If $$i_{1}=3\sin\omega t$$ and $$i_{2}=4\cos\omega t$$, then $$i_{3}$$ is

$$\displaystyle5\sin \left ( \omega t+53^{\circ} \right )$$

$$\displaystyle5\sin \left ( \omega t+37^{\circ} \right )$$

$$\displaystyle5\sin \left ( \omega t+45^{\circ} \right )$$

$$\displaystyle5\cos \left ( \omega t+53^{\circ} \right )$$

MCQ Questions for Class 12 Medical Physics Current Electricity Quiz 7

Which of the following shows energy changes that takes place when an incandescent bulb glows using a battery?

0%
Chemical>Electrical>heat>Light energy
0%
Electrical>Chemical>Light>heat energy
0%
Chemical>Electrical>Heat energy
0%
Electrical>Chemical>Light energy

A potentiometer circuit shown in the figure is set up to measure emf of cell E. As the point P moves from X to Y, the galvanometer G shows deflection always in one direction, but the deflection decreases continuously until Y is reached. The balance point between X and Y maybe obtained by

0%
Decreasing the resistance R and decreasing V
0%
Decreasing the resistance R and increasing V
0%
Increasing the resistance R and increasing V
0%
Increasing the resistance R and decreasing V

In specific resistance measurement of a wire using a meter bridge, the key k in the main circuit is kept open when we are not taking readings. The reason is

0%
the emf of cell will decrease.
0%
the value of resistance will change due to joule heating effect.
0%
the galvanometer will stop working.
0%
none of these.

$i_{1}=3\sin\omega t$ and

$i_{2}=4\cos\omega t$ , then

$i_{3}$ is

0%
$\displaystyle5\sin \left ( \omega t+53^{\circ} \right )$
0%
$\displaystyle5\sin \left ( \omega t+37^{\circ} \right )$
0%
$\displaystyle5\sin \left ( \omega t+45^{\circ} \right )$
0%
$\displaystyle5\cos \left ( \omega t+53^{\circ} \right )$

Explanation

From Kirchoff's current law,

$\displaystyle i_{3}=i_{1}+i_{2}=3\sin wt + 4\sin\left ( wt+90^{0} \right )$

$\displaystyle =\sqrt{3^{2}+4^{2}+2(3)(4)cos90^{0}}\sin (wt+\phi)$
where

$\displaystyle \tan \phi=\frac{4\sin90^{0}}{3+4\cos 90^{0}}=\frac{4}{3}$

$\displaystyle \therefore i_{3}=5\sin \left ( wt+53^{0} \right )$

Steady current flows in a metallic conductor of non-uniform cross-section. Which of the following quantities is a constant along the conductor ?

0%
Current
0%
Current density
0%
Electric field
0%
Drift speed

A voltmeter having a resistance of

$998\Omega$ is connected to a cell of emf

$2$ V and internal resistance

$2\Omega$ . The error in the measurement of emf will be.

0%
$4\times 10^{-1}$ V
0%
$2\times 10^{-3}$ V
0%
$4\times 10^{-3}$ V
0%
$2\times 10^{-1}$ V

Explanation

$I=\displaystyle \frac{2}{99}=\frac{2}{1000}=2\times 10^{-3}$ A

$=2$ mA
Internal resistance is

$2\Omega$
Hence, the measure of emf will be

$E=2-I\times 2$

$=2-2mA\times 2$
Therefore, error in measured value

$=4\times 10^{-3}$ V.

In a potentiometer of wire length

$l$ , a cell of emf V is balanced at a length

$\dfrac{l}{3}$ from the positive of the wire. For another cell of emf

$1.5 V$ , the balancing length becomes

0%
$\dfrac{l}{6}$
0%
$\dfrac{l}{2}$
0%
$\dfrac{l}{3}$
0%
$\dfrac{2l}{3}$
0%
$\dfrac{4l}{3}$

