Maths-
Period of 
Maths-General
Answer:The correct answer is:
Book A Free Demo
+91 
Grade*
Select Grade
Related Questions to study
physics-
In a network as shown in the figure, the potential difference across the resistance 2R is (the cell has an emf of E volt and has no ingternal resistance)
In the given circuit, resistors 4R and 2R are connected in parallel while resistance R is connected in series to it.
Hence, equivalent resistance is
Given, emf is E volts, therefore
Potential difference across R is
Potential difference across 2R is
Hence, equivalent resistance is
Given, emf is E volts, therefore
Potential difference across R is
Potential difference across 2R is
In a network as shown in the figure, the potential difference across the resistance 2R is (the cell has an emf of E volt and has no ingternal resistance)
physics-General
In the given circuit, resistors 4R and 2R are connected in parallel while resistance R is connected in series to it.
Hence, equivalent resistance is
Given, emf is E volts, therefore
Potential difference across R is
Potential difference across 2R is
Hence, equivalent resistance is
Given, emf is E volts, therefore
Potential difference across R is
Potential difference across 2R is
physics-
In the adjoining figure the equivalent resistance between A and B is
Equivalent circuit of the given circuit is
Between points C and D resistors 2
, 2 and 2 are in series, therefore, their equivalent resistance,
Resistors R’ and 6 are in parallel, therefore their equivalent resistance is given by
Now between points A and B 1, 3
and 1 are in series.
Therefore, resultant resistance is
R=1+3+1=5
Between points C and D resistors 2
, 2 and 2 are in series, therefore, their equivalent resistance,Resistors R’ and 6
are in parallel, therefore their equivalent resistance is given byNow between points A and B 1
, 3 and 1 are in series.Therefore, resultant resistance is
R=1+3+1=5
In the adjoining figure the equivalent resistance between A and B is
physics-General
Equivalent circuit of the given circuit is
Between points C and D resistors 2, 2 and 2 are in series, therefore, their equivalent resistance,
Resistors R’ and 6 are in parallel, therefore their equivalent resistance is given by
Now between points A and B 1, 3 and 1 are in series.
Therefore, resultant resistance is
R=1+3+1=5
Between points C and D resistors 2
, 2 and 2 are in series, therefore, their equivalent resistance,Resistors R’ and 6
are in parallel, therefore their equivalent resistance is given byNow between points A and B 1
, 3 and 1 are in series.Therefore, resultant resistance is
R=1+3+1=5
maths-
Period of 
 is
Period of  is
maths-General
physics-
The charge on the capacitor of capacitance 
shown in the figure below will be
The charge on the capacitor of capacitance shown in the figure below will be
physics-General
physics-
Each resistance shown in figure is 2. The equivalent resistance between A and B is
Given circuit is a balanced Wheatstone bridge. So, diagonal resistance of 2 will be ineffective.
Equivalent resistance of upper arms
=2+2=4
Equivalent resistance of lower arms
=2+2=4
will be ineffective.Equivalent resistance of upper arms
=2+2=4
Equivalent resistance of lower arms
=2+2=4
Each resistance shown in figure is 2. The equivalent resistance between A and B is
physics-General
Given circuit is a balanced Wheatstone bridge. So, diagonal resistance of 2 will be ineffective.
Equivalent resistance of upper arms
=2+2=4
Equivalent resistance of lower arms
=2+2=4
will be ineffective.Equivalent resistance of upper arms
=2+2=4
Equivalent resistance of lower arms
=2+2=4
physics-
The equivalent resistance across A and B is
The equivalent circuit can be redrawn as
we have,
So, the given circuit is a balanced Wheatstone’s bridge.
Hence, the equivalent resistance
we have,
So, the given circuit is a balanced Wheatstone’s bridge.
Hence, the equivalent resistance
The equivalent resistance across A and B is
physics-General
The equivalent circuit can be redrawn as
we have,
So, the given circuit is a balanced Wheatstone’s bridge.
Hence, the equivalent resistance
we have,
So, the given circuit is a balanced Wheatstone’s bridge.
