Question 513 Marks
Express unified atomic mass unit in kg.
AnswerThe unified atomic mass unit is the standard unit that is used for indicating mass on an atomic or molecular scale (atomic mass). One unified atomic mass unit is approximately the mass of one nucleon (either a single proton or neutron) and is numerically equivalent to 1g/mol. It is defined as one- twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state.
View full question & answer→Question 523 Marks
A small error in the measurement of the quantity having the highest power (in a given formula) will contribute maximum percentage error in the value of the physical quantity to whom it is related. Explain why?
Answer$\text{Let}\text{ Z}=\text{A}^{\text{m}}\times\text{B}^{\text{n}}\times\text{C}^{\text{l}}$ Where $\text{m}>\text{n}>\text{l}$ $\therefore$ Maximum fractional error in Z is given by: $\frac{\Delta\text{Z}}{\text{Z}}=\text{m}.\frac{\Delta\text{A}}{\text{A}}+\text{n}.\frac{\Delta\text{B}}{\text{B}}+\text{l}.\frac{\Delta\text{C}}{\text{C}}$ As $\text{m}>\text{n}>\text{l}$ $\therefore\text{m}\times\frac{\Delta\text{A}}{\text{A}}$ will contribute the maximum percentage error in the value of A.
View full question & answer→Question 533 Marks
The displacement of a progressive wave is represented by $\text{y} = \text{A} \sin(\omega \text{t} – \text{k x} ),$ where $x$ is distance and $t$ is time. Write the dimensional formula of $(i) \omega$ and $(ii) k.$
AnswerWe have to apply principle of homogeneity to solve this problem. Principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be same, i.e., dimensions of $\text{LHS}$ and $\text{RHS}$ should be equal. According to the problem, $\text{y}=\text{A}\sin(\omega\text{t}-\text{kx})$ Here $y = [L]$ hence $\text{A}\sin(\omega\text{t}-\text{kx})=[\text{L}]$ Here $A = [L]$, which peak value of $y$ So, $\omega\text{t}-\text{kx}$ Should be dimensionless,
- $[\omega\text{t}]=\text{constant}$
$\Rightarrow[\omega]=[\text{T}^{-1}]$
- $[\text{Kx}] = \text{Constant}$
$\Rightarrow[\text{k}]=[\text{L}^{-1}]$ View full question & answer→Question 543 Marks
The number of particles crossing per unit area perpendicular to x-axis in unit time N is given by: $\text{N}=-\text{D}\Big(\frac{\text{n}_2-\text{n}_1}{\text{x}_2-\text{x}_1}\Big),$ where $n_1$ and $n_2$ are the number of particles per unit volume at $x_1$ and $x_2$ respectively. Deduce the dimensional formula for D.
Answer$\text{D}=-\text{N}\Big(\frac{\text{x}_2-\text{x}_1}{\text{n}_2-\text{n}_1}\Big)[\text{N}]=\text{L}^{-2}\text{T}^{-1}$ $[\text{D}]=\frac{\text{L}^{-2}\text{T}^{-1}\text{L}}{\text{L}^{-3}}$ $=\text{L}^2\text{T}^{-1},[\text{x}_2]=[\text{x}_1]=\text{L}$ $[\text{n}_2]=[\text{n}_1]=\frac{\text{N}_0}{\text{L}_3}=\text{L}^{-3}$
View full question & answer→Question 553 Marks
The length, breadth and thickness of a rectangular sheet of metal are $4.234m, 1.005m$, and $2.01cm$ respectively. Give the area and volume of the sheet to correct significant figures.
AnswerGiven that, length, $\mathrm{I}=4.234 \mathrm{~m}$ breadth, $\mathrm{b}=1.005 \mathrm{~m}$ thickness, $\mathrm{t}=2.01 \mathrm{~cm}=2.01 \times 10^{-2} \mathrm{~m}$ Area of the sheet $=2(\mathrm{I} \times 0$ $+\mathrm{b} \times \mathrm{t}+\mathrm{t} \times \mathrm{l})=2(4.234 \times 1.005+1.005 \times 0.0201+0.0201 \times 4.234)=2(4.3604739)=8.7209478 \mathrm{~m} 2$ As area can contain a maximum of three significant digits, therefore, rounding off, we get Area $=8.72 \mathrm{~m}^2$ Also, volume $=1 \times \mathrm{b} \times \mathrm{t}$ $V=4.234 \times 1.005 \times 0.0201=0.0855289=0.0855 \mathrm{~m}^3($ Significant Figures $=3)$
View full question & answer→Question 563 Marks
Density $\rho$ or a piece of an object of mass m and volume V is given by the formula $\rho=\frac{\text{m}}{\text{v}}$ If $\text{m}=(375.32\pm0.01)\text{g}$ and $\text{V}=(136.41\pm0.01)\text{cm}^3,$ find percentage error in $\rho.$
Answer$\rho=\frac{\text{m}}{\text{v}},$ density $'\rho'$ of a piece $\text{m}=(375.32\pm0.01)\text{g}$ $\text{V}=(136.41\pm0.01)\text{cm}^3$ $\frac{\text{dm}}{\text{m}}=\frac{0.01}{375.32}=0.0000266$ $\frac{\text{dV}}{\text{V}}=\frac{0.01}{136.41}=0.0000733$ $\frac{\Delta\rho}{\rho}=\frac{\Delta\text{m}}{\text{m}}+\frac{\Delta\text{V}}{\text{V}}$ $=0.0000266+0.0000733$ $=0.0000999$ Percentage error $=\frac{\Delta\rho}{\rho}\times100\%=0.01\%$
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Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity): The mass of an elephant.
AnswerConsider a ship of known base area floating in the sea. Measure its depth in sea $\left(\right.$ say $\left.\mathrm{d}_1\right)$. Volume of water displaced by the ship, $\mathrm{Vb}=\mathrm{Ad}_1$ Now, move an elephant on the ship and measure the depth of the ship ( $\mathrm{d}_2$ ) in this case. Volume of water displaced by the ship with the elephant on board, $\mathrm{V}_{\mathrm{be}}=\mathrm{Ad}_2$ Volume of water displaced by the elephant $=A d_2-A d_1$ Density of water $=D$ Mass of elephant $=A D\left(d_2-d_1\right)$
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Find the value of $60W$ on a system having $100g, 20cm$ and $1$ min as the fundamental units.
Answer$n_1 = 60W$, power is $[M^1L^2T^{-3}]$ In first system, $M_1 = 1kg, L_1 = 1m$, and $T_1 = 1s$ In second system $M_2 = 100g, L^2 = 20m$, And $T_2 = 1 min = 60s $$\text{So, }\text{n}_2=\text{n}_1\Big[\frac{\text{M}_1}{\text{M}_2}\Big]\Big[\frac{\text{L}_1}{\text{L}_2}\Big]^1\Big[\frac{\text{T}_1}{\text{T}_2}\Big]^{-3}$
$=60\Big[\frac{1000\text{g}}{100\text{g}}\Big]\Big[\frac{100\text{cm}}{20\text{cm}}\Big]\Big[\frac{1\text{s}}{60\text{s}}\Big]^{-3}$
$=60\times\frac{1000}{100}\times\frac{100}{20}\times\frac{100}{20}\times60\times60\times60$
$=3.24\times10^9\text{ units}$
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If $x = at^2 + bt + c$, where x is displacement as a function of time. Write the dimensions of a b and c.
