Question 13 Marks
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 total mass of rain-bearing clouds over India during the Monsoon.
AnswerDuring monsoons, a metrologist records about 215 cm of rainfall in India i.e., the height of water column, $\mathrm{h}=215 \mathrm{~cm}$ $=2.15 \mathrm{~m}$ Area of country, $\mathrm{A}=3.3 \times 10^{12} \mathrm{~m}^2$
Hence, volume of rain water, $\mathrm{V}=\mathrm{A} \times \mathrm{h}=7.09 \times 10^{12} \mathrm{~m}^3$ Density of water, $\mathrm{p}=1 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$
Hence, mass of rain water $=\mathrm{p} \times \mathrm{V}=7.09 \times 10^{15} \mathrm{~kg}$ Hence, the total mass of rain-bearing clouds over India is approximately $7.09 \times 10^{15} \mathrm{~kg}$
View full question & answer→Question 23 Marks
Explain this common observation clearly: If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).
AnswerLine of sight is defined as an imaginary line joining an object and an observer’s eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly. On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.
View full question & answer→Question 33 Marks
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 number of air molecules in your classroom.
AnswerLet the volume of the room be V. One mole of air at NTP occupies 22.4 li.e., $22.4 \times 10^{-3} \mathrm{~m}^3$ volume. Number of molecules in one mole $=6.023 \times 10^{23} \therefore$ Number of molecules in room of volume V
$=6.023 \times 10^{23} / 22.4 \times 10^{-3} \mathrm{~V}=134.915 \times 10^{26} \mathrm{~V}=1.35 \times 10^{28} \mathrm{~V}$
View full question & answer→Question 43 Marks
The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107K, and its outer surface at a temperature of about 6000K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data : mass of the Sun = 2.0 × 1030kg, radius of the Sun = 7.0 × 108m.
AnswerMass of the Sun, $M = 2.0 \times 10^{30}kg$ Radius of the Sun, $R = 7.0 \times 10^8m$
Density, $\rho= ?$
$\rho=\frac{\text{mass}}{\text{volume}}=\frac{\text{M}}{\frac{4}{3}\pi\text{R}^3}=\frac{3\text{M}}{4\pi\text{R}^3}=\frac{3\times2.0\times10^{30}}{4\times3.14(7\times10^8)^3}$
$=1.392\times10^3\text{Kg/m}^3$
The density of the Sun is in the density range of solids and liquids. This high density is attributed to the intense gravitational attraction of the inner layers on the outer layer of the Sun.
View full question & answer→Question 53 Marks
A calorie is a unit of heat (energy in transit) and it equals about $4.2 J$ where $1J = 1kg m^2 s^{–2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha\text{ kg},$ the unit of length equals $\beta\text{ m},$ the unit of time is $\gamma\text{ s}.$ Show that a calorie has a magnitude $4.2\ \alpha^{-1}\ \beta^{-2}\ \gamma^2$ in terms of the new units.
AnswerGiven that, 1 Calorie =$4.2 J = 4.2Kg m^2 s^{-2} …… (i)$ As new unit of mass $\alpha\text{ kg}$$\therefore\ 1\text{kg}=1/\alpha$ new unit of mass
Similarly, $1\text{m}=\beta^{-1}$ new unit of length$1\text{s}=\gamma^{-1}$ new unit of time
Putting these values in (i), we get 1 calorie = 4.2 ($\alpha^{-1}$ new unit of mass)($\beta^{-1}$ new unit of length)$^2$($\gamma^{-1}$ new unit of time)$^{-2}=4.2\ \alpha^{-1}\beta^{-2}\ \gamma^{2}$ new unit of energy (Proved)
View full question & answer→Question 63 Marks
A physical quantity P is related to four observables a, b, c and d as follows:$\text{P}=\text{a}^3\text{b}^3/\big(\sqrt{\text{c}}\text{ d}\big)$
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?
Answer$\text{P}=\frac{\text{a}^3\text{b}^2}{(\sqrt{\text{c}}\text{ d})}$Maximum fractional error in P is given by
$\frac{\Delta\text{P}}{\text{P}}=\pm\Big[3\frac{\Delta\text{a}}{\text{a}}+2\frac{\Delta\text{b}}{\text{b}}+\frac{1}{2}\frac{\Delta\text{c}}{\text{c}}+\frac{\Delta\text{d}}{\text{d}}\Big]\\=\pm\Big[3\Big(\frac{1}{100}\Big)+2\Big(\frac{3}{100}\Big)+\frac{1}{2}\Big(\frac{4}{100}\Big)+\frac{2}{100}\Big]$
$=\pm\frac{13}{100}=\pm0.13$
Percentage error in $\text{P}=\frac{\Delta\text{P}}{\text{P}}\times100=\pm0.13\times100=\pm13\%$
Percentage error in P = 13%
Value of P is given as 3.763.
By rounding off the given value to the first decimal place, we get P = 3.8.
