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Find the dimensions of
(a) linear momentum,
(b) frequency and
(c) pressure.
a) Linear momentum \(\quad: \mathrm{mv} \quad=\left[M L T^{-1}\right]\)
b) Frequency : \(\frac{1}{T}=\left[M^{0} L^{0} T^{-1}\right]\)
c) Pressure: \(\frac{\text { Force }}{\text { Area }}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^{2}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)
Find the dimensions of
(a) angular speed \(\omega\),
(b) angular acceleration \(\alpha\),
(c) torque \(\Gamma\) and
(d) moment of interia \(I\).
Some of the equations involving these quantities are
\(\omega=\frac{\theta_{2}-\theta_{1}}{t_{2}-t_{1}}, \quad \alpha=\frac{\omega_{2}-\omega_{1}}{t_{2}-t_{1}}, \quad \Gamma=F . r\) and \(I=m r^{2} .\)
The symbols have standard meanings.
a) Angular speed \(\omega=\theta / t=\left[M^{0} L^{0} T^{-1}\right]\)
b) Angular acceleration \(\alpha=\frac{\omega}{t}=\frac{M^{0} L^{0} T^{-2}}{T}=\left[M^{0} L^{0} T^{-2}\right]\)
c) Torque \(\tau=F r=\left[\mathrm{MLT}^{-2}\right][L]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\)
d) Moment of inertia \(=\mathrm{Mr} r^{2}=[\mathrm{M}]\left[\mathrm{L}^{2}\right]=\left[\mathrm{ML}^{2} \mathrm{~T}^{0}\right]\)
Find the dimensions of
(a) electric field \(E\),
(b) magnetic field \(B\) and
(c) magnetic permeability \(\mu_{0}\).
The relevant equations are
\(F=q E, F=q v B\), and \(B=\frac{\mu_{0} I}{2 \pi a}\);
where F is force, q is charge, v is speed, I is current, and a is distance.
a) Electric field \(E=F / q=\frac{M L T^{-2}}{[I T]}=\left[{M L T}^{-3} \mathrm{I}^{-1}\right]\)
b) Magnetic field \(B=\frac{F}{q v}=\frac{M L T^{-2}}{[I T]\left[L T^{-1}\right]}=\left[{M T}^{-2}\mathrm{I}^{-1}\right]\)
c) Magnetic permeability \(\mu_{0}=\frac{\mathrm{B} \times 2 \pi \mathrm{a}}{\mathrm{I}}=\frac{\left.\mathrm{MT}^{-2} \mathrm{I}^{-1}\right] \times[\mathrm{L}]}{[\mathrm{l}]}=\left[\mathrm{MLT}^{-2} \mathrm{I}^{-2}\right]\)
Find the dimensions of
(a) electric dipole moment \(p\) and
(b) magnetic dipole moment \(M\).
The defining equations are \(p=q . d\) and \(M=I A\); where \(d\) is distance, \(A\) is area, \(q\) is charge and \(I\) is current.
Find the dimensions of Planck’s constant \(h\) from the equation \(E=h v\) where \(E\) is the energy and \(v\) is the frequency.
\(\mathrm{E}=\mathrm{h} v\) where \(\mathrm{E}=\) energy and \(v=\) frequency.
\(\mathrm{h}=\frac{\mathrm{E}}{v}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{T}^{-1}\right]}\) \(=\left[\mathrm{M} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right]\)
Find the dimensions of
(a) the specific heat capacity \(c\),
(b) the coefficient of linear expansion \(\alpha\) and
(c) the gas constant \(R\).
Some of the equations involving these quantities are \(Q=m c\left(T_{2}-T_{1}\right), l_{t}=l_{0}\left[1+\alpha\left(T_{2}-T_{1}\right)\right]\) and \(P V=n R T\)
a) Specific heat capacity \(=C=\frac{Q}{m \Delta T}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{M}][\mathrm{K}]}=\left[L^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]\)
b) Coefficient of linear expansion \(=\alpha=\frac{\mathrm{L}_{1}-\mathrm{L}_{2}}{\mathrm{~L}_{0} \Delta \mathrm{T}}=\frac{[\mathrm{L}]}{[\mathrm{L}][\mathrm{R}]}=\left[\mathrm{K}^{-1}\right]\)
c) Gas constant \(=\mathrm{R}=\frac{\mathrm{PV}}{\mathrm{nT}}=\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^{3}\right]}{[(\mathrm{mol})][\mathrm{K}]}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1}(\mathrm{~mol})^{-1}\right]\)
Taking force, length, and time to be the fundamental quantities find the dimensions of
(a) density,
(b) pressure,
(c) momentum and
(d) energy.
