book 0sample integral cal

8/3/2019 Book 0Sample Integral Cal

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INTRODUCTION :

Meaning of Integration is to 'join' or 'combine'. The process of Integration is just the reverse of

differentiation.

Integration given scalar sum, it gives the area under the x y curve. Area represents that physical quantityobtained by the product of x & y.

A = y.dxRemember :

Area of Velocity time curve DisplacementArea of Acceleration time curve change in Velocity

y

dx

Area undercurve

Area of Force time curve Impulse(change in linear momentum)

Area of Force displacement curve WorkArea of Power time curve WorkArea of Pressure volume curve WorkArea of Torque time curve change in Angular momentum

Points to be remembered while using Integration in Physics :

(i) Before doing integration ensure that the given physical quantity is scalar or vector.

(ii) It the given physical quantity is scalar, then integrate directly and if it is vector quantity then first see the

direction. If direction remains constant, then integrate directly and if direction varies then first find the

components and then integrate.

(iii) When doing integration there should not be more than two variables, each separated on different sides.

(iv) If there are more than two variables then first convert them into two variables.

(v) Be careful while deciding limits and use relative limits.

(vi) Always put limits of that variable only with respect to which integration is being done.

(vii) In physics every relation should be known with conditions. If condition is variable then use calculus in

physics.

APPLICATIONS IN PHYSICS

In physics, there are numerous situations in which we have to calculate the integral of a given function.

For example,

(i) Velocity functionrv of a particle is the integral of its acceleration function

ra .

That is, v(t) adt= rr

(ii) Displacement functionrx of a particle is the integral function of its velocity function

rv .

That is x vdt= r r

APPLICATIONS OF THE

INTEGRAL IN PHYSICS

CHAPTER

6


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INTEGRAL CALCULUS378

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(iii) Impulse I is defined as the integral of force with respect to time.

That is I Fdt= r r

(iv) Work W is defined as the integral of force with respect to displacement.

W F.ds= r r

Mathematically, definite integral is a number unlike the indefinite integral which is a function. In physics,

definite integral is not merely a number but a physical concept or physical quantity with certain units.

(v) The integral of acceleration function between two time limits gives the change in velocity between the

time interval.

v21 = v2 v1 =

2

1

ad t

(vi) The integral of velocity function between two time limits gives the change in position (i.e. displacement)

between the given time interval

x21 = x2 x1 =

2

1

vdt

Integration can be used for calculating 'v' from 'a' & position from 'v'. a v x.(vii) The integral of force within a given time interval is equal to the change in momentum of the particle during

that time interval.

p21

= p2

p1

=

2

1

Fdt

(viii) The integral of force within the specific space is equal to the change in kinetic energy of the particle.

K21 = K2 K1 =

2

1

Fdx

Example 1 :

A particle moves with a constant acceleration a = 2 ms 2 . along a straight line. If it starts with an initial

velocity of 5 ms 1 ., then obtain an expression for its instantaneous velocity.

Sol. Using the relation between velocity and acceleration, we get

v = ad t = 2 dt = 2 t + cwhere c is the constant of integration. Its value can be obtained by using the initial condition.

That is at t = 0; v = 5 ms 1

Thus, 5 = 2 (0) + c c = 5 ms 1

Therefore, v = 2t + 5 is the required expression for instantaneous velocity.

Example 2 :

In the above example if the particle occupies a position x = 7 m at t = 1s, then obtain an expression for the

instantaneous displacement of the particle.

Sol. Using the relation between displacement and velocity, we get

x = vdt = (2t 5) dt+ or x = t2 + 5t + cwhere c is the constant of integration. Its value can be determined by using the given conditions. That is,

at t = 1 s ; x = 7m

7 = (1)2 + 5 (1) + c c = 1m Thus, x = t2 + 5t + 1


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APPLICATIONSOFTH EINTEGRALIN PHYSICS 379

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Example 3 :

If a particle is moving with velocity u and a resistive force acts on it which is proportional to v2, then find

the velocity of the particle after travelling distance x.

Sol. As given in the question, F v2 or F = K v2

or m.a = Kv2 or m.v.dv

dx= Kv2

orv

u

1

v . dv =

K

m

x

0

dx or [ ]ve ulog v = K

m[ ]

x

0x

or logev logeu =

K

m (x 0) or loge

v

u =

K

m (x)

or

Kx

mv

eu

= orKx

mv ue=

Example 4 :

If a force F is being applied on a block of mass m, which is proportional to displacement x, then find the

work done in displacing the block by distance d.