View full question & answer→Question 73 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{l} \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 83 Marks
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 $\left(\mathrm{d}_2\right)$ 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)$
View full question & answer→Question 93 Marks
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 $=1450 \mathrm{~m} / \mathrm{s}$ Time lag between transmission and reception of Sonar waves $=77 \mathrm{~s}$ 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$ $\times 77=38.5 \mathrm{~s}$
$\therefore$ Distance between the ship and the submarine $(S)=1450 \times 38.5=55825 \mathrm{~m}=55.8 \mathrm{~km}$
View full question & answer→Question 103 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 10^6km = 824.7 \times 10^9m$ Angular diameter $= 35.72^“ = 35.72 \times 4.874 \times 10^{-6}$ rad Diameter of Jupiter = 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}$
View full question & answer→Question 113 Marks
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 123 Marks
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 133 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 143 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}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 be1" $=$ $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^{\prime \prime} \therefore$ We have $\theta=\frac{\mathrm{I}}{\mathrm{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 153 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}\times10^4}=94.11$
View full question & answer→Question 163 Marks
Find the dimensions of the following quantities
- Acceleration.
- Angle.
- Density.
- Kinetic energy.
- Gravitational constant.
- Permeability.
Answer
- Acceleration $=\frac{\text{Velocity}}{\text{Time}}$
$\therefore[\text{Acceleration}]=\frac{[\text{Velocity}]}{[\text{Time}]}=\frac{[\text{LT}^{-1}]}{[\text{T}]}=[\text{LT}^{-2}]$
- $\text{Angle}=\frac{\text{Distance}}{\text{Distance}}$
$\therefore$ angle is dimensionless
- $\text{Density}=\frac{\text{Mass}}{\text{Volume}}$
$\therefore[\text{Density}]=\frac{[\text{Mass}]}{[\text{Volume}]}$
$\therefore[\text{Density}]=\frac{[\text{Mass}]}{[\text{Volume}]}=\frac{[\text{M}]}{[\text{L}^3]}=[\text{ML}^{-3}]$
- Kinetic energy $=\frac{1}{2}\text{Mass}\times\text{Velocity}^2$
$\therefore$ [Kinetic energy] $=[\text{Mass}]\times[\text{Velocity}]^2=[\text{ML}^2\text{T}^{-2}]$
- Constant of gravitation occurs in Newton's law of gravitation:
$\text{F}=\text{G}\frac{\text{m}_1\text{m}_2}{\text{d}^2}$
$\therefore[\text{G}]=\frac{[\text{F}][\text{d}^2]}{[\text{m}_1][\text{m}_2]}$
$=\frac{\text{MLT}^{-2}\text{L}^2}{[\text{M}][\text{M}]}=[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$
- Permeability occurs in Ampere's law of force:
$\Delta\text{F}=\mu\frac{(\text{i}_1\Delta\text{l}_1)(\text{i}_2\Delta\text{l}_2)\sin\theta}{\text{r}^2}$
$\therefore[\mu]=\frac{[\Delta\text{F}][\text{r}^2]}{[\text{i}_1\Delta\text{l}_1][\text{i}_2\Delta\text{l}_2]}$
$=\frac{[\text{MLT}^{-2}]}{[\text{AL}][\text{AL}]}=[\text{MLT}^{-2}\text{A}^{-2}]$ View full question & answer→Question 173 Marks
Why length, mass and time are chosen as base quantities in mechanics?
AnswerIn mechanics, length, mass and time are chosen as the base quantities because
- There is nothing simpler to length, mass and time.
- All other quantities in mechanics can be expressed in terms of length, mass and time.
- Length, mass and time cannot be derived from one another.
View full question & answer→Question 183 Marks
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 total mass of rain-bearing clouds over India during the Monsoon.
AnswerDuring monsoons, a metrologist records about 215 cm of rainfall in India i.e., the height of water column, $\mathrm{h}=215 \mathrm{~cm}$ $=2.15 \mathrm{~m}$ Area of country, $\mathrm{A}=3.3 \times 10^{12} \mathrm{~m}^2$ Hence, volume of rain water, $\mathrm{V}=\mathrm{A} \times \mathrm{h}=7.09 \times 10^{12} \mathrm{~m}^3$ Density of water, $\mathrm{p}=1 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ Hence, mass of rain water $=\mathrm{p} \times \mathrm{V}=7.09 \times 10^{15} \mathrm{~kg}$ Hence, the total mass of rain-bearing clouds over India is approximately $7.09 \times 10^{15} \mathrm{~kg}$
View full question & answer→Question 193 Marks
Express the average distance of earth from the sun in:
- Light year.
- Par sec.
AnswerAverage distance of earth from the sun is $(r) \frac{1.496\times10^{11}}{9.46\times10^{15}}=1.58\times10^{-5}\text{ ly}$ Also, $\text{r}=\frac{1.496\times10^{11}}{3.08\times10^{16}}\text{per sec}$ $=4.86\times10^{-6}\text{per sec}$
View full question & answer→Question 203 Marks
A physical quantity Q is given by$\text{Q}= \frac{\text{A}^2\text{B}^\frac{3}{2}}{\text{C}^{+4}\text{D}^\frac{1}{2}}$
The percentage error in A, B, C, D are 1%, 2%, 4%, 2% respectively. Find the percentage error in Q.
