MCQ Questions for Class 12 Medical Physics Current Electricity Quiz 11

In the circuit shown below, the cell is ideal, with

$emf=2V$ . The resistance of the coil of the galvanometer

$G$ is

$1 \Omega$ . The steady current through the galvanometer is

0%
$0.2A$
0%
$0.1A$
0%
$0.4A$
0%
Zero

In an electrolyte

$3.2\times 10^{18}$ bivalent positive ions drift to the right per second while

$3.6\times 10^{18}$ monovalent negative ions drift to the left per second. Then the current is

0%
$1.6$ amp to the left
0%
$1.6$ amp to the right
0%
$0.45$ amp to the right
0%
$0.45$ amp to the left

Explanation

Net current

$i_{net}=i_{(+)}+i_{(-)}$

$=\dfrac {n_{(+)} q_{(+)}}{t}+\dfrac {n_{(-)} q_{(-)}}{t}$

$=\dfrac {n_{(+)} }{t}\times 2e+\dfrac {n_{(-)} }{t}\times e$

$=3.2\times 10^{18}\times 2\times 1.6\times 10^{-19}+3.6\times 1.6\times 10^{18}\times 1.6\times 10^{-19}$

$=1.6\ A$ (towards right)

1731446_1815208_ans_eb5ef10b694c4a669937752a1b019aff.png

Variation of current and voltage in a conductor has been shown in the diagram below. The resistance of the conductor is.

0%
$4$ ohm
0%
$2$ ohm
0%
$3$ ohm
0%
$1$ ohm

Explanation

from ohm's law

$R=\dfrac VI$

also, slope of given graph

$\dfrac VI$

$R=$ slope of graph

$s=\dfrac{y_2-y_1}{x_2-x_1}$

$s=\dfrac{4-1}{4-1}=1$

Resistance = 1 ohm

Which is a wrong statement

0%
The Wheatstone bridge is most sensitive when all the four resistance are of the same order
0%
In a balanced Wheatstone bridge, interchanging the position of galvanometer and cell affects the balance of the bridge
0%
Kirchhoff's first law (for currents meeting at a junction in a electric circuit) expresses the conversion of change
0%
The rheostat can be used a a potential divider

A potential difference of

$V$ is applied at the ends of a copper wire of length

$l$ and sdiameter

$d$ . On doubling only

$d$ , drift velocity

0%
Becomes two times
0%
Becomes half
0%
Does not change
0%
Becomes one fourth

Explanation

Drift velocity

$v_d=\dfrac {V}{\rho l n e}; v_d$ does not depend upon diameter.

A solenoid is at potential of difference

$60\ V$ and current flows through it is

$15$ ampere, then the resistance of coil will be

0%
$4\ \Omega$
0%
$8\ \Omega$
0%
$0.25\ \Omega$
0%
$2\ \Omega$

Explanation

Given,

Potential difference

$V=60\ V$

Current

$i=15\ A$

Resistance

$R=?$

Resistance

$=\dfrac {Potential\ difference}{Current}=\dfrac{60V}{15A}=4\Omega$

Consider the circuit given here with the following parameters E.M.F. of the cell

$=12\ V$ . Internal resistance of the cell

$=2\omega$ . Resistance

$R=4\Omega$
Which one of the following statements in true

0%
Rate of energy loss in the source is $=8\, W$
0%
Rate of energy conversion in the source is $16\ W$
0%
Power output in is $=8\ W$
0%
Potential drop across $R$ is $=16\ V$

Explanation

From ohm's Law

$i=\dfrac{V}{R}$

wher

$R=4+2=6\Omega$

$i=\dfrac{12}{(4+2)}=2\ A$

Energy loss inside the source

$=i^2r=(2)^2\times 2=8\ W$

Kirchhoff

$I$ law and

$II$ law of current, proves the

0%
Conservation of charge and energy
0%
Conservation of current and energy
0%
Conservation of mass and charge
0%
None of thee

