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

Question 1

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

$\frac{r}{R} = \frac{l_{1} - l_{2}}{l_{2}}$
here R=10 ohm
$l_{1}$ = 560 cm
$l_{2}$ =560-60=500 cm
$r = R \times (\frac{l_{1} - l_{2}}{l_{2}})$
$= 10 \times (\frac{560 - 500}{500})$
=1.2 ohm

Question 2

Eels are able to generate current with biological cells called electroplaques. The electroplaques in an eel are arranged in 100 rows, each row stretching horizontally along the body of the fish containing 5000 electroplaques. The arrangement is suggestively shown below. Each electroplaques has an emf of 0.15 V and internal resistance of 0.25 $Ω$ . The water surrounding the eel completes a circuit between the head and its tail. If the water surrounding it has a resistance of 500 ohm, the current an eel can produce in water is about

A. 1.5 A

B. 3.0 A

C. 15 A

D. 30 A

SOLUTION

Solution : A

Given problem is the case of mixed grouping of cells

So total current produced $i = \frac{n E}{R + \frac{n r}{m}}$
Here m = 100, n = 5000, R = 500 $Ω$
E = 0.15V and r = 0.25 $Ω$

$\Rightarrow i = \frac{5000 \times 0.15}{500 + \frac{5000 \times 0.25}{100}} = \frac{750}{512.5} \approx 1.5 A$

Question 3

In Wheatstone's bridge P=9ohm, Q=11ohm, R=4ohm and S=6ohm. How much resistance must be put in parallel to the resistance S to balance the bridge

A. 24 ohm

\frac{44}{9}

ohm

C. 26.4 ohm

D. 18.7 ohm

SOLUTION

Solution : C

$\frac{P}{Q} = \frac{R}{S^{'}}$ (For balancing bridge)
$\Rightarrow S^{'} = \frac{4 \times 11}{9} = \frac{44}{9}$
$\Rightarrow \frac{1}{S^{'}} = \frac{1}{r} + \frac{1}{6}$
$\Rightarrow \frac{9}{44} - \frac{1}{6} = \frac{1}{r}$
$\Rightarrow r = \frac{132}{5} = 26.4 Ω$

Question 4

For comparing the e.m.f.'s of two cells with a potentiometer, a standard cell is used to develop a potential gradient along the wires. Which of the following possibilities would make the experiment unsuccessful.

A. The e.m.f. of the standard cell is larger than the E e.m.f.'s of the two cells

B. The diameter of the wires is the same and uniform throughout

C. The number of wires is ten

D. The e.m.f. of the standard cell is smaller than the e.m.f.'s of the two cells

SOLUTION

Solution : D

The emf of the standard cell must be greater than that of experimental cells, otherwise balance point is not obtained.

Question 5

In the circuit shown below $E_{1} = 4.0 V, R_{1} = 2 Ω, E_{2} = 6.0 V, R_{2} = 4 Ω$ and $R_{3} = 2 Ω$ . The current $I_{1}$ is

A. 1.6 A

B. 1.8 A

C. 1.25 A

D. 1.0 A

SOLUTION

Solution : B

Applying Kirchhoff's law for the loops (1) and (2) as shown in figure For loop(1)

$- 2 i_{1} - 2 (i_{1} - i_{2}) + 4 = 0 \Rightarrow 2 i_{1} - i_{2} = 2 . . . . (i)$
For loop (2)
$- 2 (i_{1} - i_{2}) + 4 i_{2} - 6 = 0 \Rightarrow - i_{1} + 3 i_{2} = 3 . . . (i i)$
On solving equation (i) and (ii) $i_{1} = 1.8 A .$

Question 6

The figure shows a circuit diagram of a ‘Wheatstone Bridge’ to measure the resistance G of the galvanometer. The relation $\frac{P}{Q} = \frac{R}{G}$ will be satisfied only when

A. The galvanometer shows a deflection when switch S is closed

B. The galvanometer shows a deflection when switch S is open

C. The galvanometer shows no change in deflection whether S is open or closed

D. The galvanometer shows no deflection

SOLUTION

Solution : C

In balance condition, no current will flow through the branch containing S.