Sol. F x or F = kxNow small work done in displacing the body by a small distance dx is

dw = F.dx

So total work done in displacing the body by a

distance d is

dw =d

0

Kx.dx

w = K

d

0

xdx or w = Kd

2

0

x

2

= K

2d

02

or w = K

2d

2

Example 5 :

Find the work done in giving charge Q to a spherical shell of radius R.

Sol. Small work done in giving small charge dq to the spherical shell is given by

dw = dq. V; Now as V =KQ

R

dw =KQ

R

.dq

Total work down in giving charge Q is

Q Q

o 0

KQ Kdw .dq Q.dq

R R= = or w =

Q2 2

0

K Q KQ

R 2 2R

=


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Again ,dv 10

v.ds v 1

=+

(v2 + v) dv = 10 ds

Integrating,

3 2v v

10s C3 2

+ = +

Using v = 0 at s = 0 ; C ' = 0

3 2v v

10s3 2

+ =

Putting v = 15 ; s = 123.75m

Example 8 :

A particle moving along a straight line, starts from origin with an initial speed of u. It is subjected to

retardation which is given by kv where k is a constant and v is its velocity. Obtain the displacement-timerelation.

Sol. Given :dv

kvdt

= + (The minus stands for retardation)

ordv

kdtv

=

Integrating both sides, we get log v = kt + C1Using boundary conditions, v = u, at t = 0 C1 = log u

The equation reduces to ev

kt logu

=

Ktv eu

= [Q if an = b, then logab = n (remember)]

Ktv ue= Ktds ue dt= [Particle stop whenkt

uv 0 , t

e= = ]

Integrating once again we getkt

2

us e C

k

= +

(where C2 is a constant)

At t = 0, s = 0 2 2u u

0 (1) C Ck k

= + =

The equation of motion becomesktus (1 e )

k

=

[Particle stop when t u

s finite distance travelled in infinite time why!k

= = ]

Example 9 :

Figure given below shows the variation of rate of

change in acceleration with time, for a particle moving

along a straight line. (Assume the body starts from

rest) Find the following :

(i) Acceleration at instant (a) t = 2 sec and (b) t = 3 sec.

1 2 3

1(m/s)

dt

da

t (sec)

(ii) Velocity at instant (a) t = 1 sec (b) t = 5 sec.

(iii) Displacement at the instant t = 2 sec.


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Sol. (i) (a)Between t = 0 and t = 2 sec, it is evident thatda

1 da dtdt

= =

Integrating both sides, we have

a t

0 0

da dt a t= = [Q Initial acceleration is 0]

At t = 2 sec, acceleration = 2 m/sAlternative : Change in acceleration = Area below the graph,

a 0 = 2 1 = 2 m/s

(b) At t = 3 sec.

Change in acceleration a 0 = Area below the graph = 2 1 + 1 1 = 2.5 a = 2.5 m/s(ii) (a) Velocity at t = 1 sec.

Fromdv

adt

= , we have dv = a dt

1v 1 2

0 0 0

tdv dt v 0.5m/s

2

= = =

(b) Now, acceleration at time t between t = 2 sec. and t = 3 sec, can be obtained as follows :

From ANP and AMB (fig.) we have ANP and AMB are similar triangles Ratio of corresponding sides should be equal

1 2 3

1

dt

da

t (sec)

A

P

B

N

M

i.e., AN AM

NP MB=

da1

1 dadt3 t da (3 t)dt

t 2 1 dt

= = =

Integrating, we have

a t

2 2

da (3 t)dt= t

2

2

ta 2 3t2

=

213(t 2) (t 4)2

2 2

t t3t 4 a 3t 2

2 2 = +

Acceleration at t = 3 sec. a (3) = 2.5 m/s

Velocity at t = 5 sec.v 5

0 0

dv a dt = 2 3 5

0 2 3

a d t a dt adt + +

2 3 52

0 2 3

t 5t d t 3t 2 dt dt

2 2

+ + +

[ ]

2 32 3 2

5

30 2

t r 3t 52t t

2 6 2 2

+ + +

3

2

19 3 52 (5) 2(1) (2) 9.33m/s

6 2 2

+ + + (iii) Displacement at t = 2 sec, From, ds = v dt.

