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CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 6 - MCQExams.com
CBSE
Class 12 Medical Physics
Current Electricity
Quiz 6
A potential difference of 20V is needed to make a current of 0.05A flow through a resistor. What potential difference is needed to make a current of 300 mA flow through the same resistor?
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60V
0%
120V
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40V
0%
150V
Explanation
Answer is B.
According to Ohm's law, $$V = IR$$. That is, $$R = \dfrac VI$$.
In this case, the voltage is $$20\ V$$ and the current is $$0.05\ A$$.
So, the resistance $$R = 20/0.05 = 400$$ ohms.
Now, to make a current of 300 mA flow through the same resistor of 400 ohms, the potential difference to be applied is calculated as follows.
Here, $$I = 300 mA = 0.3\ A$$ and $$R = 400$$ ohms.
Therefore, $$V = 0.3\times 400 = 120\ V$$.
Hence, to make a current of 300 mA flow through the same resistor of 400 ohms a potential difference of 120 V is applied.
Kilowatt hour (kWh) represents the unit of :
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0%
power
0%
impulse
0%
momentum
0%
energy
Explanation
1 kW h = 1 kW ×1 h = 1000 W × 3600 s = 3600000 J =
3.6 $$\ast$$ 10$$^{6}$$ J
The energy used in households, industries and commercial establishments are usually expressed in kilowatt hour.
A cell of emf $$5$$ V can supply a total energy of $$9000$$ J, then the total charge that can be obtained from the cell would be _____ C
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$$180$$
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$$18000$$
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$$1800$$
0%
$$18$$
Explanation
EMF = 5V
Energy = 9000J
Power = Voltage $$\times$$ current
P = VI ($$\varepsilon $$ = energy )(t = time )(t = time)(it = q)
$$\varepsilon =VIt\\ \varepsilon =Vq\\ \dfrac { 9000 }{ 5 } = 1800C\\ q=1800c$$
Power of an electric heater is $$1000$$ W and it is running for $$1$$ hour. Calculate the energy consumed by it
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$$3.6$$ MJ
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$$1.2$$ MJ
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$$1.8$$ MJ
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$$4.2$$ MJ
Explanation
Power$$=P=1000W$$
$$t=1h=3600s$$
Power is energy consumed per unit time.
$$P=\dfrac{E}{t}$$
$$\implies E=Pt$$
$$\implies E=3600\times 1000$$
$$\implies E=3.6\times 10^6J$$
Energy$$=E=3.6MJ$$
Answer-(A)
Electric current cannot flow through
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Silver
0%
Wood
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Copper
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None of these
Explanation
Wood is an insulator and hence does not allow electric current to pass through it.
Answer-(B)
State whether given statement is True or False
Silk threads are very good conductors.
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True
0%
False
Explanation
Silk threads are bad conductor electricity. so it allows less amount of current at high voltage applied.
Given statement is $$False$$.
The amount of work done (or, energy spent) by a source of power $$1kW$$ in $$1h$$ time is:
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$$1J$$
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$$1Wh$$
0%
$$1kWh$$
0%
$$1calorie$$
Explanation
$$Energy$$ $$spent$$ = $$Power \times time$$
$$E=1kWh \times 1h$$ =
$$1kWh$$
$$1kWh$$ is the amount of work done (or, energy spent) by a source of $$1kW$$power in $$1h$$ time.
Convert 1 kWh to SI unit of energy.
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$$3.6\times 10^6J$$
0%
$$3.6\times 10^8J$$
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$$1.8\times 10^6J$$
0%
$$1.8\times 10^8J$$
Explanation
We know that,
$$1\ watt = 1\ joule/sec$$ and $$1\ kW=1000\ watt$$
Therefore,
$$1\ kWh$$ $$ = 1000\ watt\times 1\ hour = 1000\ joule/sec \times 3600\ sec$$ $$ = 3.6\times 10^6$$ $$J$$
Calculate the electric current in the circuit shown.