There is a current of

$20$ amperes in a copper wire of

$10^{-6}$ square meter area of cross-section. If the number of free electrons per cubic meter is

$10^{29}$ , then the drift velocity is

0%
$125\times 10^{-3}m/sec$
0%
$12.5\times 10^{-3}m/sec$
0%
$1.25\times 10^{-3}m/sec$
0%
$1.25\times 10^{-4}m/sec$

Explanation

we know that drift velocity

$v_d =\dfrac {i}{nAe}$

According to the question:

$v_d=\dfrac {20}{10^{29}\times 10^{-6}\times 1.6\times 10^{-19}}=1.25\times 10^{-3}m/s$

Which of the following statement is correct

0%
Liquid obey fully the ohm's law
0%
Liquid obey partially the ohm's law
0%
There is no relation between current and $p.d$ for liquids
0%
None of the above

Explanation

Liquid is non-ohomic conductor.

$VI$ graph, we will not get a straight line in case of liquids.

The drift velocity of free electrons in a conductor is

$'v'$ when a current

$'i'$ is flowing in it. If both the radius and current are doubled then drift velocity will be

0%
$v$
0%
$\dfrac v2$
0%
$\dfrac v4$
0%
$\dfrac v8$

Explanation

We know that

$v_d =\dfrac {i}{ne \pi r^2}\Rightarrow v_d \propto \dfrac {i}{r^2}\Rightarrow \dfrac {v}{v'}=\dfrac {i_1}{i_2}\times \left(\dfrac {r_2}{r_1}\right)^2 \Rightarrow v' =\dfrac {v}{2}$

A current

$I$ is passing through a wire having two sections

$P$ and

$Q$ of uniform diameter

$d$ and

$d/2$ respectively. If the mean drift velocity in sections

$P$ and

$Q$ is denoted by

$v_P$ and

$v_Q$ respectively then

0%
$v_P =v_Q$
0%
$v_P=\dfrac 12 v_Q$
0%
$v_P=\dfrac 14 v_Q$
0%
$v_P =2 v_Q$

Explanation

Drift velocity

$v_d=\dfrac {i}{neA}\Rightarrow v_d \propto \dfrac {1}{A}$ or

$v_d \propto \dfrac {1}{d^2}$

$\Rightarrow \ \dfrac {v_P}{v_Q}=\left(\dfrac {d_Q}{d_P}\right)^2=\left(\dfrac {d/2}{d}\right)^2=\dfrac 14\Rightarrow v_P =\dfrac 14 v_Q$

In a balanced Wheatstone's network, the resistance in the arms

$Q$ and

$S$ are interchanged. As a result of this

0%
Network is not balanced
0%
Network is still balanced
0%
Galvanometer shows zero deflection
0%
Galvanometer and the cell must be interchanged to balance

Explanation

We know that in In a balanced Wheatstone's network,

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

$Q$ is replaced by

$S$ then ratio becomes

$\dfrac{P}{S} = \dfrac{R}{Q}$ which is not equal to the previous ratio. Hence, network is not balanced.

1736938_1815728_ans_de88bf72aac5415ab5329658fb2b9f86.png

A copper wire has a square cross-section

$2.0\ mm$ on a side. It carries a current of

$8\ A$ and the density of the electrons is

$8\times 10^{28}m^{-3}$ . The drift speed of electrons is equal to

0%
$0.156\times 10^{-3}\ m.s$
0%
$0.156\times 10^{-2}\ m.s$
0%
$3.12\times 10^{-3}\ m.s$
0%
$3.12\times 10^{-2}\ m.s$

Explanation

$v_d =\dfrac {i}{nAe}=\dfrac {8}{8\times 10^{28}\times (2\times 10^{-3})^2 \times 1.6\times 10^{-19}}$

$=0.156\times 10^{-3}m/sec$

The resistance of an incadescent lamp is

0%
Greater when switched off
0%
Smaller when switched on
0%
Greater when switched on
0%
The same whether it is switched off or switched on

Explanation

$R\propto \dfrac {1}{\tau}$ , where

$\tau =$ Relaxation time.