AnswerAll the terms should have the same dimension:
$\therefore[\text{a}]=\Big[\frac{\text{x}}{\text{t}^2}\Big]\text{s}=[\text{LT}^{-2}]$
$[\text{b}]=\Big[\frac{\text{x}}{\text{t}}\Big]\text{s}=[\text{LT}^{-1}]$
$[\text{c}]=[\text{x}]=[\text{L}]$
View full question & answer→Question 603 Marks
Experiments show that the frequency (n) of a tuning fork depends upon the length (I) of the prong, the density (d) and the Young's modulus (Y) of its material. From dimensional considerations, find a possible relation for the frequency of the tuning fork.
Answer$\text{Let n }= \text{Kl}^\text{a}\text{d}^\text{b}\text{y}^\text{c}, $ Where $\text{K}$ Substituting the dimension of all the quantities involved We have: $[\text{T}^{-1}]= [\text{L}]^\text{a}[\text{ML}^{-3}]^\text{b}[\text{ML}^{-1}\text{T}^{-2}]^\text{c}$ $ [\text{M}^0\text{L}^0\text{T}^{-1}]= [\text{M}] ^{(\text{b+c})}[\text{L}]^{\text{a-3b-c}}[\text{T}] ^{-2\text{c}}$ Comparing powers of M, L and T we get $\text{b+c = 0}$ $\text{a}-3\text{b}-\text{c}=0$ $-2\text{c}= -1$ Or $\text{a = }-1, \text{b}= \frac{-1}{2} \text{ and c} =\frac{1}{2}$ This gives $\text{n}= \text{Kl}^{-1}\text{d}^\frac{-1}{2}\text{y}\frac{1}{2}$ Or $\text{n}= \frac{\text{K}}{1}\sqrt{\frac{\text{y}}{\text{d}}}$
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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 generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be $77.0s$. What is the distance of the enemy submarine? (Speed of sound in water = $1450m s^{–1}$).
AnswerLet the distance between the ship and the enemy submarine be ‘S’. Speed of sound in water = $1450m/s$ Time lag between transmission and reception of Sonar waves = 77s In this time lag, sound waves travel a distance which is twice the distance between the ship and the submarine (2S). Time taken for the sound to reach the submarine = $1/2 × 77 = 38.5s$
$\therefore$ Distance between the ship and the submarine (S) = $1450 × 38.5 = 55825m = 55.8km$
View full question & answer→Question 623 Marks
The density p of a piece of metal of a mass m and volume V is given by the formula = $\frac{\text{M}}{\text{V}}.,$ If $\text{m} = 375.32 \pm0.01\text{g},$ and $\text{V} = 136.41\pm 0.01\text{cm}^3$ Find % error in $\rho$
AnswerGiven $dm = 0.01g$ and $d = 0.01cm^3$ The error in mass, $\frac{\text{dm}}{\text{m}}=\frac{0.01}{375.32}$ $=0.0000266$ and error in volume $\frac{\text{dv}}{\text{V}}=\frac{0.1}{136.41}$
$=0.0000733$ As density is a function of both mass and volume the error in its value should be sum of the errors in the mass and volume.
$\therefore\frac{\text{dm}}{\text{m}}+\frac{\text{dv}}{\text{V}}$
$=(0.0000266+0.0000733)$ % error in $\rho=\frac{\text{dp}}{\text{p}}\times100\%$ $=0.0000999\times100\%$ $=0.0099\%$
View full question & answer→Question 633 Marks
In the expression $P = El^2 m^{-5} G^{-2}, E, m, l$ and $G$ denote energy, mass, angular momentum and gravitational constant, respectively. Show that $P$ is a dimensionless quantity.
AnswerAccording to the problem, expression is $\text{P}=\text{El}^2\text{m}^{-5}\text{G}^{-2}$ where E is energy $[\text{E}]=[\text{ML}^2\text{T}^{-2}],$ m is mass [m] = [M], L is angular momentum $[\text{L}] = [\text{ML}^2 \text{T}^{-1}],$ G is gravitational constant $[\text{G}] = [\text{M}^{-1}\text{L}^2\text{T}^{-2}]$ Substituting dimensions of each physical quantity in the given expression, $[\text{P}]=[\text{ML}^2\text{T}^{-2}]\times[\text{ML}^2\text{T}^{-1}]^2\times[\text{M}]^{-5}\times[\text{M}^{-1}\text{L}^3\text{T}^{-2}]^{-2}$ $=[\text{M}^{1+2-5+2}\text{L}^{2+4-6}\text{T}^{-2-2+4}]$ $=[\text{M}^0\text{L}^0\text{T}^0]$ This shows that P is a dimensionless quantity.
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The distance of the Sun from the Earth is $1.496 \times 10^{11}m$ (i.e., 1 A.U.). If the angular diameter of the Sun is 2000”, find the diameter of the Sun.
Answer
Here, $\theta=2000''=\frac{2000}{3600}\times\frac{\pi}{180}\text{rad}$
$=9.7\times10^{-3}\text{rad}$
$\text{d}=1.496\times10^{11}\text{m}$
From the figure,
$\theta=\frac{\text{D}}{\text{d}}$
$\therefore\text{D}=\theta\text{d}$
$=9.7\times10^{-3}\times1.496\times{10}^{11}$
$=1.45\times10^{9}\text{m}$ View full question & answer→Question 653 Marks
The radius of a sphere is measured as $(2.1 \pm 0.5) \text{cm}$ calculate its surface area with error limits.
AnswerRadius of the sphere $=(2.1 \pm 0.5) \text{cm}$ $\therefore\text{r}=2.1\text{ and }\Delta\text{r}=\pm0.5$ $\text{S.A.}=4\pi\text{r}^2=4\times3.14\times2.1\times2.1$ $=55.4\text{cm}^2$ As per the principle of error $\frac{\Delta\text{s}}{\text{s}}=\pm2.\frac{\Delta\text{r}}{\text{r}}$ $\frac{\Delta\text{s}}{55.4}=\pm\frac{2\times0.5}{2.1}$ $\therefore\Delta\text{s}=\pm26.4\text{cm}$ $\therefore$ Error limits are $\pm26.4\text{cm}$ $\therefore$ Surface area of the sphere $=(55.4\pm26.4)\text{cm}^2$
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Reynold's number $N_R$(a dimensionless quantity) determines the condition of laminar flow of a viscous liquid through a pipe. $N_R$ is a function of the density of the liquid 'r', its average speed is 'y' and the coefficient of viscosity of the liquid is 'h'. If N, is given directly proportional to 'd' (the diameter of the pipe), show from dimensional consideration that $\text{N}_\text{R}\propto \frac{\text{dp}\rho}{\eta}$ the unit of '$\eta$' in SI system is kg $m^{-1}s^{-1}$?