Question 7

What will be the equivalent resistance of circuit shown in figure between points A and D

10 Ω

20 Ω

30 Ω

40 Ω

SOLUTION

Solution : C

The equivalent circuit of above fig between A and D can be drawn as

So $R_{e q} = 10 + 10 + 10 = 30 Ω$

Question 8

In the network shown in the figure each of resistance is equal to $2 Ω$ . The resistance between A and B is

1 Ω

2 Ω

3 Ω

4 Ω

SOLUTION

Solution : B

Taking the portion COD is figure to outside the triangle (left), the above circuit will be now as resistance of each is $2 Ω$ the circuit will behaves as a balanced wheatstone bridge and no current flows through CD. Hence $R_{A B} = 2 Ω$

Question 9

A primary cell has an emf of 1.5 volts, when short-circuited it gives a current of 3 amperes. The internal resistance of the cell is

A. 4.5 ohm

B. 2 ohm

C. 0.5 ohm

\frac{1}{4.5}

ohm

SOLUTION

Solution : C

$i_{S C} = \frac{E}{r} \Rightarrow 3 = \frac{1.5}{r} \Rightarrow r = 0.5 Ω$

Question 10

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 11

A torch bulb rated as 4.5 W, 1.5 V is connected as shown in the figure. The emf of the cell needed to make the bulb glow at full intensity is

A. 4.5 V

B. 1.5 V

C. 2.67 V

D. 13.5 V

SOLUTION

Solution : D

When bulb glows with full intensity, potential difference across it is 1.5 V. So current through the bulb and resistance of $1 Ω$ are 3 A and 1.5 A respectively. So main current from the cell i = 3 + 1.5 = 4.5 A. By using $E = V + i r \Rightarrow E = 1.5 + 4.5 \times 2.67 = 13.5 V$ .

Question 12

A group of N cells whose emf varies directly with the internal resistance as per the equation $E_{N} = 1.5 r_{N}$ are connected as shown in the following figure. The current i in the circuit is

A. 0.51 amp

B. 5.1 amp

C. 0.15 amp

D. 1.5 amp

SOLUTION

Solution : D

$i = \frac{E_{e q}}{r_{e q}} = \frac{1.5 r_{1} + 1.5 r_{2} + 1.5 r_{3} + . . . . . .}{r_{1} + r_{2} + r_{3} + . . . . . .} = 1.5$ amp.

Question 13

In the circuit shown here $E_{1} = E_{2} = E_{3} = 2 V$ and $R_{1} = R_{2} = 4 Ω$ . The current flowing between point A and B through battery $E_{2}$ is

A. Zero

B. 2 A from A to B

C. 2 A from B to A

D. None of these

SOLUTION

Solution : B

The equivalent circuit can be drawn as since $E_{1}$ and $E_{3}$ are parallely connected

So current $i = \frac{2 + 2}{2} = 2 A m p$ from A to B

Question 14

The magnitude and direction of the current in the circuit shown will be

\frac{7}{3}

A from a to b through e

\frac{7}{3}

A from b and a through e

C. 1.0 A from b to a through e

D. 1.0 A from a to b through e

SOLUTION

Solution : D

Current $i = \frac{10 - 4}{3 + 2 + 1} = 1 A$ from a to b via e

Question 15

A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to this current, the temperature of the wire is raised by $Δ T$ in time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. The temperature of wire is raised by same amount $Δ T$ in the same time t. The value of N is

A. 4

B. 6

C. 8

D. 9

SOLUTION

Solution : B

Heat $= m S Δ T = i^{2} R t$
Case I: Length (L) $\Rightarrow$ Resistance = R and mass = m
Case II: Length (2L) $\Rightarrow$ Resistance = 2R and mass = 2m
So $\frac{m_{1} S_{1} Δ T_{1}}{m_{2} S_{2} Δ T_{2}} = \frac{i_{1}^{2} R_{1} t_{1}}{i_{2}^{2} R_{2} t_{2}} \Rightarrow \frac{m S Δ T}{2 m S Δ T} = \frac{i_{1}^{2} R t}{i_{2}^{2} 2 R t} \Rightarrow i_{1} = i_{2} \Rightarrow \frac{(3 E)}{R} = \frac{(N E)}{2 R} \Rightarrow N = 6$

Free Objective Test 02 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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