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Multiple choice questions on EMF and Electric measurements(Electric Circuits)






Multiple Choice Questions

Question 1
For accurate measurements, the resistance of a voltmeter should be
  1. infinite
  2. as small as possible
  3. as large a possible
  4. equal to the resistance across which potential difference is to be measured

Answer

Voltmeter when connected across any circuit element should not draw any current from it. If the voltmeter resistance is not high as compared to the resistance of the circuit element, across which it is connected, the measured voltage becomes less. Thus resistance of the voltmeter should be as large as possible.
Hence (c) is the correct answer


Paragraph Type Questions

(A)Use figure given below for questions 2 to 4
Consider the circuit given below
Multiple choice questions on EMF and Electric measurements for JEE Main and Advanced
Question 2
The rate at which energy is dissipated is 20Ω resistor
(a) 14.2 Watt
(b) 8.97 Watt
(c) 97.7 Watt
(d) 47.2 Watt
Question 3
Current through 10Ω resistor is
  1. 1.54 A
  2. 0.67 A
  3. 1.22 A
  4. 2.21 A
Question 4
Power of battery having emf E1=20V is
  1. 44.2 Watt
  2. 30.8 Watt
  3. 13.4 Watt
  4. 20 Watt

Answer(2-4)

(i) Consider the circuit given in the question
Multiple choice questions on EMF and Electric measurements for JEE Main and Advanced
Rate at which energy is dissipated in 20 Ω is
P=I2R
So we need to calculate the current through 20 Ω resistor. Applying Kirchoff’s loop rule in ABCDA we get equation
E1 -6I1 +10 I2 =0
or
20 -6I1 +10 I2 =0  ----(1)
Applying Kirchoff’s loop rule in CDEFC we get equation
E2 -10I2 -20I3  =0
24- 10I2 -20I3  =0        ---(2)
From Kirchoff’s junction rule at point C
I1 + I2 = I3
putting this in equation 2 we find
24-30I2 -20I1 =0       -----(3)
Solving equation 1 and 3 for the value of I1 we get
I1=2.21A
Again from equation 1 we find
I2=-0.67A here consider reversing the direction of current from the assumed one.
And finally
I3=1.54A
Rate at which energy is dissipated in 20 Ω is
P=I2R=(1.54)2x20=47.2 Watt
Hence (d) is the correct answer
(ii) I2 is the current through 10Ω resistor
Hence (b) is the correct answer
(iii) Power of battery is
Pemf=iE
Power supplied by battery E1 = I1E1=44.2 Watt
Hence (a)  the correct answer


Multiple Choice Questions

Question 5
Consider the figure given below
Multiple choice questions on EMF and Electric measurements for JEE Main and Advanced
If ammeter shows a reading of 10A and voltmeter having internal resistance 3000 Ω measures a voltage of 200V, the resistance R is given by
  1. 201.3 Ω
  2. 2013 Ω
  3. 20.13 Ω
  4. 20.00 Ω

Answer

Applying Ohm’s law in section PQ
V=IR
The resistances, R and internal resistance of the voltmeter are connected in parallel
$\frac {I}{V} =\frac {1}{R}  + \frac {1}{R_v}$
or
$\frac {1}{R} = \frac {I}{V} - \frac {1}{R_v} = \frac {R_v I -V}{VR_v}$
Thus
$R=\frac {V}{I - \frac {V}{R_v}}$
putting the respective values and calculating we find R=20.13Ω
Hence (c) the correct answer


Question 6
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if
(a) both length and radius of wire are halved
(b) both length and radius of wire are doubled
(c) the radius of wire id doubled
(d) length of wire is doubled

Answer

$H=\frac {V^2 t}{R}$
but
$R=\frac {\rho L}{A} =\frac {\rho L}{\pi r^2}$
Hence
$H=\frac {V^2 t r^2 \pi}{\rho L}
Hence (b) the correct answer


Question 7
If a wire of resistance 2Ω is covered with ice and a voltage of 210V is applied across the wire , then rate of melting ice would be
  1. 0.85g/s
  2. 1.92g/s
  3. 6.56g/s
  4. none of the above

Answer

Rate at which heat is produced in the wire is
$\frac {heat}{time} =\frac {V^2}{R} = 2205 Js^-1$
Latent heat of ice=80c/gm=80x4.2J/gm
Hence rate of melting ice would be =$\frac {2205}{80X4.2} =6.56g/s$
Hence (c) the correct answer


