Multiple Choice Questions
Question 1
For accurate measurements, the resistance of a voltmeter should be
- infinite
- as small as possible
- as large a possible
- equal to the resistance across which potential difference is to be measured
Solution
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
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
- 1.54 A
- 0.67 A
- 1.22 A
- 2.21 A
Question 4
Power of battery having emf E
1=20V is
- 44.2 Watt
- 30.8 Watt
- 13.4 Watt
- 20 Watt
Solution 2, 3, 4
(i) Consider the circuit given in the question
Rate at which energy is dissipated in 20 Ω is
P=I
2R
So we need to calculate the current through 20 Ω resistor. Applying Kirchoff’s loop rule in ABCDA we get equation
E
1 -6I
1 +10 I
2 =0
or
20 -6I
1 +10 I
2 =0 ----(1)
Applying Kirchoff’s loop rule in CDEFC we get equation
E
2 -10I
2 -20I
3 =0
24- 10I
2 -20I
3 =0 ---(2)
From Kirchoff’s junction rule at point C
I
1 + I
2 = I
3
putting this in equation 2 we find
24-30I
2 -20I
1 =0 -----(3)
Solving equation 1 and 3 for the value of I
1 we get
I
1=2.21A
Again from equation 1 we find
I
2=-0.67A here consider reversing the direction of current from the assumed one.
And finally
I
3=1.54A
Rate at which energy is dissipated in 20 Ω is
P=I
2R=(1.54)
2x20=47.2 Watt
Hence (d) is the correct answer
(ii) I
2 is the current through 10Ω resistor
Hence (b) is the correct answer
(iii) Power of battery is
P
emf=iE
Power supplied by battery E
1 = I
1E
1=44.2 Watt
Hence (a) the correct answer
Multiple Choice Questions
Question 5
Consider the figure given below
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
- 201.3 Ω
- 2013 Ω
- 20.13 Ω
- 20.00 Ω
Solution
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
Solution
$H=\frac {V^2 t}{R}$
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
- 0.85g/s
- 1.92g/s
- 6.56g/s
- none of the above
Solution
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?
- Both Peltier and Joule effects are reversible
- Both Peltier and Joule effects are irreversible
- Joule effect is reversible and Peltier effect is irreversible
- Joule effect is irreversible and Peltier effect is reversible
Solution
(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?
- x/2
- x/4
- 2x
- x
Solution
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
- Both assertion and reason are true and reason is correct explanation of assertion
- Both assertion and reason are true and reason is not the correct explanation of assertion
- Assertion is true but reason is false
- 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.
Solution 10-13
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
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
Solution
Given that current I enter the cube at point A and leaves at point G as shown below in the figure
Here we have
I
1=2I
2
And also from point rule we have I=3I
1
If E is the emf connected across the system then
E=IR
eq
Applying Kirchhoff’s law to any path between A and G , say ABFG , we get
I
1R+I
2R+I
1R=V=IR
eq
Putting I
1=2I
2 in above equation we get
5I
2R=IR
eq
Since I=3I
1 so we have I=6I
2
From this we have
5I
2R=6I
2R
eq
Or, R
eq=5R/6
Linked comprehension type question
(B) Consider the figure given below
Here AB is a wire of total resistance R
0 . A jockey connected at point C can divide the wire into resistors of resistance pR
0 and (p-1)R
0 . Assuming that the batteries are identical and have zero internal resistance
Question 15
Current through ideal ammeter for any 0<p<1 is
Solution
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
Solution
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
Question 17
Total resistance in the circuit is
(a) 5Ω
(b) 6Ω
(c) (11/6)Ω
(d) 3Ω
Solution
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
Solution
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
Solution
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
Solution
$V_E$ 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
Solution
Power supplied is P=VI = 24W
Matrix match type question
Question 22
Consider the two circuits given below
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
|
Solution
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.
The value of resistance R is
(a) 800Ω
(b) 900Ω
(c) 1200Ω
(d) 1000Ω
Solution
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 q
0 on the capacitor is
Question 25
How long would it take for capacitor to be charged to (1/2)q
0 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 0.80q
0 is
Solution 24-26
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
|
Solution
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
Solution
(b), (d)
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Class 12 Maths
Class 12 Physics