AnswerAs the Reynold's Number $N_R$ depends on density p, average speed v and coefficient ofviscosity I, then let us say.$\text{N}_\text{R}\propto \rho^\text{a}\text{v}^b\eta^\text{c}$
Again is proportional to the diameter of the pipe , combining the two we have . $\text{N}_\text{R}\propto \rho ^\text{a}\text{v}^\text{b}\eta^\text{c}\text{d}$
$\text{N}_\text{R }= \text{ K} \rho^\text{a} \text{v}^\text{b}\eta^\text{c}\text{d}$
We write ,$[\text{N}_\text{R}]= [\text{M}^0\text{L}^0\text{T}^0]$
$[\rho]= [\text{MK}^{-3}]$
$[\text{v}]= [\text{LT}^{-1}]$
$[\eta] = [ \text{ML}^{-1} \text{T}^{-1}]$
$[\text{d}]= [\text{L}]$
Syubstituting the dimension in (i), we have ,$[\text{M}^0\text{L}^0\text{T}] = [\text{ML}^{-3}]^\text{a}[\text{LT}^{-1}]^\text{b}[\text{ML}^{-1}\text{T}^{-1}]^\text{c}[\text{L}]$
$= \text{M}^{(\text{a+c})} \text{L }^{(-3\text{a+b+c+1})}\text{T}^{(\text{-b-c})}$
Comparing the dimensions of M,L and T, we have,$\text{a+c}=0$
$-3\text{a+b-c+1=0}$
$-\text{b}-\text{c} =0$
On simplifying, we get$\text{c}= -1$
$\text{b}=1$
$\text{a}=1$
Therefore, the relation(i) becomes$\text{N}_\text{R} =\text{K}\rho^1 \text{v}^1\eta^{-1}\text{d}$
$\text{N}_\text{R}=\text{K}\rho^1\text{v}^1\eta^{-1}\text{d}$
$\text{N}_\text{R} = \text{K}\rho\frac{\text{vd}}{\eta}$
$\text{N}_\text{R} = \propto\rho\frac{\text{vd}}{\eta}$
View full question & answer→Question 673 Marks
The number of particles crossing per unit area perpendicular to x-axis in unit time N is given by: $\text{N}=-\text{D}\Big(\frac{\text{n}_2-\text{n}_1}{\text{x}_2-\text{x}_1}\Big)\text{s}$ where $n_1$ and $n_2$ are the number of particles per unit volume at $x_1$ and $x_2$ respectively. Deduce the dimensional formula for D.
Answer$\text{D}=-\text{N}\Big(\frac{\text{x}_2-\text{x}_1}{\text{n}_2-\text{n}_1}\Big)\text{s}$
$[\text{N}]=\frac{\text{N}_0}{[\text{L}^2\text{T}]}=[\text{L}^{-2}\text{T}^{-1}]$
$[\text{D}]=\frac{[\text{L}^{-2}\text{T}^{-1}\text{L}]}{[\text{L}^{-3}]}=[\text{L}^2\text{T}^{-1}]$
$[\text{x}_2]=[\text{x}_1]=[\text{L}]$ And $[\text{n}_2]=[\text{n}_1]=\frac{\text{N}_0}{[\text{L}^{-3}]}$
View full question & answer→Question 683 Marks
When the planet Jupiter is at a distance of $824.7$ million kilometers from the Earth, its angular diameter is measured to be $35.72$” of arc. Calculate the diameter of Jupiter.
AnswerDistance of Jupiter from the Earth, $D=824.7 \times 10^6 \mathrm{~km}=824.7 \times 10^9 \mathrm{~m}$ Angular diameter $=35.72^{\prime \prime}=35.72 \times 4.874 \times$ $10^{-6} \mathrm{rad}$ Diameter of Jupiter $=\mathrm{d}$ Using the relation,
$\theta=\frac{\text{d}}{\text{D}}$ $\text{d}=\theta\text{ D}=824.7\times10^9\times35.72\times4.872\times10^{-6}$ $= 143520.76 \times 10^3\text{m} = 1.435 \times 10^5\text{Km}$
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The radius of curvature of a concave mirror, measured by a spherometer is given by: $\text{R}=\frac{\text{l}^2}{6\text{h}}+\frac{\text{h}}{2}$ The value of l and h are 4.0cm and 0.065cm respectively where l is measured by a metre scale and h by the spherometer. Find the relative error in the measurement of R.
AnswerGiven that I - 4cm and Al = 0.1cm (least count of the metre scale) here l is the distance between the legs of the spherometer. As $\text{R}=\frac{\text{l}^2}{6\text{h}}+\frac{\text{h}}{2}$ $\therefore\frac{\Delta\text{R}}{\text{R}}=\frac{2\Delta\text{l}}{\text{l}}+\Big(-\frac{\Delta\text{h}}{\text{h}}\Big)+\frac{\Delta\text{h}}{\text{h}}$ $\Rightarrow\frac{\Delta\text{R}}{\text{R}}=2\frac{\Delta\text{l}}{\text{l}}+\frac{\Delta\text{h}}{\text{h}}+\frac{\Delta\text{h}}{\text{h}}$ (Considering the magnitude only) $=2\Big(\frac{\Delta\text{l}}{\text{l}}+\frac{\Delta\text{h}}{\text{h}}\Big)=2\Big(\frac{0.1}{4}\Big)+2\times\Big(\frac{0.001}{0.065}\Big)$ $=0.05+0.03=0.08$
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A body travels uniformly a distance of $(13.8 \pm 0.2)\text{m}$ in a time $(4.0 \pm 0.3) \text{s}.$ What is the velocity of the body within error limits?
AnswerHere, $\text{S}=(13.8\pm0.2)\text{cm},\text{t}=(4.0\pm0.3)\text{s}$ $\therefore\text{V}=\frac{13.8}{4.0}=3.45\text{ms}^{-1}$ Also $\frac{\Delta\text{V}}{\text{V}}=\pm\Big(\frac{\Delta\text{S}}{\text{S}}+\frac{\Delta\text{t}}{\text{t}}\Big)$ $=\pm\Big(\frac{0.2}{13.8}+\frac{0.3}{4.0}\Big)=\pm0.0895$ $\Delta\text{V}=\pm0.3$ (rounding off to one place of decimal) $\text{V}=(3.45\pm0.3)\text{ms}^{-1}$
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A laser signal is beamed towards the planet Venus from Earth and its echo is received 8.2 minutes later. Calculate the distance of Venus from the Earth at that time.
