(2) Characteristics of transistors : A transistor can be connected in a circuit in the following three different
configurations.
(i) Common base (CB) (ii) Common emitter (CE) (iii) Common collector (CC)
(i) CB characteristics : The graphs between voltages and currents when base of a transistor is common to input
and output circuits are known as CB characteristic of a transistor.
PNP Ic VCB = – 10 V
Ie
E C Ie (mA) VCB = – 20 V VCB = 0 Ic (mA) Ie = 15 mA
+ Ie = 10 mA
B VCB = output Ie = 5 mA
– Ib RL Ie = 0 mA
VEB = input –
VCB (in volt)
+
VEB (in volt)
(ii) CE characteristics : The graphs between voltages an d currents when emitter of a transistor is common to
input and output circuits are known as CE characteristics of a transistor.
PNP C Ic
Ie B Ib (µA) VCE = – 0.1 V VCE = – 0.2 V
– E Ic (mA) Ib = 30 mA
+ Ie RL Ib = 20 mA
VCE = output Ib = 10 mA
VEB = input – Ib = 0 mA
+
VEB (in volt) Knee voltage VCE (in volt)
(3) Transistor as an amplifier : A device which increases the amplitude of the input signal is called
amplifier.
Amplifier
Input signal Output amplified signal
The transistor can be used as an amplifier in the following three configuration
(i) CB amplifier (ii) CE amplifier (iii) CC amplifier
PNP Ie B PNP Ic
Ie Ic C
EC
Input ~ B VC RL Output signal Input ~ E Output
signal Ib – signal Ie RL – signal
+
VEB VCB VEB VCE
+–
+
CB amplifier –+ CE amplifier
(4) Parameters of CE/CB amplifiers
Transistor as C.E. amplifier Transistor as C.B. amplifier
(i) Current gain (α) (i) Current gain (β)
(a) α ac = Small change in collector current (∆ic ) ; (a) β ac = ∆ic VCE = constant
Small change in collector current (∆ie ) ∆ib
VB (constant)
(b) α dc (orα) = Collector current(ic ) (b) β dc = ic
Emitter current (ie ) ib
valve of αdc lies between 0.95 to 0.99 valve of βac lies between 15 and 20
(ii) Voltage gain (ii) Voltage gain
Av = Change in output voltage(∆Vo ) Av = ∆Vo = β ac × Resistance gain
Change in input voltage (∆Vi ) ∆Vi
⇒ Av = αac × Resistance gain
(iii) Power gain = Change in output power(∆Po ) (iii) Power gain = ∆Po = β 2 × Resistance gain
Change in input power(∆Pc ) ∆Pi ac
⇒ Power gain = α 2 × Resistance gain Note : ≅ Trans conductance (gm) : The ratio
ac
of the change in collector current to the change in
emitter base voltage is called trans conductance. i.e.
gm = ∆ic . Also gm = AV ; RL = Load resistance
∆VEB RL
(5) Relation between α and β : β = 1 α or α = 1 β β
−α +
(6) Comparison between CB, CE and CC amplifier
S.No. Characteristic Amplifier
CB CE CC
≈ 150 – 800 kΩ high
(i) Input resistance (Ri) ≈ 50 to 200 Ω low ≈ 1 to 2 kΩ medium ≈ kΩ low
(ii) Output resistance (Ro) ≈ 1 – 2 kΩ high ≈ 50 kΩ medium 20 – 200 high
(iii) Current gain 0.8 – 0.9 low 20 – 200 high
Low
(iv) Voltage gain Medium High Low
Zero
(v) Power gain Medium High
Voltage
(vi) Phase difference between Zero 180o
input and output voltages
(vii) Used as amplifier for current Power
Example
Example: 1 What is the coordination number of sodium ions in the case of sodium chloride structure [CBSE 1988]
Solution : (a)
Example: 2 (a) 6 (b) 8 (c) 4 (d) 12
Solution : (a) In NaCl crystal Na+ ion is surrounded by 6 Cl − ion, therefore coordination number of Na+ is 6.
Example: 3
A Ge specimen is doped with Al. The concentration of acceptor atoms is ~1021 atoms/m3. Given that the
Solution : (c) intrinsic concentration of electron hole pairs is ~ 1019 / m3 , the concentration of electrons in the specimen is
Example: 4 [AIIMS 2004]
Solution : (c) (a) 1017 / m3 (b) 1015 / m3 (c) 104 / m3 (d) 102 / m3
Example: 5
ni2 = nhne ⇒ (1019 )2 = 10 21 × ne ⇒ ne = 1017 / m3 .
Solution : (a)
Example: 6 A silicon specimen is made into a P-type semi-conductor by doping, on an average, one Indium atom per
5 ×107 silicon atoms. If the number density of atoms in the silicon specimen is 5 ×1028 atoms/m3, then the
number of acceptor atoms in silicon will be [MP PMT 1993, 2003]
(a) 2.5 ×1030 atoms/cm3 (b) 1.0 ×1013 atoms/cm3 (c) 1.0 ×1015 atoms/cm3 (d) 2.5 ×1036 atoms/cm3
Number density of atoms in silicon specimen = 5 × 1028 atom/m3 = 5 × 1022 atom/cm3
Since one atom of indium is doped in 5 × 107 Si atom. So number of indium atoms doped per cm-3 of silicon.
n= 5 × 10 22 = 1 × 1015 atom / cm3 .
5 × 107
A P-type semiconductor has acceptor levels 57 meV above the valence band. The maximum wavelength of
light required to create a hole is (Planck’s constant h = 6.6 ×10−34 J-s) [MP PET 1995]
(a) 57 Å (b) 57 ×10−3 Å (c) 217100 Å (d) 11.61×10−33 Å
E = hc ⇒ λ = hc = 6.6 × 10 −34 × 3 × 108 = 217100 Å.
λ E 57 × 10 −3 × 1.6 × 10 −19
A potential barrier of 0.50V exists across a P-N junction. If the depletion region is 5.0 ×10−7 m wide, the
intensity of the electric field in this region is [UPSEAT 2002]
(a) 1.0 ×106 V / m (b) 1.0 ×105 V / m (c) 2.0 ×105 V / m (d) 2.0 ×106 V / m
E = V = 0.50 = 1 × 106 V/m.
d 5 × 10 −7
A 2V battery is connected across the points A and B as shown in the figure given below. Assuming that the
resistance of each diode is zero in forward bias and infinity in reverse bias, the current supplied by the battery
when its positive terminal is connected to A is 10 Ω [UPSEAT 2002]
(a) 0.2 A 10 Ω
(b) 0.4 A AB
(c) Zero
(d) 0.1 A
Solution : (a) Since diode in upper branch is forward biased and in lower branch is reversed biased. So current through
Example: 7
circuit i = V ; here rd = diode resistance in forward biasing = 0
Solution : (b) R + rd
Example: 8
So i = V = 2 = 0.2A .
Solution : (a) R 10
Example: 9
Current in the circuit will be [CBSE PMT 2001]
(a) 5 A 20
40
(b) 5 A 30
50
(c) 5 A i
10 20
5V
5
(d) 20 A
The diode in lower branch is forward biased and diode in upper branch is reverse biased
∴ i = 5 = 5 A
20 + 30 50
Find the magnitude of current in the following circuit [RPMT 2001]
(a) 0 –4V 3Ω – 1 V
(b) 1 amp
(c) 0.1 amp
(d) 0.2 amp
Diode is reverse biased. Therefore no current will flow through the circuit.
The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 V at all currents and a
maximum power rating of 100 milliwatts. What should be the value of the resistor R, connected in series with
the diode for obtaining maximum current [CBSE PMT 1997]
(a) 1.5 Ω R 0.5 V
(b) 5 Ω
(c) 6.67 Ω
(d) 200 Ω
The current through circuit i= P 100 × 10 −3 = 0.2A 1.5 V
V 0.5
Solution : (b) =
Example: 10
Solution : (d) ∴ voltage drop across resistance = 1.5 – 0.5 = 1 V ⇒ R = 1 = 5Ω
0.2
For a transistor amplifier in common emitter configuration for load impedance of 1 kΩ (hfe = 50 and hoe = 25)
the current gain is
[AIEEE 2004]
(a) – 5.2 (b) – 15.7 (c) – 24.8 (d) – 48.78
In common emitter configuration current gain Ai = 1 −h fe = 1 + 25 − 50 × 10 3 = – 48.78.
+ hoe RL × 10 −6
Example: 11 In the following common emitter configuration an NPN transistor with current gain β = 100 is used. The
output voltage of the amplifier will be
[AIIMS 2003]
(a) 10 mV 1KΩ 10KΩ
(b) 0.1 V Vout
(c) 1.0 V 1mV
(d) 10 V
Solution : (c) Voltage gain = Output voltage ⇒ Vout = Vin × Voltage gain
Input voltage
Example: 12
⇒ Vout = Vin × Current gain × Resistance gain = Vin × β × RL = 10 −3 × 100 × 10 = 1V.
Solution : (a) RBE 1
Example: 13
While a collector to emitter voltage is constant in a transistor, the collector current changes by 8.2 mA when
Solution : (d)
Example: 14 the emitter current changes by 8.3 mA. The value of forward current ratio hfe is [KCET 2002]
Solution : (a)
(a) 82 (b) 83 (c) 8.2 (d) 8.3
h fe = ∆ic Vce = 8.2 = 82
∆ib 8.3 − 8.2
The transfer ratio of a transistor is 50. The input resistance of the transistor when used in the common-emitter
configuration is 1 KΩ. The peak value for an ac input voltage of 0.01 V peak is [CBSE PMT 1998]
(a) 100 µA (b) 0.01 mA (c) 0.25 mA (d) 500 µ A
ic = βib = β × Vi = 50 × 0.01 = 500 × 10 −6 A = 500µ A
Ri 1000
In a common base amplifier circuit, calculate the change in base current if that in the emitter current is 2 mA
and α = 0.98 [BHU 1995]
(a) 0.04 mA (b) 1.96 mA (c) 0.98 mA (d) 2 mA
∆ic = α ∆ie = 0.98 × 2 = 196 mA
∴ ∆ib = ∆ie − ∆ic = 2 − 1.96 = 0.04 mA .
Friction
Introduction.
If we slide or try to slide a body over a surface the motion is resisted by a bonding between the body and the
surface. This resistance is represented by a single force and is called friction.
The force of friction is parallel to the surface and opposite to the direction of intended motion.
Types of Friction.
(1) Static friction : The opposing force that comes into play when one body tends to move over the surface
of another, but the actual motion has yet not started is called static friction.
(i) If applied force is P and the body remains at rest then static friction F = P. R
(ii) If a body is at rest and no pulling force is acting on it, force of friction on it is zero. P
F
(iii) Static friction is a self-adjusting force because it changes itself in accordance mg
with the applied force.
(2) Limiting friction : If the applied force is increased the force of static friction also increases. If the applied
force exceeds a certain (maximum) value, the body starts moving. This maximum value of static friction upto which
body does not move is called limiting friction.
(i) The magnitude of limiting friction between any two bodies in contact is directly proportional to the normal
reaction between them.
Fl ∝ R or Fl = µ s R
(ii) Direction of the force of limiting friction is always opposite to the direction in which one body is at the
verge of moving over the other
(iii) Coefficient of static friction : (a) µs is called coefficient of static friction and defined as the ratio of force of
limiting friction and normal reaction µs = F
R
(b) Dimension : [M 0 L0T 0 ]
(c) Unit : It has no unit.
(d) Value of µs lies in between 0 and 1
(e) Value of µ depends on material and nature of surfaces in contact that means whether dry or wet ; rough
or smooth polished or non-polished.
(f) Value of µ does not depend upon apparent area of contact.
(3) Kinetic or dynamic friction : If the applied force is increased further and sets the body in motion, the
friction opposing the motion is called kinetic friction.
(i) Kinetic friction depends upon the normal reaction.