When lamp is switched on, temperature of filament increase.

hence

$\tau$ decrease so

$R$ increase.

If you are provided three resistances

$2 \Omega, 3 \Omega$ and

$6 \Omega$ . How will you connect them so as to obtain the equivalent resistance of

$4\Omega$

0%
0%
0%
0%
None of these

The material of wire of potentiometer is

0%
Copper
0%
Steel
0%
Mangamint
0%
Aluminum

At room temperature copper has free electrons density of

$8.4\times 10^{28}$ per

$m^3$ . The copper conductor has a cross-section of

$10^{-6}m$ and carries a current of

$5.4\ A$ , The electrons drift velocity in copper is

0%
$400\ m/s$
0%
$0.4\ m/s$
0%
$0.4\ mm/s$
0%
$72\ m/s$

Explanation

$V_d =\dfrac {i}{neA}=\dfrac {5.4}{8.4\times 10^{28}\times 10^{-6}\times 1.6\times 10^{-19}}$

$=0.4\times 10^{-3}m/sec =0.4\ mm/sec$ .

Consider the circuit below. All bulbs are identical. Choose the CORRECT option

0%
If the bulb A fuse,then bulb C stays lighted, bulb B bums brightly
0%
If the bulb B fuse, then bulb C goes off and D stays lighted
0%
If the bulb C fuse, bulbs A and B goes off
0%
If the bulb d fuse, the event is unnoticable, and bulbs A,B,C, and stays lighted

Ina potentiometer experiment two cells of e.m.f and

$E$ are used in series and in conjunction and the balancing length is found to be

$58\ cm$ of the wire. If the polarity of

$E$ is revered then the balancing length becomes

$29\ cm$ . The ratio

$\dfrac{E_1}{E_2}$ of the e.m.f of the two cell is

0%
$1:1$
0%
$2:1$
0%
$3:1$
0%
$4:1$

Explanation

In potentiometer we know that

$E (emf) \propto l (length)$

$\dfrac{E_1}{E_2}=\dfrac{l_1+l_2}{l_1-l_2}=\dfrac{58+29}{58-29}=\dfrac{3}{1}$

The internal resistance of a primary cell is 4 ohm. It generates a current of 0 .2 amp in an external resistance of 21 ohm. The rate of which chemical energy is consumed providing the current is

0%
$0 . 42$ J/s
0%
$0 . 84$ J/s
0%
$5$ J/s
0%
$1$ J/s

Explanation

The rate of consumption of chemical energy = Rate of generation of electrical energy.

The energy generated is calculated from the formula P = VI. Substituting V=IR (Ohm's law) in the above equation we get,

$P=V^2/R$

In this case, there are external and internal resistances. So, the equation is rewritten as I

$P=I^2\times (R+r)$

Therefore, Chemical energy =

$0.2^2(21+4)=1\,J$

Hence, The rate at which chemical energy is consumed in providing the current is 1 J/s

The resistor of resistance R is connected to 25 V supply and heat produced in it 25 J sec. The value of R is

0%
$225 \Omega$
0%
$1 \Omega$
0%
$25 \Omega$
0%
$50 \Omega$

Explanation

The power dissipated across the resistance is given by

$P = \dfrac{V^{2}}{R} \Rightarrow R = \dfrac{V^{2}}{P} = \dfrac{25\times25}{25} = 25\Omega$

Thus 25 ohm is the resistance if the resistor.