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$$1.5A$$
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$$0.5A$$
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$$2.5A$$
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$$2A$$
Explanation
Any current that passes through the $$10 \Omega$$ resistor also passes through $$5 \Omega$$, i.e., $$10\Omega$$ and $$5 \Omega$$ are connected in series. Their equivalent resistance is $$R_{eq} = 10 \Omega + 5 \Omega = 15 \Omega$$.
The equivalent circuit is shown in the above figure.
The current $$\displaystyle i = \frac {V}{R_{eq}}=\frac {7.5 V}{15 \Omega}=0.5A$$
Some matters do not allow the electricity to pass through it. What is called a matter which does not allow the electricity to pass though it?
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Conductor
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Insulator
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Semiconductor
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All of these
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None of these
Explanation
Substances which do not allow electricity to pass through them are called insulators.Examples: Mica,Plastic.
Two heater coils $$A$$ and $$B$$ made of the same material are connected in parallel across the same mains. The length and diameter of the wire used in $$A$$ are doubled as that of the wire used in $$B$$. If $$H_1$$ and $$H_2$$ are the quantities of heat liberated per second in $$A$$ and $$B$$, respectively, then $$H_1 : H_2 =$$
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$$1 : 4$$
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$$2 : 1$$
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$$1 : 3$$
0%
$$1 : 1$$
Explanation
Since the coils are in parallel across the same mains, $$V$$ is the same for both
$$\displaystyle \frac {H_1}{H_2} = \frac {R_2}{R_1} \left ( \because H= \frac {V^2}{R}\right)$$
$$\displaystyle \frac {H_1}{H_2} = \frac {2}{1} $$
The substance which allows electric current to flow through it is called
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Conductor
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Insulator
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Semiconductor
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None of these
Explanation
The substance which allows electric current to pass through it are called $$conductor.$$
Answer-(A)
The current in the given circuit is
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$$0.3\;\text{amp}$$
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$$0.4\;\text{amp}$$
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$$0.1\;\text{amp}$$
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$$0.2\;\text{amp}$$
Explanation
Two batteries are joined with opposite polarity so, total e.m.f.=$$5-2=3\;V$$
Total resistance = $$10+20=30\Omega$$
Current = $$\displaystyle\frac{3}{30}=0.1\;\text{A}$$
Watt hour represents:
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Electric energy
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Current
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Voltage
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Power
Explanation
The watt-hour (Wh) is a unit of electric energy equivalent to one watt (1 W) of power expended for one hour (1 h) of time. Watt-hour is commonly used in electrical applications.
$$1\ Wh = 1\times 60\times 60 = 3600\ J$$
Unit used in selling electrical energy to consumer is:
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$$Volt-Ampere$$
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$$Kilowatt-hour$$
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$$Volt/second$$
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None of these
Explanation
The commercial unit of electrical energy is $$Kilowatt-hour\ (kWh)$$.
$$1\ kWh$$ is equal to $$1\ unit$$ of energy.
An electric bulb of resistance $$20 \Omega$$ draws a current of $$0.04 A$$. Calculate the potential
difference at the ends.
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$$0.8 V$$
0%
$$2 V$$
0%
$$0.032 V$$
0%
$$5 V$$
Explanation
Given,
Resistance of resistor $$R = 20 \Omega$$
Current draws $$ I = 0.04 A$$
According to ohm's law
Potential difference, $$V = IR = 0.04 \times 20 = 0.8 V$$.
Hence Option A
Number of KWh in 1 Joule.
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$$\displaystyle 3.6\times { 10 }^{ 6 }\ KWh$$
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$$\displaystyle 2.77\times { 10 }^{ -7 }\ KWh$$
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$$\displaystyle 600\ KWh$$
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$$\displaystyle 1.6\times { 10 }^{ -19 }\ KWh$$
Explanation
We know that $$KWh$$ is kilo watt-hour-
In SI unit-
$$K = 10^3$$
$$hour = 3600\ s$$
$$1\ kWh = 3.6\times 10^6\ J$$
$$\therefore$$ $$1$$ $$J = \dfrac{1}{3.6\times 10^6} = 2.77\times 10^{-7}$$ $$kWh$$
Which of these units of energy is the largest?