Explanation

$\dfrac{E_1}{L_1} = \dfrac{E_2}{L_2}$

$\dfrac{V}{l/3} = \dfrac{1.5 V} {x}$

$\therefore x = l/2$

The resistance of a bulb filament is

$100\Omega$ at a temperature at

$100^{o}C$ . If its temperature coefficient of resistance be

$0.005/^{o}C$ , then its resistance will become

$200\Omega$ at a temperature of

0%
$300^{o}C$
0%
$400^{o}C$
0%
$500^{o}C$
0%
$200^{o}C$

Explanation

$\textbf{Given : }$ $R_{100} = 100\Omega$ , $T_1 = 100^oC$ , $R' = 200\Omega$ , $\alpha = 0.005$ per $^oC$

$\textbf{Step 1: Relation between resistance and temperature }$

$R = R_{0}(1+\alpha (T_2-T_1))$

where $\alpha$ is the temperature coefficient of resistance.

So,

$R' = R_{100}(1+\alpha (T_2-T_1))$

$....(1)$

$\textbf{Step-2: Final temperature }$

Putting values of

$R_{100},T_1,R',\alpha$ in equation

$(1)$ , we get

$200 = 100(1+0.005 (T_2-100))$

Or,

$T_2 - 100 = 200$

$\Rightarrow\ \ \ T_2 = 300^oC$

Hence, Option A is correct.

Figure below shows a portion of an electric circuit with the currents in amperes and their directions. The magnitude and direction of the current in the portion

$PQ$ is

0%
$0 A$
0%
$3 A$ from $P$ to $Q$
0%
$4 A$ from $Q$ to $P$
0%
$6 A$ from $Q$ to $P$

Explanation

At junction the net current arriving should be equal to the net current leaving

The net current

$8+2=4+2+1+x$ ,

$x=3A$

Now at junction Q the net current arriving is

$3+3=6A$ . This will flow toward P.

The best instrument for accurate measurement of emf of a cell.

0%
Potentiometer
0%
Wheatstone Bridge
0%
Voltmeter
0%
Ammeter

In Wheatstone's bridge, three resistors P, Q, R are connected in three arms in order and

$4^{th}$ arm is formed by two resistors

$s_1$ and

$s_2$ connected in parallel. The condition for bridge to be balanced is

$\displaystyle\frac{P}{Q}=$ .

0%
$\displaystyle\frac{R(s_1+s_2)}{s_1s_2}$
0%
$\displaystyle\frac{s_1s_2}{R(s_1+s_2)}$
0%
$\displaystyle\frac{Rs_1s_2}{(s_1+s_2)}$
0%
$\displaystyle\frac{(s_1+s_2)}{Rs_1s_2}$

Explanation

Let the equivalent resistance of the resistors

$s_1$ and

$s_2$ connected in parallel be

$S$ .

Equivalent resistance in parallel combination

$\dfrac{1}{S} = \dfrac{1}{s_1}+\dfrac{1}{s_2}$

$\implies$

$S = \dfrac{s_1s_1}{s_1+s_2}$

Wheatstone bridge is balanced if

$\dfrac{P}{Q} = \dfrac{R}{S}$

$\dfrac{P}{Q} = \dfrac{R}{\dfrac{s_1s_2}{s_1+s_2}}$

$\implies$

$\dfrac{P}{Q} = \dfrac{R(s_1+s_2)}{s_1s_2}$

An electron in potentiometer wire experiences a force

$2.4\times 10^{-19}$ N. The length of potentiometer wire is

$6m$ . The e.m.f. of the battery connected across the wire is (electronic charge

$=1.6\times 10^{-19}$ C).

0%
$6$ V
0%
$9$ V
0%
$12$ V
0%
$15$ V

Explanation

Electric field in the wire

$E = \dfrac{F}{e}$

$\therefore$

$E = \dfrac{2.4\times 10^{-19}}{1.6\times 10^{-19}} = 1.5$

$N/C$

Length of the wire

$l = 6$ m

Thus e.m.f of the battery

$\mathcal{E} = El$

$\therefore$

$\mathcal{E} = 1.5\times 6 = 9$ V

A potentiometer wire of length

$10 m$ is connected in series with a battery. The e.m.f. of a cell balances against

$250 cm$ length of wire. If length of potentiometer wire is increased by

$1 m$ , the new balancing length of wire will be

0%
$2.00 m$
0%
$2.25 m$
0%
$2.50 m$
0%
$2.75 m$

Explanation

$l=10m$

Balancing length

$=250cm=2.5m$

Lets, voltage drop across

$10m=V\quad Volt$

$\implies$ Voltage drop across

$1m=\dfrac V{10}=0.v$

$\implies$ Voltage drop across

$2.5m=0.25V$

Now, we know the e.m.f of cell=

$0.25V$

Case=

$II$ When

$l=11m$

Lets, voltage drop across

$11m=V\quad Volt$

To drop V potential we use

$11m$ long wire.

