In this page we have dimensional analysis practice problems . Hope you like them and do not forget to like , social share
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Question 1 The air bubble formed by explosion inside water perform oscillations with time period T which depends on pressure (p), density (ρ) and on energy due to explosion (E). Establish relation between T, p, E and ρ. Answer
As per the question,time period depends on Pressure,density and energy
$T=C P^a \rho ^b E^c$ where, symbols have their usual meaning and C is a constant
Now we need to find the values of a,b,c by using dimensional analysis
Question 2 The velocity v of a particle depends upon the time ‘t’ according to the equation
Determine the units of a, b, c and d. What physical quantities they represent. All have SI units. Answer
The magnitudes of physical quantities can be added together or subtracted from one another only if they have the same dimensions. So, the dimensions of d will be [T]
Now Velocity has the dimension $[LT^{-1}]$, Now each of the quantity which are being added on the RHS should also have same dimension
so bt should also have $[LT^{-1}]$ dimension, which means b dimension is $[LT^{-2}]$
Now $\frac {c}{d+t}$ should also $[LT^{-1}]$ dimension, which means c dimension is $[L]$
Now $\sqrt {ab}$ should have $[LT^{-1}]$ dimension,which means a dimension is [L]
So a represent length, b represent acceleration, c represent length and d represent time
Question 3 Mass of the Sun is $2.0 \times 10^{30}$ Kg and radius of sun is $7.0 \times 10^{8}$m. Express the density of sun. Also explain why the order of magnitude of density of sun is in the range of solids and not in the range of liquids and gases? Answer
Given that, \(M=2\times 10^{30}kg\,\, R=7\times 10^8m\)
Lets calculate the density based on the given data
\begin{align*}
Density, \,\, \rho \,\,&=\frac{mass}{volume}=\frac{M}{\frac{4}{3}\pi R^3}\\
&=\frac{3M}{4\pi R^3}\\
&=\frac{3\times 2\times 10^{30}}{4\times 3.14\times \left( 7\times 10^8 \right) ^3}\\
&=\text{1.392 }\times 10^3kg/m^3
\end{align*}
Which is the order of density of solids and liquids not gases.High density of sun is because of inward gravitational attractions or the outer layers of the sun.
Question 4 A physical quantity X is connected from $X=\frac{a{{b}^{2}}}{C}$ . Calculate the percentage error in X, when % error in a, b, c are 4, 2, and 3 respectively. Answer
$\frac {\Delta X}{X} =\frac {4}{100} + 2 \frac {2}{100} + \frac {3}{100} $
$\frac {\Delta X}{X} =\frac {11}{100} $
So percentage error is X is 11%
Question 5 Percentage error in the measurement of height and radius of cylinder are $x$ and $y$ respectively. Find the percentage error in the measurement of volume. Which of the two measurements height or radius needs more attention? Answer
Volume is given by
$V= \pi r^2 H$
Error in V is given by
$ \frac {\Delta V}{V} = 2\frac {\Delta r}{r} + \frac {\Delta h}{h}$
Question 6 Orbital radius of mercury around the sun is 0.38 A.U. Find the angle of maximum elongation for mercury and its distance from earth when the elongation is maximum. Answer
A.U stands for Astronomical unit. 1 A.U =$1.496 \times 10^{11} m$
Some Theory around the question. This question is based on Copernicus method.Some point to remember
a. The planets which are closer to the Sun than the Earth are called inferior planets.
b.The Mercury and the Venus are two inferior planets.
c. Copernicus assumed the orbits of inferior planets as perfectly circular.
d. The angle formed at the Earth between the Earth planet direction and the Earth-Sun Direction is called the planet elongation and is denoted by symbol $\theta$
e. The planets elongation changes continuously as planets are moving.
f. When elongation acquires maximum value it appears to be farthest from the sun, At this time the Sun and the Earth subtends an angle of 90° at the planet.
g. Average Distance between Sun and earth is 1 A.U
Question 7 Convert 4.29 light years into parsecs. Calculate the parallax of a star at this distance when viewed from two locations of earth six months apart in its orbit around the sun. Answer
Lets first calculate the light year
Now we know that distance travelled by light in 1 year is called 1 light year
Speed of light = $ 3 \times 10^8$ m/s and $1 year =365 \times 24 \times 60 \times 60$ s
So 1 light year = $3 \times 10^8 \times 365 \times 24 \times 60 \times 60 $ m
\( 1light\,\,year=9.46\times 10^{15}m\)
\begin{align*}
x=4.29\,light\,\,year &=4.29\times 9.46\times 10^{15}m\\
&=\frac{4.29\times 9.46\times 10^{15}}{3.08\times 10^{16}}par\sec =1.323 parsec
\end{align*}
In an orbit around the sun, the distance between six months apart locations is diameter of the Earth orbit itself which is $3 \times 10^{11}m$.