AnswerWe know that speed of laser light $\text{c}=3\times10^8\text{m}/\text{ s}$ Time of echo, t = 8.2 minutes = 8.2 × 60 seconds If distance of venus be d, then $\text{t}=\frac{2\text{d}}{\text{c}}$ $\text{d}=\frac{1}{2}\text{ct}=\frac{1}{2}\times3\times10^8\times8.2\times60\text{m}$ $=7.38\times10^{10}\text{m}=7.4\times10^{10}\text{m}.$
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Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.
AnswerIt is indeed very true that precise measurements of physical quantities are essential for the development of science. For example, ultra-shot laser pulses (time interval ∼ $10^{–15}\ s$) are used to measure time intervals in several physical and chemical processes. X-ray spectroscopy is used to determine the inter-atomic separation or inter-planer spacing. The development of mass spectrometer makes it possible to measure the mass of atoms precisely.
View full question & answer→Question 733 Marks
If two resistances of values $\text{R}_1=(2.0\pm0. 1)12$ and $\text{R}_2=(12.3\pm0.2)\Omega$ are put (i) in parallel and (ii) in series, find the error in the equivalent resistance.
Answer$\text{R}_1=2,\Delta\text{R}_1=0.1,$ $\text{R}_\text{s}=\text{R}_1+\text{R}_2=14.3$ $\text{R}_2=12.3,\Delta\text{R}_2=0.2,$ $\Delta(\text{R}_1+\text{R}_2)=\Delta\text{R}_\text{s}=0.3$ In series$\text{R}_\text{s}=(\text{R}_1+\text{R}_2)\pm\Delta(\text{R}_1+\text{R}_2)$ $=(14.3\pm0.3)\Omega$ In parallel,$\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}=\frac{\text{R}_1+\text{R}_2}{\text{R}_1\text{R}_2}$ $\therefore\text{R}=\frac{\text{R}_1\text{R}_2}{\text{R}_1+\text{R}_2}=\frac{\text{R}_1\text{R}_2}{\text{R}_\text{s}}$ $\frac{\Delta\text{R}}{\text{R}}=\frac{\Delta\text{R}_1}{\text{R}_1}+\frac{\Delta\text{R}_2}{\text{R}_2}+\frac{\Delta\text{R}_\text{s}}{\text{R}_\text{s}}$ $\Delta\text{R}=\text{R}\bigg[\frac{\Delta\text{R}_1}{\text{R}_1}+\frac{\Delta\text{R}_2}{\text{R}_2}+\frac{\Delta\text{R}_\text{s}}{\text{R}_\text{s}}\bigg]$ $\Rightarrow\text{R}=\frac{24.6}{14.3}=1.72$ $\Delta\text{R}=\text{R}(\frac{0.1}{2}+\frac{0.2}{12.3}+\frac{0.3}{14.3})$ $=\text{R(0.05+0.016+0.020)}$ $=1.72(0.08)=0.13$ R (in parallel)$=(1.72\pm0.13)\Omega$
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To find the value of 'g' by using a simple pendulum, the following observations were made: Length of thread $\text{l} = (100 \pm 0.1)\text{cm}$ Time period of oscillation $\text{T} = (2 ± 0.1)\text{sec}$ Calculate the maximum permissible error in measurement of 'g'. Which quantity should be measured more accurately and why?
AnswerHere, $\text{l}=(100\pm0.1)\text{cm}$ Time period of oscillation $\text{T}=(2\pm0.1)\text{sec}$ $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$ $\text{T}^2=4\pi^2\times\frac{\text{l}}{\text{g}}$ $\therefore\text{g}=\frac{4\pi^2\text{l}}{\text{T}^2}$ As per the principle of error, $\frac{\Delta\text{g}}{\text{g}}=\pm\Big(\frac{\Delta\text{l}}{\text{l}}+2\frac{\Delta\text{T}}{\text{T}}\Big)$ $\frac{\Delta\text{g}}{\text{g}}=\pm\Big(\frac{\Delta\text{0.1}}{100}+2\times\frac{0.1}{2} \Big)$ $\Delta\text{g}=\pm9.8\times0.101=\pm0.99$ $\therefore$ Maximum permissible error in the measurement of $\text{g}=\pm0.99$ . Time period of the pendulum should be measured more accurately as $\text{g}\propto\frac{1}{\text{T}^2}.$
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A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8min and 20s to cover this distance?
AnswerDistance between the Sun and the Earth = Speed of light × Time taken by light to cover the distance Given that in the new unit, speed of light = 1 unit Time taken, t = 8min 20s = 500s $\therefore$ Distance between the Sun and the Earth = 1 × 500 = 500 units
View full question & answer→Question 763 Marks
In an experiment, refractive index of glass was observed to be $1.45, 1.56, 1.54, 1.44, 1.54$ and $1.53$. Calculate:
- Mean value of refractive index.
- Mean absolute error.
- Fractional error.
- Percentage error.
Express the result in terms of absolute error and percentage error. Answer
- Mean value of $\mu$
$=\frac{1.45+1.56+1.54+1.44+1.54+1.53}{6}$
$\mu=1.51$
- Mean absolute error $=\frac{\text{sum of absolute error}}{6}$
$\Delta\overline{\mu}=\frac{0.06+0.05+0.03+0.07+0.03+0.02}{6}$
$=\frac{0.26}{6}=0.0433=0.04$
- Fractional error $=\frac{\Delta\overline{\mu}}{\mu}=\frac{0.04}{1.51}$
$=0.02649=0.03$
- Percentage error $=\frac{\Delta\overline{\mu}}{\mu}\times100=3\%$
$\mu=1.51\pm0.04$ in terms of absolute error.
Also, $\mu=1.51\pm3\%$ in terms of $\%$ error. View full question & answer→Question 773 Marks
It is known that the period T of a magnet of magnetic moment M vibrating in a uniform magnetic field of intensity H depends upon M, H and I where I is the moment of inertia of the magnet about its axis of oscillations. Show that T = $2\pi\sqrt{\frac{\text{I}}{\text{MH}}}.$
AnswerLet us first write the dimensions of various physical quantities involved, Moment of Inertia, $I = [ML^2]$ Magnetic moment has the units $Am^2$, therefore, $M = [M^0L^2A]$ Magnetic field intensity, H has the units newton per ampere metre. $\text{[H]}=\frac{\text{netwon}}{\text{A-M}}=\text{MLT}^{-2}\text{L}^{-1}\text{A}^{-1}$
$\text{H}=\text{[MT}^{-2}\text{A}^{-1}]$
$\text{T}=\text{KI}^\text{a}\text{M}^\text{b}\text{H}^\text{c}.$ Where K is constant of proportionality, $\text{[T]}=\text{[ML}^2]^\text{a}\text{[L}^2\text{A}]^\text{b}\text{[MT}^{-2}\text{A}^{-1}]^\text{c}$
$=\text{[M}^\text{a+b}\text{L}^\text{2a+2b}\text{T}^{-2\text{c}}\text{A}^\text{+b-c}]$ Equating powers of M. L and T and A on both the sides, we have, $\text{a+c}=0$
$2(\text{a+b)}=0$
$\Rightarrow\text{a+b}=0,-2\text{c}=1$ and $-\text{c+b}=0$ Solving these equations ,we get, $\text{a}=\frac{1}{2},\text{b}=\frac{-1}{2},\text{c}=\frac{-1}{2}$
$\therefore \text{T}=\text{KI}^\frac{1}{2}\text{M}^{\frac{-1}{2}}\text{H}^{\frac{-1}{2}}$ It can be proved experimentally that $\text{K}=2\pi$
$\therefore \text{T}=2\pi\sqrt{\frac{\text{I}}{\text{MH}}}$
View full question & answer→Question 783 Marks
Write the dimensions of:
- Linear density.