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$$1\ Calorie$$
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$$1\ Joule$$
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$$1\ Erg$$
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$$1\ Kilowatt-hour$$
Explanation
Energy is the ability to do work.
$$1\ {calorie} = 4.186\ J$$
$$1\ erg= 1\times 10^{-7}\ J$$
$$1\ kWh=3.6\times 10^{6}$$ $$J$$
$$1\ kilowatt-hour$$ is the largest unit of energy.
Smallest commercial unit of energy is
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Kilowatt hour
0%
Watt second
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Watt hour
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Watt minutes
Explanation
Smallest commercial unit of energy is called Watt hour $$(Wh)$$.
One watt hour is the energy consumed when $$1 \ watt$$ of power is used for $$1 \ hour.$$
Energy consumed in our house has a unit of
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watts
0%
horse power
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kilowatt-hours
0%
none
Commercial unit of energy is .......................
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Joule
0%
Volt
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Kilowatt hour
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Watt second
Explanation
Commercial unit of electrical energy is kilowatt hour $$\left(kWh\right)$$
$$1 kwh = 3.6 \times {10}^{6} J$$
Calculate the monthly bill if a heater of 100 watt is used at the rate of rs. 1 per unit for 1 hour daily.
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Rs. 200
0%
Rs. 6000
0%
Rs. 100
0%
Rs. 3
Explanation
Given : $$P=0.1$$ kW $$t = 1$$ hour per day
Energy consumed per day $$E = Pt = 0.1\times 1 = 0.1$$ kWh per day
Total energy consumed per month $$E_T = 30\times 0.1 = 3$$ kWh per month
Cost of electricity $$ =Re$$ $$ 1 $$ per unit i.e per kWh
$$\therefore$$ Monthly bill $$ = Re.$$ $$1\times 3 =$$Rs. $$3$$ per month
An electron in potentiometer with experiences a force $$2.4 \times {10}^{-19} N$$. The length of potentiometer wire is $$6 m$$. The emf of the battery connected across the wire is (electronic charge $$= 1.6 \times {10}^{-19} C$$)
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$$6 V$$
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$$9 V$$
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$$12 V$$
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$$15 V$$
Explanation
Force experienced by electron $$F = qE$$ where $$E =$$ Electric field intensity
$$\therefore 2.4 \times {10}^{-19} = 1.6 \times {10}^{-19} E \Rightarrow E = 1.5 {N}/{m}$$
Moreover $$E =$$ potential gradient
$$E = \dfrac{dV}{dl}$$
$$\therefore E \times l = V \Rightarrow V = 1.5 \times 6 = 9 V$$
Copper and Carbon wires are connected in series and the combined resistor is kept at $$\displaystyle { 0 }^{ \circ }C$$. Assuming the combined resistance does not vary with temperature, the ratio of the resistances of Carbon and Copper wires at $$\displaystyle { 0 }^{ \circ }C$$ is:
(Temperature coefficients of resistivity of Copper and Carbon respectively are $$\displaystyle 4\times { { 10 }^{ -3 } }/{ ^{ \circ }{ C } }$$ and $$\displaystyle -0.5\times { { 10 }^{ -3 } }/{ ^{ \circ }{ C } }$$)
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$$8$$
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$$6$$
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$$2$$
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$$4$$
Explanation
It is given that combined series resistance does not change with temperature.