Fk ∝ R or Fk = µk R where µk is called the coefficient of kinetic friction
(ii) Value of µk depends upon the nature of surface in contact.
(iii) Kinetic friction is always lesser than limiting friction Fk < Fl ∴ µk < µ s
i.e. coefficient of kinetic friction is always less than coefficient of static friction. Thus we require more force to
start a motion than to maintain it against friction. This is because once the motion starts actually ; inertia of rest has
been overcome. Also when motion has actually started, irregularities of one surface have little time to get locked
again into the irregularities of the other surface.
(iv) Types of kinetic friction
(a) Sliding friction : The opposing force that comes into play when one body is actually sliding over the
surface of the other body is called sliding friction. e.g. A flat block is moving over a horizontal table.
(b) Rolling friction : When objects such as a wheel (disc or ring), sphere or a cylinder rolls over a surface,
the force of friction comes into play is called rolling friction.
≅ Rolling friction is directly proportional to the normal reaction (R) and inversely proportional to the radius (r)
of the rolling cylinder or wheel.
Frolling = µr R
r
µr is called coefficient of rolling friction. It would have the dimensions of length and would be measured in metre.
≅Rolling friction is often quite small as compared to the sliding friction. That is why heavy loads are
transported by placing them on carts with wheels.
≅In rolling the surfaces at contact do not rub each other.
≅The velocity of point of contact with respect to the surface remains zero all the times although the centre of
the wheel moves forward.
Graph Between Applied Force and Force of Friction.
(1) Part OA of the curve represents static friction (Fs ) . Its value Force of friction A C
increases linearly with the applied force B
(2) At point A the static friction is maximum. This represent limiting Fs Fk
friction (Fl ). Fl
(3) Beyond A, the force of friction is seen to decrease slightly. The O Applied force
portion BC of the curve therefore represents the kinetic friction (Fk ) .
(4) As the portion BC of the curve is parallel to x-axis therefore kinetic friction does not change with the
applied force, it remains constant, whatever be the applied force.
Friction is a Cause of Motion.
It is a general misconception that friction always opposes the motion. No doubt friction opposes the motion of
a moving body but in many cases it is also the cause of motion. For example :
(1) In moving, a person or vehicle pushes the ground backwards (action) and the rough surface of ground
reacts and exerts a forward force due to friction which causes the motion. If there had been no friction there will be
slipping and no motion.
Friction
θ
Action
(2) In cycling, the rear wheel moves by the force communicated to it by pedalling while front wheel moves by
itself. So, when pedalling a bicycle, the force exerted by rear wheel on ground makes force of friction act on it in the
forward direction (like walking). Front wheel moving by itself experience force of friction in backward direction (like
rolling of a ball). [However, if pedalling is stopped both wheels move by themselves and so experience force of
friction in backward direction.]
While pedalling Pedalling is stoped
(3) If a body is placed in a vehicle which is accelerating, the force of friction is the cause of motion of the body
along with the vehicle (i.e., the body will remain at rest in the accelerating vehicle until ma < µ smg). If there had
been no friction between body and vehicle the body will not move along with the vehicle.
a ma
µsmg
From these examples it is clear that without friction motion cannot be started, stopped or transferred from one body
to the other.
Problem 1. If a ladder weighing 250N is placed against a smooth vertical wall having coefficient of friction between it and
floor is 0.3, then what is the maximum force of friction available at the point of contact between the ladder
and the floor [AIIMS 2002]
(a) 75 N (b) 50 N (c) 35 N (d) 25 N
Solution : (a) Maximum force of friction Fl = µsR = 0.3 × 250 = 75N
Problem 2. On the horizontal surface of a truck (µ = 0.6), a block of mass 1 kg is placed. If the truck is accelerating at the
rate of 5m/sec2 then frictional force on the block will be [CBSE PMT 2001]
(a) 5 N (b) 6 N (c) 5.88 N (d) 8 N
Solution : (a) Limiting friction = µsR = µsmg = 0.6 ×1× 9.8 = 5.88 N
Problem 3. When truck accelerates in forward direction at the rate of 5m / s 2 a pseudo force (ma) of 5N works on block
Solution : (a) in back ward direction. Here the magnitude of pseudo force is less than limiting friction So, static friction works
in between the block and the surface of the truck and as we know, static friction = Applied force = 5N.
Problem 4.
A block of mass 2 kg is kept on the floor. The coefficient of static friction is 0.4. If a force F of 2.5 N is applied
Solution : (c) on the block as shown in the figure, the frictional force between the block and the floor will be [MP PET 2000]
Problem 5. (a) 2.5 N F
(b) 5 N
Solution : (a)
(c) 7.84 N
Problem 6.
(d) 10 N
Solution : (d)
Applied force = 2.5 N and limiting friction = µmg = 0.4 × 2 × 9.8 = 7.84 N
As applied force is less than limiting friction. So, for the given condition static friction will work.
Static friction on a body = Applied force = 2.5 N.
A block A with mass 100 kg is resting on another block B of mass 200 kg. As shown in figure a horizontal rope
tied to a wall holds it. The coefficient of friction between A and B is 0.2 while coefficient of friction between B
and the ground is 0.3. The minimum required force F to start moving B will be [RPET 1999]
(a) 900 N A
(b) 100 N BF
(c) 1100 N
(d) 1200 N
Two frictional force will work on block B. fAB A
F = fAB + fBG = µABmag + µBG(mA + mB)g fBG BF
= 0.2 × 100 × 10 + 0.3 (300) × 10 Ground
= 200 + 900 = 1100N. (This is the required minimum force)
A 20 kg block is initially at rest on a rough horizontal surface. A horizontal force of 75 N is required to set the
block in motion. After it is in motion, a horizontal force of 60 N is required to keep the block moving with
constant speed. The coefficient of static friction is [AMU 1999]
(a) 0.38 (b) 0.44 (c) 0.52 (d) 0.60
Coefficient of static friction µS = Fl = 75 = 0.38 .
R 20 × 9.8
A block of mass M is placed on a rough floor of a lift. The coefficient of friction between the block and the
floor is µ. When the lift falls freely, the block is pulled horizontally on the floor. What will be the force of friction
(a) µ Mg (b) µ Mg/2 (c) 2µ Mg (d) None of these
When the lift moves down ward with acceleration 'a' then effective acceleration due to gravity
g' = g – a
∴g' = g − g = 0 [As the lift falls freely, so a = g]
So force of friction = µmg' = 0
Advantages and Disadvantages of Friction.
(1) Advantages of friction
(i) Walking is possible due to friction.
(ii) Two body sticks together due to friction.
(iii) Brake works on the basis of friction.
(iv) Writing is not possible without friction.
(v) The transfer of motion from one part of a machine to other part through belts is possible by friction.
(2) Disadvantages of friction
(i) Friction always opposes the relative motion between any two bodies in contact. Therefore extra energy has
to be spent in over coming friction. This reduces the efficiency of machine.
(ii) Friction causes wear and tear of the parts of machinery in contact. Thus their lifetime reduces.
(iii) Frictional force result in the production of heat, which causes damage to the machinery.
Methods of Changing Friction.
We can reduce friction
(1) By polishing.
(2) By lubrication.
(3) By proper selection of material.
(4) By streamlining the shape of the body.
(5) By using ball bearing.
Also we can increase friction by throwing some sand on slippery ground. In the manufacturing of tyres,
synthetic rubber is preferred because its coefficient of friction with the road is larger.
Angle of Friction.
Angle of friction may be defined as the angle which the resultant of limiting SR
friction and normal reaction makes with the normal reaction. θ
By definition angle θ is called the angle of friction F P
tanθ = F
R
mg
∴ tan θ = µ [As we know F =µ ]
R
or θ = tan−1(µ)
Hence coefficient of limiting friction is equal to tangent of the angle of friction.
Resultant Force Exerted by Surface on Block.
In the above figure resultant force S = F 2 + R 2
S = (µmg)2 + (mg)2
S = mg µ 2 + 1
when there is no friction (µ = 0) S will be minimum i.e., S = mg
Hence the range of S can be given by, mg ≤ S ≤ mg µ 2 + 1
Angle of Repose.
Angle of repose is defined as the angle of the inclined plane with horizontal such that a body placed on it is
just begins to slide. RF
By definition α is called the angle of repose.
In limiting condition F = mg sinα mg sin α
α
and R = mg cosα α mg cos α
mg
F
So R = tanα
∴ F = µ = tanθ = tanα [As we know F =µ = tanθ ]
R R
Thus the coefficient of limiting friction is equal to the tangent of angle of repose.
As well as α = θ i.e. angle of repose = angle of friction.
Problem 7. A body of 5 kg weight kept on a rough inclined plane of angle 30o starts sliding with a constant velocity. Then the
coefficient of friction is (assume g = 10 m/s2) [JIPMER 2002]
(a) 1 / 3 (b) 2 / 3 (c) 3 (d) 2 3
Solution : (a) Here the given angle is called the angle of repose So, µ = tan 30o = 1
3
Problem 8.
The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half is rough. A body
starting from the rest at top comes back to rest at the bottom if the coefficient of friction for the lower half is
given [Pb PMT 2000]
(a) µ = sin θ (b) µ = cot θ (c) µ = 2 cos θ (d) µ = 2 tan θ
Solution : (d) For upper half by the equation of motion v2 = u2 + 2as
v2 = 02 + 2(g sin θ )l / 2 = gl sinθ [As u = 0, s = l / 2, a = g sin θ ] Smooth
For lower half Rough
0 = u2 + 2g(sin θ − µ cos θ ) l /2 [As v = 0, s = l / 2, a = g (sinθ − µ cosθ ) ]
⇒ 0 = gl sin θ + gl(sin θ − µ cos θ ) [As final velocity of upper half will be equal to the initial velocity of lower half]
⇒ 2 sinθ = µ cosθ ⇒ µ = 2 tanθ
Calculation of Necessary Force in Different Conditions.
If W = weight of the body, θ = angle of friction, µ = tanθ = coefficient of friction
then we can calculate necessary force for different condition in the following manner :
(1) Minimum pulling force P at an angle α from the horizontal P
By resolving P in horizontal and vertical direction (as shown in figure) α
For the condition of equilibrium
F = P cosα and R = W − P sinα
By substituting these value in F = µR
P cosα = µ (W − P sinα) R
P sinα
⇒ P cosα = sinθ (W − P sinα) [As µ = tanθ ] F P cosα
cosθ
W
W sinθ
⇒ P = cos (α − θ )
(2) Minimum pushing force P at an angle α from the horizontal P
By Resolving P in horizontal and vertical direction (as shown in the figure) α
For the condition of equilibrium
R
F = P cosα and R = W + P sinα
By substituting these value in F = µR
⇒ P cosα = µ (W + P sinα)
⇒ P cosα = sinθ (W + P sin α ) [As µ = tanθ ] F P cosα
cosθ P sinα
W
⇒ P = W sinθ
cos (α + θ )
(3) Minimum pulling force P to move the body up an inclined plane
By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the figure)
For the condition of equilibrium
R + P sinα = W cos λ P R + P sinα
∴ R = W cos λ − P sinα α P cosα
and F + W sin λ = P cosα λ F + W sinλ λ W cosλ
∴ F = P cosα − W sin λ λ
By substituting these values in F = µR and solving we get
P = W sin (θ + λ)
cos (α − θ )
(4) Minimum force on body in downward direction along the surface of inclined plane to start its motion
By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the figure)
For the condition of equilibrium P R + P sinα
R + P sinα = W cos λ α
F
∴ R = W cos λ − P sinα λ
and F = P cosα + W sin λ P cosα
By substituting these values in F = µR and +
solving we get
W sinλ
λ W cosλ
λW
P = W sin(θ − λ)
cos (α − θ )
(5) Minimum force to avoid sliding a body down an inclined plane
By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the figure)
For the condition of equilibrium P R + P sinα F + P cosα
R + P sinα = W cos λ
α
∴ R = W cos λ − P sinα λ W sinλ λ W cosλ
and P cosα + F = W sin λ
∴ F = W sin λ − P cosα λW
By substituting these values in F = µR and solving we get
P = W sin (λ −θ)
+ α)
cos (θ P
α
(6) Minimum force for motion and its direction
Let the force P be applied at an angleα with the horizontal.
