Physics-
The graph between the displacement 
and 
for a particle moving in a straight line is shown in figure. During the interval 
and 
the acceleration of the particle is
, 
Physics-General
- - 0 + 0
- + 0 + +
- + 0 - +
- - 0 - 0
Answer:The correct answer is: - 0 + 0Region 
shows that graph bending toward time axis 
acceleration is negative.
Region 
shows that graph is parallel to time axis 
velocity is zero. Hence acceleration is zero.
Region 
shows that graph is bending towards displacement axis acceleration is positive.
Region 
shows that graph having constant slope velocity is constant. Hence acceleration is zero
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maths-
The minimum and maximum values of 
3x are
The minimum and maximum values of 3x are
maths-General
physics-
The displacement-time graphs of two moving particles make angles of 
with the 
axis. The ratio of their velocities is
Slope of displacement time-graph is velocity
The displacement-time graphs of two moving particles make angles of with the axis. The ratio of their velocities is
physics-General
Slope of displacement time-graph is velocity
maths-
Extreme Values:The range of 
Extreme Values:The range of
maths-General
physics-
The given graph shows the variation of velocity with displacement. Which one of the graph given below correctly represents the variation of acceleration with displacement?
The 
equation from the given graph can be written as,
Substituting
from Eq. (i), we get
Thus,
graph is a straight line with positive slope and negative intercept.
equation from the given graph can be written as,Substituting
from Eq. (i), we getThus,
graph is a straight line with positive slope and negative intercept.The given graph shows the variation of velocity with displacement. Which one of the graph given below correctly represents the variation of acceleration with displacement?
physics-General
The equation from the given graph can be written as,
Substituting from Eq. (i), we get
Thus, graph is a straight line with positive slope and negative intercept.
equation from the given graph can be written as,Substituting
from Eq. (i), we getThus,
graph is a straight line with positive slope and negative intercept.physics-
What is the equivalent resistance between 
in the given circuit?
and 
are in series
are in parallel
and 
are in series
and 
are in parallel
and 
are in series
Now,
and 
are in parallel
What is the equivalent resistance between in the given circuit?
physics-General
and are in series are in paralleland are in series and are in parallel and are in seriesNow,
and are in parallelphysics-
The current in the 1
resistor shown in the circuit is
In the given circuit 4 resistors are connected in parallel, this combination is connected in series with 1 resistance.
Also, R’’=2 +1 =3
From Ohm’s law,
resistors are connected in parallel, this combination is connected in series with 1 resistance.Also, R’’=2
+1 =3From Ohm’s law,
The current in the 1 resistor shown in the circuit is
physics-General
In the given circuit 4 resistors are connected in parallel, this combination is connected in series with 1 resistance.
Also, R’’=2 +1 =3
From Ohm’s law,
resistors are connected in parallel, this combination is connected in series with 1 resistance.Also, R’’=2
+1 =3From Ohm’s law,
physics-
The current in the 1 resistor shown in the circuit is
In the given circuit 4 resistors are connected in parallel, this combination is connected in series with 1 resistance.
Also, R’’=2 +1 =3
From Ohm’s law,
resistors are connected in parallel, this combination is connected in series with 1 resistance.Also, R’’=2
+1 =3From Ohm’s law,
The current in the 1 resistor shown in the circuit is
physics-General
In the given circuit 4 resistors are connected in parallel, this combination is connected in series with 1 resistance.
Also, R’’=2 +1 =3
From Ohm’s law,
resistors are connected in parallel, this combination is connected in series with 1 resistance.Also, R’’=2
+1 =3From Ohm’s law,
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,
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-
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-
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-
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-
In the circuit shown the value of I in ampere is
We can simplify the network as shown
So, net resistance,
R=2.4+1.6=4.0
Therefore, current from the battery.
Now, from the circuit (b),
4I’ =6I
But
=I+I’
So, net resistance,
R=2.4+1.6=4.0
Therefore, current from the battery.
Now, from the circuit (b),
4I’ =6I
But
=I+I’In the circuit shown the value of I in ampere is
physics-General
We can simplify the network as shown
So, net resistance,
R=2.4+1.6=4.0
Therefore, current from the battery.
Now, from the circuit (b),
4I’ =6I
But=I+I’
So, net resistance,
R=2.4+1.6=4.0
Therefore, current from the battery.
Now, from the circuit (b),
4I’ =6I
But
=I+I’physics-
Six resistors, each of value 3 are connected as shown in the figure. A cell of emf 3V is connected across 
The effective resistance across and the current through the arm will be
The equivalent circuit is shown as
We can emit the resistance in the arm DF as balance condition is satisfied.
Therefore, the 3 resistances in arm CD and DE are in series.
Similarly, for arms CF and FE, R’’=6
are in parallel
R’’’=3
Now, R’’’ and 3 resistances are in parallel
Moreover, V across AB=3V and resistance in the arm=3
∴ Current through the arm will be
We can emit the resistance in the arm DF as balance condition is satisfied.
Therefore, the 3
resistances in arm CD and DE are in series.Similarly, for arms CF and FE, R’’=6
are in parallelR’’’=3
Now, R’’’ and 3
resistances are in parallelMoreover, V across AB=3V and resistance in the arm=3
∴ Current through the arm will be
Six resistors, each of value 3 are connected as shown in the figure. A cell of emf 3V is connected across The effective resistance across and the current through the arm will be
physics-General
The equivalent circuit is shown as
We can emit the resistance in the arm DF as balance condition is satisfied.
Therefore, the 3 resistances in arm CD and DE are in series.
Similarly, for arms CF and FE, R’’=6
are in parallel
R’’’=3
Now, R’’’ and 3 resistances are in parallel
Moreover, V across AB=3V and resistance in the arm=3
∴ Current through the arm will be
We can emit the resistance in the arm DF as balance condition is satisfied.
Therefore, the 3
resistances in arm CD and DE are in series.Similarly, for arms CF and FE, R’’=6
are in parallelR’’’=3
Now, R’’’ and 3
resistances are in parallelMoreover, V across AB=3V and resistance in the arm=3
∴ Current through the arm will be
physics-
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- graph