Silver and copper voltameter are connected in parallel with a battery of e.m.f. 12 V. In 30 minutes, 1 gm of silver and 1.8 gm of copper are liberated. The power supplied by the battery is

0%
24.13 J/sec
0%
2.413 J/13
0%
0.2413 J/sec
0%
2413 J/sec

Explanation

The current taken by the silver voltameter

$I_{1} = \dfrac{m}{Zt}=\dfrac{1}{11.2\times 10^{-4}\times 30\times 60}=0.496 A$ and by copper voltameter

$I_{2}=\dfrac{1.8}{6.6\times 10^{-4}\times 30\times 60}=1.515 A$

Total current

$I = (I_{1}+I_{2})=2.011A$

Power

$P = IV = 2.011\times 12 = 24.132 J / sec$

Two bulbs of

$100 W$ and

$200 W$ working at

$220$ volt are joined in series with 220 volt supply. Total power consumed wll be approximately.

0%
$65$ watt
0%
$33$ watt
0%
$300$ watt
0%
$100$ watt

Explanation

The two bulbs of 100W and 200W are working at 220V are joined in series with 220V supply

Thus ,

$P_{s}=\dfrac{P_{1}P_{2}}{P_{1}+P_{2}}=\dfrac{100\times 200}{100+200}=\dfrac{200}{3}\approx 65watt$

$60 \ Watt$ bulb operates on

$220 \ V$ supply. The current flowing through the bulb is

0%
$\dfrac{11}{3} \ amp$
0%
$\dfrac{3}{11} \ amp$
0%
$3 \ amp$
0%
$6 \ amp$

Explanation

The power rated on the bulb

$P = 60 \ W$ ,it operates on

$V = 220 \ V$ supply .

Power is given by

$P=Vi$

Hence,

$\Rightarrow i = \dfrac{P}{V} = \dfrac{60}{220} = \dfrac{3}{11} \ amp$

1723369_1816967_ans_62d819f3b5cd490d94ec0283db286960.png

A hot electric iron has a resistance of

$80 \Omega$ and is used on a 200 V source. The electrical energy if it is used for hours, will be

0%
$8000 Wh$
0%
$2000 Wh$
0%
$1000 Wh$
0%
$800 Wh$

Explanation

The hot electric iron has resistance of 80 Ohm and it is across 200V source. So the electrical energy is given by Energy

$\dfrac{V^{2}}{R} t = \dfrac{200 \times 200 \times 2}{80} = 1000 Wh$

In an electric heater 4 amp current passes for 1 minute at potential difference of 250 volt, the power of heater and energy consumed will be respectively

0%
1 kW, 60 kJ
0%
0.5 kW, 30 kJ
0%
10 kW, 600 kJ
0%
None of thesee

Explanation

A 4 amp of current passes for 1 minutes at a potential difference of 250 volt.

The power generated is given by

$P = VI$

$P=250 \times 4=1kW$

Heat generation is given by

$H=VIt$

$H=60kW$

If a 2 kW boiler is used everyday for hour, then electrical energy consumed by boiler in thirty days is

0%
$15$ unit
0%
$60$ unit
0%
$120$ unit
0%
$240$ unit

Explanation

A 2kW boiler is used everyday for an hour, Then electric current consumed by the boiler in thirty days is given by

Energy =

$P \times t = 2 \times 1 \times 30 = 60 k WH = 60 unit$

For ensuring dissipation of same energy in all three resistors

$(R_{1},R_{2},R_{3})$ connected as shown in figure, their values must be related as]

0%
$R_{1}=R_{2}=R_{3}$
0%
$R_{2}=R_{3}andR_{1}=4R_{2}$
0%
$R_{2}=R_{3}andR_{1}=\dfrac{1}{4}R_{2}$
0%
$R_{1}=R_{2}+R_{3}$

Explanation

The voltage in

$R_2$ and

$R_3$ is same therefore according to

$H=\dfrac{V^2}{R} t$

$R_2=R_3$

Also the energy in all resister is same

$i^2Rt=i_1^2Rt$ ....(i)