Question 8
Which of the following statements is/are correct?
  1. Both Peltier and Joule effects are reversible
  2. Both Peltier and Joule effects are irreversible
  3. Joule effect is reversible and Peltier effect is irreversible
  4. Joule effect is irreversible and Peltier effect is reversible

Answer

(d) is the correct answer



Question 9
The arrangement shown below is of the meter bridge experiment. Here AC=x corresponds to the null deflection in the galvanometer. What will be the value of AC if the radius of the wire AB is doubled?
Multiple choice questions on EMF and Electric measurements for JEE Main and Advanced
  1. x/2
  2. x/4
  3. 2x
  4. x

Answer

When radius of the wire is changed , it will not affect the ratio of resistance of two arms AC and BC of the slide wire bridge.
Hence (d) is the correct answer


Assertion Reason type questions

  1. Both assertion and reason are true and reason is correct explanation of assertion
  2. Both assertion and reason are true and reason is not the correct explanation of assertion
  3. Assertion is true but reason is false
  4. Both assertion and reason are false.
 
Question 10
Assertion
The distribution of currents within the network for a given amount of current always takes place in a way that leads to minimum total power distribution in the circuit
Reason
For maximum output current by a source the source resistance must be equal to external load resistance.
 
Question 11
Assertion
When temperature of a cold junction of a thermocouple is lowered , the value of neutral temperature of this thermocouple rises
Reason
More thermo emf is produced when temperature difference between two junctions of thermocouple is raised.
 
Question 12
Assertion
A domestic appliance, working on a three pin , will continue working even if the top pin is removed
Reason
Third pin is used as a safety device.
 
Question 13
Assertion
A potentiometer of larger length is used for accurate measurements
Reason
Potential gradient for potentiometer of larger length with a given battery becomes small.

Answer

10 (b)
11.(d)
12 (a)
13 (a)
 


Multiple Choice Questions

Question 14
12 wires each of resistance RΩ are connected in the form of a skeleton cube as shown below in the figure
Practice Questions on EMF and Electric measurements for JEE Main and Advanced
If current I enter at point A and leaves at diagonally opposite point G then the equivalent resistance of the cube would be
(a)12R/5
(b)12R
(c) 6R/5
(d) 5R/6

Answer

Given that current I enter the cube at point A and leaves at point G as shown below in the figure

Here we have
I1=2I2
And also from point rule we have I=3I1
If E is the emf connected across the system then
E=IReq
Applying Kirchhoff’s law to any path between A and G , say ABFG , we get
I1R+I2R+I1R=V=IReq
Putting I1=2I2 in above equation we get
5I2R=IReq
Since I=3I1 so we have I=6I2
From this we have
5I2R=6I2Req
Or, Req=5R/6


Linked comprehension type question

(B) Consider the figure given below
Practice Questions on EMF and Electric measurements for JEE Main and Advanced
Here AB is a wire of total resistance R0 . A jockey connected at point C can divide the wire into resistors of resistance pR0 and (p-1)R0 . Assuming that the batteries are identical and have zero internal resistance
Question 15
Current through ideal ammeter for any 0<p<1 is
Practice Questions on EMF and Electric measurements for JEE Main and Advanced

Answer

Let I be the amount of current through the ammeter, I1 be the current due to left hand EMF and $I_2$ be the current due to right hand emf. Applying Kirchhoff's law we find
$E-Ir-I_1pR_0=0$
$E-Ir-I_2(1-p)R_0=0$
and
$I=I_1+I_2$
Solving for I we get
$I=\frac{E}{\left(r+R_0p-R_0p^2\right)} $


Question 16
Value of p for which ammeter reads maximum value is
(a) 0
(b) 1
(c) 2
(d) 3

Answer

For maximum I denominator should be smallest which is for two values of p i.e., for p=0,1.