- Power.
- Impulse.
- Velocity gradien.
- Mass per unit area.
- Kinetic energy.
- Angular acceleration.
- Couple.
- Moment of force.
- Work done.
Answer$i. \left[M L^{-1} T^0\right)$
$ii. \left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right]$
$iii. \left[M L T^{-1}\right)$
$iv. \left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]$
$v. \left[\mathrm{ML}^{-2} \mathrm{~T}^0\right)$
$vi. \left[M^1 L^2 T^{-2}\right]$
$vii. \left[M^0 L^0 \mathrm{~T}^{-2}\right]$
$viii. \left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$
$ix. \left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$
$x. \left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$.
View full question & answer→Question 793 Marks
Find an expression for viscous force $F$ acting on a tiny steel ball of radius $r$ moving in a viscous liquid of viscosity $n$ with a constant speed $v$ by the method of dimensional analysis.
AnswerIt is given that viscous force $F$ depends on:
- Radius $r$ of steel ball.
- Coefficient of viscosity $n$ of viscous liquid.
- Speed $v$ of the ball.
$\text{i.e}.,\text{F}=\text{k}\text{r}^{\text{a}}\eta^{\text{b}}\text{v}^{\text{c}}$ where $k$ is dimensionless constant
Dimensional formula of force
$\text{F}=[\text{MLT}^{-2}],\text{r}=[\text{L}]$
$\eta=[\text{M}^1\text{L}^{-1}\text{T}^{-1}]\text{ and }\text{v}=[\text{LT}^{-1}]$
We have:
$[\text{MLT}^{-2}]=[\text{L}^{\text{a}}][\text{M}^1\text{L}^{-1}\text{T}^{-1}]^{\text{b}}[\text{LT}^{-1}]^{\text{c}}$
$=[\text{M}^{\text{a}}\text{L}^{\text{a}-\text{b}+\text{c}}\text{T}^{-\text{b}-\text{c}}]$
Comparing powers of $M, L$ and $T$ on either side of equation,
We get:
$\text{a}=1$
$\text{a}-\text{b}+\text{c}=1$
$-\text{b}-\text{c}=2$
On solving, these above equations,
We get,
$\text{a}=1,\text{b}=1$ and $\text{c}=1$
Hence, the relation becomes
$\text{F}=\text{kr}\eta\text{v}$ View full question & answer→Question 803 Marks
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?
AnswerWe know $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$ Or $\text{T}^2=4\pi^2\frac{\text{l}}{\text{g}}$ $\text{g}=4\pi^2\frac{\text{l}}{\text{T}^2}$ $\therefore\frac{\Delta\text{g}}{\text{g}}=\frac{\Delta\text{l}}{\text{l}}+2\frac{\Delta\text{T}}{\text{T}}$ % error in g = 1% + 2 × 2% = 5%
View full question & answer→Question 813 Marks
The radius of the Earth is $6.37 \times 10^6\ m$ and its mass is $5.975 \times 10^{24}\ kg$. Find the Earth's average density to appropriate significant figures.
AnswerRadius of the earth $(R) = 6.37 \times 10^6m$ Volume of the earth $(\text{V})=\frac{4}{3}\pi\text{R}^3\text{m}^3$
$=\frac{4}{3}\times(3.142)\times(6.37\times10^6)^3\text{m}^3$
Average density $(\text{D})=\frac{\text{Mass}}{\text{Volume}}=\frac{\text{M}}{\text{V}}=0.005517\times10^{6}\text{kg}\text{m}^{-3}$
The density is accurate only up to three significant figures which is the accuracy of the least accurate factor, namely, the radius of the earth.
View full question & answer→Question 823 Marks
Why has second been defined in terms of periods of radiations from cesium$-133$?
AnswerSecond has been defined in terms of periods of radiation, because
- This period is accurately defined.
- This period is not affected by change of physical conditions like temperature, pressure and volume etc.
- The unit is easily reproducible in any good laboratory.
View full question & answer→Question 833 Marks
A capacitor of capacitance $\text{C}=(2.0\pm0.1)\mu\text{F}$ is charged to a voltage$\text{V}=(20\pm0.5)\text{V}$ Calculate the charge Q with error limits.
Answer$\text{Q}=\text{CV}=2.0\times20=40\mu\text{C}$ $=40\times10^{-6}\text{C}$ $\frac{\Delta\text{Q}}{\text{Q}}=\frac{\Delta\text{C}}{\text{C}}+\frac{\Delta\text{V}}{\text{V}}$ $=\frac{0.1}{2.0}+\frac{0.5}{20}=\frac{3}{40}$ $\Delta\text{Q}=\frac{3}{40}\times\text{Q}=\frac{3}{40}\times10^{-6}=3\mu\text{C}$ Hence, $\text{Q}=(40\pm3.0)\times10^{-6}\text{C}$
View full question & answer→Question 843 Marks
The orbital velocity v of a satellite may depend on its mass m, distance r from the centre of Earth and acceleration due to gravity g. Obtain an expression for orbital velocity.
AnswerLet orbital velocity of satellite be given by the relation $v = k\ m^ar^bg^c$ where k is a dimensionless constant and a, b and c are the unknown powers. Writing dimensions on two sides of equation, we have:$[\text{M}^0\text{L}^1\text{T}^{-1}=[\text{M}]^{\text{a}}[\text{L}]^{\text{b}}[\text{LT}^{-2}]^{\text{c}}$
$=[\text{M}^{a}\text{L}^{\text{b}+\text{c}}\text{T}^{-2\text{c}}]$
Applying principle of homogeneity of dimensional equation,
We find that:
$\text{a}=0$
$\text{b}+\text{c}=1$
$-2\text{c}=-1$
On solving these equations,
We find that:
$\text{a}=0,\text{b}=+\frac{1}{2}\text{ and }\text{c}=+\frac{1}{2}$
$\therefore\text{v}=\text{k}\text{r}^{\frac{1}{2}}\text{g}^{\frac{1}{2}}$
Or $\text{v}=\text{k}\sqrt{\text{rg}}.$
View full question & answer→Question 853 Marks
The farthest objects in our Universe discovered by modern astronomers are so distant that 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?