$i_1=\dfrac{R_3}{R_2+R_3} i$

$I_1=i_2$

Thus

$i_1^2R_1t=|dfrac{i^2 R_2 t}{4}$

$R_1=R_2/4$

1737863_1817403_ans_56408915a19f4e67ac9d0d168295415e.png

An electric is designed to draw power P at voltage V.If the voltage is V it draws a power P. Then

0%
$P = \left ( \dfrac{V_{0}}{V} \right )^{2}P_{0}$
0%
$P = \left ( \dfrac{V}{V_{0}} \right )^{2}P_{0}$
0%
$P = \left ( \dfrac{V}{V_{0}} \right )P_{0}$
0%
$P = \left ( \dfrac{V_{0}}{V} \right )P_{0}$

Explanation

Electric is designed to draw power p at the given voltage V , then the power developed is given by

$P \alpha V^{2} \Rightarrow \dfrac{P}{P_{0}} = \left ( \dfrac{V}{V_{0}} \right )^{2} \Rightarrow P = \left ( \dfrac{V}{V_{0}} \right )^{2} P_{0}$

Thus $P = \left ( \dfrac{V}{V_{0}} \right )^{2} P_{0}$ is the final answer.

An electric bulb marked 40 W and 200 V, is used in a circuit of supply voltage 100 v, Now its power is

0%
$100 W$
0%
$40 W$
0%
$20 W$
0%
$10 W$

Explanation

The power genrated is given by

$P=\dfrac{V^{2}}{R}\Rightarrow \dfrac{P_{2}}{P_{1}}=\dfrac{V^{2}_{1}}{V^{2}_{1}}(\because R is constant )\Rightarrow \dfrac{P_{2}}{P_{1}}=\left ( \dfrac{100}{200} \right )^{2}=\dfrac{1}{4}\Rightarrow P_{2}=\frac{P_{1}}{4}=\dfrac{40}{4}=10W$

If a

$5 \,V$ battery is provoding

$2 \,A$ current in a conductor then what is the resistance of conductor ?

0%
$3$ Ohm
0%
$2.5$ Ohm
0%
$10$ Ohm
0%
$2$ Ohm

Explanation

Given that:

Battery supplies a voltage of

$V=5 \ V$

Current provided by the battery

$I=2 \ A$

Resistance= ?

By Ohm's law,

$V=IR$

$\Rightarrow R=\dfrac{V}{I}=\dfrac{5}{2}=2.5 \Omega$

In the figure, the potentiometer wire AB of length L and resistance

$9r$ is joined to the cell D of emf

$\varepsilon$ and internal resistance r. The cell Cs emf is

$\epsilon /2$ and its internal resistance is

$2r$ . The galvanometer G will show no deflection when the length AJ is

0%
$\dfrac{4L}{9}$
0%
$\dfrac{5L}{9}$
0%
$\dfrac{7L}{18}$
0%
$\dfrac{11L}{18}$

What happens to the current in each resistor when resistors of different values in parallel combination are connected to a source of electricity?

0%
Values of current and potential difference are different
0%
Values of current and potential difference are same
0%
Values of current are different but value of potential difference is same
0%
Values of current are the same but values of potential difference are different

Kilowatt hour is unit of

0%
Energy
0%
Power
0%
Impulse
0%
Force

Kirchhoffs first and second laws are based on

0%
Charge and Energy Conservation Laws
0%
Current and Energy Conservation Laws
0%
Mass and Charge Conservation Laws
0%
None of these

Potentiometer is a type of instrument by which potential difference can be measured

0%
Zero
0%
Infinite
0%
Uncertain
0%
Depends on external resistance

In a Wheatstone bridge, the position of battery and galvanometer is interchanged, then the new balanced position will be

0%
Unchanged
0%
Will changed
0%
Can't say anything
0%
May change or may not be, it depends upon the resistance of galvanometer and battery

Five resistances of R

$\Omega$ were taken. First three resistances are connected in parallel combination and rest two are connected in series combination, then the equivalent resistance is :

