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NCERT Exemplar Class 11 Physics Solutions Chapter 2 Units and Measurement

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NCERT Exemplar Class 11 Physics Solutions Chapter 2 Units and Measurement

Edited By Safeer PP | Updated on Aug 08, 2022 06:13 PM IST

NCERT Exemplar Class 11 Physics solutions Chapter 2 offers reliable and comprehensive answers to all the questions and queries related to measurement and the units used to measure quantities. NCERT Exemplar Class 11 Physics chapter 2 solutions have been curated by experts and throw light on the internationally accepted measurement standards. The chapter deals with different types of units, the procedure to go about measuring an abstract or a physical quantity, and the plausible errors in a measuring instrument.

NCERT Exemplar Class 11 Physics solutions chapter 2 PDF Download could be availed by the students for a better understanding of concepts and help revise the chapter for a flawless academic performance in final exams as well as competitive exams.

NCERT Exemplar Class 11 Physics Solutions Chapter 2: MCQ I

Question:1

The number of significant figures in 0.06900 is:
a) 5
b) 4
c) 2
d) 3

Answer:

The answer is the option (b) 4
Explanation: The two zeros before 6 are insignificant in the number 0.06900. Thus, there are 4 significant numbers.

Question:2

The sun of the numbers 436.32, 227.2,and 0.301 in appropriate significant figures is:
a)663.821
b)664
c)663.8
d)663.82

Answer:

The answer is the option (c) 663.8
Explanation: After addition, the sum is rounded off to the minimum number of decimal places of the numbers that are added. Hence, option (c) is correct

Question:3

The mass and volume of a body are 4.237\; g and 2.5\; cm^{3} , respectively. The density of the material of the body in correct significant figures is:
a) 1.6048 \; g/cm^{3}
b) 1.69 \; g/cm^{3}
c) 1.7 \; g/cm^{3}
d) 1.695 \; g/cm^{3}

Answer:

The answer to a multiplication or division is rounded off to the same number of significant figures as possessed by the least precise term used in the calculation. The final result should retain as many significant figures as are there in the original number with the least significant figures. In the given question, density should be reported to two significant figures .
\begin{aligned} &\text { Density }=\frac{4.237 \mathrm{g}}{2.5 \mathrm{cm}^{3}}=1.6948\\ \end{aligned}
After rounding off the number, we get density =1.7

Question:3

The mass and volume of a body are 4.237\; g and 2.5\; cm^{3} , respectively. The density of the material of the body in correct significant figures is:
a) 1.6048 \; g/cm^{3}
b) 1.69 \; g/cm^{3}
c) 1.7 \; g/cm^{3}
d) 1.695 \; g/cm^{3}

Answer:

The answer is the option (c) 1.7 \; g\; cm^{-3}
Explanation: When we are performing multiplication or division, we should keep in mind to retain as many significant figures in the original no. with the least significant figures.
We know that Density = mass/ volume
=4.237/2.5
=1.7\; g\; cm^{-3}
So, we get the above no. after rounding up 2 significant figures.

Question:4

The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give:
a) 2.75 and 2.74
b) 2.74 and 2.73
c) 2.75 and 2.73
d) 2.74 and 2.74

Answer:

The answer is the option (d) 2.74 \; and\; 2.74
Explanation:(i) 2.745 \rightarrow 2.74
Here the fourth digit is five and is preceded by, viz., an even no. Hence, it becomes 2.74.
(ii) 2.745 \rightarrow 2.74
Here the fourth digit is 5, but the preceding digit is 3, viz., an odd no. Hence, it is increased by 1, and we get 2.74.

Question:5

The length and breadth of a rectangular sheet are 16.2 \; cm and 10.1 \; cm respectively. The area of the sheet in appropriate significant figures and error is:
a) 164\pm 3\; cm^{2}
b) 163.62\pm 2.6\; cm^{2}
c) 163.6\pm 2.6\; cm^{2}
d) 163.62\pm 3\; cm^{2}

Answer:

The answer is the option (i) 164\pm 3\; cm^{2}
Explanation: Given :
I=16.2\; cm,\; and \; \Delta I=0.1
b=10.1\; cm,\; \Delta b=0.1
I=16.1\pm 0.1\; and\; b=10.1\pm 0.1
Now, area (a) =I\times b
=16.2\times 10.1
=163.62\; cm^{2}
=164\; cm^{2}
\Delta \frac{A}{A}=\frac{\Delta I}{I}+\frac{\Delta b}{b}
=\frac{0.1}{16.2}+\frac{0.1}{10.1}
\frac{\Delta A}{164}=(\frac{10.1\times0.1+16.2\times0.1}{16.2\times10.1})
Thus, \Delta A=2.63\; cm^{2}
We get \Delta A= 3\; cm^{2} by rounding off \Delta I\; and\; \Delta b
Thus, \Delta A=(163\pm 3)cm^{2}

Question:6

Which of the following pairs of physical quantities does not have same dimensional formula?
a) work and torque
b) angular momentum and Planck’s constant
c) tension and surface tension
d) impulse and linear momentum

Answer:

The answer is the option (c) Tension & surface tension
Explanation : We know that,
Work = force \times displacement
=[ML^{2}T^{-2}]
& Torque = force \times distance
=[ML^{2}T^{-2}]
Hence, their dimensions are the same.
(b) Dimensions of-
Angular momentum =[ML^{2}T^{-1}]
& Planck’s constant =[ML^{2}T^{-1}]
Hence, their dimensions are the same.
(c) Dimensions of-
Tension =[MLT^{-2}]
Surface tension=[ML^{0}T^{-2}]
Their dimensions are not the same, hence opt (c).
(d) Dimensions of-
Impulse =[MLT^{-1}]
Momentum =[MLT^{-1}]
Thus, their dimensions are also the same.