To drop

$1V$ of potential we will find the balancing length at

$=\dfrac{}{V}m$

To drop

$0.25V$ balancing length will be

$=\dfrac11{V}\times 0.25V$

$=2.75m$

A student's

$9.0V, 7.5W$ portable radio was left on from

$9:00\ P.M.$ until

$3:00 A.M.$ How much charge passed through the wires?

0%
$18000C$
0%
$24000C$
0%
$6000C$
0%
$12000C$

Explanation

As we know,
Power,

$P = V\cdot I$

$I = \dfrac {P}{V}$

$\Rightarrow I = \dfrac {Q}{t} = \dfrac {P}{V}$

$\Rightarrow Q = \dfrac {P}{V} \times t = \dfrac {7.5}{9}\times 3600\times 6$

$= 18000\ C$

$[\because t = 9\ PM$ to

$3\ AM = 6\ hr = 6\times 3600s]$

A potentiometer is an ideal device of measuring potential difference because

0%
It uses a sensitive galvanometer
0%
It is an elaborate arrangement
0%
It has a long wire hence heat developed is quickly radiated
0%
It does not disturb the potential difference it measures

To make a battery of two cells, the negative terminal of one cell is connected to the negative terminal of the other cell.

0%
True
0%
False

In a potention meter of one metre length an unknown emf voltage source is balanced at 60 cm length of potention meter wire, while a 3v battery is balanced at 45 cm length. Then, the emf of the unknown voltage source is :

0%
3 V
0%
2.25 V
0%
4 V
0%
4.5 V

Explanation

Let

$l_1$ and

$l_2$ be the balancing length with the cells

$E_1$ and

$E_2$ respectively, then

$E_1 = xl_1$ and

$E_2 = xl_2$

$\dfrac {E_1}{E_2} = \dfrac {l_1}{l_2}$

$\dfrac {E_1}{3} = \dfrac {60}{35}$

$E_1 = 4V$

A cylindrical conductor is placed near other positively charged conductor. The net charge acquired by the cylindrical conductor will be :

0%
Positive only
0%
Negative only
0%
Zero
0%
Either positive or negative

In potentiometer experiment, null point is obtained at a particular point for a cell on potentiometer wire x cm long. If the length of the potentiometer wire is increased without changing the cell, the balancing length will (Driving source is not changed)

0%
increase
0%
decrease
0%
not change
0%
becomes zero

Explanation

Hint: The potential drop across a piece of uniform cross-section wire carrying a constant current is directly proportionate to its length, according to the principle of a potentiometer.

Explanation:
Step 1: Concept used:
according to the principle of a potentiometer, the potential drop across a piece of uniform cross-section wire carrying a constant current is directly proportionate to its length.

Step Finding value of length by given condition:

Let driving source

$=E_p$ , Resistance and length of potentiometer wire

$= R$ and

$L$ , Balancing length

$=l$ , cell voltage

$= V$

So at balancing point-

$\dfrac{V}{E_p}=\dfrac{l}{L}$ ...........(1)

After changing length of potentiometer wire

$=L_1$ , balancing length

$=l_1$

Now

$\dfrac{V}{E_p}=\dfrac{l_1}{L_1}$ .........(2)

Step 3: Conclusion:

By solving both equations, we get:

$l_1= l(\dfrac{L_1}{L})$

$L_1 >L \implies l_1>l$
Hence option A is correct.

Five cells each of internal resistance

$0.2 \Omega$ and emf

$2 V$ are connected in series with a resistance of

$4\Omega$ .The current through the external resistance is

0%
$4 A$
0%
$2 A$
0%
$1 A$
0%
$0.5 A$

Explanation

Since the five batteries and their internal resistances are connected in series to one another, the net emf will be

$nE$ and the internal resistance is

$nr$ .

