Units and Measurements Grade 11 Solutions
Short Discussions Questions
Long Discussion Questions
| Basic units | Derived units |
| a) They are the units of basic or fundamental physical quantities. | a) They are the units of derived physical quantities. |
| b) They have their own identity. | b) They do not have their own identity. |
| c) They can be expressed individually. | c) They depend upon the basic units for their existence. |
| d) There are only seven basic units in SI system of units. | d) There are hundreds of basic units in SI system of units. |
| e) Example: metre, kilogram, second, etc. | e) Example: pascal, newton, joule, etc. |
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It can be used to check the correctness of a physical relation.
Example: $v = u + at$
Writing the given equation in dimensional form, we get,
$[LT^{-1}] = [LT^{-1}] + [LT^{-2}]*[T]$
$[LT^{-1}] = [LT^{-1}] + [LT^{-1}]$
Since, it follows principle of homogeneity of dimensions, the given physical relation is correct dimensionally. -
It can be used to convert a physical quantity from one system of units to
another.
Example: See numerical number ..... -
It can be used to obtain relation between different physical quantities.
Having given certain physical quantities, we can express them in such a
way that the it satisfies the principle of homogeneity of dimensions and
also gives us a relation between the considered quantities.
Example: See numerical number ...
$[P] = \dfrac{MLT^{-2}}{L^2} = [ML^{-1}T^{-2}]$
(i) Momentum and Impulse
Solution:
We know,
Momentum
= mass * velocity
$[p] = M * LT^{-1} = [MLT^{-1}]$
Also,
Impulse
= Force * time
$[J] = MLT^{-2} * T = [MLT^{-1}]$
Hence,
It
is shown that the given two physical quantities have identical dimension
i.e. $[MLT^{-1}]$
(ii) Torque and Energy
Solution:
We know,
Torque
= Force * effort arm
$[T] = MLT^{-2} * L = [ML^2T^{-2}]$
Also,
Energy
= Work = Force * displacement
$[E] = MLT^{-2} * L = [ML^2T^{-2}]$
Hence,
It
is shown that the given two physical quantities have identical dimension
i.e. $[ML^2T^{-2}]$
(iii) Energy and Work
Solution:
We know,
Energy
= Joule = Nm
$[E] = MLT^{-2} * L = [ML^2T^{-2}]$
Also,
Work
= Force * displacement
$[E] = MLT^{-2} * L = [ML^2T^{-2}]$
Hence,
It is shown that the given two physical quantities have
identical dimension i.e. [ML^2T^{-2}]$
Problems
1. Find the dimensional formula of the following physical quantities.
(i) Young's modulus
Solution:
We know,
Young's modulus = Stress/Strain
So,
$[Y]
= \dfrac{Force}{Area * Strain} = \dfrac{MLT^{-2}}{L^2} = [ML^{-1}T^{-2}]$
Remember: Strain is a dimensionless physical quantity.
(ii) Coefficient of viscosity
Solution:
We know,
$F = 6 \pi \eta r v$
$or, \eta =
\dfrac{F}{6 \pi r v}$
So,
$[\eta] = \dfrac{MLT^{-2}}{L *
LT^{-1}} = [ML^{-1}T^{-1}]$
(iii) Specific heat capacity
Solution:
We know,
$Q = msdt$
$or, s = \dfrac{Q}{mdt}$
So,
$[s]
= \dfrac{ML^2T^{-2}}{M * K} = [L^2T^{-2}K^{-1}]$
2. In Van der Waal's equation of state (P +a/V2)(V-b) = RT, where P is the pressure, V is the volume, and T is the temperature of a gas and R is a gas constant. What are the dimensions of the constants a and b?
Solution:
Given,
$(P + \frac{a}{V^2} ) ( V -b) = RT$
According to
principle of homogeneity of dimensions,
$[P] = [\frac{a}{V^2}]$ ---
(i)
Also,
$[V] = [b]$ --- (ii)
Solving equation (i), we get,
$a = P * V^2$
So,
$[a] =
\dfrac{Force* Volume^2}{Area} = \dfrac{MLT^{-2}* L^6}{L^2} = [ML^5T^{-2}]$
Solving equation (ii), we get,
$b = V$
$[b] = [L^3]$
Hence, the required dimensions of a and b are $ [ML^5T^{-2}]$ and $[L^3]$, respectively.