Question:7

Measure of two quantities along with the precision of respective measuring instrument is
A=2.5\; m/s\pm 0.5\; m/s
B= 0.10\; s\; \pm 0.01\;s
The value of AB will be
a) (0.25 \pm 0.08) \; m
b) (0.25 \pm 0.5) m
c) (0.25 \pm 0.05) m
d) (0.25 \pm 0.135) m

Answer:

The answer is the option (a) (0.25\pm 0.08)m
Explanation : By using the given values of A & B,
X=AB=2.5\times 0.10=0.25\; m
\frac{\Delta x}{x}=\frac{\Delta A}{A}+\frac{\Delta B}{B}
=\frac{0.5}{2.5}+\frac{0.01}{0.10}
=\frac{0.075}{0.25}
Thus, \Delta x=0.075\cong 0.08
Thus, AB=(0.25\pm 0.08)m

Question:8

You measure two quantities as A = 1.0 \; m \pm 0.2 \; m, B = 2.0 \; m \pm 0.2 \; m. We should report correct value for √ AB as:
a) 1.4 \; m \pm 0.4 \; m
b) 1.41 \; m \pm 0.15\; m
c) 1.4 \; m \pm 0.3 \; m
d) 1.4 \; m \pm 0.2\; m

Answer:

The answer is the option (d) 1.4\pm 0.2\; m
There are two significant figures in 1.0 & 2.0
\sqrt{AB}=\sqrt{1.0\times 2.0}=\sqrt{2}
=1.414\; m
After roundig up 1.0 &2.0 we het two sigificat features, i.e.
X=\sqrt{AB}
=1.4
Now, \frac{\Delta X}{X}=\frac{1}{2}[\frac{\Delta A}{A}+\frac{\Delta B}{B}]
\frac{\Delta X}{1.4}=0.1\left [ \frac{2.0+1.0}{1.0\times 2.0} \right ]
\Delta X=0.1(\frac{3}{2})((1.4)
=0.21\ m after rounding up
Thus, option (d) is the right answer.

Question:9

Which of the following measurements is most precise?
a)5.00 mm
b) 5.00 cm
c) 5.00 m
d) 5.00 km

Answer:

The answer is the option (a)5.00 mm
Explanation: The least unit is mm, and all the measurements are up to two decimal places. Hence, 5.00 mm is the most precise.

Question:10

The mean length of an object is 5\; cm. Which of the following measurements is most accurate?
a) 4.9\; cm
b) 4.805\; cm
c) 5.25\; cm
d) 5.4\; cm

Answer:

The answer is the option (a) 4.9 cm
Explanation : \left | \Delta a_{1} \right |=\left | 5-4.9 \right |=0.1cm,\left |\Delta a_{2} \right |=\left | 5-4.805 \right |=0.195\; cm
\left | \Delta a_{3} \right |=\left | 5-5.25 \right |=0.25cm,\left |\Delta a_{4} \right |=\left | 5-5.4\right |=0.4\; cm
Thus,\left | \Delta a_{1} \right | it is the mmost acurate.

Question:11

Young’s modulus of steel is 1.9 \times 10^{11} N/m^{2}. When expressed in CGS units of dynes/cm2, it will be equal to:
a) 1.9 \times 10^{10}
b) 1.9 \times 10^{11}
c) 1.9 \times 10^{12}
d) 1.9 \times 10^{13}

Answer:

The answer is the option (c) 1.9\times 10^{12}
Explanation : Young's modulus (Y) =1.9\times 10^{11}N/m
Converting into dyne/cm2-
Y=\frac{1.9\times 10^{11}\times 10^{5}dynes}{10^{4}cm^{2}}
=1.9\times 10^{11+5-4}
=1.9\times 10^{12}dyne/cm^{2}

Question:12

If momentum (P), area (A), and time (T) are taken to be fundamental quantities, then energy has the dimensional formula
a)(P^{1}A^{-1}T^{1})
b) (P^{2}A^{1}T^{1})
c) (P^{1}A^{-1/2}T^{1})
d) (P^{1}A^{1/2}T^{-1})

Answer:

The answer is the option d) (P^{1}A^{1/2}T^{-1})
Explanation: Let us consider [P^{a}A^{b}T^{c}] as the formula for energy for fundamental quantities P, A & T.
Thus, dimensional formula of-
Energy(E) = [P^{a}A^{b}T^{c}]
Momentum (P) = [MLT^{-1}]
Area (A) =[L^{2}]
Time(T) = [T^{1}]
Now, E = f.s
= [MLT^{-2}L].[ ML^{2}T^{-2}]
= [MLT^{-1}]^{a}[L^{2}]^{b}[T]^{c} . [M^{a}L^{2+2b}T^{-a+c}]
Now, let us compare the powers,
a=1\; \; \; \; \; \; \; \; a+2b=2\; \; \; \; \; \; \; \; \; -a+c=-2
\; \; \; \; \; \; \; \; 1+2b=2\; \; \; \; \; \; \; \; \; -1+c=-2
\; \; \; \; \; \; \; \; 2b=2-1\; \; \; \; \; \; \; \; \; c=-2+1
\; \; \; \; \; \; \; \; b=\frac{1}{2}\; \; \; \; \; \; \; \; \; c=-1
Thus, [P^{1}A^{1/2}T^{-1}] is the dimensional formula of energy.

