference across a resistance in the circuit is balanced at 1.25 m. If a voltmeter is connected across the same resistance, it reads 0.6V. The error in the voltmeter is

A. 0.1V

B. 0.025V

C. 0.235V

D. 0.18V

SOLUTION

Solution : A

$i ρ = \frac{E}{l} = \frac{1.5}{3.75} \frac{V}{m} V^{1} = (i ρ) l = \frac{1.5}{3.75} \times 1.25 V^{1} = 0.5 V E r r o r = V - V^{1} = 0.6 - 0.5 = 0.1 V$

Question 14

A potentiometer wire of length 1 m and resistance $10 Ω$ is connected in series with a cell of e.m.f 2 volt with internal resistance $1 Ω$ and a resistance box including a resistance R. If the potential difference between the ends of the wire is 1 mV, then the value of R will be

9989 Ω

10000 Ω

19989 Ω

20000 Ω

SOLUTION

Solution : C

$i = \frac{V_{2}}{R_{1}} = \frac{1 \times 10^{- 3}}{10} = 10^{- 4} A E = i [R + r + R_{L}] 2 = 10^{- 4} [R + 1 + 10] 20000 = R + 11 R = 20000 - 11 = 19989 Ω$

Question 15

A $1 Ω$ resistance is in series with an ammeter which is balanced by 75 cm of potentiometer wire. A standard cell of 1.02 V is balanced by 50 cm. The ammeter shows a reading of 1.5A. The error in the ammeter reading

A. 0.002 A

B. 0.03 A

C. 1.01 A

D. no error

SOLUTION

Solution : B

$i p = \frac{V}{l} = \frac{1.02}{50}$
Voltage across ammeter = $\frac{V}{l} l_{1}$
$= \frac{1.02}{50} \times 75 = \frac{1.02}{2} \times 3 = 0.51 \times 3$
The true current $I_{A}^{1} = \frac{V_{A}}{R} = \frac{1.53}{1} I_{A}^{1} = 1.53 A E r r o r = I_{A}^{1} - I_{A} = 1.53 - 1.5 = 0.03 A$

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

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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