By resolving P in horizontal and vertical direction (as shown in figure)
For vertical equilibrium
R + P sinα = mg
∴ R = mg − P sinα ….(i)
….(ii)
and for horizontal motion ….(iii)
P cosα ≥ F
i.e. P cosα ≥ µR R + P sinα
F P cosα
Substituting value of R from (i) in (ii)
mg
P cosα ≥ µ (mg − P sinα)
P ≥ µ mg
cosα + µ sinα
For the force P to be minimum (cosα + µ sinα) must be maximum i.e.
d [cosα + µ sin α ] = 0 ⇒ − sinα + µ cosα = 0 1+ µ2 µ
dα
α
∴ tanα = µ 1
or α = tan −1(µ) = angle of friction
i.e. For minimum value of P its angle from the horizontal should be equal to angle of friction
As tanα = µ so from the figure sinα = µ and cosα = 1
1+ µ2 1+ µ2
By substituting these value in equation (iii)
P≥ µ mg ≥ µ mg ∴ Pmin = µmg
1 + µ2 1+ µ2 1+ µ2
1+ µ2 1+ µ2
Problem 9. What is the maximum value of the force F such that the block shown in the arrangement, does not move
Solution : (a) (µ = 1/2 3 ) [IIT-JEE (Screening) 2003]
(a) 20 N F m = √3kg
(b) 10 N 60o
(c) 12 N
(d) 15 N
Frictional force f = µR
⇒ F cos 60 = µ(W + F sin 60) R
( )⇒ 1 f F cos 60°
F cos 60 = 23 3g + F sin 60
⇒ F = 20N . W+F sin 60°
Problem 10. A block of mass m rests on a rough horizontal surface as shown in the figure. Coefficient of friction between
the block and the surface is µ. A force F = mg acting at angle θ with the vertical side of the block pulls it. In
which of the following cases the block can be pulled along the surface
(a) tanθ ≥ µ mg = F
(b) cot θ ≥ µ θ
(c) tan θ / 2 ≥ µ
m
(d) cot θ / 2 ≥ µ
Solution : (d) For pulling of block P ≥ f
⇒ mg sinθ ≥ µR ⇒ mg sinθ ≥ µ (mg − mg cosθ ) R+mg cosθ
⇒ sinθ ≥ µ (1 − cosθ ) f mg sin θ =p
⇒ 2 sin θ cos θ ≥ µ 2 sin 2 θ ⇒ cot θ ≥ µ mg
2 2 2 2
Acceleration of a Block Against Friction.
(1) Acceleration of a block on horizontal surface
When body is moving under application of force P, then kinetic friction opposes its motion.
Let a is the net acceleration of the body R
From the figure ma
ma = P − Fk Fk P
∴ a = P − Fk
m
mg
(2) Acceleration of a block down a rough inclined plane
When angle of inclined plane is more than angle of repose, the body placed on the inclined plane slides down
with an acceleration a.
From the figure ma = mg sinθ − F R F
ma
⇒ ma = mg sinθ − µR
⇒ ma = mg sinθ − µ mg cosθ mg sinθ θ mg cosθ
∴ Acceleration a = g [sinθ − µ cosθ ] mg
θ
Note : ≅ For frictionless inclined plane µ = 0 ∴ a = g sinθ .
(3) Retardation of a block up a rough inclined plane
When angle of inclined plane is less than angle of repose, then for the upward motion
ma = mg sinθ + F
ma = mg sinθ + µ mg cosθ
Retardation a = g [sinθ + µ cosθ ] R ma
Note : ≅ For frictionless inclined plane µ = 0 ∴ a = g sinθ
mg sinθ + F θ mg cosθ
θ mg
Problem 11. A body of mass 10 kg is lying on a rough plane inclined at an angle of 30o to the horizontal and the coefficient
of friction is 0.5. The minimum force required to pull the body up the plane is [JIPMER 2000]
(a) 914 N (b) 91.4 N (c) 9.14 N (d) 0.914 N
Solution : (b) F = mg(sinθ + µ cosθ ) = 10 × 9.8 (sin 30 + 0.5 cos 30) = 91.4 N
Problem 12. A block of mass 10 kg is placed on a rough horizontal surface having coefficient of friction µ = 0.5. If a
horizontal force of 100 N is acting on it, then acceleration of the block will be [AIIMS 2002]
(a) 0.5 m/s2 (b) 5 m/s2 (c) 10 m/s2 (d) 15 m/s2
Solution : (b) a= Applied force – kinetic friction = 100 − 0.5 × 10 × 10 = 5m/s2.
mass 10
Problem 13.
A body of weight 64 N is pushed with just enough force to start it moving across a horizontal floor and the same
force continues to act afterwards. If the coefficients of static and dynamic friction are 0.6 and 0.4 respectively, the
acceleration of the body will be (Acceleration due to gravity = g) [EAMCET 2001]
(a) g (b) 0.64 g (c) g (d) 0.2 g
6.4 32
Solution : (d) Limiting friction = Fl = µ s R ⇒ 64 = 0.6 m g ⇒ m = 64 .
0.6g
Acceleration = Applied force – Kinetic friction = 64 − µKmg = 64 − 0.4 × 64 = 0.2g
Mass of the body m 64 0.6
0.6g
Problem 14. If a block moving up at θ = 30o with a velocity 5 m/s, stops after 0.5 sec, then what is µ [CPMT 1995]
(a) 0.5 (b) 1.25 (c) 0.6 (d) None of these
Solution : (c) From v = u − at ⇒ 0 = u − at ∴t = u
a
for upward motion on an inclined plane a = g(sinθ + µ cosθ ) ∴t = g(sin θ u cosθ )
+µ
Substituting the value of θ = 30o , t = 0.5 sec and u = 5m / s , we get µ = 0.6
Work Done Against Friction.
(1) Work done over a rough inclined surface
If a body of mass m is moved up on a rough inclined plane through distance s, then
Work done = force × distance R ma
= ma × s
= mg [sinθ + µ cosθ ]s s
= mg s [sinθ + µ cosθ ]
mg sinθ + F θ mg cosθ
(2) Work done over a horizontal surface θ mg
In the above expression if we put θ = 0 then
Work done = force × distance R
FP
=F×s
= µ mg s mg s
It is clear that work done depends upon
(i) Weight of the body.
(ii) Material and nature of surface in contact.
(iii) Distance moved.
Problem 15. A body of mass 5kg rests on a rough horizontal surface of coefficient of friction 0.2. The body is pulled
through a distance of 10m by a horizontal force of 25 N. The kinetic energy acquired by it is (g = 10 ms2)
(a) 330 J (b) 150 J (c) 100 J [EAMCET (Med.) 2000]
(d) 50 J
Solution : (b) Kinetic energy acquired by body = Total work done on the body – Work done against friction
= F × S − µmgS = 25 × 10 – 0.2 × 5 × 10 ×10 = 250 – 100 = 150 J.
Problem 16. 300 Joule of work is done in sliding a 2 kg. block up an inclined plane to a height of 10 meters. Taking value
of acceleration due to gravity ‘g’ to be 10 m/s2, work done against friction is [MP PMT 2002]
(a) 100 J (b) 200 J (c) 300 J (d) Zero
Solution : (a) Work done against gravity = mgh = 2 × 10 × 10 = 200J
Work done against friction = Total work done – Work done against gravity = 300 – 200 = 100J.
Problem 17. A block of mass 1 kg slides down on a rough inclined plane of inclination 60o starting from its top. If the
coefficient of kinetic friction is 0.5 and length of the plane is 1 m, then work done against friction is (Take
g = 9.8 m/s2) [AFMC 2000; KCET (Engg./Med.) 2001]
(a) 9.82 J (b) 4.94 J (c) 2.45J (d) 1.96 J
Solution : (c) W = µmg cosθ .S = 0.5 × 1 × 9.8 × 1 = 2.45 J.
2
Problem 18.
A block of mass 50 kg slides over a horizontal distance of 1 m. If the coefficient of friction between their
surfaces is 0.2, then work done against friction is [CBSE PMT 1999, 2000; AIIMS 2000; BHU 2001]
(a) 98 J (b) 72J (c) 56 J (d) 34 J
Solution : (a) W = µmgS = 0.2 × 50 × 9.8 × 1 = 98J .
Motion of Two Bodies One Resting on the Other.
When a body A of mass m is resting on a body B of mass M then two conditions are possible
(1) A force F is applied to the upper body, (2) A force F is applied to the lower body
We will discuss above two cases one by one in the following manner :
(1) A force F is applied to the upper body, then following four situations are possible
(i) When there is no friction
(a) The body A will move on body B with acceleration (F/m). mA F
M L
aA = F /m B
(b) The body B will remain at rest
aB = 0
(c) If L is the length of B as shown in figure A will fall from B after time t
t= 2L 2mL As s = 1 at2 and a = F/m
a= F 2
(ii) If friction is present between A and B only and applied force is less than limiting friction (F < Fl)
(F = Applied force on the upper body, Fl = limiting friction between A and B, Fk = Kinetic friction between A and B)
(a) The body A will not slide on body B till F < Fl i.e. F < µ smg
(b) Combined system (m + M) will move together with common acceleration aA = aB = F
M +m
(iii) If friction is present between A and B only and applied force is greater than limiting friction (F > Fl)
In this condition the two bodies will move in the same direction (i.e. of applied force) but with different
acceleration. Here force of kinetic friction µk mg will oppose the motion of A while will cause the motion of B.
F − Fk = m a A Free body diagram of A Fk = M aB Free body diagram of B
i.e. aA = F − Fk maA i.e. aB = Fk MaB
m M
AF B FK
aA = (F − µkmg) Fk ∴ aB = µk mg
m M
Note : ≅ As both the bodies are moving in the same direction.
Acceleration of body A relative to B will be a = aA − aB = MF − µk mg (m + M)
mM
So, A will fall from B after time t = 2L 2 m ML
a= MF − µk mg (m + M)
(iv) If there is friction between B and floor
(where Fl′ = µ ′(M + m) g = limiting friction between B and floor, Fk = kinetic friction between A and B)
B will move only if Fk > Fl′ and then Fk − Fl′ = M aB MaB FK
B
However if B does not move then static friction will work (not limiting
friction) between body B and the floor i.e. friction force = applied force (= Fk) Fl′
not Fl′ .
(2) A force F is applied to the lower body, then following four situations are possible
(i) When there is no friction
(a) B will move with acceleration (F/M) while A will remain at rest (relative M L Am
to ground) as there is no pulling force on A. B F
aB = F and aA =0
M
(b) As relative to B, A will move backwards with acceleration (F/M) and so will fall from it in time t.
∴ t= 2L 2ML
a= F
(ii) If friction is present between A and B only and F ′< Fl
(where F′ = Pseudo force on body A and Fl = limiting friction between body A and B)
(a) Both the body will move together with common acceleration a = F m
M+
(b) Pseudo force on the body A, F′ = ma = mF and Fl = µ smg
m+ M
(c) F ′ < Fl ⇒ mF < µ s mg ⇒ F < µ s (m + M) g
m+ M
So both bodies will move together with acceleration aA = aB = m F M if F < µ s [m + M] g
+
(iii) If friction is present between A and B only and F > Fl′
(where Fl′ = µs (m + M)g = limiting friction between body B and surface)
Both the body will move with different acceleration. Here force of kinetic friction µk mg will oppose the
motion of B while will cause the motion of A.
maA = µkmg Free body diagram of A F − Fk = MaB Free body diagram of B
i.e. a A = µk g
A maA i.e. aB = [F − µkmg] MaB
Fk M FK
F
B
Note : ≅ As both the bodies are moving in the same direction
Acceleration of body A relative to B will be
a = aA − aB = − F − µk g(m + M)
M
Negative sign implies that relative to B, A will move backwards and will fall it after time
t= 2L = 2ML
a F − µk g(m + M)
(iv) If there is friction between B and floor : The system will move only if F > Fl' then replacing F by
F − Fl′ . The entire case (iii) will be valid.