Current through external resistance

$i=\dfrac{nE}{nr+R}=\dfrac{5\times 2}{5\times 0.2+4}$

$=2 A$

Ten identical batteries each of emf 2 V are connected in series to a

$8 \Omega$ resistor. If the current in the circuit is 2 A, then the internal resistance of each battery is

0%
$0.2 \Omega$
0%
$0.3 \Omega$
0%
$0.4 \Omega$
0%
$0.5 \Omega$
0%
$1 \Omega$

Explanation

$R_{eq} = 8 + 10r$ since they are connected in series.

Also effective voltage will become

$10 \times 2 V = 20V$

Now,

$i = \dfrac{V}{10r + 8}$

$2 = \dfrac{20}{8 + 10r}$

$8 + 10r = 10$

$r = 0.2 \ \ ohm$

In a potentiometer arrangement, a cell of emf

$1.5V$ gives a balance point at

$27cm$ length of wire. If the cell is replaced by another cell and balance point shifts to

$54cm$ , the emf of the second cell is

0%
$3V$
0%
$1.5V$
0%
$0.75V$
0%
$2.25V$

Kirchhoff's first law ie,

$\displaystyle \Sigma i=0$ at a junction is based on the law of conservation of

0%
Charge
0%
Energy
0%
Momentum
0%
Angular momentum

A potentiometer wire of length

$L$ and resistance

$10\Omega$ is connected in series with a battery of e.m.f.

$2.5\ V$ and a resistance in its primary circuit. The null point corresponding to a cell of e.m.f.

$1\ V$ is obtained at a distance

$L/2$ . If the resistance in the primary circuit is doubled then the position of new null point will be

0%
$0.4\ L$
0%
$0.5\ L$
0%
$0.6\ L$
0%
$0.8\ L$

Explanation

Initial current in the circuit

$= \dfrac {2.5V}{R + 10\Omega}$
Potential difference across

$\dfrac {L}{2}$ or

$5\Omega = \dfrac {2.5}{R + 10}\times 5 = 1V$
because a cell of

$1\ volt$ balances.
From this

$R = 2.5\Omega$

$\therefore$ When

$R$ is doubled,

$i = \dfrac {2.5}{5 + 10} A$ If this balances at a length

$x, \dfrac {2.5}{15} \times R' = 1\ V$ where

$R'$ is the resistance corresponding to the length

$x$ .

$\Rightarrow R' = \dfrac {15\times 1}{5/2} = 6\Omega$ i.e.,

$x = \dfrac {L\times 6}{10} = 0.6\ L$ .

If specific resistance of a potentiometer wire is

$10^{-7}\Omega m$ and current flowing through it is

$0.1\ amp$ ., cross-sectional area of wire is

$10^{-6}m^{2}$ then potential gradient will be

0%
$10^{-2}volt/m$
0%
$10^{-4}volt/ m$
0%
$10^{-6}volt/ m$
0%
$10^{-8}volt/ m$

Explanation

Resistance of the potentiometer wire

$R = \rho\dfrac{l}{A}$
Potential difference across the wire

$V = Il$
So. potential gradient

$\dfrac {V}{l} = \dfrac {IR}{l} = \dfrac {I\rho l}{Al}$
Or

$\dfrac{V}{l} = \dfrac{I\rho }{A}= \dfrac {0.1\times 10^{-7}}{10^{-6}} = 0.01 = 10^{-2}V/m$ .

36 identical cell each having emf 1.5 volt and internal resistance

$0.5 \Omega$ are connected in series with an external resistance of

$12 \Omega$ .If 6 cells are wrongly connected then current through the circuit will be

0%
1.2 A
0%
1 A
0%
2 A
0%
4 A

Explanation

Total resistance in the circuit is,

$R_T=36\times0.5+12=30\Omega$

Net EMF in the circuit is,

$E_T=(30-6)\times1.5=36V$

So, current in the circuit is,

$I=\dfrac{E_T}{R_T}=\dfrac{36}{30}=1.2A$

A potentiometer wire of length

$1m$ has a resistance of

$100\Omega$ . It is connected in series with a resistance and a battery of emf

$2V$ of negligible resistance. A source of emf

$10mV$ is balanced against a length of

$40cm$ of the potentiometer wire. What is the value of the external resistance.