Answer$\% \text{ error in} \text{ Q}=2 \Big(\frac{\text{dA}}{A}\times100\Big)+\frac{3}{2}\Big(\frac{\text{dB}}{\text{B}}\times100\Big)$ $+4\Big(\frac{\text{dC}}{\text{C}}\times100\Big)+\frac{1}{2}\Big(\frac{\text{dB}}{\text{D}}\times100\Big) $ $= 2\times1+\frac{3}{2}\times2+4\times4+\frac{1}{2}\times2$ $=2+3+16+1= 22\%$
View full question & answer→Question 213 Marks
E, m, I and G denote energy, mass, angular momentum and gravitational constant respectively. Determine the dimensions of $\frac{\text{El}^2}{\text{m}^5\text{G}^2}$
AnswerDimensions of $\text{E}=[\text{M}\text{L}^2\text{T}^{-2}]$ Dimensions of $\text{l}=[\text{ML}^2\text{T}^{-2}]$ Dimensions of $\text{m}=[\text{M}]$ Dimensions of $\text{G}=[\text{M}^{-1}\text{L}^{3}\text{T}^{-2}]$ $\therefore$ Dimensions of $\frac{\text{El}^2}{\text{m}^5\text{G}^2}$ $=\frac{[\text{ML}^2\text{T}^{-2}]{[\text{ML}^2\text{T}^{-2}]^2}}{[\text{M}]^5[\text{M}^{-1}\text{L}^{3}\text{T}^{2}]^{-2}}=1$ Thus $\frac{\text{El}^2}{\text{m}^5\text{G}^2}$is dimensionless.
View full question & answer→Question 223 Marks
How many astronomical units (A.U.) make 1 parsec?
AnswerAccording to the definition, 1 parsec is equal to the distance at which 1AU long arc subtends an angle of 1s. $\text{But}\ \ 1''=\frac{1}{3600}\times\frac{\pi}{180}\text{rad}$ $\therefore1\text{parsec}=\frac{3600\times180}{\pi}\text{AU}$ $=206265\text{AU}\approx2\times10^5\text{AU}$
View full question & answer→Question 233 Marks
Why do we have different units for the same physical quantity?
AnswerBecause, bodies differ in order of magnitude significantly in respect to the same physical quantity. For example, interatomic distances are of the order of angstroms, inter-city distances are of the order of km, and interstellar distances are of the order of light year.
View full question & answer→Question 243 Marks
Explain this common observation clearly: If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).
AnswerLine of sight is defined as an imaginary line joining an object and an observer’s eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly. On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.
View full question & answer→Question 253 Marks
The wavelength 1 associated with a moving particle depends upon its mass m, its velocity v and Planck's constant h. Show dimensional relation between them.
AnswerSuppose wavelength à associated with a moving particle depends upon (i) its mass (m), (ii) its velocity (v) and (iii) Planck's constant (r)where, k is a dimensionless constant.
Writing dimensions of various terms,
We get:
$[\text{M}^0\text{L}^1\text{T}^0]=[\text{M}]^{\text{a}}[\text{LT}^{-1}]^{\text{b}}[\text{ML}^2\text{T}^{-1}]^{\text{c}}$
$=\text{M}^{\text{a}+\text{c}}\text{L}^{\text{b}+\text{2c}}\text{T}^{-\text{b}-\text{c}}$
Comparing power of M, L and T on two sides of equation,
We have:
$\text{a}+\text{c}=0,\text{b}+\text{2c}=1,-\text{b}-\text{c}=0$
We get:
$\text{a}=-1,\text{b}=-1,\text{c}=+1$
Hence, the relation becomes $\lambda=\frac{\text{kh}}{\text{mv}}$
View full question & answer→Question 263 Marks
The density of a cylindrical rod was measured by using the formula: $\rho=\frac{4\text{m}}{\pi\text{D}^2\text{l}}$
AnswerThe percentage errors in m, D and l are 1%, 1.5% and 0.5%. Calculate the percentage error in the calculated value of density. $\text{Density }\rho=\frac{4\text{m}}{\pi\text{D}^2\text{l}}$ $\therefore\Big(\frac{\Delta\rho}{\rho}\Big)_{\text{max}}=\frac{\Delta\text{m}}{\text{m}}+2\frac{\Delta\text{D}}{\text{D}}+\frac{\Delta\text{l}}{\text{l}}$ But, $\frac{\Delta\text{m}}{\text{m}}=1\%,\frac{\Delta\text{D}}{\text{D}}=1.5\%\text{ and }\frac{\Delta\text{l}}{\text{l}}=0.5\%$ $\therefore$ Maximum percentage error in calculated value of density $\Big(\frac{\Delta\rho}{\rho}\Big)_{\text{max}}=1\%+2\times1.5\%+0.5\%$ $=(1+3+0.5)\%=4.5\%$
View full question & answer→Question 273 Marks
The radius of curvature of a concave mirror measured by spherometer is given by $\text{R}=\frac{\text{I}^2}{6\text{h}}+\frac{\text{h}}{2}.$ The values of land h are 4cm and 0.065cm respectively. Compute the error in measurement of radius of curvature.
AnswerHere l = 4cm, $\Delta\text{I}=0.1\text{cm}$ (least count of the metre scale) Here l is the distance between the legs of the spherometer. h = 0.065cm, Ah = 0.001cm = (least count of the spherometer) $\text{As }\text{R}=\frac{\text{I}^2}{6\text{h}}+\frac{\text{h}}{2}$ Considering the magnitude only, We get: $\frac{\Delta\text{R}}{\text{R}}=2\frac{\Delta\text{I}}{\text{I}}+\frac{\Delta\text{h}}{\text{h}}+\frac{\Delta\text{h}}{\text{h}}$ $=2(\frac{\Delta\text{I}}{\text{I}}+\frac{\Delta\text{h}}{\text{h}})$ $=2\times\frac{0.1}{4}+\frac{2\times0.001}{0.065}$ $=0.05+0.03=0.08\%$
View full question & answer→Question 283 Marks
The sides of a rectangle are $(10.5\pm0.2)\text{cm}$ and $(52\pm0.1)\text{m}$ Calculate its perimeter with error limits.