However if F < Fl′ the system will not move and friction between B and floor will be F while between A and B
is zero.
Problem 19. A 4 kg block A is placed on the top of a 8 kg block B which rests on a smooth table. A just slips on B when a
force of 12 N is applied on A. Then the maximum horizontal force on B to make both A and B move together, is
(a) 12 N (b) 24 N (c) 36 N (d) 48 N
Solution : (c) Maximum friction i.e. limiting friction between A and B, Fl = 12 N.
If F is the maximum value of force applied on lower body such that both body move together
It means Pseudo force on upper body is just equal to limiting friction
F' = Fl ⇒ m m F M = 4 4 8 F = 12 ∴F = 36N.
+ +
Problem 20. A body A of mass 1 kg rests on a smooth surface. Another body B of mass 0.2 kg is placed over A as shown.
The coefficient of static friction between A and B is 0.15. B will begin to slide on A if A is pulled with a force
greater than
(a) 1.764 N B
(b) 0.1764 N A
(c) 0.3 N
(d) It will not slide for any F
Solution : (a) B will begin to slide on A if Pseudo force is more than limiting friction
F' > Fl ⇒ m m F M > µs R ⇒ m m F M > 0.15mg ∴F > 1.764 N
+ +
Problem 21. A block A of mass 2 kg rests on another block B of mass 8 kg which rests on a horizontal floor. The coefficient
of friction between A and B is 0.2, while that between B and floor is 0.5. When a horizontal force of 25 N is
applied on the block B, the force of friction between A and B is [IIT-JEE 1993]
(a) Zero (b) 3.9 N (c) 5.0 N (d) 49 N
Solution : (a) Limiting friction between the block B and the surface A 2kg
B 8kg
FBS = µBS.R = 0.5(m + M ) g = 0.5(2 + 8)10 = 50N 25
Surface
but the applied force is 25 N so the lower block will not move i.e. there is no
pseudo force on upper block A. Hence there will be no force of friction between A and B.
Motion of an Insect in the Rough Bowl.
The insect crawl up the bowl up to a certain height h only till the component of its weight along the bowl is
balanced by limiting frictional force.
Let m = mass of the insect, r = radius of the bowl, µ = coefficient of friction r O
for limiting condition at point A Fl R θ
y
R = mg cosθ ......(i) and Fl = mg sinθ ......(ii) A h
Dividing (ii) by (i) mg sinθ
Fl mg cosθ mg
R
tanθ = = µ [As Fl = µR]
∴ r2 − y2 =µ or y = r
y 1+ µ2
So 1 1
Problem 22. h = r − y = r 1 − , ∴ h = r 1 −
1 + µ 2 1 + µ 2
An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the
insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an
angle α with the vertical, the maximum possible value of α is given by
[IIT-JEE (Screening) 2001]
(a) cot α = 3 α
(b) tanα = 3
(c) sec α = 3
(d) cosec α = 3
Solution : (a) From the above expression, for the equilibrium R = mg cosα and F = mg sinα .
Substituting these value in F = µR we get tan α = µ or cot α = 1 = 3 .
µ
Minimum Mass Hung From the String to Just Start the Motion.
(1) When a mass m1 placed on a rough horizontal plane : Another mass m2 hung from the string
connected by pulley, the tension (T) produced in string will try to start the motion of mass m1 .
At limiting condition
T = Fl R
T
⇒ m2 g = µR
F1 m1
⇒ m2 g = µ m1g
m1g T
∴ m2 = µm1 this is the minimum value of m2 to start the motion. m2
Note : ≅ In the above condition Coefficient of friction µ = m2
m1
(2) When a mass m1 placed on a rough inclined plane : Another mass m2 hung from the string
connected by pulley, the tension (T) produced in string will try to start the motion of mass m1 .
At limiting condition
For m2 T = m2 g ...... (i) R T
For m1 T = m1g sinθ + F ⇒ T = m1g sinθ + µR m1
⇒ T = m1 g sinθ + µm1 g cosθ T
......(ii) m2
From equation (i) and (ii) m2 = m1[sinθ + µ cosθ ] m1g sinθ + F θ m1g cosθ m2g
this is the minimum value of m2 to start the motion m1g
Note : ≅ In the above condition Coefficient of friction
µ = m2 θ − tanθ
m1 cos
Problem 23. Two blocks of mass M1 and M2 are connected with a string passing over a pulley as shown in the figure. The
block M1 lies on a horizontal surface. The coefficient of friction between the block M1 and horizontal surface is
Solution : (b) µ. The system accelerates. What additional mass m should be placed on the block M1 so that the system does
not accelerate
(a) M2 − M1 m
µ M1
(b) M2 − M1
µ
M1 M2
µ
(c) M2 −
(d) (M2 − M1)µ
By comparing the given condition with general expression
µ = M2 ⇒ m+ M1 = M2 ⇒ m = M2 − M1
m + M1 µ µ
Problem 24. The coefficient of kinetic friction is 0.03 in the diagram where mass m2 = 20 kg and m1 = 4 kg . The
acceleration of the block shall be (g = 10ms−2) 20 kg m2 T
(a) 1.8 ms−2
(b) 0.8 ms−2 T
(c) 1.4 ms−2 m1 4 kg
Solution : (c) (d) 0.4 ms−2 m2a
Let the acceleration of the system is a F m2 T
From the F.B.D. of m2
T − F = m2a ⇒ T − µm2g = m2a .....(i) T
⇒ T − 0.03 × 20 × 10 = 20a ⇒ T − 6 = 20a ....(ii) m1 m1 a
From the FBD of m1
m1g − T = m1a Fm1g
⇒ 4 × 10 − T = 4a ⇒ 40 − T = 4a
Solving (i) and (ii) a = 1.4m / s 2 .
Maximum Length of Hung Chain .
A uniform chain of length l is placed on the table in such a manner that its l' part is hanging over the edge of
table with out sliding. Since the chain have uniform linear density therefore the ratio of mass or ratio of length for
any part of the chain will be equal.
We know µ = m2 = mass hanging from the table [From article 5.15]
m1 mass lying on the table
∴ For this expression we can rewrite above expression in the following manner
µ= length hanging from the table [As chain have uniform linear density] ( l – l′ )
length lying on the table
∴ µ = l′ l′
− l′
l
by solving l′ = µl
(µ + 1)
Problem 25. A heavy uniform chain lies on a horizontal table top. If the coefficient of friction between the chain and the
table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the
table is
[CBSE PMT 1990]
(a) 20% (b) 25% (c) 35% (d) 15%
Solution : (a) From the expression l' = µ µ 1 l = 0.25 1 l [As µ = 0.25]
+ 0.25 +
⇒ l' = 0.25 l = l = 20% of the length of the chain.
1.25 5
Coefficient of Friction Between Body and Wedge .
A body slides on a smooth wedge of angle θ and its time of descent is t.
S S
Smooth wedge Rough wedge
θ θ
If the same wedge made rough then time taken by it to come down becomes n times more (i.e. nt)
The length of path in both the cases are same.
For smooth wedge For rough wedge
S = u t + 1 at 2 S = u t + 1 at 2
2 2
S = 1 (g sin θ ) t 2 .....(i) S = 1 g (sinθ − µ cosθ )(nt)2 .....(ii)
2 2
[As u = 0 and a = g sinθ ] [As u = 0 and a = g (sinθ − µ cosθ )]
From equation (i) and (ii) 1 (g sin θ ) t 2 = 1 g (sin θ − µ cosθ )(nt)2
2 2
⇒ sinθ = (sinθ − µ cosθ )n2
⇒ µ = tanθ 1 − 1
n2
Problem 26. A body takes just twice the time as long to slide down a plane inclined at 30o to the horizontal as if the plane
were frictionless. The coefficient of friction between the body and the plane is [JIPMER 1999]
(a) 3 (b) 3 (c) 4 (d) 3
4 3 4
Solution : (a) µ = tan θ 1 − 1 = tan 30 1 − 1 = 3 .
n2 22 4
Stopping of Block Due to Friction.
(1) On horizontal road
(i) Distance travelled before coming to rest : A block of mass m is moving initially with velocity u on a
rough surface and due to friction it comes to rest after covering a distance S.
Retarding force F = ma = µR
⇒ ma = µ mg
∴ a = µg .
From
v 2 = u2 − 2aS ⇒ 0 = u2 − 2µ g S [As v = 0, a = µg] S v=0
∴ u
S = u2
2µg
or S = P2 [As momentum P = mu]
2µm2 g [As v = 0, a = µ g]
(ii) Time taken to come to rest
From equation v = u − a t ⇒ 0 = u − µ g t
∴t = u
µg
(iii) Force of friction acting on the body
We know, F = ma
So, F = m (v − u)
t
F = mu [As v = 0]
t
F = µ mg As t = u
µg
(2) On inclined road : When block starts with velocity u its kinetic energy will be converted into potential
energy and some part of it goes against friction and after travelling distance S it comes to rest i.e. v = 0.
And we know that retardation a = g [sinθ + µ cosθ ]
By substituting the value of v and a in the following equation v=0
v 2 = u2 − 2a S S
u
⇒ 0 = u2 − 2g [sinθ + µ cosθ ]S
θ
∴S = 2g u2 µ cosθ )
(sinθ +
Problem 27. A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10s. Then the
coefficient of friction is [AIEEE 2003]
(a) 0.01 (b) 0.02 (c) 0.03 (d) 0.06
Solution : (d) v = u − at = u − µg t = 0
∴ µ = u = 6 = 0.06 .
gt 10 × 10
Problem 28. A 2 kg mass starts from rest on an inclined smooth surface with inclination 30o and length 2 m. How much
will it travel before coming to rest on a surface with coefficient of friction 0.25 [UPSEAT 2003]
(a) 4 m (b) 6 m (c) 8 m (d) 2 m
Solution : (a) v 2 = u 2 + 2aS = 0 + 2 × g sin 30 × 2
v = 20
Let it travel distance ‘S’ before coming to rest
S = v2 = 2× 20 = 4m.
2µg 0.25 × 10
Stopping of Two Blocks Due to Friction.
When two masses compressed towards each other and suddenly released then energy acquired by each block
will be dissipated against friction and finally block comes to rest
i.e., F × S = E [Where F = Friction, S = Distance covered by block, E = Initial kinetic energy of the block]
⇒ F × S = P2 [Where P = momentum of block]
2m
A B
⇒ µmg × S = P2 [As F = µ mg] m1 m1 m2 m2
2m
S1 S2
⇒ S = P2
2µm2 g
In a given condition P and µ are same for both the blocks.
1 S1 m2 2
m2 S2
So S ∝ ∴ = m1
Velocity at the Bottom of Rough Wedge.
A body of mass m which is placed at the top of the wedge (of height h) starts moving downward on a rough
inclined plane.
Loss of energy due to friction = FL (Work against friction)
PE at point A = mgh u=0
Am
KE at point B = 1 2 L
2 h
mv
By the law of conservation of energy mB
v
i.e. 1 mv 2 = mgh − FL
2
v= 2 (mgh − FL)
m
Sticking of a Block With Accelerated Cart.
When a cart moves with some acceleration toward right then a pseudo force (ma) acts on block toward left.
This force (ma) is action force by a block on cart.
Now block will remain static w.r.t. block. If friction force µR ≥ mg a
⇒ µma ≥ mg [As R = ma] F F Mm
CART
⇒ a≥ g
µ ma m R
∴ a min = g mg
µ
This is the minimum acceleration of the cart so that block does not fall.
and the minimum force to hold the block together
Fmin = (M + m) amin
Fmin = (M + m) g
µ
Sticking of a Person With the Wall of Rotor.