AnswerTime taken by quasar light to reach Earth $=3$ billion years $=3 \times 10^9$ years $=3 \times 10^9 \times 365 \times 24 \times 60 \times 60$ s Speed of light $=3 \times 10^8 \mathrm{~m} / \mathrm{s}$ Distance between the Earth and quasar $=\left(3 \times 10^8\right) \times\left(3 \times 10^9 \times 365 \times 24 \times 60 \times 60\right)=283824 \times$ $10^{20} \mathrm{~m}=2.8 \times 10^{22} \mathrm{~km}$
View full question & answer→Question 863 Marks
In the relation $\text{p}=\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{e}^{-\Big(\frac{\text{az}}{\theta}\Big)}},\text{p}$ is the pressure, Z is the distance, and is the temperature. What is the dimensional formula of p?
AnswerSince, ${\text{e}^{-\Big(\frac{\text{aZ}}{\theta}\Big)}}$is dimensionless, We have $\frac{\text{aZ}}{\theta}=1$ Or $\text{a}=\frac{\theta}{\text{Z}}=\frac{\text{K}}{\text{L}}=[\text{L}^{-1}\text{K}]$ We find that $\frac{\text{a}}{\text{b}}=$ dimensions of p and $\text{b} = [\text{ML}^{-1}\text{T}^{-2}].$ Therefore, dimensional formula of p is obtained as $\text{p}=\frac{\text{a}}{[\text{ML}^{-1}\text{T}^{-2}]}=\frac{[\text{L}^{-1}\text{K}]}{[\text{ML}^{-1}\text{T}^{-2}]}$ $=[\text{M}^{-1}\text{L}^0\text{T}^2\text{K}]$
View full question & answer→Question 873 Marks
The radius of the Earth is $6.37 × 106m$ and its average density is $5.517 \times 10^3kg m^{-3}$. Calculate the mass of earth to correct significant figures.
Answer$\text{Mass}=\text{Volume}\times\text{density}$ Volume of earth $=\frac{4}{3}\pi\text{R}^3=\frac{4}{3}\times3.142\times(6.37\times10^6)^3\text{m}^3$ Mass of earth $=\frac{4}{3}\times3.142\times(6.37\times10^6)^3\times5.517\times10^3\text{kg}$ $=5974.01\times10^{21}\text{kg}=5.97401\times10^{24}\text{kg}$
The radius has three significant figures and the density has four. Therefore, the final result should be rounded upto three significant figures. Hence, mass of the earth = $5.97 \times 10^{24}kg$.
View full question & answer→Question 883 Marks
The measured value of length, breadth and height of a block of wood along with maximum permissible errors are expressed as follows:$\text{I}=12.80\pm0.01\text{cm},$ $\text{b}=10.12\pm0.01\text{cm}$ $\text{h}=5.26\pm0.01\text{cm}$ Calculate the percentage error in the volume of the block.
AnswerGiven: $ \text{I}\pm\Delta\text{I}=10.08\pm0.01\text{cm}$ $\text{b}\pm\Delta\text{b}=10.12\pm0.01\text{cm}$ $\text{h}\pm\Delta\text{h}=5.62\pm0.01\text{cm}$ Volume of the block (V) $=\text{I}\times\text{b}\times\text{h}$ $\therefore$ Percentage error in the volume is given by $=\frac{\Delta\text{V}}{\text{V}}=(\frac{\Delta\text{I}}{\text{I}}+\frac{\Delta\text{b}}{\text{b}}+\frac{\Delta\text{h}}{\text{h}})\times100\%$ $=(\frac{0.01}{12.08}+\frac{0.01}{10.12}+\frac{0.01}{5.62})\times100\%$ $=(0.000827+0.000988+0.00178)\times100\%$ $=0.003585\times100\%=0.3585\%$
View full question & answer→Question 893 Marks
The principle of ‘parallax’ in section $2.3.1$ is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit $\approx 3 \times 10^{11} \mathrm{~m}$. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of $1”$ (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of $1”$ (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?
AnswerDiameter of Earth's orbit $=3 \times 10^{11} \mathrm{~m}$ Radius of Earth's orbit, $\mathrm{r}=1.5 \times 10^{11} \mathrm{~m}$ Let the distance parallax angle be $1^{\prime \prime}=$ $4.847 \times 10^{-6} \mathrm{rad}$. Let the distance of the star be D. Parsec is defined as the distance at which the average radius of the Earth's orbit subtends an angle of 1”.
$\therefore$ We have $\theta=\frac{\text{r}}{\text{D}}$ $\text{D}=\frac{\text{r}}{\theta}=\frac{1.5\times10^{11}}{4.847\times10^{-6}}$ $=0.309\times10^{-6}\approx3.09\times10^{16}\text{m}$ Hence, 1 parsec $\approx3.09\times10^{16}\text{m}.$
View full question & answer→Question 903 Marks
An experiment measured quantities a, b, c and then x is calculated by using the relation $\text{x}=\frac{\text{ab}^2}{\text{c}^3}$ If the percentage errors in measurements of a, b and care $\pm1\%, \pm2\% \text{ and } \pm 1.5\%$ respectively, then calculate the maximum percentage error in value of x obtained.