NCERT Exemplar Class 11 Physics Solutions Chapter 2: MCQII

Question:13

On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct:
a) y=a\; \sin \; \frac{2\pi t}{T}
b) y=a\; \sin \; vt
c) y=\frac{a}{T}\sin (\frac{t}{a})
d) y=a\sqrt{2}[\sin (\frac{2\pi t}{t})-\cos (\frac{2\pi t}{T})]

Answer:

The answer is the option (b) y=a\; \sin \; vt and (c) y=\frac{a}{T}\sin (\frac{t}{a})
Explanation: (b) Here, v.t is an angle, whose dimensions are – [LT^{-1}] [T] = [L]
Thus, it is not true that sin vt is dimensionless.
(c) Here the dimension of amplitude a/T on the R.H.S. is equal to \frac{[L]}{[T]} = [LT^{-1}], viz., not equal to the dimensions of y and the angle \frac{t}{a} = [LT^{-1}], which means that they are not dimensionless.

Question:14

If P, Q, R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity?
a) \frac{(P - Q)}{R}
b) PQ - R
c)\frac{ PQ}{R}
d)\frac{ (PR-Q^{2})}{R}
e) \frac{(R + Q)}{P}

Answer:

The answer is the option a) \frac{(P - Q)}{R} and e) \frac{(R + Q)}{P}
Explanation:
(i) The different physical quantities (P-Q) & (R+Q) in option a & e and never be added or subtracted. Thus, they will be meaningless.
(ii) In opt (d), dimensions of PR & Q^{2} may be equal.
(iii) in opt (b), the dimension of PQ may be equal to the dimension of R.
(iv) There is no addition or subtraction, which gives the possibility of opt (c).
Hence, opt (a) & (e).

Question:15

Photon is quantum of radiation with energy E = hv where v is frequency and h is Planck’s constant. The dimensions of h are the same as that of:
a) linear impulse
b) angular impulse
c) linear momentum
d) angular momentum

Answer:

The answer is the option (b) Angular Impulse and (d) Angular momentum
Explanation: It is given that E = hv
Hence, h (Planck’s constant) =\frac{E}{v}
=\frac{[ML^{2}T^{-2}]}{[T^{-1}]}
=[ML^{2}T^{-1}]
Now, Linear Impulse = F.t
=\frac{dp}{dt}.dt=dp
=mv=[MLT^{-1}]
& Angular Impulse = = \tau.dt= \frac{dL}{dt}.dt
= dL= mvr
= [M][LT^{-1}][L]
= [ML^{2}T^{-1}]
Now, Linear momentum = mv
= [MLT^{-1}]
& Angular momentum (L) = mvr

= [ML^{2}T^{-1}]
Thus, the dimensional formulae of Planck’s constant, Angular Impulse and Angular Momentum are the same.

Question:16

If Planck’s constant (h) and speed of light in vacuum (c) are taken as two fundamental quantities, which of the following can in addition be taken to express length, mass, and time in terms of the three chosen fundamental quantities?
a) mass of electron (me)
b) universal gravitational constant (G)
c) charge of electron (e)
d) mass of proton (mp)

Answer:

The answer is the option (a) Mass of electron (me) and (b) Universal gravitational constant (G) and (d) Mass of proton (mp)
Explanation: Dimensions of –
h (Planck’s constant)
= [ML^{2}T^{-1}]
c (Speed of light in vacuum)
= \frac{s}{t}
= [LT^{-1}]
Thus, dimension of hc
= [ML^{2}T^{-1}][LT^{-1}]
= [ML^{3}T^{-2}]
G = \frac{Fr^{2}}{M_{1}M_{2}}
= \frac{[ML^{3}T^{-2}]}{[M][M]}
= [M^{-1}L^{3}T^{-2}]
Now,
\frac{hc}{G} = [ML^{3}T^{-2}]/[M^{-1}L^{3}T^{-2}]
=[M^{2}]
Thus,
M= \sqrt{\frac{hc}{G}}
= [h^{\frac{1}{2}} c^{\frac{1}{2}}G^{\frac{1}{2}}]
Now,
\frac{h}{c} = [ML^{2}T^{-1}]/[LT^{-1}]
= [ML]
=\sqrt{\frac{hc}{G}}\times L
Now,
L=\frac{h}{c}\times \sqrt{\frac{G}{hc}}
=\frac{\sqrt{Gh}}{\frac{c^{3}}{2}}
= [G^{\frac{1}{2}} h^{\frac{1}{2}} c^{\frac{-3}{2}}]
Also, c = [LT^{-1}]
= [G^{\frac{1}{2}} h^{\frac{1}{2}} c^{\frac{-3}{2}}T^{-1}]
& T= [G^{\frac{1}{2}} h^{\frac{1}{2}} c^{\frac{-3}{2}-1}]
T= [G^{\frac{1}{2}} h^{\frac{1}{2}} c^{\frac{-5}{2}}]
Therefore, in terms of chosen fundamental quantities, the physical quantities a, b & d can be used to represent L, M & T.

Question:17

Which of the following ratios express pressure?
a) Force/area
b) Energy/volume
c) Energy/area
d) Force/volume

Answer:

The answer is the option (a) Force/Area and (d) Force/Volume
Explanation: Let us start with calculating the dimension of pressure,
We know that Pressure(P) =\frac{F}{A}
= \frac{[MLT^{-2}]}{[L^{2}]}
..... this also verifies the opt (a).
Now let us compare it with other options-
(b)\frac{E}{V}=\frac{[ML^{2}T^{-2}]}{[L^{3}]}
=[ML^{-1}T^{-2}]
= Dimesions of P
Thus, opt (b) is verified as well.
(c)\frac{E}{A}=\frac{[ML^{2}T^{-2}]}{[L^{3}]}
=[ML^{0}T^{-2}] ...............viz., not equal to Dimensions of P
(d)\frac{F}{V}=\frac{[MLT^{-2}]}{[L^{3}]}
=[ML^{-2}T^{-2}] .................viz., again not equal to dimensions of P
Thus, opt (c) & (d) are discarded.