(C) Consider the circuit given below
Practice Questions on EMF and Electric measurements for JEE Main and Advanced
Question 17
Total resistance in the circuit is
(a) 5Ω
(b) 6Ω
(c) (11/6)Ω
(d) 3Ω

Answer

From the figure in the question we see that all the resistors are connected in the series combination. So equivalent resistance of the circuit is
$R=R_1+R_2+R_3 = 6$ ω


Question 18
Current in the circuit is
(a) 4A
(b) 12A
(c) 6A
(d) 2A

Answer

Total current in the circuit is I=V/R = 12/6 =2A


Question 19
Potential difference between points A and E is
(a) 12V
(b) 6V
(c) 10V
(d) 5V

Answer

For finding potential difference between points A and E we first need to find the effective resistance between these two points.
$R_{AE} = 2+3=5$ ω
$V_{AE} = IR_{AE} = 10$ V


Question 20
Potential at point E is
(a) 10 V above the ground
(b) 10 V below the ground
(c) 2 V above the ground
(d) 2 V below the ground

Answer

$V_E$ is the voltage across the resistor $R_3$ . So,
$V_E=R_3I = +2$ V above the ground


Question 21
Power supplied by the battery is
(a) 12 W
(b) 10W
(c)22W
(d)24W

Answer

Power supplied is P=VI = 24W


Matrix match type question

Question 22
Consider the two circuits given below
Practice Questions on EMF and Electric measurements for JEE Main and Advanced
Column A
Column B
(P) Point of lowest potential in circuit A has its value equal to
(U) 0.15 A
 
(Q) Circuit current in circuit B is
(V) 4V
(R) Circuit current in circuit A is
(W) 15V
(S) Potential of point B in circuit B is
(X) 90V
(T) Potential at point B in circuit A is 
(Y) .5A

Answer

First consider circuit A for finding value of lowest potential. Point of lower potential in circuit A is the point of maximum negative potential and its value is
V=IR
Where
I= 24/160 = .15 A is the current in the circuit.
So,
Lowest potential is
V=-0.15 × 100 = -15 V
That is 15 V below the ground.
Now potential at point B in circuit A = 0.15 × 60 =+9V

Now we find circuit current in circuit B
Net voltage= 12-6 = 6V
Total resistance in the circuit = 4+8 = 12 ω
And circuit current is, I =6/12 = .5A
Now potential at point B in circuit B = $V_A-(4 \times 0.5)$
Again potential at point A in circuit B =12-6=+6V which is positive w.r.t. ground
$V_B = 6-2= +4V$
P -> W ; Q -> Y ; R - >U ; S ->V ; T- >X



Multiple Choice Questions

Question 23
In the circuit shown below the reading of ammeter is the same when both switches open as both switches close.
Practice Questions on EMF and Electric measurements for JEE Main and Advanced
The value of resistance R is
(a) 800Ω
(b) 900Ω
(c) 1200Ω
(d) 1000Ω

Answer

When both the switches are open resistances $R_1$, $R_2$ and $R_3$ are in series combination,So current through ammeter would be
$I_A= \frac {E}{R_1 + R_2 + R_3}$

Again with both Switches closed

Voltage at points a and b would be same and no current passes through resistance $R_3$. Now current through E and $R_2$
$I'=\frac {E}{R_2 + \frac {RR_1}{R_1 + R}}$
Current through Ammeter would be the current through resistor $R_1$. If i is current through resistor R then I' -i would be the current through $R_1$ and ammeter. So from Kirchoff's law
$(I' -i)R_1 = iR$
or
$i= \frac {I' R_1}{R+ R_1}$
So current through ammeter
$=I' -i = I' - \frac {I' R_1}{R+ R_1}= \frac {I'R}{R+R_1}$
or
$=\frac {RE}{(R+R_1)R_2 + RR_1}$
Now since Ammeter reports same reading before and after
$\frac {RE}{(R+R_1)R_2 + RR_1}=\frac {E}{R_1 + R_2 + R_3}$
$R(R_1 + R_2 + R_3)=(R+R_1)R_2 + RR_1$
$RR_1 + RR_2 + RR_3 = RR_2 + R_1 R_2 + RR_1$
or
$R= \frac {R_1}{R_2}{R_3} = \frac {200 \times 600}{100} = 1200$ ω


Linked comprehension type question

(D) For questions from 24 to 26 consider the statement given below
A 20µF capacitor which is initially uncharged is connected to a 6V battery through a resistor with resistance equal to 200Ω
Question 24
Magnitude of final charge q0 on the capacitor is
Practice Questions on EMF and Electric measurements for JEE Main and Advanced
Question 25
How long would it take for capacitor to be charged to (1/2)q0  after it is connected to the battery
(a) 3.8×10-3s
(b) 2.8×10-3s
(c) 2.8×10-4s
(d) 3.8×10-4s
Question 26
The time taken by the capacitor to be charged to 0.80q0 is
Practice Questions on EMF and Electric measurements for JEE Main and Advanced