AnswerGiven: $\text{x}=\frac{\text{ab}^2}{\text{c}^3}$ $\therefore\Big(\frac{\Delta\text{x}}{\text{x}}\Big)_{\text{max}}=\frac{\Delta\text{a}}{\text{a}}+2\frac{\Delta\text{b}}{\text{b}}+3\frac{\Delta\text{c}}{\text{c}}$ But $\frac{\Delta\text{a}}{\text{a}}=\pm1\%,\frac{\Delta\text{b}}{\text{b}}=\pm2\%$ And $\frac{\Delta\text{c}}{\text{c}}=\pm1.5\%$ $\therefore\Big(\frac{\Delta\text{x}}{\text{x}}\Big)_{\text{max}}=1\%+2\times2\%+3\times1.5\%$ $=(1+4+4.5)\%=9.5\%$
View full question & answer→Question 913 Marks
It is known that the period T of a magnet of magnetic moment M vibrating in a uniform magnetic field of intensity B depends upon M, B and I where I is the moment of inertia of the magnet about its axis of oscillations. Show that: $\text{T}=2\pi\sqrt{\frac{\text{I}}{\text{MB}}}$
AnswerWe first note that the dimension of I are [$ML^2$]. Also the magnetic moment has the units $Am^2$ so that its dimensions can be written as [$AL^2$] where A stands for the dimensions of the electric current. Finally the magnetic field vector B has the units newton (per ampere metre) so that its dimensions can be written as: $[\text{B}]=\frac{[\text{MLT}^{-2}]}{[\text{A}][\text{L}]}=[\text{MT}^{-2}\text{A}^{1}]$ We know assume that: $\text{T}=\text{k}\text{ l}^{\text{a}}\text{ M}^{\text{b}}\text{ B}^{\text{c}}$ Substituting dimensions of all the quantities involved, We have: $[\text{T}]=[\text{ML}^2]^{\text{a}}[\text{AL}^2]^{\text{b}}[\text{MT}^{-2}\text{A}^{-1}]^{\text{c}}$
$=[\text{M}^{\text{a}+\text{c}}\text{L}^{\text{2a}+\text{2b}\text{T}^{-2\text{c}}}\text{A}^{\text{b}-\text{c}}]$ Equating powers of M, L, T and A on both sides, we have a + c = 0, 2(a + b) = 0, -2c = 1, b - c = 0. From the first three equations we get $\text{c}=\frac{-1}{2}, \text{a}=\frac{1}{2}$ and $\text{b}=\frac{-1}{2}$ These values are consistent with the fourth equations. Thus: $\text{a}=\frac{1}{2},\text{b}=\frac{-1}{2}$ and $\text{c}=\frac{-1}{2}$
$\therefore\text{T}=\text{k }\text{l}^{\frac{1}{2}}\text{ M}^{\frac{-1}{2}}=\text{k}\sqrt{\frac{\text{I}}{\text{MB}}}$ Experiments show that $\text{k}=2\pi$ Therefore $\text{T}=2\pi\sqrt{\frac{\text{I}}{\text{MB}}}$
View full question & answer→Question 923 Marks
Show dimensionally that the frequency n of transverse waves in a string of length / and mass, per unit length z under a tension T is given by n = $\frac{\text{K}}{l}\sqrt{\frac{\text{T}}{\text{m}}}.$
AnswerLet us suppose that: $\text{n}= \text{Kl}^\text{a}\text{T}^\text{b}\text{m}^\text{c}$ where K is a dimensionless constant and a, b, c arc unknown powers to be found. Writing the dimension of all the quantities involved, we get $[\text{T}^{-1}] = [\text{L}] ^\text{a}[\text{MLT}^{-2}]^\text{b}[\text{ML}^{-1}]^\text{c}$ Or $[\text{T}^{-1}]= $ L, M and T on both sides we have $\text{a + b} -\text{c}=0,\text{b + c = 0 } \text{and }\text{a }= -1$ On simplifying , we get $\text{b}= \frac{1}{2}, \text{c}= -\frac{1}{2}, \text{a}= -\text{1}$ $\therefore \text{n}= \text{Kl}^{-1}\text{T}\frac{1}{2}\text{m}\frac{-1}{2}$ $\text{or }\text{n}= \frac{\text{K}}{\text{l}}\sqrt{\frac{\text{T}}{\text{m}}}$
View full question & answer→Question 933 Marks
The frequency $'f\ '$ of vibration of a stretched string depends upon:
- Its length
- The mass per unit length $'m\ '$
- The Tension $'T\ '$ in the string.
- Obtain dimensionally an expression for frequency $'f\ ’.$
AnswerLet us suppose that $\mathrm{f}=\mathrm{Kl}^{\mathrm{a}} \mathrm{T}^{\mathrm{b}} \mathrm{m}^{\mathrm{c}}$ where $K$ is a dimensionless constant and $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are unknown powers to be determined. Writing the dimension of all the quantities involved, We get : Comparing powers $\mathrm{L}, \mathrm{M}$ and $T$ on both sides, we have $a+b-c=0, b+c=0$ and $-2 b=-1$ On simplifying,
We get: $\text{B}=\frac{1}{2},\text{c}=-\frac{1}{2},\text{a}=-1$
$\therefore\text{f}=\text{Kl}^{-1}\text{T}^{\frac{1}{2}}\text{m}^{\frac{-1}{2}}$ $\Rightarrow\text{f}=\frac{\text{K}}{\text{L}}\sqrt{\frac{\text{T}}{\text{m}}}$
View full question & answer→Question 943 Marks
A function $\text{f}(\theta)$ is defined as: $\text{f}(\theta)=1-\theta+\frac{\theta^2}{2!}-\frac{\theta^3}{3!}+\frac{\theta^4}{4!}$ Why is it necessary for q to be a dimensionless quantity?
Answer$\theta$ is represented by angle which is equal to $\frac{\text{arc}}{\text{radius}}$ so angle $\theta$ is dimensionless physical quantity. First term is 1 which is dimensionless, next term contain only powers of $\theta$, as $\theta$ is dimensionless so their powers will also be dimensionless. Hence, each term in R.H.S. expression are dimensionless so left hand side $\text{f}(\theta)$ must be dimensionless.
View full question & answer→Question 953 Marks
Consider a sunlike star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be $\Big(\frac{1}{2}\Big)^0$ from the earth. Due to atmospheric fluctuations, eye can’t resolve objects smaller than 1 arc minute.
AnswerSun's angular diameter from the earth is $\Big(\frac{1}{2}\Big)^\circ$ at 1AU. Angular diameter of the sun like star at a distance of 2 parsecs $=\frac{\Big(\frac{1}{2}\Big)^\circ}{2\times2\times10^5}=\Big(\frac{1}{8}\times10^{-5}\Big)^\circ$ $=\Big(\frac{1}{8}\times10^{-5}\Big)^\circ\times60'=7.5\times10^{-5}\text{arcmin}$ When the sun like star is seen through a telescope with magnification 100, the angular diameter of the star. $=100\times7.5\times10^{-5}=7.5\times10^{-3}\ \text{arcmin}$ But eye cannot resolve smaller than 1 arcmin due to atmospheric fluctuations. So angular size of sun like star appears as 1 arcmin.
View full question & answer→Question 963 Marks
The speed of light in air is $3.00 \times 10^8 \mathrm{~ms}^{-1}$. The distance travelled by light in one year (i.e., 365 days $=3.154 \times 10^7 \mathrm{~s}$ ) is known as light year. A student calculates one light year $=9.462 \times 10^{15} \mathrm{~m}$. Do you agree with the student? If not, write the correct value of one light year.
AnswerOne light year $=$ speed $\times$ time $=9.462 \times 10^{15} \mathrm{~m}$. When two physical quantities are multiplied, the significant figures retained in the final result should not be greater than the least number of significant figures in any of the two quantities. Since, in this case significant figures in one quantity $\left(3.00 \times 10^8 \mathrm{~ms}^{-1}\right)$ are 3 and the significant figures in the other quantity ( $\left.3.154 \times 10^7 \mathrm{~s}\right)$ are 4 , therefore, the final result should have 3 significant figures. Thus, the correct value of one light year $=9.46 \times 10^{15} \mathrm{~m}$.
View full question & answer→Question 973 Marks
The diameter of a wire as measured by a screw gauge was found to be $1.328, 1.330, 1.325, 1.334$ and $1.336\ cm$. Calculate
- Mean value of diameter.
- Absolute error in each measurement.
- Mean absolute error.
- Fractional error.
- Percentage error.