0%
$890\Omega$
0%
$7900\Omega$
0%
$680\Omega$
0%
$740\Omega$

Explanation

Given: A potentiometer wire of length

$1m$ has a resistance of

$100\Omega$ . It is connected in series with a resistance and a battery of emf

$2V$ of negligible resistance. A source of emf

$10mV$ is balanced against a length of

$40cm$ of the potentiometer wire.

To find the value of the external resistance

Solution:

$J$ is current through the potentiometer wire then

$J=\dfrac E{R+10}=\dfrac 2{R+10}$

As the source of e.m.f

$E'=10mV=10\times10^{-3}V$ is balanced by a length of

$40cm$ of the potentiometer wire, it follows that

$10\times10^{-3}=J\times$ resistance of

$40cm$ of the potentiometer wire.

Now resistance of

$40cm$ of the potentiometer wire

$=\dfrac {10}{100}\times 40 =4\Omega$

So,

$10\times10^{-3}=\dfrac 2{R+10}\times 4 \\\implies R=790\Omega$

is the value of the external resistance.

Observe the given figure and answer the following questions. The bulb will glow when a ______ is placed in between the probes.

0%
Crayon
0%
Comb
0%
Key
0%
Piece of stone

Ravi connected three bulbs with the cells and a switch as shown. When switch is moved to ON position.

0%
The bulb X will glow first.
0%
The bulb Y will glow first.
0%
The bulbs Z and X will glow first.
0%
All the bulbs will glow simultaneously.

Which of the following circuit diagram represents the given torch?

What is the value of unknown resistance R, of galvanometer shows null deflection in the given meter bridge set up shown in the fig?

0%
$97.50\Omega$
0%
$105\Omega$
0%
$220\Omega$
0%
$110\Omega$

Explanation

Given that, the galvanometer shows null deflection. Hence the given combination is Wheatstone bridge.

Let

$k$ be the resistance per unit length of the meter bridge wire.

Applying Wheatstone bridge principle we get:-

$\dfrac{20k}{55}=\dfrac{80k}{R}$

$\implies R= \dfrac{80\times 55}{20} =220\ \Omega$

1514204_941984_ans_e631a2ccf7ef4415aa9fa21ea62cd081.png

An electric heater is connected to the voltage supply. After a few seconds, current get its steady value then its initial current will be?

0%
Equal to its steady current
0%
Slightly higher than its steady current
0%
Slightly less than its steady current
0%
Zero

In Fig.,value of

${ I }_{ x }and\quad { I }_{ y }$ are respectively

0%
1A,1A
0%
1.2A,1.2A
0%
0.8A,0.8A
0%
1A,1.2A

In a meter bridge experiment, the ratio of the left gap resistance to right gap resistance is

$2 : 3$ , the balance point from left is?

0%
$60$ cm
0%
$50$ cm
0%
$40$ cm
0%
$20$ cm

Explanation

Hint:

If $R_1$ , $R_2$ , $R_3$ and $R_4$ resistors are connected in cyclic order in Wheatstone bridge, then for balance point we have

$\dfrac{R_1}{R_4} = \dfrac{R_3}{R_4}$

Step 1: Note all the values given.

It is given that ratio of left gap resistance to right gap resistance is

$2:3$ . That is, if resistance in left gap is

$2R$ , then resistance in right gap will be

$3R$ . We have

$R_1 = 2R$

$R_2 = 3R$

Step 2: Use Principle of Wheatstone bridge in meter bridge.