We have

2t

ds dt2

=


s 2 2

0 0

tds dt

2=

23

0

1 ts 1.33

2 3

= =


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2-D MOTIONExample 10 :

A particle moving on a X-Y plane, always has its acceleration directed along Y-axis, of constant magnitude

a, if its starts from origin with an initial velocity in a direction of 4/ with X-axis and magnitude 2/1times that of acceleration. Find the equation of path of the particle.

Sol. Given : ax = 0 and ay = a [constant]

1x Cv = and y 2v adt at C= = +Initially vx = C1 and vy = C2

Given,2 21 2

1C C a

2

+ =

and 1

2

Ctan 1

C 4

= =

221 1 2

a a2C C C

2 2

= = =

Again x 3adt a

x v dt t C2 2

= = = + and2

y 4

a t ay v dt at dt a t C

2 2 2

= = = +

But at t = 0, x = y = 0 C3 = C4 = 02a ax t and y [t t]

2 2 = =

Eliminating t from above questions

2a 2x 2x a 2x 2x a

y2 a a 2 a a

= =

xy (2x a)

a = or

22x

y xa

=

Example 11 :

Find the length of path described by a particle moving along the curve x = t,

3t

y t3

= , between t = 0 and

t 3= .

Sol. Given, x = t2 and

3t

y t3

= dx

2tdt

= and 2dy

1 tdt

=

From

2 2dx dy

ds dtdt dt

= + ( )2 2

ds (dx) | (dy)= +

We have2 4 2 2ds 4t 1 t 2t dt (1 t )dt= + + = +

The required length to path

33 32

0 0

ts (1 t )dt t 3 3 2 3units

3

= + = + = + =

LAWS OF MOTIONExample 12 :

A rocket moves in the absence of a gravitational field, the exhaust gases being escaping with a constant

velocity u relative to the rocket. Assuming the initial velocity to be v0, determine the velocity after a time

when the mass of the rocket decreases to half the initial value.

Sol. Let the rockets initial mass be m0Now, the final mass of the rocket = m0/2

In this case, external force F = 0, So, from Meshchersky equation,


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We havedv dm

m F udt dt

= +

dv dm

m udt dt

= dm

dv um

=

Integrating both sides of the expression, we get

0

0 0

m /2v

v m

dmdv u

m= m /200 m0v v u[lnm] =

0v v uln(1/2) = + or 0v v u ln (2)=

Note that v0 and u being oppositely directed, the terms gets added in the right hand side of the equationand v > v0.

If the initial velocity be zero, i.e., v0 = 0

Then, v = u ln (2)

Thus, v is directed opposite to the direction of u. Moreover, it is also evident, from the foregoing

discussion that the changes in a rockets velocity at any instant depends only upon the ratio of the initial

to the final mass (as the value of u remains constant).

Example 13 :

A chain PQ of length l0 is located in a smooth horizon-

tal tube, such that h length of the tube hangs freely

and touches the surface of a smooth table. The end P

is released at a certain instant. What is the velocity

with which this end will slip out of the tube ?h

P

QSol. Let be the mass of chain per unit length.Considering the portion of chain moving inside the tube,

let its length at any instant be l, and v be its velocity.

v

FIf F be the force on it due to the overhanging mass, then from

dvm F [ u 0]

dt= =Q

dvF

dt =l ......... (1)

For the overhanging, part, net force downward is gh F.

Applying the formuladv dm

m F udt dt

= +r r r

in the vertically downward direction.

dvh gh F [ u 0]

dt = =Q ......... (2)

Adding (1) and (2), we get,

dv(h ) gh

dt + = l (h ) v d v g h d + =l l


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Integrating over the proper limits and remembering that dl is negative, being the length l decreases with

time, we get

0

v v

0 h

dvdv gh

h

= +

l

l

l 0

v2

0h

0

vgh[ln(h )

2

= +

ll

20lnv gh

2 h

=

l

0v 2ghlnh

= l

WORK ENERGY & POWERExample 14 :

A force F = a + bx acts on a particle in the x-direction where a and b are constants. Find the work done by

this force during a displacement from x1 to x2.

Sol. Let the particle at any instant be at a position x. Let, under the action of force F = a + bx, it describe a small

displacement dx. (fig.)