Answer

In charging of a capacitor we have
$q=q_0(1-e^{-\frac{t}{RC}})$ --- (1)
Where $q_0 = CV$
(a) $q_0=CV=\left(20\times{10}^{-6}F\right)\left(6V\right)=\ 120\mu C$
(b) Time constant is given by
$\tau=RC=\left(200\Omega\right)\left(20\times{10}^{-6}F\right)=4\times{10}^{-3}s$.
Now from equation 1 if charge reaches ${\frac{1}{2}q_0}$ in time t then
$ e^{-\frac{t}{RC}}=\frac{1}{2}$
Taking log on both the sides and solving for t we get
$t=t_{1/2}=2.8 \times 10^{-3} $ s
(c) $-\frac{t}{\tau}=ln 0.2$
or, $\ t=\tau\times1.6=4\times{10}^{-3}\times1.6=6.4\times{10}^{-3}\ $s


Matrix match type question

Question 27
In the figure given below the circuit contains four capacitors having same capacitance of 10µF and a battery providing a voltage of 90V
Match the columns given below
 
Column A
Column B
(P) When switch S2 is open and S1 is closed and then opened after C1 , C2 and C3 are fully charged. Electric potential difference across each capacitor is
(W) 0 V
(Q) Now after opening S1 if switch S2 is closed, the electric potential difference across each capacitor is
(X) V1=V3=36V , V2=V4=18V
(R) In another case if switch S1 is open and switch S2 is first closed then electric potential difference across each capacitor is
(Y) V=30 V
(S) Now if switch S1 is closed , the potential difference across each capacitor is
(z) V1=V3=30V , V2=V4=15V

Answer

First when switch $S_2$ is open and $S_1$ is closed the capacitor $C_4$ is not included in the circuit and the effective circuit is shown below in the figure.


If battery provides a voltage of 90V then $V_1+V_2+V_3=V$ where $V_1$, $V_2$, and $V_3$ are voltages across the capacitors $C_1$, $C_2$, $C_3$ respectively.
Since all the capacitors are in series combination so charge on all the capacitor plates is same say, Q. Also given that
$C_1=C_2=C_3=C=10 \mu F$
$Q=C_1V_1= C_2V_2= C_3V_3$
$V_1=V_2=V_3=V'$
which gives ,
$3V'=V$ or $V'=90/3 = 30$V
Once the capacitors are charged this way, the potential difference across each capacitor will remain the same even after opening the switch $S_1$.

If we now open switch $S_1$ then battery is excluded from the circuit after charging of the capacitors $C_1$, $C_2$ and $C_3$. Now closing switch $S_2$ will bring $C_4$ in the circuit and effective circuit would be


Capacitors $C_1$ and $C_3$ will still carry the same charge as before and opening of switch $S_1$ leaves them isolated from any flow of charge. Charge Q originally on capacitor $C_2$ is distributed between capacitor $C_2$ and $C_4$. As $C_2=C_4$ so charge on each capacitor equals Q/2 and voltage across $C_2$ and $C_4$ would be equal to half the original voltage across C2. Thus
$V_1=V_3=30V$ , $V_2=V_4=15V$

Again with capacitors initially uncharged and switch $S_1$ open , closing $S_2$ does not provide any potential difference across any capacitor as battery is effectively disconnected.

Now we close $S_1$ with $S_2$ closed previously and circuit arrangement is as shown below in the figure.

Effective capacitance of parallel combination between capacitors $C_2$ and $C_4$ is
$C_{24} = C+C = 2C$
And the entire system has equivalent capacitance
$\frac{1}{C_{eq}}=\left(\frac{1}{C_1}+\frac{1}{C_{24}}+\frac{1}{C_3}\right)$
Or,
$C_{eq}=\frac{2}{5}C$
Thus, charge drawn from the battery onto each of $C_1$, $C_24$ and $C_3$ is,
$Q'=C_{eq}V={\frac{2}{5}CV}$
$V_1=V_3=\frac{Q'}{C}=\frac{2}{5}CV=36V$
$V_2=V_4=\frac{Q'}{2C}=\frac{1}{5}CV=18V$
Hence P ->Y ; Q -> Z ; R -> W ; S ->X


Multiple Choice Questions

Question 28
Kirchhoff's junction rule is a reflection of
(a) conservation of current density vector.
(b) conservation of charge.
(c) the fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction.
(d) the fact that there is no accumulation of charges at a junction

Answer

(b), (d)










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