Question:18

Which of the following are not a unit of time?
a) second
b) parsec
c) year
d) light year

Answer:

The answer is the option (b) Parsec and (d) Light year
Explanation: Parsec & Light year are units to measure distance while second & Year are units to measure time.

NCERT Exemplar Class 11 Physics Solutions Chapter 2: Very Short Answer

Question:19

Why do we have different units for the same physical quantity?

Answer:

(i) Different orders of quantity are measured by the same physical quantities.
(ii)The radius of a nucleus is measured in fermi. The distance between stars is measured in light-years. The length of cloth is measured in meter while the distance between two cities is measured in Kilometer or miles. Depending upon the physical quantity to be measured the units also vary.

Question:21

Name the device used for measuring the mass of atoms and molecules.

Answer:

The mass spectrograph is the device used for measuring the mass of atoms and molecules.

Name the device used for measuring the mass of atoms and molecules.

Question:22

Express unified atomic mass unit in kg.

Answer:

We know that the unified atomic mass unit or amu or u= \frac{1}{12} the mass of one c^{12} atom
Mass of 6.023\times10^{23} carbon atoms (_{6}C^{12})= 12\; gm
Thus, the mass of one _{6}C^{12} atom = \frac{12}{6.023}\times 10^{-23}gm
Now , 1 amu
= \frac{1}{12}\times \frac{12}{6.023}\times 10^{-23}\; gm
=1.66\times 10^{-24}\; gm
Thus, 1 amu = 1.66\times10^{-27}kg

Question:23

A function f(\theta ) is defined as
f(\theta )= 1-\theta +\frac{\theta ^{2}}{2!}-\frac{\theta ^{3}}{3!}+\frac{\theta ^{4}}{4!}+.........
Why is it necessary for \theta to be a dimensionless quantity ?

Answer:

(i) \theta is a dimensionless physical quantity since it is represented by an angle viz. equal to arc/radius.
(ii) As \theta is dimensionless, all its powers will also be dimensionless. Here the first term is 1 viz. also dimensionless.
Hence, the L.H.S. f(\theta) must be dimensionless so that all the terms in R.H.S. are also dimensionless.

Question:24

Why length, mass and time are chosen as base quantities in mechanics?

Answer:

Length, mass & time is chosen as base quantities in mechanics because unlike other physical quantities, they cannot be derived from any other physical quantities.

NCERT Exemplar Class 11 Physics Solutions Chapter 2: Short Answer

Question:25

(a) 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?
(b) Moon is seen to be of \left ( \frac{1}{2} \right )^{o} diameter from the earth. What must be the relative size compared to the earth?
(c) From parallax measurement, the sun is found to be at a distance of about 400 times the earth-moon distance. Estimate the ratio of sun-earth diameters.

Answer:

(a) Here we know that, \theta = arc/radius
= \frac{R_{e}}{60R_{e}}
= \frac{1}{60}rad

Now, the angle from the moon to the diameter of the earth will be-
2\theta = 2\left ( \frac{1}{60} \right )\left ( \frac{180}{\pi } \right )
= \frac{6^{o}}{\pi }
\cong \frac{6}{3.14}\cong 2^{o}
(b) If the moon is seen from the earth diametrically, the angle will be \frac{1}{2^{o}}
& if the earth is seen from the moon, the angle will be 2^{o}
Thus, the size of the moon/ size of the earth = \frac{\frac{1}{2^{o}}}{2^{o}}= \frac{1}{4}
Therefore, as compared to the earth, the size of the moon is \frac{1}{4} the size(diameter) of the earth.
(c)Let rem = x m be the distance between the earth and the moon
Now, the distance between the sun and the earth (rse) = 400xm
On a solar eclipse, the sun is completely covered by the moon; also the angle formed by the diameter of the sun, moon and the earth are equal.
Thus, \theta _{m}= \theta _{s}
......... ( the angle from the earth to the moon is \theta {_{m}}^{o} & angle from sun to earth is \theta {_{s}}^{o}
Now since \theta _{m}= \theta _{s}
\frac{D_{m}}{r_{em}}= \frac{D_{s}}{r_{x}}
\frac{D_{m}}{x}= \frac{D_{s}}{400 x}
\frac{D_{s}}{D_{M}}= 400
Thus, 4D_{m}= D_{e}or\; D_{m}= \frac{D_{e}}{4}
Thus, \frac{D_{s}}{\frac{D_{e}}{4}}= 400
\frac{4D_{s}}{D_{e}}= 400
\frac{D_{s}}{D_{e}}= \frac{400}{4}= 100
Therefore, D_{s}= 100\; D_{e}

Question:26

Which of the following time measuring devices is most precise?
(a) A wall clock.
(b) A stopwatch.
(c) A digital watch.
(d) An atomic clock.
Give reason for your answer.

Answer:

The least counts of the devices are as follows-
Wall clock = 1 sec
Stopwatch = \frac{1}{10} sec
Digital watch = \frac{1}{100} sec &
Atomic clock = \frac{1}{10^{13}} sec.
Hence, the atomic clock is the most precise.

Question:20

The radius of atom is of the order of 1\; \AA and radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus?

Answer:

Let us convert these different units into a common one, i.e., ‘m’.
Thus, Radius of an atom (R)= 1\AA
= 10^{-10}m
& Radius of a nucleus = 1 fermi
= 10^{-15}m
Now, let us calculate the ratio of the volumes of an atom to a nucleus-
\frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^{3}}= R^{3}/r^{3}
= (10^{-10}/10^{-15})^3
= (10^{5})^{3}= 10^{15}

Question:27

The distance of a galaxy is of the order of 1025 m. Calculate the order of magnitude of time taken by light to reach us from the galaxy.