Taking force, length, and time as the fundamental quantity
a) Density \(=\frac{m}{V}=\frac{(\text { forcelacceleration })}{\text { Volume }}=\frac{\left[F / L T^{-2}\right]}{\left[L^{2}\right]}=\frac{F}{L^{4} T^{-2}}=\left[FL^{-4} T^{2}\right]\)
b) Pressure \(=F / A=F / L^{2}=\left[F L^{-2}\right]\)
c) Momentum \(=m v(Force / acceleration) \times Velocity =\left[\mathrm{F} / \mathrm{LT}^{-2}\right] \times\left[\mathrm{LT}^{-1}\right]=[\mathrm{FT}]\)
d) Energy \(=\frac{1}{2} m v^{2}=\frac{\text { Force }}{\text { acceleration }} \times(\text { velocity })^{2}\)
\(
=\left[\frac{F}{L T^{-2}}\right] \times\left[L T^{-1}\right]^{2}=\left[\frac{F}{\left.L T^{-2}\right]}\right] \times\left[L^{2} T^{-2}\right]=[F L]
\)
Suppose the acceleration due to gravity at a place is \(10 \mathrm{~m} / \mathrm{s}^{2}\). Find its value in \(\mathrm{cm} /(\text { minute })^{2}\).
The average speed of a snail is \(0.020 \mathrm{miles} /\) hour and that of a leopard is \(70 \mathrm{miles} /\) hour. Convert these speeds in SI units.
The average speed of a snail is \(0.02 \mathrm{mile} / \mathrm{hr}\)
Converting to S.I. units, \(\frac{0.02 \times 1.6 \times 1000}{3600} \mathrm{~m} / \mathrm{sec}[1 mile =1.6 \mathrm{~km}=1600 \mathrm{~m}]=0.0089 \mathrm{~ms}^{-1}\)
The average speed of leopard \(=70 \mathrm{miles} / \mathrm{hr}\)
In SI units \(=70\) miles \(/\) hour \(=\frac{70 \times 1.6 \times 1000}{3600}=31 \mathrm{~m} / \mathrm{s}\)
The height of mercury column in a barometer in a Calcutta laboratory was recorded to be \(75 \mathrm{~cm}\). Calculate this pressure in SI and CGS units using the following data : Specific gravity of mercury \(=13 \cdot 6\), Density of water \(=10^{3} \mathrm{~kg} / \mathrm{m}^{3}, g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) at Calcutta. Pressure \(=h \rho g\) in usual symbols.
Height \(\mathrm{h}=75 \mathrm{~cm}\), Density of mercury \(=13600 \mathrm{~kg} / \mathrm{m}^{3}, \mathrm{~g}=9.8 \mathrm{~ms}^{-2}\) then
Pressure \(=h \rho g =10 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}\) (approximately)
In C.G.S. Units, \(P=10 \times 10^{5} dyne / \mathrm{cm}^{2}\)
Express the power of a 100 watt bulb in CGS unit.
In S.I. unit 100 watt \(=100\) Joule/sec and In C.G.S. Unit \(=10^{9} \mathrm{erg} / \mathrm{sec}\)
The normal duration of I.Sc. Physics practical period in Indian colleges is 100 minutes. Express this period in microcenturies. 1 microcentury \(=10^{-6} \times 100\) years. How many microcenturies did you sleep yesterday?
1 micro century \(=10^{4} \times 100\) years \(=10^{-4} \times 365 \times 24 \times 60 \mathrm{~min}\); So, \(100 \min =10^{5} / 52560=1.9\) microcentury
The surface tension of water is 72 dyne/cm. In SI unit its value is
Surface tension of water \(=72\) dyne/cm; In S.I. Unit, 72 dyne \(/ \mathrm{cm}=0.072 \mathrm{~N} / \mathrm{m}\)
The kinetic energy \(K\) of a rotating body depends on its moment of inertia \(I\) and its angular speed \(\omega\). Assuming the relation to be \(K=k I^{a} \omega^{b}\) where \(k\) is a dimensionless constant, find \(a\) and \(b\). Moment of inertia of a sphere about its diameter is \(\frac{2}{5} M r^{2}\).