Thus, $$R_{Cu}+R_{C}=R_{Cu}(1+\alpha_{Cu}\Delta T)+R_C(1+\alpha_C\Delta T)$$
for all values of $$\Delta T$$
Thus, $$R_{Cu}\alpha_{Cu}=-R_{C}\alpha_{C}$$
$$\implies \dfrac{R_C}{R_{Cu}}=-\dfrac{\alpha_{Cu}}{\alpha_C}$$
$$=-\dfrac{4\times 10^{-3}}{-0.5\times 10^{-3}}=8$$
The electric current $$i$$ in the circuit shown is:
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$$6\ A$$
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$$2\ A$$
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$$3\ A$$
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$$4\ A$$
Explanation
Using Kirchhoff's junction law at A : $$3 + 2 = i_1$$
$$\implies i_1 = 5$$ A
Using Kirchhoff's junction law at B : $$ i_1 = 2 + i_2$$
$$\therefore$$ $$ 5 = 2 + i_2$$ $$\implies i_2 = 3$$ A
Using Kirchhoff's junction law at C : $$ i_2 + 1 = i$$
$$\therefore$$ $$ 3 + 1 = i$$ $$\implies i = 4$$ A
This question relates to the DC circuit shown below.
The values for current, voltage, and resistance in the circuit are graphed from point A through point G in the graphs directly below.
Which of the graphs shows the voltage from point A to point G?
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A
0%
B
0%
C
0%
D
0%
E
Explanation
Here the current through entire circuit is constant and the resistance is increasing from A to G. So by Ohm's law, the voltages throughout the circuit will decrease
as the distance from point A increases. Thus, e
ach increase in the resistance produces a corresponding decrease in voltage. So graph A will be the correct.
A steady current flow in a metallic conductor of non-uniform cross-section. The quantity/quantities remaining constant along the whole length of the conductor is/are:
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current, electric field and drift speed
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drift speed only
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current and drift speed
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current only
Explanation
When a steady current flows in a metallic conductor of non-uniform cross-section then the drift speed is $$V_d=I/neA$$ and the electric field $$E=I/ \rho A$$
$$\Rightarrow V_d\propto \dfrac{1}{A}$$ and $$E\propto \dfrac{1}{A}$$
$$\Rightarrow$$ Only current remains constant.
In a metallic conductor of non-uniform cross-section, only the current remains constant along the entire length of the conductor.
Two cells A and B of emf 2V and 1.5 V respectively, are connected as shown in figure through an external resistance $$10\Omega$$. the internal resistance of each cell is $$5\Omega$$. The potential difference $$E_A$$ and $$E_B$$ across the terminals of the cells A and B respectively are:
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$$E_A = 2.0 V. E_B = 1.5 V$$
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$$E_A = 2.125 V, E_B = 1.375 V$$
0%
$$E_A = 1.875 V, E_B = 1.625 V$$
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$$E_A = 1.875 V, E_B = 1.375 V$$
Explanation
$$R_{eq} = 10 + 5 + 5 = 20 \Omega$$
$$V_{eq} = 2 - 1.5 = 0.5V$$
$$I = \dfrac{V_{eq}}{R_{eq}} \\ \\ I = \dfrac{0.5}{20} \\ I = 0.025 A$$
$$E_A = 2 - R_AI \\ E_A = 2 - 5 * 0.025 = 2 - 0.125 = 0.875 V $$
$$E_B = 1.5 + R_BI \\ E_B = 1.5 + 5 * 0.025 = 1.5 + 0.125 = 1.625V $$
Hence option 'C' is correct.
A meter bridge shown in fig.is balanced.what is the value of resistance X.
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$$ 20 \Omega $$
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$$ 10 \Omega $$
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$$ 5 \Omega $$
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$$ 2.5 \Omega $$
Explanation
Since the meter bridge is balanced,
$$\dfrac{10}{0.5} = \dfrac{X}{0.5}$$
$$X = 10 \Omega$$
In shown diagram, if voltage of battery is doubled and resistance kept constant then what is the new Power dissipated at resistor(Consider $$P$$ was the initial power dissipation) ?
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0%
$$\dfrac{P}{4}$$
0%
$$\dfrac{P}{2}$$
0%
P
0%
2P
0%
4P
Explanation
Power, $$P=\dfrac{V^2}{R}$$
As R is kept constant so $$P \propto V^2$$
$$P'$$is new power dissipation.
