Physics Assignment Help Service

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About Physics Assignment: The purpose of physics is to

formulate laws that govern the functioning of physical world.

Physical laws are more often than not based on mathematical

language. Physics is regarded as the most fundamental subject

of natural sciences. It is divided into many sub areas like

electronics, quantum physics, astronomy, and biophysics. We

offer Physics Assignment Help to scholars of all related

fields of physics. The primary highlights of our service includes

prompt response on assignment quotes, high quality plagiarism free solution, on time

delivery and query handling at no additional cost. We don’t believe in hiring a lot of physics

assignment tutors available for you irrespective of stress on quality. Rather we believe on

hiring specialized tutors with good experience in teaching and solving assignments.

Sample of Physics Assignment Illustrations and Solutions:

Question-1: Show that the following relation for the time period of a body executing

S.H.M. is correct.

T = 2π 1

accelaration / displacement

Solution: T = 2π displacement

acceleration

Dimension of L.H.S. quantity = [M0L0T-1]

Dimensions of R.H.S. quantity

= (dimension of length)1/2/(dimension of acceleration)1/2

=[L]1/2/[LT-2]1/2 = 1/[T-1] or [M0L0T1]

Since the dimension of L.H.S. quantity is equal to the dimension of R.H.S. quantity, the

relation is dimensionally correct.

(c) To derive a relationship between different physical quantities. The concept of the

homogeneity of dimensions can also be used to find a possible relationship between

different physical quantities provided we have some preliminary idea of the dependence of

various quantities on each other. This will be clear from the following examples.

Potential difference required, V = IR = 15 × 112 = 180 volts.

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Question-2: If the frequency of a stretched string depends upon the length of the string.

the tension in the string and the mass per unit length of the string. Establish a relation for

the frequency using the concept of dimension.

Solution: Let us assume that frequency is directly proportional to Ix, Ty and d2. The possible

relation for frequency (n) will be

n ∝ lx Ty d2

The dimension of L.H.S. quantity

= [M0L0T-1]

The dimension of L.H.S. quantity

=[L]x × [MLT-2]Y × [M/L]z

∴ [M0L0T-1]

= [L]x × [My] × [Ly] × [T-2y] × [Mz] × [L-z]

= [My+z] × [L x+y-z] × [T-2y]

Equating the powers, we have

y + z = 0, y + x – z = 0, we have

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y + z = 0, y + x –z = 0, and -2u = -1

From these equations we get x = -1, y = ½

Therefore lx Ty dz = 1/1 T

d ,

Hence n ∝ 1/1 T

d = K 1/1

T

d

Question-3: On the basis of dimensional analysis obtain an expression for the

(i)distance travelled by a body in time t if its initial velocity be u and acceleration f.

(ii) the acceleration of a particle moving with a uniform speed u in a circle of radius r,

(iii) time period of a small drop of liquid vibrating under surface tension S, if it depends

upon thee density d, radius r and surface tension.

Solution: (i) Let the distance travelled be given by the following relation

S = K ua fb tc ∴ [S] = [ua fb tc]

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i.e., in the dimensional form

[L M0T0] = [LT-1]a [LT-2]b [T]c

or L M0T0 = La+bM0TI-a-2b+c

By equating the indices of basic quantities, we get a + b = 1 and a+2b = c

As these two equations are not enough to find the three unknown numbers a, b and c. We

make use of the fact that in case acceleration is zero, then

S = k1uabtc and in dimensional form

[L M0T0] = [LT-2]b [T]c or [LM0T0] = LbTc-a

Equating indices in the above relation we have a = 1 and c = 1 i.e., S = k1 ut.

Again if body starts from rest i.e. S = k2fbtc and in dimensional form

[L M0T0] = [LT-2]b [T]c or [LM0T0] = LbTc-2b

Equating indices in the above equation we have b = 1 and c = 2 i.e., S = K2ft2

Now if the body possesses both initial velocity and acceleration, the equation will have both

terms k1 ut and k2ft2, so that

S = k1 ut+k2ft2

It is observed experimentally that constant k1 = 1 and k2 = ½

∴ S = ut + ½ ft2.

(ii) As the acceleration of the particle moving in a circle depends upon the velocity of the

particle (v) and the radius of the circle (r) we can say acceleration = kvarb.

Dimensionally [LT-2] = [LT -1]a [L]b

or LT-2 = La+b T-a

Equating indices we have, a + b = 1 and a = 2.

Hence we have b = -1 and a = 2.

Therefore we have acceleration = kv2/r

(iii) As the time period of the drop depends upon the density of the liquid d, radius r and

surface tension S we can write for time period (T) = k da rb Sc writing these in dimensional

from we have

[M0L0T0] [L-3 M]a [L]b [MT-2]c

or M0 L0 T = L-3a+b Ma+c T-2c

Equating the indices we have

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-3a + b = 0, a +c = 0 and -2c = 1

Solving these equations we have

a = ½,b = 3/2 and c = -1/2

Thus T = kd1/2r3/2 S1/2

or T = k dr 3

S

Question-4: On the basis of dimensional analysis show that the following relationship is

correct V = E/d where V is the velocity of the sound, E is the elasticity and d is the density

of the medium.

Solution: We know dimension of velocity is M0 LT-1, of density [ML-3] and of elasticity

[ML-1T-2] being same as that of pressure) in the relation v = E, d, Dimensions of the

R.H.S.

= Dimension of E

Dimension of d 1/2 = [ML-1T-2]1/2 / [ML-3 T0]½

= [ML-1 T-2 L-1/2 + 3/2 T-1 + 0 = M0 LT-1

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= M1/2-1/2L-1/2 + 3/2 T-1+0 = M0Lt-1

=The dimensions of L.H.S. (that of velocity).

Thus the above relation is correct.

Question-5: Show on the basis of dimensional analysis that the following relations are

correct :

(i) n = P r4/δvl where P is the pressure difference between the two ends of the capillary

tube; r its radius, l its length, δv the rate of flow of liquid through it (volume per second)

and n is the coefficient of viscosity.

(ii) v2 – u2 = as, where u is initial velocity, v is final velocity, a is acceleration of the body

and S is the distance moved.

(iii) p = 3g/4r G where p is the density of earth, G is the gravitational constant, r is the

radius of the earth and g is acceleration due to gravity.

Solution: (i) From relation n = Pr4/δvl

We have for the dimension of R.H.S.

= [dimension of P] [dimension of r]4 / [dimension of b v/t] [dimension of l]

= [ML-1 T-2] [L]4 / [L3 T-1] [L] = ML3 T-2 / L4 T-1 = ML-1 T-1

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Now the dimension of L.H.S. n

= [Force ]

Area × dv

dy

= MLT −2

L2LT −1

L = ML-1T-1

As the dimensions of L.H.S. are equal to the dimensions of R.H.S. the above relation is

correct.

(ii) From relation v2 – u2 = aS

We have for the dimension of L.H.S.

= [dimension of v]2 – [dimension of u]2 = L2T-2

The dimensions of R.H.S.

= [dimension of a] [dimensions of S]

= [LT-2] [L] = T2T-2

As the dimension of L.H.S. = dimension of R.H.S. the above relation is correct.

(iii)From relation ρ = 3g/4r G

We have for the dimension of R.H.S. quantity.

= [dimension of g]/ [dimension of r] [dimension of G]

= LT-2/[L] [M-1L3T ] = ML-3

∴ Dimension of density = Dimension of R.H.S.

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