A person with a mass m stands in contact against the wall of a cylindrical drum (rotor). The coefficient of
friction between the wall and the clothing is µ.
If Rotor starts rotating about its axis, then person thrown away from the centre due to centrifugal force at a
particular speed w, the person stuck to the wall even the floor is removed, because friction force balances its weight
in this condition.
From the figure. F
Friction force (F) = weight of person (mg)
⇒ µR = mg R FC
mg
⇒ µ Fc = mg [Here, Fc= centrifugal force]
⇒ µmω 2 r = mg
min
∴ ω min = g
µr
Problem 29. A motorcycle is travelling on a curved track of radius 500m if the coefficient of friction between road and tyres is
0.5. The speed avoiding skidding will be [MH CET (Med.) 2001]
(a) 50 m/s (b) 75 m/s (c) 25 m/s (d) 35 m/s
Solution : (a) v = µrg = 0.5 × 500 × 10 = 50 m/s.
Problem 30. A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction
between the block and the wall is 0.2. The weight of the block is [AIEEE 2003]
(a) 2 N
(b) 20 N 10 N
(c) 50 N
(d) 100 N
Solution : (a) For equilibrium F
Weight (W) = Force of friction (F) R 10 N
W = µR = 0.2 × 10 = 2N W
Problem 31. A body of mass 2 kg is kept by pressing to a vertical wall by a force of 100 N. The friction between wall and
body is 0.3. Then the frictional force is equal to [Orissa JEE 2003]
(a) 6 N (b) 20 N (c) 600 N (d) 700 N
Solution : (b) For the given condition Static friction = Applied force = Weight of body = 2 × 10 = 20 N.
Problem 32. A fireman of mass 60kg slides down a pole. He is pressing the pole with a force of 600 N. The
coefficient of friction between the hands and the pole is 0.5, with what acceleration will the fireman
slide down (g = 10 m/s2) [Pb. PMT 2002]
(a) 1 m/s2 (b) 2.5 m/s2 (c) 10 m/s2 (d) 5 m/s2
Solution : (d) Friction = µR = 0.5 × 600 = 300 N, Weight = 600 N
ma = W – F ⇒ a = W−F = 600 − 300 F
m 60 R
600 N
W
∴ a = 5 m/s 2
Problem 33. The system shown in the figure is in equilibrium. The maximum value of W, so that the maximum value of
Solution : (c) static frictional force on 100 kg. body is 450 N, will be 100 kg 45o
(a) 100 N
(b) 250 N W
(c) 450 N
(d) 1000 N
For vertical equilibrium T1 sin 45o = W ∴ T1 = W F T1 sin 45o T1
sin 45o T2 45o
For horizontal equilibrium T2 = T1 cos 45o = W cos 45o = W T1 cos 45o
sin 45o
W
and for the critical condition T2 = F
∴ W = T2 = F = 450 N
Surface tension
Intermolecular Force.
The force of attraction or repulsion acting between the molecules are known as intermolecular force. The
nature of intermolecular force is electromagnetic.
The intermolecular forces of attraction may be classified into two types.
Cohesive force Adhesive force
The force of attraction between molecules of same The force of attraction between the molecules of the
substance is called the force of cohesion. This force is lesser different substances is called the force of adhesion.
in liquids and least in gases.
Ex. (i) Two drops of a liquid coalesce into one when Ex. (i) Adhesive force enables us to write on the blackboard
brought in mutual contact. with a chalk.
(ii) It is difficult to separate two sticky plates of glass welded (ii) A piece of paper sticks to another due to large force of
with water. adhesion between the paper and gum molecules.
(iii) It is difficult to break a drop of mercury into small (iii) Water wets the glass surface due to force of adhesion.
droplets because of large cohesive force between the
mercury molecules.
Note : ≅ Cohesive or adhesive forces are inversely proportional to the eighth power of distance between the
molecules.
Surface Tension.
The property of a liquid due to which its free surface tries to have minimum surface area and behaves as if it
were under tension some what like a stretched elastic membrane is called surface
tension. A small liquid drop has spherical shape, as due to surface tension the Imaginary line
liquid surface tries to have minimum surface area and for a given volume, the
sphere has minimum surface area.
Surface tension of a liquid is measured by the force acting per unit length
on either side of an imaginary line drawn on the free surface of liquid, the
direction of this force being perpendicular to the line and tangential to the free surface of liquid. So if F is the force
acting on one side of imaginary line of length L, then T = (F/L)
(1) It depends only on the nature of liquid and is independent of the area of surface or length of line considered.
(2) It is a scalar as it has a unique direction which is not to be specified.
(3) Dimension : [MT – 2]. (Similar to force constant)
(4) Units : N/m (S.I.) and Dyne/cm [C.G.S.]
(5) It is a molecular phenomenon and its root cause is the electromagnetic forces.
Force Due to Surface Tension.
If a body of weight W is placed on the liquid surface, whose surface tension is T. If F is the minimum force
required to pull it away from the water then value of F for different bodies can be calculated by the following table.
Body Figure Force
Needle F = 2l T + W
(Length = l ) F F = 2π (r1 + r2)T + W
TT
Hollow disc F = 2π (r + r)T + W
(Inner radius = r1 F F = 4πrT + W
Outer radius = r2) F = 2πrT + W
F
Thin ring F = 8l T + W
(Radius = r) F
F = 4l T + W
Circular plate or disc F
(Radius = r)
F
Square frame
(Side = l )
Square plate
Examples of Surface Tension.
(1) When mercury is split on a clean glass plate, it forms (2) When a greased iron needle is placed gently on the
globules. Tiny globules are spherical on the account of surface of water at rest, so that it does not prick the water
surface tension because force of gravity is negligible. The surface, the needle floats on the surface of
bigger globules get flattened from the middle but have water despite it being heavier because T T
round shape near the edges, figure
the weight of needle is balanced by the
vertical components of the forces of
surface tension. If the water surface is
pricked by one end of the needle, the mg
needle sinks down.
(3) When a molten metal is poured into water from a (4) Take a frame of wire and dip it in soap solution and take
suitable height, the falling stream of metal breaks up and it out, a soap film will be formed in the frame. Place a loop
the detached portion of the liquid in small quantity acquire of wet thread gently on the film. It will remain in the form,
the spherical shape. we place it on the film according to
Molten metal figure. Now, piercing the film with a
Water pin at any point inside the loop, It Thread
immediately takes the circular form loop
as shown in figure.
(5) Hair of shaving brush/painting brush when dipped in (6) If a small irregular piece of camphor is floated on the
water spread out, but as soon as it is taken out, its hair stick surface of pure water, it does not remain steady but dances
together. about on the surface. This is because, irregular shaped
camphor dissolves unequally and decreases the surface
tension of the water locally. The unbalanced forces make it
move haphazardly in different directions.
(7) Rain drops are spherical in shape because each drop (8) Oil drop spreads on cold water. Whereas it may remain
tends to acquire minimum surface area due to surface as a drop on hot water. This is due to the fact that the
tension, and for a given volume, the surface area of sphere surface tension of oil is less than that of cold water and is
is minimum. more than that of hot water.
Factors Affecting Surface Tension.
(1) Temperature : The surface tension of liquid decreases with rise of temperature. The surface tension of
liquid is zero at its boiling point and it vanishes at critical temperature. At critical temperature, intermolecular forces
for liquid and gases becomes equal and liquid can expand without any restriction. For small temperature
differences, the variation in surface tension with temperature is linear and is given by the relation
Tt = T0 (1 − α t)
where Tt , T0 are the surface tensions at t oC and 0o C respectively and α is the temperature coefficient of
surface tension.
Examples : (i) Hot soup tastes better than the cold soup.
(ii) Machinery parts get jammed in winter.
(2) Impurities : The presence of impurities either on the liquid surface or dissolved in it, considerably affect
the force of surface tension, depending upon the degree of contamination. A highly soluble substance like sodium
chloride when dissolved in water, increases the surface tension of water. But the sparingly soluble substances like
phenol when dissolved in water, decreases the surface tension of water.
Applications of Surface Tension.
(1) The oil and grease spots on clothes cannot be removed by pure water. On the other hand, when
detergents (like soap) are added in water, the surface tension of water decreases. As a result of this, wetting power
of soap solution increases. Also the force of adhesion between soap solution and oil or grease on the clothes
increases. Thus, oil, grease and dirt particles get mixed with soap solution easily. Hence clothes are washed easily.
(2) The antiseptics have very low value of surface tension. The low value of surface tension prevents the
formation of drops that may otherwise block the entrance to skin or a wound. Due to low surface tension, the
antiseptics spreads properly over wound.
(3) Surface tension of all lubricating oils and paints is kept low so that they spread over a large area.
(4) Oil spreads over the surface of water because the surface tension of oil is less than the surface tension of cold water.
(5) A rough sea can be calmed by pouring oil on its surface.
(6) In soldering, addition of ‘flux’ reduces the surface tension of molten tin, hence, it spreads.
Molecular Theory of Surface Tension.
The maximum distance upto which the force of attraction between two molecules is appreciable is called
molecular range (≈ 10−9 m) . A sphere with a molecule as centre and radius equal to molecular range is called the
sphere of influence. The liquid enclosed between free surface (PQ) of the liquid and an imaginary plane (RS) at a
distance r (equal to molecular range) from the free surface of the liquid form a liquid film.
To understand the tension acting on the free surface of a liquid, let us consider four liquid molecules like A, B,
C and D. Their sphere of influence are shown in the figure.
(1) Molecule A is well within the liquid, so it is attracted equally in all directions. Hence the net force on this
molecule is zero and it moves freely inside the liquid. P D C BQ
RS
(2) Molecule B is little below the free surface of the liquid and it is also
attracted equally in all directions. Hence the resultant force on it is also zero.
(3) Molecule C is just below the upper surface of the liquid film and the A
part of its sphere of influence is outside the free liquid surface. So the number of
molecules in the upper half (attracting the molecules upward) is less than the number of molecule in the lower half
(attracting the molecule downward). Thus the molecule C experiences a net downward force.
(4) Molecule D is just on the free surface of the liquid. The upper half of the sphere of influence has no liquid
molecule. Hence the molecule D experiences a maximum downward force.
Thus all molecules lying in surface film experiences a net downward force. Therefore, free surface of the liquid
behaves like a stretched membrane.