\(\mathrm{K}=\mathrm{kl}^{\mathrm{a}} \omega^{\mathrm{b}}\) where \(\mathrm{k}=\) Kinetic energy of rotating body and \(\mathrm{k}=\) dimensionless constant Dimensions of left side are,
\(
\mathrm{K}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]
\)
Dimensions of right side are,
\(
\mathrm{I}^{\mathrm{a}}=\left[\mathrm{ML}^{2}\right]^{\mathrm{a}}, \omega^{\mathrm{b}}=\left[\mathrm{T}^{-1}\right]^{\mathrm{b}}
\)
According to the principle of homogeneity of dimension,
\(
\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]=\left[\mathrm{ML}^{2}\right]^{\mathrm{a}} \left[\mathrm{T}^{-1}\right]^{\mathrm{b}}
\)
Equating the dimension of both sides,
\(
2=2 a \text { and }-2=-b \Rightarrow a=1 \text { and } b=2
\)
The theory of relativity reveals that mass can be converted into energy. The energy \(E\) so obtained is proportional to certain powers of mass \(m\) and the speed \(c\) of light. Guess a relation among the quantities using the method of dimensions.
Let energy \(E \propto M^{a} C^{b}\) where M=Mass, C= speed of light \(\Rightarrow E=K M^{a} C^{b}(K= proportionality constant )\)
Dimension of the left side
\(E=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\)
Dimension of right side
\(
\begin{aligned}
&M^{a}=[M]^{a},[C]^{b}=\left[L T^{-1}\right]^{b} \\
&\therefore\left[M L^{2} T^{-2}\right]=[M]^{a}\left[L T^{-1}\right]^{b} \\
&\Rightarrow a=1 ; b=2
\end{aligned}
\)
So, the relation is \(\mathrm{E}=\mathrm{KMC}^{2}\)
Let I= current through a conductor, R= its resistance and V= potential difference across its ends. According to Ohm’s law, product of two of these quantities equals the third. Obtain Ohm’s law from dimensional analysis. Dimensional formulae for R and V are \(\mathrm{ML}^{2} \mathrm{I}^{-2} \mathrm{~T}^{-3}\) and \(\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{I}^{-1}\) respectively.
Dimensional formulae of \(\mathrm{R}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{I}^{-2}\right]\)
Dimensional formulae of \(\mathrm{V}=\left[\mathrm{ML}^{2} \mathrm{~T}^{3} \mathrm{I}^{-1}\right]\)
Dimensional formulae of \(I=[I]\)
\(
\begin{aligned}
&\therefore\left[\mathrm{ML}^{2} \mathrm{~T}^{3} \mathrm{I}^{-1}\right]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{I}^{-2}\right][\mathrm{I}] \\
&\Rightarrow \mathrm{V}=\mathrm{IR}
\end{aligned}
\)
The frequency of vibration of a string depends on the length \(L\) between the nodes, the tension \(F\) in the string, and its mass per unit length \(m\). Find the expression for its frequency from dimensional analysis.