So, $$\dfrac{P'}{P}=\dfrac{(2V)^2}{V^2}=\dfrac{4V^2}{V^2}$$ or $$P'=4P$$
Thus, option E is correct.
Potentiometer measures the potential difference more accurately than a voltmeter because :
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it has a wire of high resistance
0%
it has a wire of low resistance
0%
it does not draw current from external circuit
0%
it draws a heavy current from external circuit
Explanation
When we measure the emf of a cell by the potentiometer then no current flows in the circuit in zero-deflection condition ie, cell is in open circuit. Thus, in this condition the actual value of a cell is found. In this way, potentiometer is equivalent to an ideal voltmeter of infinite resistance.
Note. The emf in the potentiometer is measured by null method in which zero deflection position is found on the wire.
A $$2V$$ battery, a$$990 \Omega$$ resistor and a potentiometer of $$2m$$ length, all are connected in series of the resistance of potentiometer wire is $$10\Omega$$, then the potential gradient of the potentiometer wire is
Report Question
0%
$$0.05Vm^{-1}$$
0%
$$0.5Vm^{-1}$$
0%
$$0.01Vm^{-1}$$
0%
$$0.001Vm^{-1}$$
Explanation
Potential gradient $$x=\displaystyle\frac{e}{(R+R_{h}+r)}\cdot \frac{R}{L}$$
$$\Rightarrow \displaystyle x=\frac{2}{(990+10)}\times \frac{10}{2}=0.01Vm^{-1}$$
The circuit shows a battery and a variable resistor supplying power to a light circuit. As the variable resistor is moved towards left, the bulbs dim a little. What happens if the variable resistor is moved all the way to the right?
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0%
The lights brighten considerably
0%
The total circuit resistance increases
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The total circuit current decreases
0%
The total applied voltage increases
0%
The total power used in the circuit increases
Explanation
When the resistor is moved towards left , total resistance of the circuit increases and hence current in circuit decreases , so bulbs dim a little . The reverse process will happen when the resistor is moved towards right , total resistance decreases and current increases therefore bulbs will light brighten considerably and also the total power used will increase.
If current in a source of e.m.f. is in the direction of e.m.f., the energy of the source
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0%
Increase
0%
Decrease
0%
Remains constant
0%
Zero
Explanation
This is charging of source . The current is stored in the form of chemical energy in battery . Higher potential is applied across the battery to overcome the emf of battery.
The five different colored graphs above represent different possible relationship between voltage across a resistor and current through the resistor.
Which of the graphs shows the correct relationship for a resistor that obeys Ohm's law?
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Red
0%
Orange
0%
Blue
0%
Purple
Explanation
Ohm's law states that if physical condition are unchanged then voltage is directly proportional to current ,
$$V\propto I$$ ,
this relation is best depicted by blue graph , therefore blue graph obeys Ohm's law .
What do you add to distilled water for making it to conduct electricity.
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0%
sulphuric acid
0%
water
0%
milk
0%
base
In a potentiometer experiment of a cell of emf 1.25 V gives balancing length of 30 cm. If the cell is replaced by another cell, balancing length is found to be 40 cm. What is the emf of second cell?
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0%
$$\simeq 1.57 V$$
0%
$$\simeq 1.67 V$$
0%
$$\simeq 1.47 V$$
0%
$$\simeq 1.37 V$$
Explanation
Given : $$E_1 = 1.25$$ V $$l_1 = 30$$ cm $$l_2 = 40$$ cm
Emf of the cell $$E\propto l$$
$$\implies$$ $$\dfrac{E_2}{E_1} = \dfrac{l_2}{l_1}$$
$$\therefore$$
$$\dfrac{E_2}{1.25} = \dfrac{40}{30}$$
$$\implies$$ $$E_2 = 1.67$$ V
Kirchoffs 1st and 2nd laws are based on conservation of
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Energy and charge respectively
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Charge and energy respectively
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Mass and charge respectively
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None of these above
Explanation
Kirchhoff's $$1$$st law states that the total amount of current coming at a junction must be equal to the total amount of current going away from it i.e. the total charge must be conserved. Thus
Kirchhoff's $$1$$st law is based on the conservation of charge.