Problem 1. A wooden stick 2m long is floating on the surface of water. The surface tension of water 0.07 N/m. By putting
Solution : (d) soap solution on one side of the sticks the surface tension is reduced to 0.06 N/m. The net force on the stick
will be [Pb. PMT 2002]
(a) 0.07 N (b) 0.06 N (c) 0.01 N (d) 0.02 N
Force on one side of the stick F1 = T1 × L = 0.07 × 2 = 0.14 N
and force on other side of the stick F2 = T2 × L = 0.06 × 2 = 0.12N
So net force on the stick = F1 − F2 = 0.14 − 0.12 = 0.02N
Problem 2. A thin metal disc of radius r floats on water surface and bends the surface downwards along the perimeter
Solution : (c) making an angle θ with vertical edge of disc. If the disc displaces a weight of water W and surface tension of
water is T, then the weight of metal disc is [AMU (Med.) 1999]
(a) 2π rT + W (b) 2π rT cosθ – W (c) 2π rT cosθ + W (d) W – 2π rT cosθ
Weight of metal disc = total upward force T T
= upthrust force + force due to surface tension θθ
= weight of displaced water + T cos θ (2π r)
= W + 2π rT cos θ
Problem 3. A 10 cm long wire is placed horizontally on the surface of water and is gently pulled up with a force of
Solution : (a) 2 × 10−2 N to keep the wire in equilibrium. The surface tension in Nm–1 of water is [AMU (Med.) 1999]
Problem 4. (a) 0.1 N/m (b) 0.2 N/m (c) 0.001 N/m (d) 0.002 N/m
Solution : (d) Force on wire due to surface tension F = T × 2l
∴T = F = 2 × 10 −2 = 0.1 N/m
2l 2 × 10 × 10 −2
There is a horizontal film of soap solution. On it a thread is placed in the form of a loop. The film is pierced
inside the loop and the thread becomes a circular loop of radius R. If the surface tension of the loop be T, then
what will be the tension in the thread [RPET 1996]
(a) πR2 / T (b) πR2T (c) 2πRT (d) 2RT
Suppose tension in thread is F, then for small part ∆l of thread 2 × T × ∆l
∆l = Rθ and 2F sinθ / 2 = 2T∆l = 2TRθ F cosθ /2
θ /2
⇒F = TRθ = TRθ = 2TR (sinθ / 2 ≈ θ / 2) F cosθ /2 ∆l F
sinθ / 2 θ /2 F θ /2 θ /2θ /2
F sinθ/2 F sinθ/2
Problem 5. A liquid is filled into a tube with semi-elliptical cross-section as shown in the figure. The ratio of the surface
tension forces on the curved part and the plane part of the tube in vertical position will be
(a) π (a + b) ba
4b
(b) 2πa
b
(c) πa
4b
(d) π (a − b)
4b
Solution : (a) From the figure Curved part = semi perimeter = π (a + b)
2
and the plane part = minor axis = 2b
∴ Force on curved part = T × π (a + b)
2
2b a π (a+b)
2
and force on plane part = T × 2b
∴ Ratio = π (a + b)
4b
Problem 6. A liquid film is formed over a frame ABCD as shown in figure. Wire CD can slide without friction. The mass to
be hung from CD to keep it in equilibrium is
(a) Tl AB
g
2Tl Liquid
g film
(b) D C
g l
2Tl
(c)
(d) T × l
Solution : (b) Weight of the body hung from wire (mg) = upward force due to surface tension (2Tl ) ⇒ m= 2Tl
g
Surface Energy.
The molecules on the liquid surface experience net downward force. So to bring a molecule from the interior
of the liquid to the free surface, some work is required to be done against the intermolecular force of attraction,
which will be stored as potential energy of the molecule on the surface. The potential energy of surface molecules
per unit area of the surface is called surface energy. C L L' D
Unit : Joule/m2 (S.I.) erg/cm2 (C.G.S.)
Dimension : [MT–2] l T × 2l F
If a rectangular wire frame ABCD, equipped with a sliding wire LM
dipped in soap solution, a film is formed over the frame. Due to the surface B M M' A
x
tension, the film will have a tendency to shrink and thereby, the sliding wire
LM will be pulled in inward direction. However, the sliding wire can be held in this position under a force F, which
is equal and opposite to the force acting on the sliding wire LM all along its length due to surface tension in the soap
film.
If T is the force due to surface tension per unit length, then F = T × 2l
Here, l is length of the sliding wire LM. The length of the sliding wire has been taken as 2l for the reason that
the film has got two free surfaces.
Suppose that the sliding wire LM is moved through a small distance x, so as to take the position L' M' . In this
process, area of the film increases by 2l × x (on the two sides) and to do so, the work done is given by
W = F × x = (T × 2l) × x = T × (2lx) = T × ∆A
∴ W = T × ∆A [∆A = Total increase in area of the film from both the sides]
If temperature of the film remains constant in this process, this work done is stored in the film as its surface energy.
From the above expression T = W or T=W [If ∆A = 1]
∆A
i.e. surface tension may be defined as the amount of work done in increasing the area of the liquid surface by
unity against the force of surface tension at constant temperature.
Work Done in Blowing a Liquid Drop or Soap Bubble.
(1) If the initial radius of liquid drop is r1 and final radius of liquid drop is r2 then
W = T × Increment in surface area
W = T × 4π [r22 − r12 ] [drop has only one free surface]
(2) In case of soap bubble
W = T × 8π [r22 − r12 ] [Bubble has two free surfaces]
Splitting of Bigger Drop.
When a drop of radius R splits into n smaller drops, (each of radius r) then surface area of liquid increases.
Hence the work is to be done against surface tension.
Since the volume of liquid remains constant therefore 4 πR 3 = n 4 πr 3 ∴ R 3 = nr 3
3 3
Work done = T × ∆A = T [Total final surface area of n drops – surface area of big drop] = T[n4πr 2 − 4πR 2 ]
Various formulae of work done
4πT[nr2 − R2] 4πR2T[n1 / 3 − 1] 4πTr 2n2 / 3[n1 / 3 − 1] 4πTR3 1 − 1 R
r R
If the work is not done by an external source then internal energy of liquid decreases, subsequently
temperature decreases. This is the reason why spraying causes cooling.
By conservation of energy, Loss in thermal energy = work done against surface tension
JQ = W
⇒ JmS∆θ = 4πTR 3 1 − 1
r R
⇒ J 4 πR 3 d S∆θ = 4πR 3 T 1 − 1 [As m=V×d= 4 π R3 × d ]
3 r R 3
∴ Decrease in temperature ∆θ = 3T 1 − 1
JSd r R
where J = mechanical equivalent of heat, S = specific heat of liquid, d = density of liquid.
Formation of Bigger Drop.
If n small drops of radius r coalesce to form a big drop of radius R then surface area of the liquid decreases.
Amount of surface energy released = Initial surface energy – final surface energy
E = n4πr 2T − 4πR 2T
Various formulae of released energy R
4πT[nr2 − R2] 4πR2T(n1 / 3 − 1) 4πTr 2n2 / 3(n1 / 3 − 1) 4πTR3 1 − 1
r R
(i) If this released energy is absorbed by a big drop, its temperature increases and rise in temperature can be
given by ∆θ = 3T 1 − 1
JSd r R
(ii) If this released energy is converted into kinetic energy of a big drop without dissipation then by the law of
conservation of energy.
1 mv 2 = 4πR 3 T 1 − 1 ⇒ 1 4 πR 3 dv 2 = 4πR 3 T 1 − 1 ⇒ v2 = 6T 1 − 1
2 r R 2 3 r R d r R
∴ v= 6T 1 − 1
Problem 7. d r R
Solution : (b) Two small drops of mercury, each of radius R, coalesce to form a single large drop. The ratio of the total
surface energies before and after the change is [AIIMS 2003]
(a) 1 : 21 / 3 (b) 21 / 3 : 1 (c) 2 : 1 (d) 1 : 2
As R = n1 / 3r = 21/ 3 r ⇒ R2 = 22 / 3 r 2 ⇒ r2 = 2−2 / 3
R2
Initial surface energy = 2(4π r 2T) = 2 r2 = 2 × 2−2 / 3 = 21/3
Final surface energy (4πR 2T) R2
Problem 8. Radius of a soap bubble is increased from R to 2R work done in this process in terms of surface tension is
[CPMT 1991; RPET 2001; BHU 2003]
(a) 24πR2S (b) 48πR2S (c) 12πR2S (d) 36πR2S
Solution : (a) ( )W = 8πT R22 − R12 = 8πS[(2R)2 − (R)2 ] = 24πR 2S
Problem 9. The work done in blowing a soap bubble of 10cm radius is (surface tension of the soap solution is 3 N /m )
100
[MP PMT 1995; MH CET 2002]
(a) 75.36 × 10 −4 J (b) 37.68 × 10 −4 J (c) 150.72 × 10 −4 J (d) 75.36 J
Solution : (a) W = 8πR 2T = 8π (10 × 10 −2 )2 3 = 75.36 × 10 −4 J
100
Problem 10. A drop of mercury of radius 2mm is split into 8 identical droplets. Find the increase in surface energy. (Surface
tension of mercury is 0.465 J/m2) [UPSEAT 2002]
(a) 23.4µJ (b) 18.5µJ (c) 26.8µJ (d) 16.8µJ
Solution : (a) Increase in surface energy = 4πR 2T(n1/ 3 − 1) = 4π (2 × 10−3 )2(0.465)(81/ 3 − 1) = 23.4 × 10 −6 J = 23.4µJ
Problem 11. The work done in increasing the size of a soap film from 10cm × 6cm to 10cm × 11cm is 3 × 10−4 J . The
surface tension of the film is [MP PET 1999; MP PMT 2000; AIIMS 2000; JIPMER 2001, 02]
(a) 1.5 × 10−2 Nm−1 (b) 3.0 × 10 −2 Nm−1 (c) 6.0 × 10 −2 Nm−1 (d) 11.0 × 10−2 Nm−1
Solution : (b) A1 = 10 × 6 = 60cm2 = 60 × 10−4 m2 , A2 = 10 × 11 = 110cm2 = 110 × 10−4 m2
As the soap film has two free surfaces ∴W = T × 2∆ A
⇒W = T × 2 × (A2 − A1) ⇒ T = W = 3 × 10−4 = 3 × 10−2 N/m
2 × 50 × 10−4 2 × 50 × 10−4
Problem 12. A film of water is formed between two straight parallel wires of length 10cm each separated by 0.5cm. If their
separation is increased by 1 mm while still maintaining their parallelism, how much work will have to be done
(Surface tension of water = 7.2 × 10−2 N / m ) [Roorkee 1986; MP PET 2001]
(a) 7.22 × 10−6 J (b) 1.44 × 10−5 J (c) 2.88 × 10−5 J (d) 5.76 × 10−5 J
Solution : (b) As film have two free surfaces W = T × 2∆A
W = T × 2l × x lF
= 7.2 × 10−2 × 2 × 0.1 × 1 × 10−3 x
= 1.44 × 10−5 J
Problem 13. If the work done in blowing a bubble of volume V is W, then the work done in blowing the bubble of volume
2V from the same soap solution will be [MP PET 1989]
(a) W/2 (b) 2 W (c) 3 2 W (d) 3 4 W
Solution : (d) As volume of the bubble V = 4 πR3 ⇒ R = 3 1 / 3 V 1 / 3 ⇒ R2 = 3 2 / 3 V 2/ 3 ⇒ R2 ∝ V 2/3
3 4π 4π
Work done in blowing a soap bubble W = 8πR2T ⇒ W ∝ R 2 ∝ V 2 / 3
W2 V2 2 / 3 2V 2 / 3
W1 V1 V
∴ = = = (2)2 / 3 = (4)1 / 3 ⇒ W2 = 34W
Problem 14. Several spherical drops of a liquid of radius r coalesce to form a single drop of radius R. If T is surface tension
and V is volume under consideration, then the release of energy is
(a) 3VT 1 + 1 (b) 3VT 1 − 1 (c) VT 1 − 1 (d) VT 1 + 1
r R r R r R r2 R2
Solution : (b) Energy released = 4πTR 3 1 − 1 = 3 4 πR 3 T 1 − 1 = 3VT 1 − 1
r R 3 r R r R
Excess Pressure.
Due to the property of surface tension a drop or bubble tries to contract and so compresses the matter
enclosed. This in turn increases the internal pressure which prevents further contraction and equilibrium is achieved.
So in equilibrium the pressure inside a bubble or drop is greater than outside and the difference of pressure
between two sides of the liquid surface is called excess pressure. In case of a drop excess pressure is provided by
hydrostatic pressure of the liquid within the drop while in case of bubble the gauge pressure of the gas confined in
the bubble provides it.
Excess pressure in different cases is given in the following table :
Plane surface Concave surface
∆P=0 ∆P
∆P = 0 ∆P = 2T
R
Convex surface Drop
∆P = 2T ∆P = 2T
R R
∆P ∆P
Bubble in air Bubble in liquid
∆P ∆P = 4T ∆P ∆P = 2T
R R
Bubble at depth h below the free surface of liquid of density d Cylindrical liquid surface
R T
R
h ∆P = 2T + hdg ∆P =
R
∆P
Liquid surface of unequal radii Liquid film of unequal radii
∆P ∆P = T 1 + 1 ∆P ∆P = 2T 1 + 1
R1 R1
R 2 R 2
Note : Excess pressure is inversely proportional to the radius of bubble
(or drop), i.e., pressure inside a smaller bubble (or drop) is higher
than inside a larger bubble (or drop). This is why when two
bubbles of different sizes are put in communication with each
other, the air will rush from smaller to larger bubble, so that the
smaller will shrink while the larger will expand till the smaller
bubble reduces to droplet.