Frequency \(f=K L^{a} F^{b} M^{c}, M=\) Mass/unit length, \(L=\) length, \(F=\) tension (force)
Dimension of \(f=\left[\mathrm{T}^{-1}\right]\)
Dimension of right side,
\(
\begin{aligned}
&L^{a}=\left[L^{a}\right], F^{b}=\left[M L T^{-2}\right]^{b}, M^{c}=\left[M L^{-1}\right]^{c} \\
&\therefore\left[T^{-1}\right]=K[L]^{a}\left[M L T^{-2}\right]^{b}\left[M L^{-1}\right]^{c} \\
&M^{0} L^{0} T^{-1}=K M^{b+c} L^{a+b-c} T^{-2 b}
\end{aligned}
\)
Equating the dimensions of both sides,
\(
\begin{array}{ll}
\therefore b+c=0 & \ldots(1) \\
-c+a+b=0 & \ldots(2) \\
-2 b=-1 & \ldots(3)
\end{array}
\)
Solving the equations we get,
\(
a=-1, b=1 / 2 \text { and } c=-1 / 2
\)
\(\therefore\) So, frequency \(f=K^{-1} F^{1 / 2} M^{-1 / 2}=\frac{K}{L} F^{1 / 2} M^{-1 / 2}=\frac{K}{L}\times\sqrt{\frac{F}{M}}\)
Which of the following equations are dimensionally correct:
(a) \(h=\frac{2 S \cos \theta}{\rho r g}\),
(b) \(v=\sqrt{\frac{P}{\rho}}\),
(c) \(V=\frac{\pi P r^{4} t}{8 \eta l}\),
(d) \(v=\frac{1}{2 \pi} \sqrt{\frac{m g l}{I}}\);
where \(h=\) height, \(S=\) surface tension, \(\rho=\) density, \(P=\) pressure, \(V=\) volume, \(\eta=\) coefficient of viscosity, \(v=\) frequency and \(I=\) moment of inertia.
a) \(h=\frac{2 s \cos \theta}{\rho r g}\)
\(\mathrm{LHS}=[\mathrm{L}]\)
\(\text { Surface tension }=S=F / I =\frac{M L T^{-2}}{L}=\left[M T^{-2}\right]\)
Density \(=\rho=M/V=\left[\mathrm{ML}^{-3} \mathrm{~T}^{0}\right]\)
Radius \(=r=[\mathrm{L}], \mathrm{g}=\left[\mathrm{LT}{ }^{-2}\right]\)
\(\mathrm{RHS}=\frac{2 \mathrm{~S} \cos \theta}{\rho r g}=\frac{\left[\mathrm{MT}^{-2}\right]}{\left[\mathrm{ML}^{-3} \mathrm{~T}^{0}\right][\mathrm{L}]\left[\mathrm{LT}^{-2}\right]}=\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}\right]=[\mathrm{L}]\)
\(\mathrm{LHS}=\mathrm{RHS}\)
So, the relation is correct
b) \(v=\sqrt{\frac{p}{\rho}}\) where \(v=\) velocity
LHS = Dimension of \(\mathrm{V}=\left[L \mathrm{~T}^{-1}\right]\)
Dimension of \(p=F / A=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)
Dimension of \(\rho=\mathrm{m} / \mathrm{V}=\left[\mathrm{ML}^{-3}\right]\)
\(\mathrm{RHS}=\sqrt{\frac{\mathrm{p}}{\rho}}=\sqrt{\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]}{\left[M L^{-3}\right]}}=\left[\mathrm{L}^{2} \mathrm{~T}^{-2}\right]^{1 / 2}=\left[L \mathrm{~T}^{-1}\right]\)
So, the relation is correct.
c) \(V=\left(\pi p r^{4} t\right) /(8 \eta l)\)
LHS = Dimension of \(\mathrm{V}=\left[\mathrm{L}^{3}\right]\)
Dimension of \(p=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right], r^{4}=\left[L^{4}\right], t=[T]\)
Coefficient of viscosity \(=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]\)
RHS \(=\frac{\pi p r^{4} t}{8 \eta \mid}=\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^{4}\right][\mathrm{T}]}{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right][L]} = \left[\mathrm{L}^{3}\right]\)
So, the relation is correct.
d) \(v=\frac{1}{2 \pi} \sqrt{(\mathrm{mgl} / \mathrm{I})}\)
LHS = dimension of \(v=\left[T^{-1}\right]\)
\(R H S=\sqrt{(m g \mid / I)}=\sqrt{\frac{[M]\left[L T^{-2}\right][L]}{\left[M L^{2}\right]}}=\left[T^{-1}\right]\)
\(\mathrm{LHS}=\mathrm{RHS}\)
So, the relation is correct.
Let \(x\) and \(a\) stand for distance. Is \(\int \frac{d x}{\sqrt{a^{2}-x^{2}}}\) \(=\frac{1}{a} \sin ^{-1} \frac{a}{x}\) dimensionally correct?