Kirchhoff's $$2$$nd law states that the sum of the potential drop across all the components in a loop must be zero. Thus
Kirchhoff's $$2$$nd law is based on the conservation of energy.
A steady electric current is flowing through a cylindrical
conductor.
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The electric field at the axis of the conductor is zero.
0%
The magnetic field at the axis of the conductor is zero.
0%
The electric field in the vicinity of the conductor is zero.
0%
The magnetic field in the vicinity of the conductor is zero.
Explanation
Electric field that should be created when a steady current is passing through a part of a circuit, then the conducting cylindrical wire will not produce an electric field but only magnetic field as there is flow of just electrons and not protons. The electric field in the vicinity,on the other hand will be zero because the conductor(here) is electrically neutral.
Human body is a
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good conductor of electricity
0%
poor conductor of electricity
0%
semiconductor
0%
good insulator
Explanation
Human body is a good conductor of electricity which allow electric current to flow.
Answer-(A)
Potentiometer measures potential more accurately because
Report Question
0%
it measures potential in the open circuit
0%
it uses sensitive galvanometer for null deflection
0%
it uses high resistance potentiometer wire
0%
it measures potential in the closed circuit
Explanation
Potentiometer measures voltage when galvanometer shows zero current rating, mean it takes zero current (open circuit) while measuring voltage across any component, that is why it is more accurate as all the current passing through that component only.
$$1$$ kilowatt hr $$=$$ __________ joules.
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0%
$$10.6 \times {10}^{6}$$
0%
$$3.6 \times {10}^{6}$$
0%
$$30.6 \times {10}^{6}$$
0%
$$3.6 \times {10}^{5}$$
Explanation
$$kWh$$ is the commercial unit of energy.
$$1\ kWh =1\times 1000\times 3600 = 3.6\times 10^6$$ joules
For driving current of $$2A$$ for $$6$$ minute in a circuit. $$1000 J$$ of work is to be done. The e.m.f. of the source in the circuit is
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0%
$$1.38V$$
0%
$$1.68V$$
0%
$$2.03V$$
0%
$$3.10V$$
Explanation
Current passing through the circuit $$I = 2 A$$
Time of operation $$t = 6$$ minute $$ = 360$$ s
Work done $$W = 1000 J$$
Using $$W = VIt$$
$$\therefore$$ $$1000 = V (2)(360)$$
$$\implies$$ $$V = 1.38$$ volts
Electromotive force represents
Report Question
0%
force
0%
energy
0%
energy per unit charge
0%
current
Explanation
Electromotive force is the potential difference (or voltage $$V$$) of the cell.
Using $$V = \dfrac{W}{q}$$
where $$W$$ is the work done in moving a charge $$q$$ from one point to another having potential difference $$V$$.
Thus electromotive force represents energy per unit charge.