Now, using parallax method
Question 8 In an experiment, on determining the density of a rectangular block, the dimensions of the block are measured with a Vernier calipers (with a least count of 0.01 cm) and its mass is measured with a beam balance of least count of 0.1 g. The measured values are Lenght(L) = 5.12 cm, Breadth(b) = 2.56 cm, thickness(t) = 0.37 cm and mass(m) = 39.39 g.How do we report our result for the density of the block. Answer
The density of the block is given by
$ \rho = \frac {mass}{volume} =\frac {m}{L \times b \times t}$
Calculating with giving values
$\rho= \frac {39.39}{ 5.12 \times 2.56 \times 0.37} = 8.1037$ g/cm^{3}
Now uncertainty given in various quantities as per the given Least count of the instruments
$m = \pm 0.01$g
$l = \pm 0.01$ cm
$b = \pm 0.01$ cm
$t = \pm 0.01$ cm
Maximum relative error, in the density value is, therefore, given by
We cannot, therefore, report the calculated value of $\rho$ (= 8.1037 g/cm^{3}) up to the fourth decimal place. Since $\Delta \rho = 0.3$g/cm^{3} the value of density can be regarded as accurate up to the first decimal place only. Hence the value of density must be rounded off as 8.1 g/cm^{3} and the result of measurements should be reported as
$\rho = (8.1 \pm 0.3)$ g/cm^{3}
Question 9 It is required to find the volume of a rectangular block. A Vernier callipers is used to measure the length, width and height of the block. The measured values are 1.37 cm, 4.11 cm, and 2.56 cm respectively. Answer
Volume of the rectangular block(V) is given by
V=LBH
Now L=1.37 cm,B=4.11 cm,H=2.56 cm
$V=4.11 \times 2.56 \times 1.37=14.41$ cm^{3}
For Vernier callipers,Least count is 0.01 cm.
If all your measurements were 0.01 cm lower, the volume would be
$V=4.10 \times 2.55 \times 1.36=14.21$ cm^{3}
If all your measurements were 0.01 cm higher, the volume would be
$V=4.12 \times 2.57 \times 1.37=14.61$ cm^{3}
We can see the maximum value is .20 higher then calculated values
So we could report answer as $(14.41 \pm 0.20)$ cm^{3}
Question 10 Two specific heat capacities of a gas are measured as C_{p}=(12.28±0.2) units and C_{V}=(3.97±0.3) units. Find the value of gas constant R. Answer
Given C_{p}=(12.28±0.2) units and C_{V}=(3.97±0.3) units.
$R=C_P-C_V = (12.28\pm 0.2) - (3.97\pm 0.3) = (8.31 \pm 0.5)$ units
Theory questions on error analysis
Note:- Write all the points you know about these questions when you write the answers. Question 1 Define error. What are different sources of errors? Question 2 How can we minimize errors? Question 3 What are the different ways of expressing an error? Question 4 What is Absolute Error? Define the term accuracy. Answer
The magnitude of the difference between the individual measurement and the true value of the quantity is called the absolute error of the measurement
The accuracy of a measurement is a measure of how close the measured value is to the true value of the quantity.
Question 5 What do you mean by precision? According to you are instrument of high precision accurate? Answer
Precision tells us to what resolution or limit the quantity is measured.
Question 6 What is relative error or percentage error? Answer
The relative error is the ratio of the mean absolute error $\Delta a_{mean}$ to the mean value $a_{mean}$ of the quantity measured.
$a_{mean} = \frac {a_1 + a_2 + a_3}{3}$
$\Delta a_{mean}= \frac {|a_1 - a_{mean}| + |a_2 - a_{mean}| + |a_3 - a_{mean}|}{3}$
Parallax error occurs when the measurement of an object's length is more or less than the true length because of your eye being positioned at an angle to the measurement markings