Answer:

It is given that the distance travelled by light from galaxy to earth = 10^{25}m.
We know that speed of light = 3\times10^{8}m/s
Thus, the required time = distance / speed
= \frac{10^{25}}{3}\times10^{-8}\; sec
= \frac{1}{3}\times 10^{17}sec
= 3.3\bar{3}\times 10^{16}sec

Question:28

The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, calculate the minimum inaccuracy in the measurement of distance.

Answer:

Let ‘n’ be the no. Of parts on the Vernier scale, n = 50
Now, the no. Of parts of M.S. coinciding with n parts of vernier scale = (n-1)
Now, Least count of an instrument = L.C. of Mainscale/n
= 0.5mm/50
Thus, the minimum accuracy will be 0.01 mm.

Question:29

During a total solar eclipse the moon almost entirely covers the sphere of the sun. Write the relation between the distances and sizes of the sun and moon.

Answer:

Thus, \Omega _{s}= \Omega _{m}
Thus, \frac{\pi (\frac{D_{s}}{2})^{2}}{r{_{s}}^{2}}= \frac{\pi (\frac{D_{m}}{2})^{2}}{r{_{m}}^{2}}
i.e., \frac{D_{s}^{2}}{4r_{s}^{2}}= \frac{D_{m}^{2}}{4r_{m}^{2}}
Thus, \frac{4R_{s}^{2}}{4r_{s}^{2}}= \frac{4R_{m}^{2}}{4r_{m}^{2}}
Taking square roots,
\frac{R_{s}}{r_{s}}=\frac{R_{m}}{r_{m}}
Or,
\frac{R_{s}}{R_{m}}=\frac{r_{s}}{r_{m}}
Therefore, the ratio of the size of the sun to the moon is equal to the distances from the sun to the moon from earth.

Question:30

If the unit of force is 100 N, unit of length is 10 m and unit of time is 100 s, what is the unit of mass in this system of units?

Answer:

The dimensions of -
Force = [M^{1}L^{1}T^{-2}]
=100 N ..........(i)
Length = [L^{1}]
= 10 m ...........(ii)
Time = [T^{1}]
= 100 sec ...........(iii)
Now let us substitute the values of eq. (ii) & (iii) in eq. (i)
We get,
M\times10\times100^{-2}=100
\frac{10M}{10000}=100
Thus, M=10^{5}kg, L=10^{1}m,F=10^{2}N \; and \; T=10^{2}seconds

Question:31

Give an example of
(a) a physical quantity which has a unit but no dimensions.
(b) a physical quantity which has neither unit nor dimensions.
(c) a constant which has a unit.
(d) a constant which has no unit.

Answer:

(a) A plane angle is an example of a physical quantity which has unit but no dimension since, plane angle = arc/radius in radian & solid angle.
(b) Relative density is a physical quantity which neither has neither units nor dimension since µr (relative density) is a ratio of same physical quantities.
(c) Gravitational constant (G) and Dielectric constant (k) are an example of constant that have a unit.
G=6.67 \times 10^{-11}N-m^{2}-kg
k=9 \times 10^{9}N-m^{2}-C^{-2}(Vacuum)
(d) Examples of the constant that has no units are Reynolds no. & Avogadro’s no.

Question:32

Calculate the length of the arc of a circle of radius 31.0 cm which \frac{\pi }{6} subtends an angle of at the centre.

Answer:

Given :radius = 31 cm
Angle = \frac{\pi }{6}
Length of arc = ?
Now, we know that
Angle = length of arc/radius of arc
\frac{\pi }{6}=\frac{x}{31}
x = 31 \times \frac{\pi }{6}
= 31 \times \frac{3.14 }{6}
x = 16.22 \; cm

Question:34

The displacement of a progressive wave is represented by y = A\; \sin (wt - k \; x ), where x is distance and t is time. Write the dimensional formula of (i) \omega and (ii) k.

Answer:

According to the principle of homogeneity dimensional formula on L.H.S. & R.H.S. is equal.
Thus, the dimension of A \; \sin (\omega t - kx) = Dimension of y
\omega t - kx has no dimensions because it is an angle of sin.
\frac{2\pi }{T}t=Kx
[M^{0}L^{0}T^{0}] = k[L]
Thus, \omega has no dimension & dimension of k=[M^{0}L^{-1}T^{0}].

Question:35

Time for 20 oscillations of a pendulum is measured as t_{1}= 39.6 \; s; t_{2} = 39.9 \; s; t_{3} = 39.5\; s. What is the precision in the measurements? What is the accuracy of the measurement?

Answer:

Given data :
t_{1}= 39.6 \; s
t_{2} = 39.9 \; s
and t_{3} = 39.5\; s
L.C. of the instrument is 0.1 sec, hence Precision (LC) = 0.1 sec.
Now,
For 20 oscillations, the mean value will be
=\frac{39.6+39.9+39.5}{3}
=\frac{119}{2}
=39.7 \; s
Now the absolute errors in measurement
\left | \Delta t_{1} \right |=\left | \bar{t}- t_{1}\right |=\left | 39.7-39.6 \right |=\left | 0.1 \right |=0.1\; s
\left | \Delta t_{2} \right |=\left | \bar{t}- t_{2}\right |=\left | 39.7-39.9 \right |=\left | 0.2 \right |=0.2\; s
\left | \Delta t_{3} \right |=\left | \bar{t}- t_{3}\right |=\left | 39.7-39.5 \right |=\left | 0.2 \right |=0.2\; s
Now, Mean absolute error =\frac{0.1+0.2+0.2}{3}
=\frac{0.5}{3}\cong 0.2\; s
& Accuracy of measurement =\pm 0.2\; s

NCERT Exemplar Class 11 Physics Solutions Chapter 2: Long Answer

Question:36

A new system of units is proposed in which unit of mass is \alpha kg, unit of length \beta m and unit of time γ s. How much will 5 J measure in this new system?