Work done during the displacement dx will be

dW = F dx = (a + bx) dx

dxx

FOTotal work done can be obtained by summing up the work done in individual element displacements

(i.e., by integrating)

x2

x1

W dW (a bx)dx= = + [Q x varies from x1 to x2]

2

1

x2

x

bxW a

2

= +

= )xx(

2

b)xx(a

21

2212 + = )]xx(ba2[

2

)xx(21

12 ++

Example 15 :

A block of mass m moves on a rough horizontal plane with a velocity v at point O. The resistive frictional

forces varies as F = ar + br, where a and b are constants and r, the distance from point O. (fig.) Find the

total distance covered by the block before it comes to rest.

Sol. The work done by friction (against the block) during a

O rsmall displacement dr

F dr = (ar + br) dr

Total work done = initial K.E. of the block

i.e.,x

2 2

0

1(ar br )dr mv

2+ = [where x is the distance covered by the block before it comes to rest]

x2 3

2

0

ar br 1mv

2 3 2

+ =

2 32ax bx 1

mv2 3 2

+ =

or x2 (3a + 2bx) = 3mv 2


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CONSERVATION OF MOMENTUMExample 16 :

A cuboidal block of mass M and cross sectional area A is moving parallel to its length along X-axis at a

uniform speed of v0. At t = 0, the front face is at x = 0. The region corresponding to x > 0 is filled with fine

dust particles of uniform density at rest. As the front face of the cylinder collides with the dust particles,they stick to its surface. Assuming that the dust particles stick only to the front face of the cylinder,

determine the distance covered by the cylinder by the end of t = t0.

Sol. Evidently, the block moves under the absence of an external force, hence the linear momentum would

remain constant. Since mass of the cylinder continuously increases, so its velocity decreases.

The initial momentum of the system (Block + dust particles) at t = 0 is

1pr

= momentum of block + momentum of dust particles

=0 0

(Mv 0 ) i Mv i+ = ........... (1)

Consider the cylinders position after time t, let the X co-ordinate

of the front face be x.

Then instantaneous velocity of the cylinder isdx v idt

=r

.X-axis

x

region x0

v

The mass of dust particles adhered to the cylinders face, is (A x).Momentum of the system after time t, is given by,

2

dx p (M A x) v (M A x) idt

= + = + r r

............. (2)

From principle of conservation of linear momentum.

1 2p p=r r

0 0dx Mv i (M A x) i Mv dt (M A x )dxdt

= + = +

Integrating both sides, and remembering that at t = 0, x = 0 and at t = t0, x = l (say)

t0

0

0 0

Mv dt (M A x)dx= + l

[ ]2

t00 0

0

A xMv t Mx

2

= +

l

22

0 0 0 0

AMv t M A 2M 2Mv t 0

2

= + + =

ll l l

Solving for l we get

20 02M 4M 8v t A

2A

+ =

l

Ignoring the ve root2

0 0

1M 2Mv t A M

A

= + l

CENTRE OF MASS

Example 17 :

Find the centre of mass of the homogeneous triangular lamina bounded by y = 3x, y = 0, and x = 2.

Sol. The lamina is sketched in figure. We have drawn a vertical area strip, so we will use x as our variable. The

strips range between x = 0 and x = 2, so the limits of integration are 0 and 2. The endpoints and midpoint

of the strip are labeled in terms of the variable x. We have used the formula for the midpoint of the line

segment between (x, 3x) and (x, 0) to determine that


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x x 3x 0 3x(x,y) , x,

2 2 2

+ + = = % %

The variable area strip has

Length = 3x 0 = 3x,

Width = dx, Area = dA = 3x dx,

x

y

(x,3x/2)(x,3x)

(x,0)

Mass = dM = dA = 3x dx

It follows that,

2

0

M dM 3x dx 6= = = ,

2 22

x 0

0 0M xdM 3xdx 3x dx 8= = = = = %

and

2

y 0

0

3xM ydM 3xdx 12

2=

= = = %

so, x 0M 8 4

xM 6 3

= = = =

andy 0M 12

y 2M 6

= = = =

We see that the point4

(x,y) ,23

= is the intersection of the medians of the triangle as we knew it

should be.

ROTATIONAL DYNAMICS

Example 18 :

A flywheel of moment of inertia 8 kg-m starts rotating from rest about its fixed central axis, such that its

angular acceleration is proportional to its angular displacement. Find the work done on the flywheel by

the time its angular displacement is 100 revolutions, if the angular acceleration is 1 102 rad/s, when its

angular displacement is 1 revolution.