Let specific resistance of wire of meter bridge is

$\rho$ . And let null point lies at distance

$x$ from left end. Then resistance of left and right end of wire will be

$R_3 = \rho x$

$R_4 = \rho (100-x)$

By principle of wheatstone stone bridge, we have

$\dfrac{R_1}{R_4} = \dfrac{R_3}{R_4}$

$\Rightarrow \dfrac{2R}{3R} = \dfrac{\rho x}{\rho (100-x)}$

$\Rightarrow \dfrac{2}{3} = \dfrac{x}{100-x}$

$\Rightarrow 200 - 2x = 3x$

$\Rightarrow 5x = 200$

$\Rightarrow x = 40cm$

Thus Balance Point from left end is 40cm away.

Option C is correct.

Which of the following I-V graph represents ohmic conductors?

Explanation

Ohm's law

$I=\dfrac{V}{R}$ is an equation of straight line. Hence I-V characteristics for ohmic conductors is also straight line and its slope gives,

$m=\dfrac{1}{resistance}$ of the conductor.

Option A

In the circuit shown in figure heat developed across

$2\Omega, 4\Omega$ and

$3\Omega$ resistances are in the ratio of.

0%
$2 : 4 : 3$
0%
$8 : 4 : 12$
0%
$4 : 8 : 27$
0%
$8 : 4 : 27$

Explanation

Let

$i_1$ current flowing through

$2 \Omega$ resistor and

$i_2$ current flowing through

$4 \Omega$ resistor

$I = i_1 + i_2$ ___(1)

$2 \Omega$ and

$4 \Omega$ resistors connected in parallel, Hence both have same potential difference.

So,

$i_1 R_1 = i_2 R_2$

$i_1 \times 2 = (I - I_1) 4$

$i_1 \times 2 = 4 I - 4i,$

$6 i_1 = 4 I$

$i_1 = \dfrac{4}{6} I = \dfrac{2}{3} I$

$i_2 = I - i , = I - \dfrac{2}{3} I = \dfrac{(3-2)I}{3} = \dfrac{I}{3}$

Heat

$= l^2 Rt ,$ time is same for all

$(Heat)_{2 \Omega} : (Heat)_{4 \Omega} : (Heat)_{3 \Omega} = \left(\dfrac{2}{3} I\right)^2 \times 2 : \left(\dfrac{I}{3} \right)^2 \times 4 : (I)^2 \times 3$

$= \dfrac{8}{9} I^2 : \dfrac{4}{9} I^2 : 3 I^2$

Ratio

$= 8 : 4 : 27$

1494235_941778_ans_e53b7e76b93247e38c6610adf7be17b1.png

The battery of a trunk has an emf of

$24$ V. If the internal resistance of the battery is

$0.8\Omega$ . What is the maximum current that can be drawn from the battery?

0%
$30$ A
0%
$32$ A
0%
$33$ A
0%
$34$ A

Explanation

Emf of the battary,

$\varepsilon=24\ volt$

Internal Resistance of the battary,

$r=0.8\ \Omega$

Assuming load resistance to be

$R$
So, current can be given as:

$I=\dfrac{\varepsilon}{R_{net}}$

$\Rightarrow I=\dfrac{24}{(0.8+R)}$

For maximizing the current to

$I_{max}, R=0$

So,

$I_{max}=\dfrac{24}{0.8}$

$\Rightarrow I_{max}=30\ A$

Option

$(A)$ is correct

1441857_941924_ans_7894abec8ad44d6f9a135713abd6d367.png

Figure (a) and figure (b) both are showing the variation of resistivity

$(\rho)$ with temperature (T) for some materials. Identify the type of these materials.

0%
Conductor and semiconductor
0%
Conductor and Insulator
0%
Insulator and semiconductor
0%
Both are conductors

In a Wheatstone bridge if the battery and galvanometer are interchanged then the deflection in galvanometer will:

0%
Change in previous direction
0%
Not change
0%
Change in opposite direction
0%
None of these

Point out the right statements about the validity of Kirchhoff's junction rule.

0%
It is based on conservation of charge
0%
Outgoing currents add up and are equal to incoming currents at a junction
0%
Bending or reorienting the wire does not change the validity of Kirchhoff's junction rule
0%
All of above

Explanation

By Kirchhoff's junction rule. Incoming current

$=$ Outgoing current

The net charge is conserved and it is based on conservation of charge.

Also bending or reorienting the wire does not invalidate the conservation of charge principle. So, option D is correct.

