NCERT Exemplar Class 11 Physics Solutions Chapter 2 Units and Measurement 2021

NCERT Exemplar Class 11 Physics Solutions Chapter 2 Units and Measurement

Edited By Vishal kumar | Updated on Apr 08, 2025 05:25 PM IST

Do you know what the distance of the moon from Earth is? It sounds interesting, right? This chapter will let you know how things we see around us—like speed, distance, and weight—are measured with accuracy using different tools and measuring kits.

The key concepts include SI units, dimensional analysis, significant figures, measurement errors, and how to check the correctness of equations using dimensions. This chapter builds a strong foundation for solving numerical problems with accuracy and precision.

NCERT Exemplar Class 11 Physics Solutions Chapter 2 gives clear and accurate answers to all the questions related to Units and Measurements. Subject experts make these solutions and explain the standard units used worldwide. This chapter helps you understand different types of units, how to measure physical and abstract quantities, and the possible errors that can occur while measuring.

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This Story also Contains

NCERT Exemplar Class 11 Physics Solutions Chapter 2: MCQ I
NCERT Exemplar Class 11 Physics Solutions Chapter 2: MCQ II
NCERT Exemplar Class 11 Physics Solutions Chapter 2: Very Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 2: Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 2: Long Answer
Class 11 Physics NCERT Exemplar Solutions Chapter 2 Topics:
What will The Students Learn in NCERT Exemplar Class 11 Physics Solutions Chapter 2?
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-Wise
NCERT Exemplar Class 11th Solutions

NCERT Exemplar Class 11 Physics Solutions Chapter 2 Units and Measurement

You can download the PDF of NCERT Exemplar Chapter 2 Solutions to understand the concepts better and revise easily. NCERT is super helpful for doing well in your final exams and competitive exams, too!

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 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 $2.5 c m^{3}$ , respectively. The density of the material of the body in correct significant figures is:
a) $1.6048 g / c m^{3}$
b) $1.69 g / c m^{3}$
c) $1.7 g / c m^{3}$
d) $1.695 g / c m^{3}$

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} Density = \frac{4.237 g}{2.5 {cm}^{3}} = 1.6948 \end{aligned}

After rounding off the number, we get density =1.7

Question:4

Answer:

The answer is the option

(c) 1.7 g c m^{- 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 c m^{- 3}

So, we get the above no. after rounding up 2 significant figures.

Question:5

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 a n d 2.74

Explanation:

(i) 2.745 \to 2.74

Here the fourth digit is five and is preceded by, viz., an even no. Hence, it becomes 2.74.

(i i) 2.745 \to 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:6

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

Answer:

The answer is the option (i)

164 \pm 3 c m^{2}

Explanation: Given :

I = 16.2 c m, a n d Δ I = 0.1

b = 10.1 c m, Δ b = 0.1

I = 16.1 \pm 0.1 a n d b = 10.1 \pm 0.1

Now, area (a) =

I \times b

= 16.2 \times 10.1

= 163.62 c m^{2}

= 164 c m^{2}

Δ \frac{A}{A} = \frac{Δ I}{I} + \frac{Δ b}{b}

= \frac{0.1}{16.2} + \frac{0.1}{10.1}

\frac{Δ A}{164} = (\frac{10.1 \times 0.1 + 16.2 \times 0.1}{16.2 \times 10.1})

Thus,

Δ A = 2.63 c m^{2}

We get

Δ A = 3 c m^{2}

by rounding off

Δ I a n d Δ b

T h u s, Δ A = (163 \pm 3) c m^{2}

Question:7

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 option (c): Tension & surface tension
Explanation: We know that,
Work = force

\times

displacement

= [M L^{2} T^{- 2}]

& Torque = force

\times

distance

= [M L^{2} T^{- 2}]

Hence, their dimensions are the same.
(b) Dimensions of-
Angular momentum

= [M L^{2} T^{- 1}]

& Planck’s constant

= [M L^{2} T^{- 1}]

Hence, their dimensions are the same.
(c) Dimensions of-
Tension

= [M L T^{- 2}]

Surface tension

= [M L^{0} T^{- 2}]

Their dimensions are not the same, hence opt (c).
(d) Dimensions of-
Impulse

= [M L T^{- 1}]

Momentum

= [M L T^{- 1}]

Thus, their dimensions are also the same.

Question:8

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 = A B = 2.5 \times 0.10 = 0.25 m

\frac{Δ x}{x} = \frac{Δ A}{A} + \frac{Δ B}{B}

= \frac{0.5}{2.5} + \frac{0.01}{0.10}

= \frac{0.075}{0.25}

Thus,

Δ x = 0.075 ≅ 0.08

Thus,

A B = (0.25 \pm 0.08) m

Question:9

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{A B} = \sqrt{1.0 \times 2.0} = \sqrt{2}

= 1.414 m

After rounding up 1.0 &2.0, we got two significant features, i.e.

X = \sqrt{A B}

= 1.4

Now,

\frac{Δ X}{X} = \frac{1}{2} [\frac{Δ A}{A} + \frac{Δ B}{B}]

\frac{Δ X}{1.4} = 0.1 [\frac{2.0 + 1.0}{1.0 \times 2.0}]

Δ X = 0.1 (\frac{3}{2}) ((1.4)

= 0.21 m

after rounding up
Thus, option (d) is the right answer.

Question:10

Which of the following measurements is most precise?
a) $5.00 m m$
b) $5.00 c m$
c) $5.00 m$
d) $5.00 k m$

Answer:

The answer is option (a),

5.00 m m

Explanation: The smallest unit is mm, and all the measurements are up to two decimal places. Hence, $5.00 mm is the most precise.