Problem 15. The pressure inside a small air bubble of radius 0.1mm situated just below the surface of water will be equal to
(Take surface tension of water 70 × 10−3 Nm−1 and atmospheric pressure = 1.013 × 105 Nm− 2 )
(a) 2.054 × 103 Pa (b) 1.027 × 103 Pa (c) 1.027 × 105 Pa [AMU (Med.) 2002]
(d) 2.054 × 105 Pa
Solution : (c) Pressure inside a bubble when it is in a liquid = Po + 2T = 1.013 × 105 + 2× 70 × 10−3 = 1.027 ×105 Pa.
R 0.1 × 10−3
Problem 16. If the radius of a soap bubble is four times that of another, then the ratio of their excess pressures will be
[AIIMS 2000]
(a) 1 : 4 (b) 4 : 1 (c) 16 : 1 (d) 1 : 16
Solution : (a) Excess pressure inside a soap bubble ∆P = 4T ⇒ ∆P1 = r2 = 1: 4
r ∆P2 r1
Problem 17. Pressure inside two soap bubbles are 1.01 and 1.02 atmospheres. Ratio between their volumes is
[MP PMT 1991]
(a) 102 : 101 (b) (102)3 : (101)3 (c) 8 : 1 (d) 2 : 1
Solution : (c) Excess pressure ∆P = Pin − Pout = 1.01atm − 1atm = 0.01atm and similarly ∆P2 = 0.02atm
and volume of air bubble V = 4 πr 3 ∴V ∝ r3 ∝ 1 [as ∆P ∝ 1 or r ∝ 1 ]
3 (∆P)3 r ∆P
V1 ∆P2 3 0.02 3 2 3 8
V2 ∆P1 0.01 1 1
∴ = = = =
Problem 18. The excess pressure inside an air bubble of radius r just below the surface of water is P1. The excess pressure
inside a drop of the same radius just outside the surface is P2. If T is surface tension then
(a) P1 = 2P2 (b) P1 = P2 (c) P2 = 2P1 (d) P2 = 0, P1 ≠ 0
Solution : (b) Excess pressure inside a bubble just below the surface of water P1 = 2T
r
and excess pressure inside a drop P2 = 2T ∴P1 = P2
r
Shape of Liquid Meniscus.
We know that a liquid assumes the shape of the vessel in which it is contained i.e. it can not oppose
permanently any force that tries to change its shape. As the effect of force is zero in a direction perpendicular to it,
the free surface of liquid at rest adjusts itself at right angles to the resultant force.
When a capillary tube is dipped in a liquid, the liquid surface becomes curved near the point of contact. This
curved surface is due to the resultant of two forces i.e. the force of cohesion and the force of adhesion. The curved
surface of the liquid is called meniscus of the liquid.
If liquid molecule A is in contact with solid (i.e. wall of capillary tube) then forces acting on molecule A are
(i) Force of adhesion Fa (acts outwards at right angle to the wall of the tube).
(ii) Force of cohesion Fc (acts at an angle 45o to the vertical).
Resultant force FN depends upon the value of Fa and Fc.
If resultant force FN make an angle α with Fa.
Then tanα = Fc sin135o = Fc
Fa + Fc cos135o 2 Fa − Fc
By knowing the direction of resultant force we can find out the shape of meniscus because the free surface of
the liquid adjust itself at right angle to this resultant force.
If Fc = 2Fa Fc < 2Fa Fc > 2Fa
tanα = ∞ ∴ α = 90o tan α = positive ∴ α is acute angle tan α = negative ∴ α is obtuse angle
i.e. the resultant force acts vertically i.e. the resultant force directed outside i.e. the resultant force directed inside
downwards. Hence the liquid meniscus the liquid. Hence the liquid meniscus the liquid. Hence the liquid meniscus
must be horizontal. must be concave upward. must be convex upward.
Fa A Fc Fa A Fc Fa A Fc
α α
α
FN 45° 45°
FN
FN
Example: Pure water in silver coated Example: Water in glass capillary tube. Example: Mercury in glass capillary tube.
capillary tube.
Angle of Contact.
Angle of contact between a liquid and a solid is defined as the angle enclosed between the tangents to the
liquid surface and the solid surface inside the liquid, both the tangents being drawn at the point of contact of the
liquid with the solid.
θ < 90o θ = 90o θ > 90o
Fa > Fc Fa = Fc θ Fa < Fc θ
2 2 2
θ
concave meniscus. plane meniscus. convex meniscus.
Liquid wets the solid surface Liquid does not wet the solid surface. Liquid does not wet the solid surface.
(i) Its value lies between 0o and 180o
θ = 0o for pure water and glass, θ = 8o for tap water and glass, θ = 90o for water and silver
θ = 138o for mercury and glass, θ = 160o for water and chromium
(ii) It is particular for a given pair of liquid and solid. Thus the angle of contact changes with the pair of solid
and liquid.
(iii) It does not depends upon the inclination of the solid in the liquid.
(iv) On increasing the temperature, angle of contact decreases.
(v) Soluble impurities increases the angle of contact.
(vi) Partially soluble impurities decreases the angle of contact.
Capillarity.
If a tube of very narrow bore (called capillary) is dipped in a liquid, it is found that the liquid in the capillary
either ascends or descends relative to the surrounding liquid. This phenomenon is called capillarity.
The root cause of capillarity is the difference in pressures on two sides of (concave and convex) curved surface
of liquid.
Examples of capillarity :
(i) Ink rises in the fine pores of blotting paper leaving the paper dry.
(ii) A towel soaks water.
(iii) Oil rises in the long narrow spaces between the threads of a wick.
(iv) Wood swells in rainy season due to rise of moisture from air in the pores.
(v) Ploughing of fields is essential for preserving moisture in the soil.
(vi) Sand is drier soil than clay. This is because holes between the sand particles are not so fine as compared
to that of clay, to draw up water by capillary action.
Ascent Formula.
When one end of capillary tube of radius r is immersed into a liquid of density d which wets the sides of the
capillary tube (water and capillary tube of glass), the shape of the liquid meniscus in the tube becomes concave upwards.
R = radius of curvature of liquid meniscus. A R
θr y
T = surface tension of liquid A(P) C(P) B h
θ
B D(P) E(P)
P = atmospheric pressure (P – 2T/R)
Pressure at point A = P, Pressure at point B = P − 2T
R
Pressure at points C and D just above and below the plane surface of liquid in the vessel is also P (atmospheric
pressure). The points B and D are in the same horizontal plane in the liquid but the pressure at these points is
different.
In order to maintain the equilibrium the liquid level rises in the capillary tube upto height h.
Pressure due to liquid column = pressure difference due to surface tension
⇒ hdg = 2T
R
∴ h = 2T = 2T cosθ As R = r
Rdg rdg cosθ
(i) The capillary rise depends on the nature of liquid and solid both i.e. on T, d, θ and R.
(ii) Capillary action for various liquid-solid pair.
Meniscus Angle of contact Level
Rises
Glass Concave θ < 90o
Water Plane θ = 90o No rise no fall
Silver
Water Convex θ > 90o Fall
Glass
Mercury
(iii) For a given liquid and solid at a given place
h ∝ 1 [As T, θ, d and g are constant]
r
i.e. lesser the radius of capillary greater will be the rise and vice-versa. This is called Jurin’s law.
(iv) If the weight of the liquid contained in the meniscus is taken into consideration then more accurate ascent
formula is given by
h = 2T cosθ − r
rdg 3
(v) In case of capillary of insufficient length, i.e., L < h, the liquid will r r'
neither overflow from the upper end like a fountain nor will it tickle along the h L
vertical sides of the tube. The liquid after reaching the upper end will increase
the radius of its meniscus without changing nature such that :
hr = Lr′ L < h ∴ r ' > r
(vi) If a capillary tube is dipped into a liquid and tilted at an angle α from R
vertical, then the vertical height of liquid column remains same whereas the h α lh
length of liquid column (l) in the capillary tube increases.
Water
h = l cosα or l= h
cosα
(vii) It is important to note that in equilibrium the height h is independent h
of the shape of capillary if the radius of meniscus remains the same. That is why
the vertical height h of a liquid column in capillaries of different shapes and sizes
will be same if the radius of meniscus remains the same.
Problem 19. Water rises to a height of 10cm in a capillary tube and mercury falls to a depth of 3.5cm in the same capillary
tube. If the density of mercury is 13.6 gm/cc and its angle of contact is 135o and density of water is 1 gm/cc
and its angle of contact is 0o , then the ratio of surface tensions of the two liquids is (cos 135o = 0.7)
(a) 1 : 14 (b) 5 : 34 (c) 1 : 5 [MP PMT 1988; EAMCET (Med.) 2003]
(d) 5 : 27
Solution : (b) h = 2T cosθ ∴ hW = TW cos θ W d Hg [as r and g are constants]
rdg hHg THg cosθ Hg dW
⇒ 10 = TW . cos 0 o 13.6 ⇒ TW = 10 × 0.7 20 = 5
3.5 THg cos 135 1 THg 3.5 × 13.6 = 136 34
Problem 20. Water rises in a vertical capillary tube upto a height of 2.0 cm. If the tube is inclined at an angle of 60o with
the vertical, then upto what length the water will rise in the tube [UPSEAT 2002]
(a) 2.0 cm (b) 4.0 cm (c) 4 cm (d) 2 2 cm
3
Solution : (b) The height upto which water will rise l= h = 2cm = 4cm . [h = vertical height, α = angle with vertical]
cosα cos 60
Problem 21. Two capillary tubes of same diameter are kept vertically one each in two liquids whose relative densities are
0.8 and 0.6 and surface tensions are 60 and 50 dyne/cm respectively. Ratio of heights of liquids in the two
tubes h1 is [MP PMT 2002]
h2
(a) 10 (b) 3 (c) 10 (d) 9
9 10 3 10
Solution : (d) h = 2T cosθ [If diameter of capillaries are same and taking value of θ same for both liquids]
rdg
∴ h1 = T1 d2 = 60 × 0.6 = 36 = 9 .
h2 T2 d1 50 0.8 40 10
Problem 22. A capillary tube of radius R is immersed in water and water rises in it to a height H. Mass of water in the
Solution : (b) capillary tube is M. If the radius of the tube is doubled, mass of water that will rise in the capillary tube will
Problem 23. now be [RPMT 1997; RPET 1999; CPMT 2002]
Solution : (a) (a) M (b) 2M (c) M/2 (d) 4M
Problem 24. Mass of the liquid in capillary tube M = Vρ = (πr2h)ρ ∴ M ∝ r 2h ∝ r [As h ∝ 1 ]
r
Solution : (d)
So if radius of the tube is doubled, mass of water will becomes 2M, which will rise in capillary tube.
Problem 25.