Answer:

Joules is a unit of work/energy, so let us first write the dimensions of energy-
Energy =[ML^{2}T^{-2}]
Let n_{1} be the SI system of unit & n_{2} be the new system of unit
Thus, n_{2}u_{2}=n_{1}u_{1}
n_{2}=n_{1}(\frac{u_{1}}{u_{2}})
=n_{1}[\frac{M_{1}}{M}_{2}]^{1}[\frac{L_{1}}{L_{2}}]^{2}[\frac{T_{1}}{T_{2}}]^{-2}
Now,
M_{1}=1\; kg & M_{2}=\alpha \; kg
L_{1}= 1 \; m & L_{2}= \beta \; m
T_{1}= 1 \; second & T_{2}= \gamma \; sec
Thus, n_{2}= 5[\frac{1\; kg}{\alpha \; kg}]^{1}[\frac{1\; m}{\beta \; m}]^{2}[\frac{1\; s}{\gamma \; s}]^{-2}
=5[\alpha ^{-1}\beta ^{-2}\gamma ^{2}]
Thus, by the new system, [\alpha ^{-1}\beta ^{-2}\gamma ^{2}].

Question:38

A physical quantity X is related to four measurable quantities a, b, c and d as follows: X = a^{2} b^{3} c^{\frac{5}{2}} d^{-2}. The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4%, respectively. What is the percentage error in quantity X ? If the value of X calculated on the basis of the above relation is 2.763, to what value should you round off the result.

Answer:

As per the information given in the question, percentage error in a = (\frac{\Delta a}{a})(100) = 1\; ^{o}/_{o}
As per the information given in the question, Percentage error in b = (\frac{\Delta b}{b})(100) = 2\; ^{o}/_{o}
As per the information given in the question, Percentage error in c = (\frac{\Delta c}{c})(100) = 3\; ^{o}/_{o}
As per the information given in the question, Percentage error in d = (\frac{\Delta d}{d})(100) = 4\; ^{o}/_{o}
Percentage error in X = (\frac{\Delta x}{x})(100)
\\\frac{\Delta X}{X}\times 100=\pm(2\times1+3\times2+2.5\times3+2\times4)\\\pm(2+6+7.5+8) =\pm 23.5\; ^{o}/_{o}
Now, let us calculate the mean absolute error
=\pm \frac{23.5}{100}
=\pm 0.235=0.24 (after rounding up)
Therefore, we get ‘2.8’ after rounding X i.e. 2.763.

Question:39

In the expression P=E\; I^{2}m^{-5}G^{-2},E,m,I\; and\; G denote energy, mass, angular momentum and gravitational constant, respectively. Show that P is a dimensionless quantity.

Answer:

The dimensional formula of –
E=[ML^{2}T^{-2}]
I=[ML^{2}T^{-1}]
G=[M^{-1}L^{3}T^{-2}]
Now let us find out the dimension of P
P=EI^{2}m^{-5}G^{-2}
[P]=\frac{[E][I^{2}]}{[m^{5}][G^{2}]}
=\frac{[ML^{2}T^{-2}][ML^{2}T^{-1}]^{2}}{[M]^{5}[M^{-1}L{3}T^{-2}]^{2}}
\ \ =\frac{M^3L^{6}T^{-4}}{M^3L^{6}T^{-4}}
=[M^{0}L^{0}T^{0}]
Hence, it is clear that P is a dimensionless quantity.

Question:40

If velocity of light c, Planck’s constant h and gravitational constant G are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

Answer:

Let us first write down the dimensions of-
h=\frac{E}{v}
= \frac{[ML^{2}T^{-2}]}{[T^{-1}]}
=[ML^{2}T^{-1}]
C=[LT^{-1}]
G = [M^{-1}L^{3}T^{-2}]
(i) Let us consider that Mass m is directly proportional to [h]^{a}[C]^{b}[G]^{c}
[M^{1}L^{0}T^{0}] = k [ML^{2}T^{-1}]^{a}[LT^{-1}]^{b}[M^{-1}L^{3}T^{-1}]^{c}
[M^{1}L^{0}T^{0}] = k [M^{a-c}L^{2a+b+3c}T^{-a-b-2c}]
Comparing the powers on R.H.S. & L.H.S.
a-c=1
Thus, a=c+1
2a+b+3c=0\; \; \; \; \; \; \; .....(i)
-a-b-2c=0 \; \; \; \; \; \; \; ..... (ii)
Substituting the value of a in eq (i) & (ii)
2(c+1) + b + 3c = 0
2c + 2 + b + 3c =0
b+5c = -2 \; \; \; \; \; \; \; \; ..... (iii)
-(c+1)-b-2c = 0
-b-3c=1 \; \; \; \; \; \; \; ...... (iv)
By adding (iii) & (iv), we get,
c=\frac{1}{2}
now, a = c+1
Thus, a = \frac{-1}{2}+1 = \frac{1}{2}
Substituting these values in eq. (iii)
b + 5(-1/2) = -2
b = -2 + \frac{2}{5}
b = \frac{1}{2}
Thus, m = kh^{\frac{1}{2}}C^{\frac{1}{2}}G^{\frac{-1}{2}}
M=k\sqrt{\frac{hc}{G}}
Now, let L\; \alpha \; C^{a}h^{b}G^{c}
Thus, [L^{1}] = k[LT^{-1}]^{a}[ML^{2}T^{-1}]^b[M^{-1}L^{3}T^{2}]^{c}
=k[M^{b-c}L^{a+2b+3c}T^{-a-b-2c}]
Equating the powers on both sides
b-c=0
Thus,b=c
a+2b+3c = 1 \; \; \; \; \; \; \; ...... (i)
-a-b-2c = 0 \; \; \; \; \; \; .... (ii)
Now, since b=c
a+2b+3b=1 becomes
a+5b=1 \; \; \; \; \; \; .....(iii)
& eq. (ii)
a+b+2c = 0 becomes
a + 3b = 0\; \; \; \; \; \; \; \; ...... (iv)
Subtracting eq (iv) from eq (iii) we get,
b=\frac{1}{2}
Thus, c=\frac{1}{2}
And eq (iv) will be
a +3(\frac{1}{2}) = 0
a=\frac{-3}{2}
Therefore, I = k\; C^{\frac{3}{2}}h^{\frac{1}{2}}G^{\frac{1}{2}}
L=k\sqrt{\frac{hG}{c^{3}}}
Let us consider T\; \alpha \; G^{a}h^{b}C^{c}
Thus, [M^{0}L^{0}T^{-1}] = k[M^{-1}L^{3} T^{-2}]^{a}[ML^{2}T^{-1}][LT^{-1}]^{c}
= k[M^{-a+b} L^{3a+2b+c}T^{-2a-b-c}]
-a+b=0 \; \; \; \; \; \; \; ... (i)
3a + 2b + c = 0 \; \; \; \; \; \; ....... (ii)
-2a-b-c = 1 \; \; \; \; \; \; \; \; .... (iii)
On adding eq (ii) & (iii) we get,
a+b=1 \; \; \; \; \; \: \: \: \: \: ..... (iv)
From eq (i) &(iv)
b=\frac{1}{2}
a=\frac{1}{2}
Thus, c=\frac{-5}{2}
T = k\; G^{\frac{1}{2}}h^{\frac{1}{2}}C^{\frac{-5}{2}}
Thus, T = k\sqrt{\frac{hG}{c^{5}}}

Question:41

An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that T=\frac{k}{R}\sqrt{\frac{r3}{g}} where k is a dimensionless constant and g is acceleration due to gravity.

Answer:

According to Kepler’s 3rd law of planetary motion,
T^{2}\; \alpha \; \; r^{3}
T\; \alpha \; \; r^{\frac{3}{2}}
Now, R & g also affects T
Thus,
T \; \; \; \alpha\; \; \; g^{a}R^{b}r^{\frac{3}{2}}
T = K[LT^{-2}]^{a}[L^{b}][L]^{\frac{3}{2}}
[T^{1}] = K[L^{a+b+\frac{3}{2}}T^{-2a}]
Now, comparing M, L & T according to the principle of homogeneity
a+b+\frac{3}{2} = 0 \; \; \; \; \; \; ..... (i)
-2a = 1 \; \; \; \; \; \; ..... (ii)
a=\frac{-1}{2}
Thus, \frac{-1}{2} + b + \frac{3}{2} = 0

b+1=0
B=-1
-2a=1
T=Kg^{\frac{-1}{2}}R^{-1}r^{\frac{3}{2}}
i.e., T=\frac{K}{R}\sqrt{\frac{r^{3}}{g}}

An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that where k is a dimensionless constant and g is acceleration due to gravity.

Question:42

In an experiment to estimate the size of a molecule of oleic acid, 1mL of oleic acid is dissolved in 19mL of alcohol. Then 1mL of this solution is diluted to 20mL by adding alcohol. Now, 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film we can calculate the thickness of the film which will give us the size of oleic acid molecule.
Read the passage carefully and answer the following questions:

a) Why do we dissolve oleic acid in alcohol?
b) What is the role of lycopodium powder?
c) What would be the volume of oleic acid in each mL of solution prepared?
d) How will you calculate the volume of n drops of this solution of oleic.
e) What will be the volume of oleic acid in one drop of this solution?

Answer:

(a) We can reduce the concentration of oleic acid by dissolving it in a proper solvent to get a molecular level. It can be dissolved in organic solvent alcohol and not ionic solvent water since it is an organic compound.
(b) Mixing of oleic acid in water is prevented by Lycopodium when a drop of oleic acid is poured on water. Thus, if we spread lycopodium powder on the water surface, the layer of oleic acid will be formed on the surface of the powder,
(c)(\frac{1}{20})(\frac{1}{20\; V}) = \frac{V}{400} ml is the concentration of oleic acid in an alcohol solution. Its required conc. in 1 ml solution is 1/400 ml.
(d) By dropping one ml of solution drop by drop in a beaker through burette and counting its no., can be used to calculate the volume of n drop solution. If there is 1ml of n drop, then its volume will be 1/n ml.
(e) If there is 1ml of n drop, then its volume will be 1/n ml. The concentration of oleic acid in one drop of the solution will be
\frac{1}{400\; V}=\frac{1}{400}.\frac{1}{n}ml
=\frac{1}{400}ml of oleic acid.

Question:43

a) How many astronomical units (AU) make 1 parsec?
b) Consider the sun like a star at a distance of 2 parsec. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be (1/2) degree from the earth. Due to atmospheric fluctuations, eye cannot resolve objects smaller than 1 arc minute.
c) Mars has approximately half of the earth’s diameter. When it is closer to the earth it is at about ½ AU from the earth. Calculate at what size it will disappear when seen through the same telescope.