3. Check the correctness of the relation:
$\rho = \dfrac{3g}{4 \pi r G}$ where the symbols have their usual meanings.
Solution:
Given,
$\rho = \dfrac{3g}{4 \pi r G}$
Dimensions
of the various physical quantities are:
$[ \rho ] = [ML^{-3}]$
$[g]
= [LT^{-2}]$
$[r] = [L]$
$[G] = [M^{-1}L^3T^{-2}]$
Taking
RHS
$[ \dfrac{3g}{4 \pi r G}] = \dfrac{LT^{-2}}{L * M^{-1}L^3T^{-2}}
= [ML^{-3}]$
Taking LHS
$[ \rho ] = [ML^{-3}]$
Since,
the dimensions of both LHS and RHS are equal, the given relation is
dimensionally correct.
4. Express 1 kg m2 s-2 in to g cm2 s-2.
Solution:
To find: $1 kg m^2s^{-2} = ? gcm^2s^{-2}$
Given,
Quantity
= Joule = $[ML^2T^{-2}]$
So, a=1, b=2, c=-2
We have,
In
SI units,
$n_1 = 1$
$M_1 = 1kg = 1000g$
$L_1 = 1m =
100cm$
$T_1 = 1 s$
In CGS units,
$n_2 = ?$
$M_2
= 1g$
$L_2 = 1cm$
$T_2 = 1s$
Using formula,
$n_2
= n1 \left [ \dfrac{M_1}{M_2} \right ] ^a \left [\dfrac{L_1}{L_2}
\right]^b \left[ \dfrac{T_1}{T_2} \right]^c$
$= 1 * \left[\dfrac{1000g}{1g} \right]^1 * \left[ \dfrac{100cm}{1cm} \right ] ^2 * \left[ \dfrac{1s}{1s} \right ]^{-2}$
$= 10^3 * 10^4 * 1$
$= 10^7$
$\therefore n_2 = 10^7$
Hence, $1 kg m^2s^{-2} = 10^7 gcm^2s^{-2}$
5. If the value of G, the gravitational constant in CGS system is 6.67*10-8, calculate its value in MKS-system using dimensional analysis.
Solution:
[G] = [M-1L3T-2]
So, a = -1, b = 3,
c = -2
In CGS system,
$n_1 = 6.67*10^{-8}$
$M_1 = 1 g$
$L_1 =
1cm$
$T_1 = 1s$
In MKS system,
$n_2 = ?$
$M_2 = 1kg =1000g$
$L_2 = 1m =
100cm$
$T_2 = 1s$
Using formula,
$n_2 = n1 \left [ \dfrac{M_1}{M_2} \right ] ^a \left
[\dfrac{L_1}{L_2} \right]^b \left[ \dfrac{T_1}{T_2} \right]^c$
$= 6.67*10^{-11} * \left[ \dfrac{1g}{1000g} \right ] ^{-1} * \left [ \dfrac{ 1cm}{100cm} \right]^3 * \left [ \dfrac{1s}{1s} \right ]^{-2}$
$= 6.67*10^{-8} * 10^3 * 10^{-6} * 1$
$= 6.67*10 * 10^{-11}$
$\therefore n_2 = 6.67*10^{-11}$
Hence, the value of G in MKS system is $6.67 * 10^{-11}$.
6. If force, length and time are taken as base quantities find the dimension of work, momentum and density.
Solution:
Let force, length and time be represented as [F], [L], and [T], respectively.
We know,
Work = Force * displacement = Force * Length
$[W] =
[FL]$
And,
Momentum = Mass * Velocity = Mass * Velocity / Time * Time =
Mass * Acceleration * Time = Force * Time
$[M] = [FT]$
Also,
Density = Mass / Volume = Force * Time * Time /
Volume-Length
$[D] = [FL^{-4}T^2]$
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