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Physics Homework Help
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Sample of Physics Homework Illustrations and Answers:
Illustration:1 Deduce the dimensional formula of : (a) Velocity, (b) acceleration, and
(c) force.
Solution :
(a) Dimension of velocity
= Dimension of length/ Dimension of time = [L]/[T] = [LT-1]
Hence the dimensional formulae for velocity will be [M0 LT-1].
(b) Dimension of acceleration
= Dimension of velocity/Dimension of time = [𝐿𝑇−1]/[T] = [LT-2]
= [M0 LT-2]
(c) [Dimension of force] = [Dimension of mass] × [Dimension of acceleration]
= [M] × [LT-2] = [MLT-2]
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Illustration:2 Deduce the dimension of (a) Pressure, (b) Work and (c) Power.
Solution :
(a) [Dimension of pressure]
= [Dimension of force]/[Dimension of area ] = [MLT-2] / [L2] = [ML-1T-2]
(b) Dimension of work.
= [Dimension of force] × [Dimension of distance]
= [MLT-2] × [L] = [ML2T-2]
(c) Dimension of power
= [Dimension of work] / [Dimension of time] = [ML2T-2]/[T]
= [ML2T-3].
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Illustration:3 Deduce the dimensional formula for (a) modulus of elasticity, and (b)
coefficient of viscosity.
Solution :
(a) Modulus of elasticity.
Deduce the dimensional formula for (a) modulus of elasticity, and (b) coefficient of
viscosity.
Solution : (a) Modulus of elasticity.
Y = Stress/Strain = Force/Area/Change in length/Original length
∴ Dimension of Y
= (Dimension of force) × (Dimension of length)/ (Dimension of area) × (Dimension of
length)
= [MLT-2] × [L]/ [L2] × [L] = [ML-1T-2]
(b) the coefficient of viscosity (n) of a liquid is defined as tangential force required per
unit area to maintain unit velocity gradient between the two layers of the liquid unit
distance apart or
n = F/A 1/(dv/dx)
= Force/ Area × (Velocity change/Distance)
Dimension of n =
(Dimension of force) × (Dimension of distance)/ (Dimension of area) × (Dimension of
velocity)
= [MLT-2] × [L]/[L2] × [LT-1] = [ML2T-2]/ [L3T-1].
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Illustration: 4 Convert one Newton into Poundal (unit of force in F.P.S. system).
Solution :
The dimensional formula of force is MLT-2. One Poundal = 1 Ib-ft/sec2 and one Newton =
1 Kg- m/sec2 1 Ib = 0.4536 kg and I ft = 0.3048 metre
∴ M units in F.P.S. system
= 0.4536 M units in M.K.S. system.
L units in F.P.S. system.
= T units in M.K.S. system.
∴ I Newton = 1 Poundal (1/0.4536) × (1/0.3048) × (1/1)-2
or 1 N = 7.233 Poundal.
Illustration: 5 If the value of G in C.G.S. system is 6.67 × 10-8 dynes cm2/g2, what
will be its value in M.K.S. system ?
Solution : The dimensional formula for G is [M-1 L3 T-2] metre = 100 cm, 1 kg = 1000 g
∴ M unit of M.K.S. system.
= 1000 M unit of C.G.S. system.
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L unit of M.K.S. system
= 100 L units of C.G.S. system.
T units of M.K.S. system,
= 100 L units of C.G.S. system
= T units of C.G.S. system.
If x is the value of G in M.K.S. system then we have
X = 6.67 × (1/1000)-1 × (1/100)3 × (1/1)-2
= 6.67 × 10-8 × 103 ×10-6 × 1
= 6.67 × 10-11 Nm2/kg2.
(b) To check the correctness of a derived relationship between physical
quantities. By making use of the fact that the dimension of the quantities on the left
hand side must be same as that of the dimensions of the quantities of the right hand
side of the dimensional equation (this principle is known as the dimensional equation
(this principle is known as the principal of homogeneity of dimension), we can check the
correctness of any relation connecting various physical quantities. Let us consider the
following examples :