Dimension of the left side \(=\int \frac{d x}{\sqrt{\left(a^{2}-x^{2}\right)}}=\int \frac{L}{\sqrt{\left(L^{2}-L^{2}\right)}}=\left[L^{0}\right]\)
Dimension of the right side \(=\frac{1}{a} \sin ^{-1}\left(\frac{a}{x}\right)=\left[L^{-1}\right]\)
So, the dimension of \(\int \frac{d x}{\sqrt{\left(a^{2}-x^{2}\right)}} \neq \frac{1}{a} \sin ^{-1}\left(\frac{a}{x}\right)\)
So, the equation is dimensionally incorrect.
A calorie is a unit of heat (energy in transit) and it equals about \(4.2 \mathrm{~J}\) where \(1 \mathrm{~J}=\) \(1 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\). Suppose we employ a system of units in which the unit of mass equals \(\alpha\) \(\mathrm{kg}\), the unit of length equals \(\beta \mathrm{m}\), the unit of time is \(\gamma \mathrm{s}\). Calculate that a calorie has a magnitude ______ in terms of the new units.
Dimensional formula of energy is \(\left[M^{1} L^{2} T^{-2}\right]\)
Comparing with \(\left[M^{a} L^{b} T^{c}\right]\), we get
\(
a=1, \quad b=2, \quad c=-2
\)
Now,
\(
\begin{aligned}
n_{2} &=n_{1}\left[\frac{M_{1}}{M_{2}}\right]^{a}\left[\frac{L_{1}}{L_{2}}\right]^{b}\left[\frac{T_{1}}{T_{2}}\right]^{c} \\
&=4.2\left[\frac{1 \mathrm{~kg}}{\alpha \mathrm{kg}}\right]^{1}\left[\frac{1 \mathrm{~m}}{\beta m}\right]^{2}\left[\frac{1 \mathrm{~s}}{\gamma \mathrm{s}}\right]^{-2}
\end{aligned}
\)
or, \(\quad n_{2}=4.2 \alpha^{-1} \beta^{-2} \gamma^{2}\)
What is the number of significant figures in the following?
(a) \(0.007 \mathrm{~m}^{2}\)
(b) \(2.64 \times 10^{24} \mathrm{~kg}\)
(c) \(0.2370 \mathrm{~g} \mathrm{~cm}^{-3}\)
(d) \(6.320 \mathrm{~J}\)
(e) \(6.032 \mathrm{~N} \mathrm{~m}^{-2}\)
(f) \(0.0006032 \mathrm{~m}^{2}\)
The mass of a box measured by a grocer’s balance is \(2.30 \mathrm{~kg}\). Two gold pieces of masses \(20.15 \mathrm{~g}\) and \(20.17 \mathrm{~g}\) are added to the box. (a) What is the total mass of the box? (b) What is the difference in the masses of the pieces to correct significant figures?
(a) Total mass of the box \(=(2.3+0.0217+0.0215) \mathrm{kg}=2.3442 \mathrm{~kg}\)
Since the least number of decimal places is 1 , therefore, the total mass of the box \(=2.3 \mathrm{~kg}\).
(b) Difference of mass \(=2.17-2.15=0.02 \mathrm{~g}\)
Since the least number of decimal places is 2 so the difference in masses to the correct significant figures is \(0.02 \mathrm{~g}\)
A famous relation in physics relates ‘moving mass’ \(m\) to the ‘rest mass’ \(m_{0}\) of a particle in terms of its speed \(v\) and the speed of light, \(c\). (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c.
\(
m=\frac{m_{0}}{\left(1-v^{2}\right)^{1 / 2}}
\). Identify the correct formula from the options.
From the given equation, \(\frac{m_{0}}{m}=\sqrt{1-v^{2}}\)
Left hand side is dimensionless.
Therefore, the right-hand side should also be dimensionless.
It is possible only when \(\sqrt{1-v^{2}}\) should be \(\sqrt{1-\frac{v^{2}}{c^{2}}}\).
Thus, the correct formula is \(m=m_{0}\left(1-\frac{v^{2}}{c^{2}}\right)^{-1 / 2}\)
A LASER is a source of very intense, monochromatic, and a unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes \(2.56 \mathrm{~s}\) to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth?