Sol. Given, angular acceleration angular displacement i.e., = k, where k is constant

when = 2 rad, 2 21 10 rad/s = 2

21 10K / s

2

=

Now work done W is given by W d= i.e., W I d Ik d= Substituting for I, k and integrating between limits

0 t o 2 (100) = = , we get

200200 2 4 2 42

0 0

8 1 10 4 10 2 10d [(200 ) 0] 8 J

2 2

= = =

Example 19 :

A uniform rod AB of length 4l and mass m is

pivoted at its centre O in such a way that it can

rotate freely in a vertical plane. (Fig.)

The rod is initially in a horizontal position when

A BO v

C

4

an insect of mass m/2 falls vertically with a speed v on the point C, and starts moving towards the end B,


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such that the rod rotates with a constant angular velocity. If the insect reaches the end B when the rod has

turned through an angle of 90, determine v.

Sol. First let us determine the constant angular velocity , with which the rod rotates.Conserving angular momentum about centre O, just before and after collision, we have

2 2m m(4 ) m 3v

v2 12 2 11

= + =

l ll

l.......... (1)

Let the collision between the rod and insect take place

at t = 0. Let after time t, the insect move through a

distance x along the rod towards B, and the rod rotate

through an angle t=

Angular momentum at time will be

2 2m(4 ) m( x) 3v

L I12 2 11

+= = +

l l

l

A

O

C

mg/2

Dx

Differentiating w.r.t. time t,

dL 3mv dx( x)

dt 11 dt= +l

l. But

dL

dt= torque

mg 3mv dx( x)cos t ( x)

2 11 dt + = +l l

lor

6vcos td t dx

11 g =

l

Integrating both sides, we have,

/ 2 1

0 0

6vcos td t dx

11 g

=

l[Q time of rotation

/ 2t

= =

]

[ ]/ 2

1

00

sin t 6vx

1 1 g

= l

6v 1 11g 11g[1 0] v

11g 16 6[3v/11 ] = =

l

2 11 11g 11v v g / 218 3

= =

ll

GRAVITATIONExample 20 :

A uniform rod of length l2 and mass density (linear) is placed along y-axis, such that the mid-point ofthe rod coincides with the origin. Find the gravitational field intensity at point P(a, 0). Hence, deduce the

result for l .Sol. The rod can be treated as a continuous mass distribution made up of infinite number of elemental lengths.

The field due to each elemental lengths at P can be determined and then their cumulative effect can be

obtained.

Consider an elemental length A, at a distance y from

the origin and of length dy. It mass dm = dy (fig.) Thefield intensity at P, due to this elemental length will be

2 2 2 2

Gdm G dydE (along PA)

(a y ) (a y )

= =

+ +

uuur

Resolving the field at P, into two components, we have

a

A

B

dy

y P


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dEcos component along the X-axis (ve)

dEsin component along the Y-axis.Consider a similar elemental length B, at a distance y below the origin. The field intensity at P, due to this

length will also be 2 2

G dydE (alongPB)

(a y )

=

+

uuur

Resolving this field, along and normal to the Y-axis, we find that, the component along Y-axis cancels, with

that due to element A. However, the components along ve X-axis add up.

Net field intensity at P due to the entire rod will be

2 2 2 2 1/2 2 2 3 / 2

0

G dy a dyE 2G a

(a y ) (a y ) (a y )

= =

+ + + l l

l

Now, let y = a tan So that dy = a sec2 d and at y = 0, = 0 and at 1y , tana

= = l

l

1 1tan ( / a ) tan ( /a )2

3 30 0

asec d G aE 2G a cos

aa sec

= =

l l

= [ ]1tan ( / a )

0 2 2

G a 2Gsin

a a a

=

+

l l

l

The above expression can be written as

1 / 22

1 /2 22

2

G 2G aE 1

aa

a 1

= = +

+

l

l

=

2 2

2 2

2G 1a a1 higher powers of .....

a 2

+

l l

Obviously as l2G

Ea

=

Example 21 :

The gravitational field due to a mass distribution is given by2

k E ix

=r

where k is a constant. Assuming

the potential to be zero at infinity, find the potential at a point x=a.

Sol. The change in potential2

kdV E.dr dx

x= =

r r[Q the variation of field is only along X-axis]

Integrating both sides, we get kV Cx

= + where C is a constant of integration

Given, V = 0 as x k k

C 0 V(x) V(a)x a

= = =

SHM

Example 22 :

A particle executes SHM whose displacement changes with time as tsinax = . Find the average speedbetween the instants when it passes its mean position and its displacement is a/2 from the first time.