Answer$\text{Given},\text{l}=(10.5\pm0.2)\text{cm},$ $\text{b}=(5.2\pm0.1)\text{cm}$ Perimeter of a rectangle $\text{p}=2(\text{l}+\text{b})$ $=2(10.5+5.2)=3.14\text{cm}$ $\Delta\text{p}=\pm(\Delta\text{l}+\Delta\text{b})$ $=\pm2(0.2+0.1)=\pm0.6$ Perimeter of a rectangle $=(31.4\pm0.6)\text{cm}$
View full question & answer→Question 293 Marks
The length, breadth and thickness of a rectangular sheet of metal are $4.234m, 1.005m$ and $2.01cm$ respectively. Calculate the surface area and volume of the sheet to correct significant figures.
AnswerLength $(l)=4.234 \mathrm{~m}$ Breadth $(b)=1.005 \mathrm{~m}$ Thickness $(\mathrm{h})=2.01 \mathrm{~cm}=0.0201 \mathrm{~m}$ Surface area of the sheet $=2(\mathrm{lb}+\mathrm{bh}+$
$\mathrm{lh})=2(4.234 \times 1.005+1.005 \times 0.0201+4.234 \times 0.0201)=8.7209478$. As the lowest significant figure in the given measurement is 3 (that of thickness) the volume and area should be expressed in 3 significant figures only. $\therefore$ Surface area $=8.72 \mathrm{~m}^2$ Volume of the sheet $=1 \times \mathrm{b} \times \mathrm{h}=4.234 \times 1.0050 .0201=0.0855289 \mathrm{~m}^3=0.086 \mathrm{~m}^3$ Length (l) $=4.234 \mathrm{~m}$ Breadth $(\mathrm{b})=1.005 \mathrm{~m}$ Thickness $(\mathrm{h})=2.01 \mathrm{~cm}=0.0201 \mathrm{~m}$ Surface area of the sheet $=2(\mathrm{lb}+\mathrm{bh}+\mathrm{lh})=$
$2(4.234 \times 1.005+1.005 \times 0.0201+4.234 \times 0.0201)=8.7209478$. As the lowest significant figure in the given measurement is 3 (that of thickness) the volume and area should be expressed in 3 significant figures only.
$\therefore$ Surface area $=8.72 \mathrm{~m}^2$ Volume of the sheet $=1 \times \mathrm{b} \times \mathrm{h}=4.234 \times 1.0050 .0201=0.0855289 \mathrm{~m}^3=0.086 \mathrm{~m}^3$
View full question & answer→Question 303 Marks
If the velocity of light (c), the constant of gravitation (G) and Planck's constant (h) be chosen as the fundamental units, find the dimensions of mass, length and time in the new system.
AnswerLet us write the dimensions of c, G and h in terms of M, L and T. $\text{Let}\text{M}=\text{Kc}^a\text{G}^b\text{h}^\text{y},$ where K is constant $\text{[M]}=\text{[LT}^{-1}]^\alpha\text{[M}^{-1}\text{[L}^3\text{T}^{-2}]^\beta\text{[ML}^2\text{T}^{-1}]^\gamma$ $=\text{[M}^{-\beta+\gamma}\text{L}^{\alpha+3\beta+2\gamma}\text{T}^{-\alpha-2\gamma}]$ Comparing the powers of M, L and T on both the sides, we have: $-\beta+\gamma=1$ $\alpha+3\beta+2\gamma=0$ $-\alpha-2\beta-\gamma=0$ Solving these equations, we get, $\alpha=\frac{1}{2},\beta=\frac{-1}{2},\gamma=\frac{1}{2}$ $\therefore \text{M}=\text{Kc}^\frac{1}{2}\text{G}^\frac{1}{2}\text{h}^\frac{1}{2}$ Taking$\text{K}=\text{I},$ we can write $\text{M}=_\text{c}^\frac{1}{2}\text{G}^{\frac{-1}{2}}\text{h}^\frac{1}{2}$ Similary,we can prove that. $\text{L}=_\text{c}^\frac{-3}{2}\text{G}^\frac{1}{2}\text{h}^\frac{1}{2}$ $\text{T}=_\text{c}^\frac{-5}{2}\text{h}^\frac{1}{2}\text{G}^\frac{1}{2}$
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For a glass prism of refracting angle 60°, the minimum angle of deviation $D_m$ is found to be 36° with a maximum error of 1.05°. When a beam of parallel light is incident on the prism, find the range of experimental value of refractive index $'μ'.$ It is known that the refractive index $'μ'$ of the material of the prism is given by: $\mu=\frac{\sin\Big(\frac{\text{A}+\text{D}_\text{m}}{2}\Big)}{\sin\frac{\text{A}}{2}}$
AnswerError in calculated $D_m$ is $\pm1.05^{\circ}$ Given $\mu=\frac{\sin\Big(\frac{\text{A}+\text{D}_\text{m}}{2}\Big)}{\sin\frac{\text{A}}{2}}$ $\mu=\frac{\sin\Big(\frac{36^{\circ}\pm1.05}{2}}{\sin\Big(\frac{60}{2}\Big)}=\frac{\sin\Big(\frac{37.05}{2}\Big)}{\sin30^{\circ}}$ $=\frac{\sin(18.525)}{\frac{1}{2}}\text{or}\frac{\sin(17.475^{\circ})}{\frac{1}{2}}$
$\Rightarrow2\times0.755\text{ or }2\times0.73$
$\Rightarrow1.51\text{ or }1.46$ Here range of $\mu$ is $1.46\leq\mu\leq1.51$
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By using the method of dimension, check the accuracy of the following formula : $ \text{T}=\frac{\text{rh}\rho\text{g}}{2\cos\theta},$where T is the surface tension, h is the height of the liquid,$\rho$ is the density of the liquid, g acceleration due to gravity $\theta$ angle of contact, and r is the radius of the tube.