We known that speed of laser light \(=c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\). If \(\mathrm{d}\) be the distance of Moon from the earth, the time taken by laser signal to return after reflection at the Moon’s surface
\(
\begin{aligned}
t &=2.56 \mathrm{~s}=\frac{2 d}{c}=\frac{2 d}{3 \times 10^{8} \mathrm{~ms}^{-1}} \\
\Rightarrow \quad d &=\frac{1}{2} \times 2.56 \times 3 \times 10^{8} \mathrm{~m}=3.84 \times 10^{8} \mathrm{~m} .
\end{aligned}
\)
A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR, the time delay between the generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be \(77.0 \mathrm{~s}\). What is the distance of the enemy submarine? (Speed of sound in water \(=1450 \mathrm{~m} \mathrm{~s}^{-1}\) ).
Here speed of sound in water \(\mathrm{v}=1450 \mathrm{~m} \mathrm{~s}^{-1}\) and time of echo \(t=77.0 \mathrm{~s}\). If distance of enemy submarine be \(d\), then \(t=2 d / v\)
\(
\therefore d=v t / 2=1450 \times 77.0 / 2=55825 \mathrm{~m}=55.8 \times 10^{3} \mathrm{~m} \text { or } 55.8 \mathrm{~km} \text {. }
\)
The farthest objects in our Universe discovered by modern astronomers are so distant that the light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes \(3.0\) billion years to reach us?
The time taken by light from the quasar to the observer \(t=3.0\) billion years \(=3.0 \times 10^{9}\) years As \(1 \mathrm{ly}=9.46 \times 10^{15} \mathrm{~m}\)
\(\therefore\) Distance of quasar from the observer \(d=3.0 \times 10^{9} \times 9.46 \times 10^{15} \mathrm{~m}\) \(=28.38 \times 10^{24} \mathrm{~m}=2.8 \times 10^{25} \mathrm{~m}\) or \(2.8 \times 10^{22} \mathrm{~km}\).
The radius of a sphere is measured as \((2.1 \pm 0.5) \mathrm{cm}\). What is its surface area with error limits?
Radius of the sphere \(=(2.1 \pm 0.5) \mathrm{cm}\)
\(\therefore r=2.1\) and \(A r=\pm 0.5\)
\(
\begin{aligned}
&\text { S.A. }=4 \pi r 2 \\
&=4 \times 3.14 \times 2.1 \times 2.1 \\
&=55.4 \mathrm{~cm} 2
\end{aligned}
\)
As per the principle of error
\(\frac{\Delta s}{s}=\pm 2 \cdot \frac{\Delta r}{r}\)
\(\frac{\Delta s}{55.4}=\pm \frac{2 \times 0.5}{2.1}\)
\(\therefore \quad \Delta s=\pm 26.4 \mathrm{~cm}\)
\(\therefore\) Error limits are \(\pm 26.4 \mathrm{~cm}\)
\(\therefore\) Surface area of the sphere \(=(55.4 \pm 26.4) \mathrm{cm}^{2}\)
The voltage across a lamp is \((6.0 \pm 0.1)\) volt and the current passing through it is (4.0 \(\pm 0.2)\) ampere. What is the power consumed by the lamp?
Power \(P=V \times I\)
\(
P=6 \times 4=24 \text { watt }
\)
Here
\(
\Delta V=\pm 0.1 \text { volts }
\)
and
\(
\Delta I=\pm 0.2 \mathrm{~A}
\)
As per the principle of error;
\(
\begin{aligned}
\therefore \quad \frac{\Delta P}{P} &=\pm\left(\frac{\Delta V}{V}+\frac{\Delta I}{I}\right) \\
\frac{\Delta P}{24} &=\pm\left(\frac{0.1}{6}+\frac{0.2}{4}\right) \\
\frac{\Delta P}{24} &=\pm \frac{0.8}{12} \\
\therefore \quad \Delta P &=\pm 1.6 \text { watt }
\end{aligned}
\)
\(\therefore\) Power with error limit is \((24 \pm 1.6)\) watt
If the length and time period of an oscillating pendulum have errors of \(1 \%\) and \(2 \%\) respectively, what is the error in the estimate of \(g\)?