Sol. Since the displacement equation is given as, x = a sin t

when x = 0, t = 0 and whena

x , t

2 6

= =


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Also, instantaneous velocity v = acos t Required value

/ 6

0/ 6

0

a cos( t)dt

v

dt

< > =

=

[ ]

/6

0

/ 6

0

sin ( t)a a sin sin(0)

6

t 06

=

=1 6 3a

u2

=

FLUID MECHANICSExample 23 :

To what height should a cylindrical vessel be filled with the liquid to make the force with which the liquid

presses on the side of the vessel equal to the force exerted by the liquid on the bottom of the vessel ?

Sol. Let r be the radius of the cylinder and h the height to which the liquid is filled in it. (fig.)

Clearly, if be the density of the liquid, then the thrust offered by the liquid at the bottom.

= pressure at the bottom area 2( gh)( r ) ......... (1)

Now, in order to find the lateral thrust, consider a thin elemental ring of thickness dx, at a depth x, below

the free surface, on the vessel. The liquid pressure at each point on the ring = gx

Lateral thrust offered by the liquid on this ring = pressure area = ( gx)(2 rdx)

The total lateral thrust can be obtained by summing up (i.e., integrating) the thrusts on each elementalrings, covering the sides of the cylinder.

Total thrust =hh 2

2

0 0

xg x 2 rdx 2 r g r gh

2

= =

........... (2)

Equating equations (1) and (2) we get h = r.

Example 24 :

A light tank, initially at rest is filled with water to a height 2.5m. A small hole is punched on the side of the

tank, close to the bottom. Find the velocity acquired by the tank, as a function of depth h of water left in

it, if the tank be placed on a smooth floor. Hence calculate the velocity acquired when the tank becomes

empty.

Sol. Let a and A be areas of the hole and that of tanks cross section respectively. Consider, the instant, when

the depth of water in the tank is h.

Mass of water = Ah = M (say) .......... (1)Velocity of efflux v 2gh= ........... (2)

Let dm be the small mass of water ejected through the hole in a further small time interval dt, then

dm = av dt ........... (3)If, as a result dh be the fall in water level, then

avdh dt

A= .......... (4)

Now, using Newtons II law, for variable mass

dmMd F v

dt= +

r r ror (Ah )a 0 ( av)v = + r r [Making use of (1) and (3)]


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dv dv dhAh ( av)v Ah . ( av)v

dt dh dt= =

r rr r

dv avAh ( av)v

dh A

=

rr

[From eq. (4)]

dv 2gh 2gh dv dh

dh h

= =

r


[ ]v h

hv

02.50 2.5

2gdv dh v 2g 2 h

h

= =

v 2 2h 2.5 h =

When the tank becomes completely empty h = 0

v 2 2 9.8 2.5 13 m /s= =

ELECTROSTATICExample 25 :

The electric field at a point (x, y, z) is given by 3 E (2xi 3j 4z k ) N / C= +r

. Find the difference in poten-

tials between two points P and Q, if P ( 1m,1m,2m) and Q (1m, 2m,0)

Sol. P QV V E.d = rr

l =

( 1,2,2)3

(1, 2 ,0)

[(2x)i 3 j (4z )k].[(dx)i (dy)j (dz)k]

+ + +

=

( 1,2,2)3

(1, 2 ,0)

(2xdx 3dy 4z dz)

+ = 2 1 1 21 2 0{[x ] 3[y] [z] } + = {(0) 3(3) 8} 1volt + =

CURRENT ELECTRICITY

Example 26 :

A potential difference applied to the ends of a wire made up of an alloy drives a current through it such

that rj += , where r is the distance of the point from the axis. If R be the radius of the wire, then findthe total current through any cross section of the wire.

Sol. Let c be the centre of the cross-section considered at any

section of the wire.If r be the radius of a circular strip concentric with the cross-

section considered and, dr be its thickness then (fig.)

current through the circular strip of thickness dr will beC

r

Rdr

di j.dS ( r)(2 rdr)cos0= = + r r

or 2di 2 ( r r ) dr= +

The total current i can be obtained by summing up the currents flowing through individual circular strips,

i.e.,

R2

0

i di 2 ( r r ) dr= = + orR R2 3 2

0 0

r r 2 Ri 2 (3 2 R)

2 3 6

= + = +

book 0sample integral cal

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