AnswerIn order to find out the accuracy of the given equation we shall compare the dimensions of T and $\frac{\text{rh}\rho\text{g}}{2\cos\theta}$ $\text{T}=\frac{\text{force}}{\text{length}}=\frac{[\text{MLT}^{-2}]}{[\text{L}]}$ Dimension of $\frac{\text{rh}\rho\text{g}}{2\cos\theta}=[\text{L][L][ML}^{-3}][\text{LT}^{-2}]$ $(2\cos\theta$ has no dimension$)$The dimensions of both the sides are the same and hence the equation is correct.
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Check by the method of dimensional analysis whether the following relations are correct. $\text{v}=\sqrt{\frac{\text{P}}{\text{D}}}$ where $v =$ velocity of sound and $P =$ pressure, $D =$ density of medium $\text{n}=\frac{1}{21}\sqrt{\frac{\text{F}}{\text{m}}},$ where $n =$ frequency of vibration $l =$ legnth of the string $F =$ Stretching force $m =$ mass per unit length of the string.
Answer
- $[\text{R.H.S}]=\sqrt{\frac{[\text{P}]}{\text{[D]}}}=\sqrt{\frac{[\text{ML}^{-1}\text{T}^{-2}]}{[\text{ML}^{-3}]}}=[\text{LT}^{-1}]$
$[\text{L.H.S}]=[\text{v}]=[\text{LT}^{-1}]$
$[\text{R.H.S.}]=[\text{L.H.S.}]$
Hence, the relation is correct.
- $[\text{R.H.S.}]=\frac{1}{[\text{l}]}\sqrt{\frac{[\text{F}]}{[\text{m}]}}$
$=\frac{1}{\text{L}}\sqrt{\frac{\text{MLT}^{-2}}{\text{ML}^{-1}}}=\frac{1}{\text{L}}[\text{LT}^{-1}]=[\text{T}^{-1}]$
$[\text{L.H.S.}]=\Big[\frac{1}{\text{Time}}\Big]=\frac{1}{\text{T}}=[\text{T}^{-1}]$
Hence, the relation is correct. View full question & answer→Question 343 Marks
A planet moves around the sun in a circular orbit. The time period of revolution $T$ of the planet depends on
- Radius of the orbit $(R)$.
- Mass of the sun $M$.
- Gravitational constant $G$.
Show dimensionally that $\text{T}^2\propto\text{R}^3$ Answer$\text{T}\propto\text{R}^\text{a}\text{M}^\text{b}\text{G}^\text{c}$ $\text{Or }\text{T}=\text{KR}^\text{a}\text{M}^\text{b}\text{G}^\text{c},$Where is constant substituting the dimension on both sides, we have $[\text{T]}=[\text{L]}^\text{a}[\text{M]}^\text{b}[\text{M}^{-1}\text{L}^3\text{T}^{-2}]^\text{c}$ $[\text{M}^o\text{L}^o\text{T}]=[\text{M}^\text{b-c}\text{L}^\text{a+3c}\text{T}^\text{-2c}]$ Comparing the power of $M, L$ and $T$, We get: $\text{b}-c=0$ $\text{a+3c=0}$ $-2\text{c=1}$ On solving, We get: $\text{a}=\frac{3}{2},\text{b}=\frac{-1}{2},\text{c}=\frac{-1}{2}$ $\text{T}=\text{KR}^\frac{3}{2}\text{M}^\frac{1}{2}\text{G}^{\frac{-1}{2}}$ $\text{T}\propto\text{R}^\frac{3}{2}$ $\text{T}^2\propto\text{R}^3$
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The volume of a liquid flowing out per second of a pipe of length l and radius r is written by a student as, $\text{v}=\frac{\pi}{8}\frac{\text{Pr}^4}{\eta\text{l}}$ where P is the pressure difference between the two ends of the pipe and $\eta$ is coefficent of viscosity of the liquid having dimensional formula $ML^{-1} T^{-1}$. Check whether the equation is dimensionally correct.
AnswerIf dimensions of LHS of an equation is equal to dimensions of RHS, then equation is said to be dimensionally correct. According to the problem, the volume of a liquid flowing out per second of a pipe is given by $\text{V}=\frac{\pi}{8}\frac{\text{pr}^2}{\eta\text{l}}$ (where, V = rate of volume of liquid per unit time) Dimension of given physical quantities, $[\text{V}]=\frac{\text{Dimension of volume}}{\text{Dimension of time}}=\frac{[\text{L}^3]}{[\text{T}]}=[\text{L}^3\text{T}^{-1}],[\text{p}]=[\text{ML}^{-1}\text{T}^{-2}],$ $[\eta]=[\text{ML}^{-1}\text{T}^{-1}],[\text{l}]=[\text{L}],[\text{r}]=[\text{L}]$
$\text{LHS}=[\text{V}]=\frac{[\text{L}^3]}{[\text{T}]}=[\text{L}^3\text{T}^{-1}]$
$\text{RHS}=\frac{[\text{ML}^{-1}\text{T}^{-2}]\times[\text{L}^4]}{[\text{ML}^{-1}\text{T}^{-1}]\times[\text{L}]}=[\text{L}^3\text{T}^{-1}]$ Dimensionally, L.H.S. = R.H.S. Therefore, equation is correct dimensionally.