A circuit has a section ABC as shown in figure. If the potentials at points A, B and C are

$V_1, V_2$ and

$V_3$ respectively. The potential at point O is?

0%
$V_1+V_2+V_3$
0%
$\left[\dfrac{V_1}{R_1}+\dfrac{V_2}{R_2}+\dfrac{V_3}{R_3}\right]\left[\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\right]^{-1}$
0%
Zero
0%
$\left[\dfrac{V_1}{R_1}+\dfrac{V_2}{R_2}+\dfrac{V_3}{R_3}\right](R_1+R_2+R_3)$

Explanation

According to kirchhoff's first rule to

$O$

$-l_1 -l_2 - l_3 = 0$ or

$l_1 + l_2+l_3 = 0$ ...(i)

$V_0$ is the electric potential at

$O$

$V_1, V_2$ and

$V_3$ be the potential at

$A, B$ and

$C$ respectively then

$V_0 - V_1 = l_1R_1$ .

$V_0 - V_2 = l_2R_2$ and

$V_0 - V_3 = l_3R_3$

This,

$l_1 = \left(\dfrac{V_0-V_1}{R_1}\right), l_2=\left(\dfrac{V_0 - V_2}{R_2}\right)$ , and

$l_3 = \left(\dfrac{V_0 - V_3}{R_3}\right)$

putting volume of

$l_1, l_2$ and

$l_3$ in (I) we get

$\dfrac{V_0 - V_1}{R_1} + \dfrac{V_0-V_2}{R_2} + \dfrac{V_0 - V_3}{R_3} = 0$

$V_0 \left[\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}\right] - \left[ \dfrac{\frac{V_1}{R_1} + \frac{V_2}{R_2} + V_3}{R_3}\right] = 0$

$V_0 = \left[ \dfrac{\frac{V_1}{R_1} + \frac{V_2}{R_2} + V_3}{R_3}\right]\left[\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}\right] ^{-1}$

When an uncharged body touches a charged body, the

0%
Uncharged body acquires a similar charge.
0%
Uncharged body acquires an equal and opposite charge
0%
Charges body loses its charge
0%
Uncharged body remains uncharged

In a potentiometer the balancing with a cell is at length of

$220$ cm. On shunting the cell with a resistance of

$3\Omega$ balance length becomes

$130$ cm. What is the internal resistance of this cell?

0%
$4.5\Omega$
0%
$7.8\Omega$
0%
$6.3\Omega$
0%
$2.08\Omega$

Explanation

Given that,

Initial balancing length of potentiometer with a cell,

$l_1= 220\ cm$

Now the cell is shunted with resistance,

$R=3\Omega$

Final balancing length,

$l_2= 130\ cm$

Internal resistance of the cell,

$r= \bigg( \dfrac{l_1-l_2}{l_2} \bigg) R$

$\implies r= \bigg( \dfrac{90}{130}\bigg) \times 3= \dfrac{270}{130}=2.08 \Omega$

Hence, option

$(D)$ is correct.

$3$ V potentiometer used for the determination of internal resistance of a

$2.4$ V cell. The balance point of the cell is open circuit is

$75.8$ cm. When a resistor of

$10.2\Omega$ is used in the external circuit of the cell the balance point shifts to

$68.3$ cm length of the potentiometer wire. The internal resistance of the cell is?

0%
$2.5\Omega$
0%
$2.25\Omega$
0%
$1.12\Omega$
0%
$3.2\Omega$

Explanation

Given:

The voltage of the potentiometer is

$3\ V$ .

The voltage of the battery whose internal resistance is to be determined is

$2.4\ V$ .

Balance point of cell in open circuit is

$75.8\ cm$ .

The new balance point of the wire is

$68.3\ cm$ .

The internal resistance of the cell is given by:

$r=R\left(\dfrac{l_1}{l_2}-1\right)$

$r=10.2\left(\dfrac{75.8}{68.3}-1\right)$

$r=1.12\Omega$

Which pair is not correct for the instrument used in the laboratory and their uses from the following?