0%
$\frac{3}{7} R \ \Omega$
0%
$\frac{7}{3} R \ \Omega$
0%
$\frac{7}{8} R \ \Omega$
0%
$\frac{8}{7} R \ \Omega$

Explanation

$R_{equi} = \frac{R}{3} + R + R = \frac{7R}{3} \ \Omega$

A conducting resistance is connected to the battery and temperature of conductor decreases by the process of cooling then the value of current will be :

0%
increased
0%
decreased
0%
remain constant
0%
zero

Explanation

The resistance of the conducting wire will decrease according to equation

$R_t = R_0(1 \pm \alpha \Delta t)$ due to which current will increase according to ohm's law

$(l = \frac{V}{R})$

Kirchhoff's first law, i.e.,

$\sum I = 0$ at a junction, deals with the conservation of

0%
charge
0%
energy
0%
momentum
0%
mass

Explanation

$Charge$
Explanation: The principle of conservation of electric charge implies that: at any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node. This is known as Kirchhoff's current law.

When the balance point is obtained in the potentiometer, a current is drawn from

0%
both the cells and auxiliary battery
0%
cell only
0%
auxiliary battery only
0%
neither cell nor auxiliary battery

The resistance of one conducting wire is

$10$ . How much electric current will flow by connecting it with a battery of

$1.5 V$ ?

0%
$0.15 mA$
0%
$1.5 mA$
0%
$15 mA$
0%
$150 mA$

Explanation

From Ohm's law,

$V = IR$

$\Rightarrow I = \dfrac{V}{R}= \dfrac{1.5}{10}= 0.15A = 150 mA$

The current in a metallic conductor is plotted against voltage at two
different temperatures

$T_{1}$ and

$T_{2}$ . Which is correct

0%
$T_{1}$ > $T_{2}$
0%
$T_{1}$ < $T_{2}$
0%
$T_{1} = T_{2}$
0%
none

In Wheat stones bridge shown in the figure, galvanometer gives no deflection on pressing the key. The balance condition for the bridge is :

0%
$\dfrac{R_{1}}{R_{2}}=\dfrac{C_{1}}{C_{2}}$
0%
$\dfrac{R_{1}}{R_{2}}=\dfrac{C_{2}}{C_{1}}$
0%
$\dfrac{R_{1}}{R_{1}+R_{2}}=\dfrac{C_{1}}{C_{1}-C_{2}}$
0%
$\dfrac{R_{1}}{R_{1}-R_{2}}=\dfrac{C_{1}}{C_{1}+C_{2}}$

Explanation

$\Rightarrow V_{B} = V_{D}$

$\Rightarrow V_{A}-V_{B} = V_{A}-V_{D}$ ..........1]

$V_{B}-V_{C} = V_{D}-V_{C}$ ..............2]

Dividing 1] by 2]

$\Rightarrow \dfrac{V_{A}-V_{B}}{V_{B}-V_{C}} = \dfrac{V_{A}-V_{D}}{V_{D}-V_{C}}$

$\Rightarrow \dfrac{IR_{1}}{IR_{2}} = \dfrac{\dfrac{Q}{C_{1}}}{\dfrac{Q}{C_{2}}}$

$\dfrac{R_{1}}{R_{2}} = \dfrac{C_{2}}{C_{1}}$

On a bulb it is written

$220\ V$ and

$60\ W$ . Find out the resistance of the bulb and the value of the current flowing through it:

0%
$806.67\ \Omega \ \text{and} \ 0.27\ A$
0%
$500\ \Omega \ \text{and} \ 2\ A$
0%
$200\ \Omega \ \text{and} \ 4\ A$
0%
$100\ \Omega \ \text{and} \ 1\ A$