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Compute the following with regards to significant figures.
- $4.6\times0.128$
- $\frac{0.9995\times1.53}{1.592}$
- $876+0.4382$
Answer
- $4.6 \times 0.128 = 0.5888 = 0.59$
The result has been rounded off to have two significant digits $($as in $4.6)$
- $\frac{0.9995\times1.53}{1.592}=0.96057=0.961$
- $876 + 0.4382 = 876.4382 = 876 .$
As, there is no decimal point in $876$, therefore, result of addition has been rounded off to no decimal point. View full question & answer→Question 373 Marks
The radius of curvature of a concave mirror measured by spherometer is $\text{R}=\frac{\text{l}^2}{6\text{h}}+\frac{\text{h}}{2}$ The values of 1 and h are 4cm and 0.065cm respectively. Compute the error in measurement of radius of curvature.
AnswerWe are given: $\text{l}=4\text{cm},\Delta\text{l}=0.1\text{cm}$ (least count of the metre scale) here l is the distance between the legs of the spherometer. $\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}}$ Considering the magnitudes only, We get: $\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}}$ $=2\Big(\frac{\Delta\text{l}}{\text{l}}+\frac{\Delta\text{h}}{\text{h}}\Big)$ $=2\times\frac{0.1}{4}+\frac{2\times0.001}{0.065}=0.05+0.03=0.08.$
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Some measurements are taken for the period of oscillations of a simple pendulum. In successive measurements, the readings turn out to be $2.63s, 2.56s, 2.42s, 2.71s$ and $2.80s$. Calculate the absolute errors, relative error and percentage error.
Answer
- Mean value of time period,
$\text{T}=\frac{2.63+2.56+2.42+2.71+2.80}{5}$
$=2.62\text{sec}$
- Absolute error in time period
$\Delta\overline{\text{T}}=\frac{0.01+0.06+0.20+0.09+0.18}{5}=0.11\text{ sec}$
- Relative error, $\frac{\Delta\overline{\text{T}}}{\overline{\text{T}}}=\frac{0.11}{2.62}=0.04$
Percentage error, $\frac{\Delta\text{T}}{\text{T}}\times100=0.04\times100=4\%$ View full question & answer→Question 393 Marks
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 number of air molecules in your classroom.
AnswerLet the volume of the room be V . One mole of air at NTP occupies 22.4 I i.e., $22.4 \times 10^{-3} \mathrm{~m}^3$ volume. Number of molecules in one mole $=6.023 \times 10^{23} \therefore$ Number of molecules in room of volume $\mathrm{V}=6.023 \times 10^{23} / 22.4 \times 10^{-3} \mathrm{~V}=$ $134.915 \times 10^{26} \mathrm{~V}=1.35 \times 10^{28} \mathrm{~V}$
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Given that the time period T of oscillation of a gas bubble from an explosion under water depends upon P, d and E where P is the static pressure, d the density of water and E is the total energy of explosion, find dimensionally a relation for T.
AnswerWe are given that: T = f(P, d, E) Assuming that $T = kP^ad^bE^c$ and substituting dimensions of all the quantities involved, We have:$[\text{T}]=[\text{M}\text{L}^{-1}\text{T}^{-2}]^{\text{a}}[\text{ML}^{-3}]^{\text{b}}[\text{M}\text{L}^2\text{T}^{-2}]^{\text{c}}$
Equating powers of M, L and T on both sides, We have:$\text{a}+\text{b}+\text{c}=0$
$-\text{a}-3\text{b}+2\text{c}=0$
$-2\text{a}-2\text{c}=1$
$\text{a}=\frac{-5}{6}\text{, b}=\frac{1}{2}\text{, c}=\frac{1}{3}$
$\therefore\text{T}=\text{k}\text{ P}^{\frac{-5}{6}}\text{ d}^{\frac{1}{2}}\text{ E}^{\frac{1}{3}}$
Or $\text{T}=\text{k}\Bigg(\frac{\text{d}^{\frac{1}{2}}\text{ E}^{\frac{1}{3}}}{\text{P}^{\frac{5}{6}}}\Bigg)$
This os the required relation for T.
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State the principle of homogeneity of dimensions. Test the dimensional homogeneity of the following equation: $\text{h}=\text{h}_0+\text{v}_0\text{t}+\frac{1}{2}\text{gt}^2.$
AnswerPrinciple of Homogenity: All terms of any physical relation must have the same dimensions. $[\text{h}] = \text{L}$$[\text{h}_0] = \text{L}$
$[\text{v}_0\text{t}] = \text{LT}^{-1} \text{T} = \text{L}$
$[\frac{1}{2}\text{gt}^2] = \text{LT}^{-2}\text{T}^2 = \text{L}$
So, the equation is dimensionally correct.
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Deduce by the method of dimensions, an expression for the energy of a body executing S.H.M. assuming that the energy of the body depends upon (a) the mass m (b) the frequency v and (c) the amplitude of vibration a.