Water rises to a height h in a capillary at the surface of earth. On the surface of the moon the height of water
Solution : (b)
column in the same capillary will be [MP PMT 2001]
(a) 6h (b) 1 h (c) h (d) Zero
6
h = 2T cosθ ∴h ∝ 1 [If other quantities remains constant]
rdg g
hmoon = g earth =6⇒ hmoon = 6h [As gearth= 6gmoon]
h earth g moon
Water rises upto a height h in a capillary on the surface of earth in stationary condition. Value of h increases if
this tube is taken [RPET 2000]
(a) On sun (b) On poles
(c) In a lift going upward with acceleration (d) In a lift going downward with acceleration
h ∝ 1 . In a lift going downward with acceleration (a), the effective acceleration decreases. So h increases.
g
If the surface tension of water is 0.06 N/m, then the capillary rise in a tube of diameter 1mm is (θ = 0o)
[AFMC 1998]
(a) 1.22 cm (b) 2.44 cm (c) 3.12 cm (d) 3.86 cm
h = 2T cosθ , [θ =0, r = 1 mm = 0.5 × 10−3 m , T = 0.06 N /m, d = 10 3 kg / m3, g = 9.8 m/s2 ]
rdg 2
h = 2 × 0.06 × cosθ = 0.0244m = 2.44cm
0.5 × 10 −3 × 10 3 × 9.8
Problem 26. Two capillaries made of same material but of different radii are dipped in a liquid. The rise of liquid in one
capillary is 2.2cm and that in the other is 6.6cm. The ratio of their radii is [MP PET 1990]
(a) 9 : 1 (b) 1 : 9 (c) 3 : 1 (d) 1 : 3
Solution : (c) As h ∝ 1 ∴ h1 = r2 or r1 = h2 = 6.6 = 3
r h2 r1 r2 h1 2.2 1
Problem 27. The lower end of a capillary tube is at a depth of 12cm and the water rises 3cm in it. The mouth pressure
required to blow an air bubble at the lower end will be X cm of water column where X is [CPMT 1989]
(a) 3 (b) 9 (c) 12 (d) 15
Solution : (d) The lower end of capillary tube is at a depth of 12 + 3 = 15 cm from the free surface of water in capillary
tube.
So, the pressure required = 15 cm of water column.
Problem 28. The lower end of a capillary tube of radius r is placed vertically in water. Then with the rise of water in the
capillary, heat evolved is
(a) + π 2r 2h2 dg (b) + πr 2h2dg (c) − πr 2h2dg (d) − πr 2h2dg
J 2J 2J J
Solution : (b) When the tube is placed vertically in water, water rises through height h given by h = 2T cosθ
rdg
Upward force = 2πr × T cosθ
Work done by this force in raising water column through height h is given by
∆W = (2πrT cosθ )h = (2πrhcosθ )T = (2πrh cosθ ) rhdg = πr 2h2dg
2 cosθ
However, the increase in potential energy ∆E p of the raised water column = mg h
2
where m is the mass of the raised column of water ∴m = πr 2hd
So, ∆E P = (πr 2hd) hg = πr 2h2dg
2 2
Further, ∆W − ∆Ep = πr 2h2dg
2
The part (∆W − ∆EP ) is used in doing work against viscous forces and frictional forces between water and
glass surface and appears as heat. So heat released = ∆W − ∆E p = πr 2h2dg
J 2J
Problem 29. Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced
Solution : (d) by 75 × 10−4 N force due to the weight of the liquid. If the surface tension of water is 6 × 10−2 N / m , the
inner circumference of the capillary must be [CPMT 1986, 88]
(a) 1.25 × 10−2 m (b) 0.50 × 10−2 m (c) 6.5 × 10−2 m (d) 12.5 × 10−2 m
Weight of liquid = upward force due to surface tension
75 × 10−4 = 2πrT
Circumference 2πr = 75 × 10−4 = 75 × 10−4 = 0.125 = 12.5 × 10−2m
T 6 × 10−2
Shape of Drops.
Whether the liquid will be in equilibrium in the form of a drop or it will Solid TLA Air
spread out; depends on the relative strength of the force due to surface tension Liquid θ TSA
at the three interfaces.
TSL O
TLA = surface tension at liquid-air interface, TSA = surface tension at
solid-air interface.
TSL = surface tension at solid-liquid interface, θ = angle of contact between the liquid and solid.
For the equilibrium of molecule
TSL + TLA cosθ = TSA or cosθ = TSA − TSL …..(i) Air
TLA
TSL θ TLA
Solid O TSA
Special Cases
TSA > TSL, cosθ is positive i.e. 0o < θ < 90o .
This condition is fulfilled when the molecules of liquid are strongly attracted to that of solid.
Example : (i) Water on glass.
(ii) Kerosene oil on any surface.
TSA < TSL, cosθ is negative i.e. 90o < θ < 180o .
This condition is fulfilled when the molecules of the liquid are strongly attracted to themselves and relatively
weakly to that of solid.
Example : (i) Mercury on glass surface.
(ii) Water on lotus leaf (or a waxy or oily surface)
(TSL + TLA cosθ) > TSA
In this condition, the molecule of liquid will not be in equilibrium and experience a net force at the interface. As a
result, the liquid spreads.
Example : (i) Water on a clean glass plate.
Useful Facts and Formulae.
(1) Formation of double bubble : If r1 and r2 are the radii of smaller and
larger bubble and P0 is the atmospheric pressure, then the pressure inside ∆P r1
P1
them will be P1 = P0 + 4T and P2 = P0 + 4T . r2 r
r1 r2
P2
Now as r1 < r2 ∴ P1 > P2
So for interface ∆P = P1 − P2 = 4 T 1 − 1 …..(i)
r1 r2
As excess pressure acts from concave to convex side, the interface will be concave towards the smaller bubble
and convex towards larger bubble and if r is the radius of interface.
∆P = 4T …..(ii)
r
From (i) and (ii) 111
r = r1 − r2
∴ Radius of the interface r = r1r2
r2 − r1
(2) Formation of a single bubble
(i) Under isothermal condition two soap bubble of radii ‘a’ and ‘b’ coalesce to form a single bubble of radius ‘c’.
If the external pressure is P0 then pressure inside bubbles
Pa P0 4T , Pb P0 4T and Pc P0 4T a
a b c
= + = + = + c
and volume of the bubbles b
Va = 4 πa 3 , Vb = 4 πb 3 , Vc = 4 πc 3
3 3 3
Now as mass is conserved µa + µb = µc ⇒ Pa Va + Pb Vb = Pc Vc As PV = µRT, i.e., µ = PV
RTa RTb RTc RT
⇒ Pa Va + Pb Vb = Pc Vc …..(i)
Substituting the value of pressure and volume [As temperature is constant, i.e., Ta = Tb = Tc ]
⇒ P0 + 4T 4 πa 3 + P0 + 4T 4 πb 3 = P0 + 4T 4 πc 3
a 3 b 3 c 3
⇒ 4T(a 2 + b 2 − c 2 ) = P0 (c 3 − a 3 − b 3 )
∴ Surface tension of the liquid T = P0 (c 3 − a 3 − b 3 )
4(a 2 + b 2 − c 2 )
(ii) If two bubble coalesce in vacuum then by substituting P0 = 0 in the above expression we get
a2 + b2 − c2 = 0 ∴ c2 = a2 + b2
Radius of new bubble = c = a 2 + b 2 or can be expressed as
r = r12 + r22 . h
h1
(3) The difference of levels of liquid column in two limbs of u-tube of
unequal radii r1 and r2 is h2
h = h1 − h2 = 2T cosθ 1 − 1
dg r2
r1
(4) A large force (F) is required to draw apart normally two glass plate enclosing a thin water film because the
thin water film formed between the two glass plates will have concave surface all around. Since on the concave side
of a liquid surface, pressure is more, work will have to be done in drawing the plates apart.
F= 2 AT where T= surface tension of water film, t= thickness of film, A = area of film.
t
(5) When a soap bubble is charged, then its size increases due to outward force on the bubble.
(6) The materials, which when coated on a surface and water does not enter through that surface are known
as water proofing agents. For example wax etc. Water proofing agent increases the angle of contact.
(7) Values of surface tension of some liquids.
Liquid Surface tension Newton/metre
Mercury 0.465
Water 0.075
Soap solution 0.030
Glycerine 0.063
Carbon tetrachloride 0.027
Ethyl alcohol 0.022
Problem 30. The radii of two soap bubbles are r1 and r2. In isothermal conditions, two meet together in vacuum. Then the
radius of the resultant bubble is given by [RPET 1999; MP PMT 2001; EAMCET 2003]
(a) R = (r1 + r2) / 2 (b) R = r1(r1r2 + r2) (c) R2 = r12 + r22 (d) R = r1 + r2
Solution : (c) Under isothermal condition surface energy remain constant ∴ 8πr12T + 8πr22T = 8πR 2T ⇒ R 2 = r12 + r22
Problem 31. Two soap bubbles of radii r1 and r2 equal to 4cm and 5cm are touching each other over a common surface
S1S2 (shown in figure). Its radius will be [MP PMT 2002]
(a) 4 cm
(b) 20 cm S1 5 cm
4 cm
(c) 5 cm S2
(d) 4.5 cm
Solution : (b) Radius of curvature of common surface of double bubble r = r2r1 = 5×4 = 20cm
r2 − r1 5−4
Problem 32. An air bubble in a water tank rises from the bottom to the top. Which of the following statements are true
[Roorkee 2000]
(a) Bubble rises upwards because pressure at the bottom is less than that at the top
(b) Bubble rises upwards because pressure at the bottom is greater than that at the top
(c) As the bubble rises, its size increases
(d) As the bubble rises, its size decreases
Solution : (b, c)
Problem 33. The radii of two soap bubbles are R1 and R2 respectively. The ratio of masses of air in them will be
R13 R23 P + 4T R13 P + 4T R23
R23 R13 R1 R2
(a) (b) (c) 4T R23 (d) 4T R13
R2 R1
P + P +
Solution : (c) From PV = µRT.
µ1 P1 V1 P + 4T 4 πR13 P + 4T R13
µ2 P2 V2 R1 3 P + R1
At a given temperature, the ratio masses of air = = = .
P 4T 4 πR23 4T R 3
+ R2 3 R2 2
Problem 34. On dipping one end of a capillary in liquid and inclining the capillary at an angles 30o and 60o with the
vertical, the lengths of liquid columns in it are found to be l1 and l2 respectively. The ratio of l1 and l2 is
(a) 1 : 3 (b) 1 : 2 (c) 2 : 1 (d) 3 : 1
Solution : (a) l1 = h and l2 = h ∴ l1 = cosα 2 = cos 60o = 1/2 = 1: 3
cos α 1 cosα 2 l2 cos α 1 cos 30o 3/2
Problem 35. A drop of water of volume V is pressed between the two glass plates so as to spread to an area A. If T is the
surface tension, the normal force required to separate the glass plates is
(a) TA2 (b) 2TA2 (c) 4TA2 (d) TA2
V V V 2V
Solution : (b) Force required to separate the glass plates F = 2 AT × A = 2TA 2 = 2TA 2 .
t A (A × t) V
Transmission of heat
Introduction.
Heat energy transfers from a body at higher temperature to a body at lower temperature. The transfer of heat
from one body to another may take place by one of the following modes.
Conduction Convection Radiation
Heat flows from hot end to cold end. Each particle absorbing heat is Heat flows without any intervening
Particles of the medium simply mobile medium in the form of electromagnetic
oscillate but do not leave their place. waves.
Medium is necessary for conduction Medium is necessary for convection Medium is not necessary for radiation
It is a slow process It is also a slow process It is a very fast process
Path of heat flow may be zig-zag Path may be zig-zag or curved Path is a straight line
Conduction takes place in solids Convection takes place in fluids Radiation takes place in gaseous
and transparent media
The temperature of the medium In this process also the temperature There is no change in the
increases through which heat flows
of medium increases temperature of the medium
Conduction.
The process of transmission of heat energy in which the heat is transferred from one particle to other particle
without dislocation of the particle from their equilibrium position is called conduction.
(i) Conduction is a process which is possible in all states of matter.
(ii) In solids only conduction takes place.
(iii) In non-metallic solids and fluids the conduction takes place only due to vibrations of molecules, therefore
they are poor conductors.
(iv) In metallic solids free electrons carry the heat energy, therefore they are good conductor of heat.
(1) Variable and steady state
When one end of a metallic rod is heated, heat flows by conduction from the hot end to the cold end.
In the process of conduction each cross-section of the rod receives heat from the adjacent cross-section
towards the hot end. A part of this heat is absorbed by the cross-section itself whose temperature increases, another
part is lost into atmosphere by convection & radiation and the rest is conducted away to the next cross-section.
Because in this state temperature of every cross-section of the rod goes on increasing, hence rod is said to exist
in variable state.
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