- Diameter of wire.
Answer
- Mean value of diameter,
$\text{D}_\text{m}=\frac{1.328+1.330+1.325+1.326+1.334+1.336}{6}$
$=\frac{7.979}{6}=13298=1.330$
$[$rounding off to three decimal places$]$
- Absolute error in different observations are:
$\Delta\text{D}_1=|1.330-1.328|=0.002\text{cm}$
$\Delta\text{D}_2=|1.330-1.330|=0\text{cm}$
$\Delta\text{D}_3=|1.330-1.325|=\pm0.005\text{cm}$
$\Delta\text{D}_4=|1.330-1.326|=0.004\text{cm}$
$\Delta\text{D}_5=|1.330-1.334|=\pm0.004\text{cm}$
$\Delta\text{D}_6=|1.330-1.336|=\pm0.006\text{cm}$
- Mean absolute error
$\Delta\text{D}_{\text{mean}}=\frac{|\Delta\text{D}_1|+|\Delta\text{D}_2|+|\Delta\text{D}_3|+|\Delta\text{D}_\text{4}|+|\Delta\text{D}_5|+|\Delta\text{D}_6|}{6}$
$=\frac{0.002+0+0.005+0.004+0.004+0.004+0.006}{6}$
$=\frac{0.021}{6}=0.0035=0.004 ($rounding off to $3$ decimal places$)$
- Fractional error $=\frac{\Delta\text{D}_{\text{mean}}}{\text{D}}=\pm\frac{0.004}{1.330}=\pm0.003$
- Percentage error $=\pm0.003\times100\%=\pm0.3\%$
- Diameter of wire $=(1.330\pm0.03)\text{cm}$
Or $\text{D}=1.330\text{cm}\pm0.3\%$ View full question & answer→Question 983 Marks
$\text{P.A.M}$. Dirac, a great physicist of $20^{th}$ century found that from the following basic constants, a number having dimensions of time can be constructed:
- Charge on electron $(e)$
- Permittivity of free space $(e)$
- Mass of electron $(m)$
- Mass of proton $(m)$
- Speed of light $(c)$
- Universal gravitational constant $(G)$
Obtain Dirac's number, given that the desired number is proportional to $\text{m}_\text{p}^{-1}$ and $\text{m}^{-2}_\text{e}$ What is the significance of this number? AnswerLet $X$ be the desired number,
Then: $\text{X}=\text{k}\text{ e}^{\text{u }}\in_0^{\text{x}}\text{ m}^{-2}_\text{e}\text{ m}^{-1}_{\text{p}}\text{ c}^{\text{y}}\text{ G}^{\text{z}}\dots(\text{i})$
Here $k$ is a dimensionless constant and $x, y, z$ and $u$ are unknowns,
whose value is to be obtained from the principle of homogeneity of dimensions.
Now $[\text{X}]=[\text{M}^0\text{L}^0\text{T}^1\text{Q}^0]$
$[\text{e}]=[\text{M}^0\text{L}^0\text{T}^0\text{Q}^1]$
$[\in_0]=[\text{M}^{-1}\text{L}^{-3}\text{T}^2\text{Q}]$
$[\text{C}]=[\text{M}^0\text{L}\text{T}^{-1}]$
$[\text{G}]=[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$
Substituting dimensions of parameters involved in equation $(i),$
We get: $[\text{M}^0\text{L}^0\text{T}^1\text{Q}^0]$
$=\text{Q}^{\text{u}}[\text{M}^{-1}\text{L}^{-3}\text{T}^2\text{Q}^2]^{\text{x}}\text{M}^{-2}\text{M}^{-1}[\text{LT}^{-1}]^{\text{y}}[\text{M}^{-1}\text{L}^3\text{T}^{-2}]\text{z}$
$=[\text{M}^{-\text{x}-3-\text{z}}\text{L}^{-3\text{x}+\text{y}+\text{3z}}\text{T}^{2\text{x}-\text{y}+\text{2z}}\text{Q}^{\text{u}+2\text{x}}]$
From the principle of homogeneity of dimensions $-\text{x} - 3 - \text{z} = 0 ... (\text{ii})$
$- 3\text{x} + \text{y} + 3\text{z} = 0 \dots (\text{iii})$
$2\text{x} - \text{y} - 2\text{z} = 1\dots(\text{iv})$
$\text{u} + 2\text{x} = 0\dots(\text{v})$ Solving equations. $(ii), (iii), (iv)$ and $(v)$
We get: $\text{u}=+4,\text{x}=-2\text{ y}=-3\text{ z}=-1$
$\therefore\text{x}=\text{ke}^{4}\in^{-2}_0\text{ m}_\text{e}^{-2}\text{ m}_{\text{p}}^{-1}\text{ c}^{-3}\text{G}^{-1}$ Or $\text{x}=\frac{\text{ke}^4}{\in^2_0\text{ m}^2_\text{e}\text{ m}_{\text{p}}\text{ c}^3\text{G}}$
Experiments show that, $\text{k}=\frac{1}{16\pi^2}$
Hence, $\text{x}=\frac{\text{e}^4}{16\pi^2\in^2_0\text{ m}^2_\text{e}\text{ m}_\text{p}\text{c}^3\text{G}}$
View full question & answer→Question 993 Marks
The photograph of a house occupies an area of $1.75cm^2$ on a $35mm$ slide. The slide is projected on to a screen, and the area of the house on the screen is $1.55m^2$. What is the linear magnification of the projector-screen arrangement.
AnswerArea of the house on the slide $=1.75 \mathrm{~cm}^2$ Area of the image of the house formed on the screen $=1.55 \mathrm{~m}^2=1.55 \times$ $10^4 \mathrm{~cm}^2$ Arial magnification, $\mathrm{m}_{\mathrm{a}}=$ Area of Image/Area of Object $=(1.55 / 1.75) \times 10^4 $
$\therefore$ Linear magnifications, $\mathrm{m}_l=$ underroot $\mathrm{m}_{\mathrm{a}}=\sqrt{\frac{1.55}{1.75} \times 10^4}=94.11$
View full question & answer→Question 1003 Marks
The earth-moon distance is about $60$ earth radius. What will be the diameter of the earth (approximately in degrees) as seen from the moon?
AnswerAs the distance between moon and earth is greater than radius of earth, then radius of earth can be treated as an arc. 
According to the problem, $R_E$ = length of arc Distance between moon and earth = $60R_E$ So, angle subtended at distance r due to an arc of length l is $\theta_\text{E}=\frac{\text{l}}{\text{r}}=\frac{2\text{R}_\text{E}}{60\text{R}_\text{E}}=\frac{1}{30}\text{rad}$ $=\frac{1}{30}\times\frac{180^\circ}{\pi}\text{degree}=\frac{6^\circ}{3.14}\text{degree}=1.9^\circ\approx2^\circ$ Hence, angle subtended by diameter of the earth $2\theta=2^\circ.$ View full question & answer→