Answer$\text{E}\propto \text{m}^\alpha\text{v}^\beta\alpha^\gamma$ Or $\text{E}= \text{Em}^\alpha\text{v}^\beta \text{a}^\gamma,$where $\text{K}$ is constant Writing the dimensions of both the sides we have $[\text{ML}^2\text{T}^2]= [\text{M}]^\alpha[\text{L}]^\gamma[\text{T}^{-1}]^\beta$. $[\text{ML}^2\text{T}^2]= [\text{M}^\alpha\text{L}^\gamma\text{T}^{-\beta}]$ Comparing the powers $\text{M, L and T,}$ We get: $\alpha = 1, \beta= 2\text{ and }\gamma= 2.$ Putting the values of $\alpha,\ \beta, \ \gamma $ We have $\text{E}=\text{K mv}^2\alpha^2$
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The length, breadth and thickness of a block of metal were measured with the help of a Vernier Callipers. The measurements are: $\text{l}=(5.250\pm0.001)\text{cm}$ $\text{b}=(3.450\pm0.001)\text{cm}$ $\text{t}=(1.740\pm0.001)\text{cm}$ Find the percentage error in volume of the block.
AnswerVolume of the block is given by: $\text{V}=\text{l b t}$ Relative error in the volume of block, $\frac{\Delta\text{V}}{\text{V}}=\frac{\Delta\text{l}}{\text{l}}+\frac{\Delta\text{b}}{\text{b}}+\frac{\Delta\text{t}}{\text{t}}$ $\Delta\text{l}=0.001\text{cm}\text{ l}=5.250\text{cm}$ $\Delta\text{b}=0.001\text{cm},\text{ b}=3.450\text{cm}$ $\Delta\text{t}=0.001\text{cm,}\text{ t}=1.740\text{cm}$ $\therefore\frac{\Delta\text{V}}{\text{V}}=\frac{0.001}{5.250}+\frac{0.001}{3450}+\frac{0.001}{1.740}$ $=0.0019+0.00289+0.00575=0.00954$ $\therefore\text{Error}=\frac{\Delta\text{V}}{\text{V}}\times100\%$ $=0.00954\times100\%=0.9\%\simeq1\%$
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Given that the amplitude of the scattered light is:
- Directly proportional to that of incident light.
- Directly proportional to the volume of the scattering dust particle.
- Inversely proportional to its distance from the scattering particle.
- Dependent upon the wavelength $(\lambda)$ of the light. Show that the intensity of scattered light varies as $1$.
Answer$\therefore$ Let us write $\text{A}_\text{s}=\text{K}\text{A}_\text{i}^\text{a}\text{V}^\text{b}\text{x}^\text{c}\lambda^\text{d}$ It is clear from the statement of the equation. $\text{a}=\text{I}$ $\text{b}=\text{I}$ $\text{c}=\text{I}$ $\therefore\text{A}_\text{s}=\text{KA}^1_\text{i}\text{V}_\text{x}^{1-1}\lambda^\text{d}$ Write the dimensions on both the sides and equatin powrs of $M,l$ and $T$, We have. $\text{[L]}=\text{[L]}\text{[L}^3]\text{[L}^{-1}]\text{[L}^\text{d}]$ $=\text{[L}^{3+\text{d}}]$ $\text{A}_\text{s}\propto\frac{1}{\lambda2}$ But intensity. $\therefore$$(\text{I}_\text{s})\propto\frac{1}{\lambda4}$
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What is meant by parallax? How can we find the distance of a nearby star by parallax method?OR
Given measurement $a_1, a_2, a_3$________ $a_n$. With the help of these, explain absolute error, relative error and percent -age error.
AnswerParallax is the name given to change in the position of an object w.r.t. fixed background, when object is seen from different positions. Suppose A is the position of planet at any time. Let $\angle\text{FAN}=\theta_1$ between direction of light from distant star 'F' and near by star ‘N’. $\angle\text{FAN}=\angle\text{ANS}=\theta_1$ after six months planet is at B, $\angle\text{FBN}=\theta_2$ again $\angle\text{FBN}=\angle\text{BNS}=\theta_2$ From figure $\angle\text{ANB}=\angle\text{ANS}+\angle\text{BNS}$ $=\theta_1+\theta_2$ This is the angle which the nearby star N subtends on orbital diameter of planet. $\text{angle}=\frac{\text{arc}}{\text{radius}}$ $\theta_1+\theta_2=\frac{\text{AB}}{\text{AN}}$ $\text{AB}=2\text{AS}=2\text{AU}$ $=2\times1.5\times10^{11}=3\times10^{11}$ If $(\theta_1+\theta_2)$ is known (radians) then AN can be calculated.OR
Absolute error: Let the physical quantity be measured 'n' times, where $a_1, a_2, ... a_n$, be the measured values. The arithmetic mean.