	Instrument	Uses
(A)	Voltmeter	To measure the electric potential difference
(B)	Ammeter	To measure the electric current
(C)	Galvanometer	To detect the presence of current
(D)	Resistor	To supply the electric energy to circuit

0%
A
0%
B
0%
C
0%
D

Explanation

The uses of different electric devices are-

$(A)Voltemeter-$ Used to measure the electric potential difference between two points in the circuit.

$(B)Ammeter-$ Used to measure electric current in any branch of the circuit.

$(C)Galvanometer-$ It is used to measure the presence of current in any branch of electric circuit.

$(D)Resistor-$ It is that device which hinders the flow of current.

The option-

$D$ is wrongly matched.

Answer-(D)

In the circuit shown, the value of currents

$I_{1}, I_{2}$ and

$I_{3}$ are

0%
$3\ A, \dfrac {-3}{2}A, \dfrac {9}{2}A$
0%
$\dfrac {9}{2}A, 3\ A, \dfrac {-3}{2}A$
0%
$5\ A, 4\ A, -3\ A$
0%
$7\ A, \dfrac {5}{4}A, \dfrac {9}{2}A$

Explanation

Applying Kirchhoff;s voltage law,
In loop

$I, -27 - 6I_{2} - 2I_{1} + 24 = 0$

$6I_{2} + 2I_{1} =- 3 .... (i)$
In loop

$II, -27 - 6I_{2} + 4I_{3} = 0$

$6I_{2} - 4I_{3} = -27 .... (ii)$
At junction

$P, I_{1} - I_{2} - I_{3} = 0 .... (iii)$
Solving equations (i), (ii) and (iii) we get

$I_{1} = 3\ A, I_{2} = -3/2\ A, I_{3} = 9/2\ A$ .

In a potentiometer a cell of emf

$1.5$ V gives a balanced point at

$32$ cm length of the wire. If the cell is replaced by another cell then the balance point shifts to

$65.0$ cm then the emf of second cell is?

0%
$3.05$ V
0%
$2.05$ V
0%
$4.05$ V
0%
$6.05$ V

Explanation

Given,

Balencing point

$l_1=32cm \$ when votage applied

$V=1.5$

Balencing point

$l_2=65cm \$

Potential gradiant

$K=\dfrac{V}{x}= \dfrac{1.5}{32}$

potential diffrence in length

$l_2=K \times l_2=V'$

$V'=\dfrac{1.5}{32} \times 65=3.05 \ Volt$

Option A

1175835_941990_ans_110f6286f1a04acc825bd5674c4e23f6.png

Two cells of emf's approximately

$5$ V and

$10$ V are to be accurately compared using a potentiometer of length

$400$ cm.

0%
The battery that runs the potentiometer should have voltage of $8$ V
0%
The battery of potentiometer can have a voltage of $15$ V and R adjusted so that the potential drop across the wire slightly exceeds $10$ V
0%
The first portion of $50$ cm of wire itself should have a potential drop of $10$ V
0%
Potentiometer is usually used for comparing resistances and not voltages

Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then,

0%
negative and distribute uniformly over the surface of the sphere
0%
negative and appears only at the sphere closest to the point charge
0%
negative and distribute non-uniformly over the entire surface of the sphere
0%
zero

Explanation

$\bullet$ As the sphere is isolated, so the total charge on it will remain same, which is zero.

$\bullet$ Presence or absence of any external charge nearby cannot change its total charge.

Hence option D is correct

$\bullet$ But, Presence of external charge will induce charges on the sphere as shown in the figure, this induction happens due to redistribution of charges within the sphere only.

$\bullet$ As positive charge(

$+q$ ) will attract negative charges of the sphere. So, the negative charge will move towards right and positive charges will move towards left.

2110550_1010566_ans_0e8daf6a0b8b4e18804226b9c94a2672.png

Which of the following statements is false?

0%
Kirchhoff's second law represents energy conservation.
0%
Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude.
0%
In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed.
0%
A rheostat can be used as a potential divider.

Explanation

The balanced condition is given by

$\dfrac{P}{Q}=\dfrac{R}{S}$ ; When battery and Galvanometer are exchanged, it become

$\dfrac{P}{R}=\dfrac{Q}{S}$ ; which is same as previous.

CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 7 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 7 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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