Answer:

(a)1 A.U. long arc subtends the angle of 1s or 1 arc sec at distance of 1 parsec.
Thus, angle 1 sec = \frac{1 A.U.}{1} parsec
Thus, 1 parsec = \frac{1 A.U.}{1} arc sec
Thus, 1 parsec= \frac{630(3600)}{11}
=206182.8
=2\times10^{5}A.U.
(b)Angle of the sun’s diameter \left ( \frac{1}{2} \right )^{o} is subtended by 1 A.U. since the distance from the sun increases angle subtended in the same ratio.
Now,
2\times10^{5}A.U. will form an angle of \theta =\left ( \frac{1}{4\times10^{5}} \right )^{o}, since the diameter is the same angle subtended on earth by 1 parsec will be same.
If the sunlike star is at 2 parsec the angle becomes half = (1.25 \times 10^{-6})^{o}
Thus, angle = 75 \times 10^{-6} min
When it is seen with a telescope that has a magnification of 100, the angle formed will be 7.5 \times 10^{-3} min, viz., less than a minute.
Hence, it can’t be observed by a telescope.

Question:44

Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass (m) to energy (E) as E = mc2, where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV where 1 MeV = 1.6 \times 10^{-13}J, the masses are measured in unified equivalent of 1u is 931.5 MeV.
a) Show that the energy equivalent of 1 u is 931.5 MeV.
b) A student writes the relation as 1 u = 931.5 MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.

Answer:

(a) Given :
m=1
u=1.67\times10^{-27}kg
c=3\times10^{8}m/s
By the formula E=mc^{2}
E=1.67\times10^{-27}\times3\times10^{8}\times3\times10^{8}
=1.67\times10^{-27+16}\times9J
=\frac{(1.67)(9)(10^{-11})}{(1.6)(10^{-13})}MeV
=939.4\; Mev
\cong 931.5\; Mev
(b) 1 u mass converted into total energy will be released by 931.5 MeV.
However, 1 amu = 931.5 MeV is dimensionally incorrect.
E=mc^{2}\rightarrow 1uc^{2}\cong 931.5\; Mev, will be dimensionally correct.

Class 11 Physics NCERT Exemplar Solutions Chapter 2 Topics:

Main Subtopics in NCERT Exemplar Class 11 Physics Solutions Chapter 2 Units and Measurement

  • 2.1 Introduction
  • 2.2 The International System Of Units
  • 2.3 Measurement Of Length
  • 2.4 Measurement Of Mass
  • 2.5 Measurement Of Time
  • 2.6 Accuracy, Precision Of Instruments And Errors In Measurement
  • 2.7 Significant Figures
  • 2.8 Dimensions Of Physical Quantities
  • 2.9 Dimensional Formulae And Dimensional Equations
  • 2.10 Dimensional Analysis And Its Applications

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What will The Students Learn in NCERT Exemplar Class 11 Physics Solutions Chapter 2?

Students would be able to understand the dynamics behind the change in measuring trends, learn about the basic units, and try to bring all that knowledge to life by using these concepts in actual measuring instruments. NCERT Exemplar Class 11 Physics solutions chapter 2 would also introduce the students to utilise the system of precise measurement to answer real-life questions related to the mass of rain-bearing clouds and ascertain the speed of an aircraft through dimensional analysis and determine the distance between two stars.

NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

Important Topics To cover For Exams in NCERT Exemplar Solutions For Class 11 Physics Chapter 2 Units and Measurement

· NCERT Exemplar Class 11 Physics solutions chapter 2 introduces the student to various measurement units across the world and the standard unit that is accepted internationally. It talks about measuring different physical quantities like length and mass and abstract quantities like time.

· The dimensional formulae and equations are also introduced in this chapter, which helps to utilise dimensional analysis in real-life applications. Students are also made aware of the importance of significant figures of measurement.

· NCERT Exemplar Class 11 Physics solutions chapter 2 also helps students understand the accuracy and precision in making measurements and the errors that one stumbles on along the way. It also highlights the different types of errors found in instruments and the measuring procedure to quickly identify and rectify them.

NCERT Exemplar Class 11 Solutions

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Frequently Asked Question (FAQs)

1. How can these NCERT Exemplar Class 11 Physics Solutions Chapter 2 help?

These solutions from Chapter 2 can help better understand the basic concepts and prepare for your academic exams. Questions of multiple choice type, short answer and long answer types are solved.

2. How to download these Solutions?

 

These Class 11 physics NCERT exemplar solutions chapter 1 can be downloaded from the website and the solution page directly in PDF format.

3. What are the essential topics of this chapter?

This chapter's essential topics are the International System of Units, Measurement of quantities, accuracy and precision, dimensional formulae, and dimensional analysis. 

Articles

Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Safety Manager

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

2 Jobs Available
Conservation Architect

A Conservation Architect is a professional responsible for conserving and restoring buildings or monuments having a historic value. He or she applies techniques to document and stabilise the object’s state without any further damage. A Conservation Architect restores the monuments and heritage buildings to bring them back to their original state.

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

2 Jobs Available
Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
Field Surveyor

Are you searching for a Field Surveyor Job Description? A Field Surveyor is a professional responsible for conducting field surveys for various places or geographical conditions. He or she collects the required data and information as per the instructions given by senior officials. 

2 Jobs Available
Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Veterinary Doctor
5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Anatomist

Are you searching for an ‘Anatomist job description’? An Anatomist is a research professional who applies the laws of biological science to determine the ability of bodies of various living organisms including animals and humans to regenerate the damaged or destroyed organs. If you want to know what does an anatomist do, then read the entire article, where we will answer all your questions.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.  

2 Jobs Available
Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Production Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

2 Jobs Available
Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
QA Manager
4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
ITSM Manager
3 Jobs Available
Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process. 

2 Jobs Available
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