Hence, the equivalent resistance
physics-
Figures i) and ii) below show the displacement-time graphs of two particles moving along the 
-axis. We can say that
Figures i) and ii) below show the displacement-time graphs of two particles moving along the -axis. We can say that
physics-General
physics-
The graph of displacement 
time is
Its corresponding velocity-time graph will be
We know that the velocity of body is given by the slope of displacement 
time graph so it is clear that initially slope of the graph is positive and after some time it becomes zero (corresponding to the peak of graph) and it will become negative
time graph so it is clear that initially slope of the graph is positive and after some time it becomes zero (corresponding to the peak of graph) and it will become negativeThe graph of displacement time is
Its corresponding velocity-time graph will be
physics-General
We know that the velocity of body is given by the slope of displacement time graph so it is clear that initially slope of the graph is positive and after some time it becomes zero (corresponding to the peak of graph) and it will become negative
time graph so it is clear that initially slope of the graph is positive and after some time it becomes zero (corresponding to the peak of graph) and it will become negativephysics-
A particle starts from rest at 
and moves in a straight line with an acceleration as shown below. The velocity of the particle at 
is
Velocity of graph 
Area of
-
graph
Area of- graphA particle starts from rest at and moves in a straight line with an acceleration as shown below. The velocity of the particle at is
physics-General
Velocity of graph Area of- graph
Area of- graphmaths-
Period of 
 is
Period of  is
maths-General
physics-
There resistances of 4 each are connected as shown in figure. If the point D divides the resistance into two equal halves, the resistance between points A and D will be
The equivalent circuit is given by
4 and 2 resistances are in series on both sides.
Then 6 and6 resistances are in parallel on both sides
R=3
4
and 2 resistances are in series on both sides.Then 6
and6 resistances are in parallel on both sidesR=3
There resistances of 4 each are connected as shown in figure. If the point D divides the resistance into two equal halves, the resistance between points A and D will be
physics-General
The equivalent circuit is given by
4 and 2 resistances are in series on both sides.
Then 6 and6 resistances are in parallel on both sides
R=3
4
and 2 resistances are in series on both sides.Then 6
and6 resistances are in parallel on both sidesR=3
physics-
The equivalent resistance between the points A and B will be (each resistance is
15 )
The circuit can be shown as given below
The equivalent resistance between D and C.
Now, between A and B, the resistance of upper part ADCB,
Between A and B, the resistance of middle part AOB
Therefore, equivalent resistance between A and B
The equivalent resistance between D and C.
Now, between A and B, the resistance of upper part ADCB,
Between A and B, the resistance of middle part AOB
Therefore, equivalent resistance between A and B
The equivalent resistance between the points A and B will be (each resistance is
15 )
physics-General
The circuit can be shown as given below
The equivalent resistance between D and C.
Now, between A and B, the resistance of upper part ADCB,
Between A and B, the resistance of middle part AOB
Therefore, equivalent resistance between A and B
The equivalent resistance between D and C.
Now, between A and B, the resistance of upper part ADCB,
Between A and B, the resistance of middle part AOB
Therefore, equivalent resistance between A and B
physics-
A 3 V battery with negligible internal resistance is connected in a circuit as shown in the figure. The current I, in the circuit will be
Resistance in the arms AC and BC are in series,
∴ R’=3+3=6
Now, R’ and 3 are in parallel,
Now, V=IR
∴ R’=3+3=6
Now, R’ and 3
are in parallel,Now, V=IR
A 3 V battery with negligible internal resistance is connected in a circuit as shown in the figure. The current I, in the circuit will be
physics-General
Resistance in the arms AC and BC are in series,
∴ R’=3+3=6
Now, R’ and 3 are in parallel,
Now, V=IR
∴ R’=3+3=6
Now, R’ and 3
are in parallel,Now, V=IR
physics-
A current of 2A flows in an electric circuit as shown in figure. The potential difference
, in volts( 
are potentials at R and S respectively) is
Current through each arm
PRQ and PSQ=1A
From Eqs. (i) and (ii), we get
PRQ and PSQ=1A
From Eqs. (i) and (ii), we get
A current of 2A flows in an electric circuit as shown in figure. The potential difference, in volts( are potentials at R and S respectively) is
physics-General
Current through each arm
PRQ and PSQ=1A
From Eqs. (i) and (ii), we get
PRQ and PSQ=1A
From Eqs. (i) and (ii), we get
physics-
The total current supplied to the given circuit by the battery is
The equivalent circuit of the given circuit is as shown
Resistances 6 and 2 are in parallel
Resistances
Resistances 3 and 3 are in parallel
The current,
Resistances 6
and 2 are in parallelResistances
Resistances 3
and 3 are in parallelThe current,
The total current supplied to the given circuit by the battery is
physics-General
The equivalent circuit of the given circuit is as shown
Resistances 6 and 2 are in parallel
Resistances
Resistances 3 and 3 are in parallel
The current,
Resistances 6
and 2 are in parallelResistances
Resistances 3
and 3 are in parallelThe current,