$\text{a}_{\text{m}}=\frac{\text{a}_1+\text{a}_2+\dots+\text{a}_{\text{n}}}{\text{n}}$ $\text{a}_{\text{m}}=\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}\text{a}_{\text{i}}$ Arithmetic mean am is taken as the best possible/true value of the quantity. By definition of absolute error:$\Delta\text{a}_1=\text{a}_\text{m}-\text{a}_1$
$\Delta\text{a}_2=\text{a}_\text{m}-\text{a}_2$
$ \ \vdots\\\Delta\text{a}_\text{n}=\text{a}_{\text{m}}-\text{a}_\text{n}$
Meanb absolute error: $\Delta\text{a}_{\text{mean}}=\frac{|\Delta\text{a}_1|+|\Delta\text{a}_2|+\dots+[\Delta\text{a}_\text{}\text{n}]}{\text{n}}$ $=\frac{1}{\text{n}}\times\sum\limits^{\text{n}}_{\text{i}=1}|\Delta\text{a}_{\text{i}}|$ Relative error: It is the ratio of mean absolute error to mean value of quantity measured. $\delta_\text{a}=\frac{\text{mean absolute error}}{\text{mean value}}=\frac{\Delta\text{a}_{\text{mean}}}{\text{a}_\text{n}}$when relative/ fractional error is expressed in percentage, we call it percentage error.
$\therefore\text{Percentage error},\delta_\text{a}=\frac{\Delta\text{a}_{\text{mean}}}{\text{a}_\text{m}}\times100\%$
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The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107K, and its outer surface at a temperature of about 6000K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data : mass of the Sun = 2.0 × 1030kg, radius of the Sun = 7.0 × 108m.
AnswerMass of the Sun, $M = 2.0 \times 10^{30}kg$ Radius of the Sun, $R = 7.0 \times 10^8m$
Density, $\rho= ?$
$\rho=\frac{\text{mass}}{\text{volume}}=\frac{\text{M}}{\frac{4}{3}\pi\text{R}^3}=\frac{3\text{M}}{4\pi\text{R}^3}=\frac{3\times2.0\times10^{30}}{4\times3.14(7\times10^8)^3}$
$=1.392\times10^3\text{Kg/m}^3$
The density of the Sun is in the density range of solids and liquids. This high density is attributed to the intense gravitational attraction of the inner layers on the outer layer of the Sun.
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A calorie is a unit of heat (energy in transit) and it equals about $4.2 J$ where $1J = 1kg m^2 s^{–2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha\text{ kg},$ the unit of length equals $\beta\text{ m},$ the unit of time is $\gamma\text{ s}.$ Show that a calorie has a magnitude $4.2\ \alpha^{-1}\ \beta^{-2}\ \gamma^2$ in terms of the new units.
AnswerGiven that, 1 Calorie = $4.2 J = 4.2Kg m^2 s^{-2}$ …… (i) As new unit of mass $\alpha\text{ kg}$ $\therefore\ 1\text{kg}=1/\alpha$ new unit of mass Similarly, $1\text{m}=\beta^{-1}$ new unit of length $1\text{s}=\gamma^{-1}$ new unit of time Putting these values in (i), we get 1 calorie = 4.2 ($\alpha^{-1}$ new unit of mass)($\beta^{-1}$ new unit of length)$^2(\gamma^{-1}$ new unit of time)$^{-2} =4.2\ \alpha^{-1}\beta^{-2}\ \gamma^{2}$ new unit of energy (Proved)
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The two specific heat capacities of a gas are measured as $\text{C}_\text{p} = (12.28 \pm 0.2) $ units and $\text{C}_\text{v} = (3.97 \pm 0.3)$ units. Find the value of the gas constant R.
AnswerHere, $\text{C}_\text{p} = (12.28 \pm 0.2) $ And $\text{C}_\text{v} = (3.97 \pm 0.3)$ We know that: $\text{C}_\text{p}-\text{C}_\text{v}=\text{R}$ $(12.28\pm0.2)\pm(3.97\pm0.3)=\text{R}$ $\Rightarrow(12.28-3.97)\pm(0.2\pm0.3)=\text{R}$ $\Rightarrow(8.31\pm0.5)=\text{R}$ Hence $\text{R}=(8.31\pm0.5)\text{ units}$
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Write the dimensional formula for the following:
- Wein’s constant.
- Planck's constant.
- Specific heat.
- Latent heat.
- Rydberg's constant.
Answer
- $[\text{M}^0\text{LT}^0\text{K}]$
- $[\text{ML}^2\text{T}^{-1}]$
- $[\text{M}^0\text{L}^{2}\text{T}^{-2}\text{K}^{-1}]$
- $[\text{M}^0\text{L}^2\text{T}^{-2}]$
- $[\text{M}^0\text{L}^{-1}\text{T}^0].$
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A physical quantity P is related to four observables a, b, c and d as follows: $\text{P}=\text{a}^3\text{b}^3/\big(\sqrt{\text{c}}\text{ d}\big)$ The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?
Answer$\text{P}=\frac{\text{a}^3\text{b}^2}{(\sqrt{\text{c}}\text{ d})}$ Maximum fractional error in P is given by $\frac{\Delta\text{P}}{\text{P}}=\pm\Big[3\frac{\Delta\text{a}}{\text{a}}+2\frac{\Delta\text{b}}{\text{b}}+\frac{1}{2}\frac{\Delta\text{c}}{\text{c}}+\frac{\Delta\text{d}}{\text{d}}\Big]\\=\pm\Big[3\Big(\frac{1}{100}\Big)+2\Big(\frac{3}{100}\Big)+\frac{1}{2}\Big(\frac{4}{100}\Big)+\frac{2}{100}\Big]$ $=\pm\frac{13}{100}=\pm0.13$ Percentage error in $\text{P}=\frac{\Delta\text{P}}{\text{P}}\times100=\pm0.13\times100=\pm13\%$ Percentage error in P = 13% Value of P is given as 3.763. By rounding off the given value to the first decimal place, we get P = 3.8.
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