Explanation

$\textbf{Step1:Resistance of bulb}$

We know,
Power, $P=\dfrac {V^2}{R}$

$\implies R=\dfrac {V^2}{P}$

$R=\dfrac {(220)^2}{60}$

$R=806.67\ \Omega$

$\textbf{Step2:Apply Ohm's law}$
According to Ohm's law,
$V=IR$

$\implies I=\dfrac {V}{R}$

$I=\dfrac {220}{806.67}$

$I=0.27\ A$

copper rod AB of length L, pivoted at one end A, rotates at constant angular velocity

$\omega$ , at right angles to a uniform magnetic field of induction B. The e.m.f developed between the mid point C of the rod and end B is

0%
$\dfrac{B\omega l^{2}}{4}$
0%
$\dfrac{B\omega l^{2}}{2}$
0%
$\dfrac{3B\omega l^{2}}{4}$
0%
$\dfrac{3B\omega l^{2}}{8}$

Two non ideal batteries are connected in parallel. Consider the following statements:
The equivalent emf is smaller than either of the two emfs.
The equivalent internal resistance is smaller than either of the two internal resistances.

0%
both 1 and 2 are correct
0%
1 is correct but 2 is wrong
0%
2 is correct but 1 is wrong
0%
Both 1 and 2 are wrong

In the circuit shown in fig., the potential difference between the points C and D is balanced against 40 cm length of potentiometer wire of total length 100 cm. In order to balance the potential difference between the points D and E, the jockey should be pressed on potentiometer wire at a distance of

0%
16 cm
0%
32 cm
0%
56 cm
0%
80 cm

Explanation

Let,

$V$ be the potential difference between the points

$C$ and

$D$ and

$l_1$ is the length of potentiometer wire balanced against it. Hence the potential gradient is

$k = \dfrac {V}{l_1} = \dfrac {IR_1}{l_1}$
where,

$I$ is the current through the wire and

$R_1$ is the resistance between points

$C$ and

$D$ .

$\therefore IR_1 = kl_1$
Let,

$l_2$ is the balancing length for potential difference between the points

$D$ and

$E$ . Hence the potential gradient is

$k = \frac {V}{l_2} = \frac {IR_2}{l_2}$
where,

$I$ is the current through the wire and

$R_2$ is the resistance between points C and D.

$\therefore IR_2 = kl_2$
comparing both equations

$\dfrac {IR_2}{IR_1} = \dfrac {kl_2}{kl_1}$

$l_2 = (\dfrac {R_2}{R_1})l_1$
Put

$R_1 = \dfrac{(10 \times 10)}{10+10} = 5 \Omega, R_2 = 4 \Omega$ and

$l_1 = 40 cm$

$\therefore l_2 = (\dfrac {4}{5})40 = 32 cm$

A battery of emf

$E_{0}$

$=$ 12 V is connected across a 4m long uniform wire having resistance 4

$\Omega$ /m. The cells of small emfs

$\varepsilon_{1}$

$=$ 2V and

$\varepsilon_{2}$

$=$ 4V having internal resistance 2

$\Omega$ and 6

$\Omega$ respectively, are connected as shown in the figure. If galvanometer shows no deflection at the point N, the distance of point N from the point A is equal to

0%
$\dfrac{1}{6}m$
0%
$\dfrac{1}{3}m$
0%
$25 cm$
0%
$50 cm$

CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 11 - MCQExams.com

$\textbf{Step1:Resistance of bulb}$

We know,
Power, $P=\dfrac {V^2}{R}$

$\implies R=\dfrac {V^2}{P}$

$R=\dfrac {(220)^2}{60}$

$R=806.67\ \Omega$

$\textbf{Step2:Apply Ohm's law}$
According to Ohm's law,
$V=IR$

$\implies I=\dfrac {V}{R}$

$I=\dfrac {220}{806.67}$

$I=0.27\ A$

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 11 - MCQExams.com

\textbf{Step1:Resistance of bulb}\textbf{Step1:Resistance of bulb}

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Practice Class 12 Medical Physics Quiz Questions and Answers

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$\textbf{Step1:Resistance of bulb}$