We know
\(
\begin{aligned}
&\quad g=4 \pi^{2} \frac{l}{T^{2}} \\
&\therefore \quad \frac{\Delta g}{g}=\frac{\Delta l}{l}+2 \frac{\Delta T}{T} \\
&\% \text { error in } g=1 \%+2 \times 2 \%=5 \%
\end{aligned}
\)
The parallactic angle subtended by a distant star is \(0.76\) on the earth’s orbital diameter ( \(1.5 \times\) \(10^{11} \mathrm{~m}\) ). What is the distance of the star from the earth?
The parallactic angle, \(\phi=0.76=\frac{0.76 \times \pi}{180 \times 60 \times 60}\) radians \(=\frac{19 \pi}{1.62 \times 10^{7}}\) radians
The orbital diameter, say \(D=1.5 \times 10^{11} \mathrm{~m}\)
\(\therefore\) The required distance, \(d=\frac{D}{\phi}\)
\(
\begin{aligned}
&=\frac{1.5 \times 10^{11} \times 1.62 \times 10^{7}}{19 \pi} \mathrm{m} \\
&=\frac{2.43 \times 10^{18}}{19 \times 3.14} \mathrm{~m} \\
&=\frac{243 \times 10^{16}}{59.66} \mathrm{~m} \\
&=4.073 \times 10^{16} \mathrm{~m}
\end{aligned}
\)
Since, 1 light year \(=9.5 \times 10^{15} \mathrm{~m}\)
\(
\begin{aligned}
d &=\frac{4.073 \times 10^{16}}{9.5 \times 10^{15}} \text { light year } \\
&=4.29 \text { light year. }
\end{aligned}
\)
The velocity of a body moving in viscous medium is given by \(v=\frac{A}{B}\left[1-e^{\frac{-t}{B}}\right].\) where \(t\) is time, \(A\) and \(B\) are constants. Then the dimensions of \(A\) are
The dimensions of entropy are
What is the dimensions of Permeability?
Permeability occurs in Ampere’s law of force
\(
\begin{aligned}
&\Delta F=\mu \frac{\left(i_{1} \Delta l_{1}\right)\left(i_{2} \Delta l_{2}\right) \sin \theta}{r^{2}} \\
&{[\mu]=\frac{[\Delta F]\left[r^{2}\right]}{\left[i_{1} \Delta l_{1}\right]\left[i_{2} \Delta l_{2}\right]}=\frac{M L T^{-2}}{A L \cdot A L}=M L T^{-2} A^{-2}}
\end{aligned}
\)
What is the dimension of the Gravitational constant?
Constant of gravitation occurs in Newton’s law of gravitation
\(
\begin{aligned}
F &=G \frac{m_{1} m_{2}}{d^{2}} \\
{[G] } &=\frac{[F]\left[d^{2}\right]}{\left[m_{1}\right]\left[m_{2}\right]}=\frac{M L T^{-2} L^{2}}{M M}=M^{-1} L^{3} T^{-2}
\end{aligned}
\)
The radius of curvature of a concave mirror measured by spherometer is given by \(R=l^{2} / 6 h+\) \(\mathrm{h} / 2\). The values of \(\mathrm{l}\) and \(\mathrm{h}\) are \(4 \mathrm{~cm}\) and \(0.065 \mathrm{~cm}\) respectively. Compute the error in measurement of radius of curvature.
We are given
\(
l=4 \mathrm{~cm}, \Delta l=0.1 \mathrm{~cm} \quad \text { (least count of the metre scale) }
\)
here \(l\) is the distance between the legs of the spherometer.
As
\(
\begin{aligned}
R &=\frac{l^{2}}{6 h}+\frac{h}{2} \\
\frac{\Delta R}{R} &=\frac{2 \Delta l}{l}+\left(-\frac{\Delta h}{h}\right)+\frac{\Delta h}{h}
\end{aligned}
\)
Considering the magnitudes only, we get
\(
\begin{aligned}
\frac{\Delta R}{R} &=2 \frac{\Delta l}{l}+\frac{\Delta h}{h}+\frac{\Delta h}{h} \\
&=2\left(\frac{\Delta l}{l}+\frac{\Delta h}{h}\right)
\end{aligned}
\)
\(
\begin{aligned}
&=2 \times \frac{0.1}{4}+\frac{2 \times 0.001}{0.065} \\
&=0.05+0.03=0.08
\end{aligned}
\)
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