Four bulbs marked $$40W, \ 250V$$ are connected in series with $$250\ V$$ mains, the total power consumed is
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0%
$$10W$$
0%
$$40W$$
0%
$$320W$$
0%
$$160W$$
Explanation
Power of each bulb,
$$P = 40 W$$
Total power consumed when four bulbs are connected in series
$$\dfrac{1}{P_s} = \dfrac{1}{P_1}+$$
$$\dfrac{1}{P_2} + \dfrac{1}{P_3}+\dfrac{1}{P_4}$$
$$\therefore$$
$$\dfrac{1}{P_s} = \dfrac{1}{P}+$$
$$\dfrac{1}{P} + \dfrac{1}{P}+\dfrac{1}{P}$$
We get
$$P_s =\dfrac{P}{4}$$
$$\Rightarrow P_s = \dfrac{40}{4}$$
$$\Rightarrow P_s =10 W $$
A potentiometer has a driving cell of negligible internal resistance. The balancing length of a Daniel cell is $$5m$$. If the driving cell had internal resistance, the balancing length of the same Denial cell would have been
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0%
more
0%
less
0%
same
0%
cannot be said from the data
Explanation
Let resistance of $$AB=R$$ and power supply has voltage $$= V$$, $$AB= L$$ and $$AC=5m$$
Current in the circuit $$I=\dfrac{V}{R}$$
So balancing voltage $$V_{Bal} = R_{AC}\times I=(R\times \dfrac{5}{L})\times \dfrac{V}{R}=\dfrac{5V}{L}$$ .... (1)
Now consider internal resistance of power supply $$= r$$ and balancing length $$= l$$
Current in the circuit $$I_1 = \dfrac{V}{R+r}$$
Now balancing voltage $$V_{Bal}=( R\times\dfrac{l}{L})\times \dfrac{V}{R+r}$$ ........(2)
By equation (1)&(2) - $$\dfrac{5V}{L} =(R\times\dfrac{l}{L})\times \dfrac{V}{R+r } \implies l = 5(1+\dfrac{r}{R}) \implies l>5m$$
Which of the following devices has a source of emf inside it?
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0%
Voltmeter
0%
Ammeter
0%
Ohm-meter
0%
Rectifier
Explanation
Ohmmeter works either by applying a voltage from an internal battery to the unknown and measuring current (analog ohmmeters) or by applying a current and measuring voltage (digital ohmmeters).
A battery of e.m.f. $$E$$ has an internal resistance '$$r$$'. A variable resistance $$R$$ is connected to the terminals of the battery. A current $$I$$ is drawn from the battery. $$V$$ is the terminal P.D. lf $$R$$ alone is gradually reduced to zero, which of the following best describes $$I$$ and $$V$$?
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0%
$$I$$ approaches $$E / r$$, $$V$$ approaches $$E$$
0%
$$I$$ approaches infinity, $$V$$ approaches $$E$$
0%
$$I$$ approaches zero, $$V $$approaches $$E$$
0%
$$I$$ approaches $$E / r$$, $$V$$ approaches zero
Explanation
$$\textbf{Step 1: Finding Current}$$
From Figure,
Total resistance of the circuit $$R_{eq} = R + r$$
So, Current in the circuit $$I = \dfrac{E}{R_{eq}} = \dfrac{E}{R+r}$$ $$....(1)$$
As $$R\rightarrow 0$$, we get $$I \rightarrow \cfrac{E}{r}$$
$$\textbf{Step 2: Terminal Potential Difference(V) across the battery}$$
From Figure:
$$V=V_{AB} = V_{CD}=$$ $$I R$$
$$\Rightarrow V = \dfrac{ER}{R+r}$$ $$($$ Using Equation $$(1))$$
From above Equation, As $$R\rightarrow 0$$, we get $$V \rightarrow 0$$.
Thus option $$D$$ is correct.
$$1 Wh$$ (Watt hour) is equal to :
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0%
$$36\times { 10 }^{ 5 }J$$
0%
$$36\times { 10 }^{ 4 }J$$
0%
$$3600 J$$
0%
$$3600 { Js }^{ -1 }$$
Explanation
Watt is a unit of power such that $$1W =1\dfrac{J}{s}$$
Also we know $$1h = 3600$$ $$s$$
$$\therefore$$ $$1Wh = 1\dfrac{J}{s} \times 3600 $$ $$s = 3600$$ $$J$$
The figure shows in apart of an electric circuit, then the current I is
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0%
$$1 A$$
0%
$$3A$$
0%
$$2 A$$
0%
$$4 A$$
Explanation
According to the Kirchhoff's first law, amount of current entering is equal to amount of current leaving.
From the circuit,
$$I+6+1=5+3+2$$
$$I+7=10$$
$$I=3A$$
0:0:1
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Practice Class 12 Medical Physics Quiz Questions and Answers
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