Question:11

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

Answer:

The answer is option (a),

4.9 c m

Explanation : $| Δ a_{1} | = | 5 - 4.9 | = 0.1 c m, | Δ a_{2} | = | 5 - 4.805 | = 0.195 c m$
$| Δ a_{3} | = | 5 - 5.25 | = 0.25 c m, | Δ a_{4} | = | 5 - 5.4 | = 0.4 c m$
Thus,

| Δ a_{1} |

it is the mmost acurate.

Question:12

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 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} d y n e s}{10^{4} c m^{2}}

= 1.9 \times 10^{11 + 5 - 4}

= 1.9 \times 10^{12} d y n e / c m^{2}

Question:13

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, the dimensional formula of-
Energy(E) =

[P^{a} A^{b} T^{c}]

Momentum (P) =

[M L T^{- 1}]

Area (A) =

[L^{2}]

Time(T) =

[T^{1}]

Now, E = f.s

= [M L T^{- 2} L] . [M L^{2} T^{- 2}]

= [M L T^{- 1}]^{a} [L^{2}]^{b} [T]^{c} . [M^{a} L^{2 + 2 b} T^{- a + c}]

Now, let us compare the powers,

a = 1 a + 2 b = 2 - a + c = - 2

1 + 2 b = 2 - 1 + c = - 2

2 b = 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: MCQ II

Question:14

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) $y = a \sin \frac{2 π t}{T}$
b) $y = a \sin v t$
c) $y = \frac{a}{T} \sin (\frac{t}{a})$
d) $y = a \sqrt{2} [\sin (\frac{2 π t}{t}) - \cos (\frac{2 π t}{T})]$

Answer:

The answer is the option (b)

y = a \sin v t

and (c)

y = \frac{a}{T} \sin (\frac{t}{a})

Explanation: (b) Here, v.t is an angle, whose dimensions are –

[L T^{- 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]} = [L T^{- 1}]

, viz., not equal to the dimensions of y and the angle

\frac{t}{a} = [L T^{- 1}]

, which means that they are not dimensionless.

Question:15

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

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:16

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{[M L^{2} T^{- 2}]}{[T^{- 1}]}

= [M L^{2} T^{- 1}]

Now, Linear Impulse = F.t

= \frac{d p}{d t} . d t = d p

= m v = [M L T^{- 1}]

& Angular Impulse =

= τ . d t = \frac{d L}{d t} . d t

= d L = m v r

= [M] [L T^{- 1}] [L]

= [M L^{2} T^{- 1}]

Now, Linear momentum = mv

= [M L T^{- 1}]

& Angular momentum (L) = mvr

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

Question:17

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 (m_e)
b) universal gravitational constant (G)
c) charge of electron (e)
d) mass of proton (m_p)

Answer:

The answer is the option (a) Mass of electron (m_e) and (b) Universal gravitational constant (G) and (d) Mass of proton (mp)
Explanation: Dimensions of –
h (Planck’s constant)

= [M L^{2} T^{- 1}]

c (Speed of light in vacuum)

= \frac{s}{t}

= [L T^{- 1}]

Thus, dimension of hc

= [M L^{2} T^{- 1}] [L T^{- 1}]

= [M L^{3} T^{- 2}]

G = \frac{F r^{2}}{M_{1} M_{2}}

= \frac{[M L^{3} T^{- 2}]}{[M] [M]}

= [M^{- 1} L^{3} T^{- 2}]

Now,

\frac{h c}{G} = [M L^{3} T^{- 2}] / [M^{- 1} L^{3} T^{- 2}]

= [M^{2}]

Thus,

M = \sqrt{\frac{h c}{G}}

= [h^{\frac{1}{2}} c^{\frac{1}{2}} G^{\frac{1}{2}}]

Now,

\frac{h}{c} = [M L^{2} T^{- 1}] / [L T^{- 1}]

= [M L]

= \sqrt{\frac{h c}{G}} \times L

Now,

L = \frac{h}{c} \times \sqrt{\frac{G}{h c}}

= \frac{\sqrt{G h}}{\frac{c^{3}}{2}}

= [G^{\frac{1}{2}} h^{\frac{1}{2}} c^{\frac{- 3}{2}}]

Also,

c = [L T^{- 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:18

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{[M L T^{- 2}]}{[L^{2}]}

..... this also verifies the opt (a).
Now let us compare it with other options-

(b) \frac{E}{V} = \frac{[M L^{2} T^{- 2}]}{[L^{3}]}

= [M L^{- 1} T^{- 2}]

= Dimesions of P
Thus, opt (b) is verified as well.

(c) \frac{E}{A} = \frac{[M L^{2} T^{- 2}]}{[L^{3}]}

= [M L^{0} T^{- 2}]

...............viz., not equal to Dimensions of P

(d) \frac{F}{V} = \frac{[M L T^{- 2}]}{[L^{3}]}

= [M L^{- 2} T^{- 2}]

.................viz., again not equal to dimensions of P
Thus, opt (c) & (d) are discarded.

Question:19

Which of the following is 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:20

The radius of atom is of the order of $1 A^{\circ}$ 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:

Radius of atom $= 1 A^{\circ} = 10^{- 10} m$
Radius of nucleus $= 1$ fermi $= 10^{- 15} m$
Volume of atom $= V_{A} = \frac{4}{3} π R_{A}^{3}$
Volume of atom $= \frac{4}{3} π R_{A}^{3}$
Volume of nucleus $V_{N} = \frac{4}{3} π R_{N}^{3}$

$\frac{V_{A}}{V_{N}} = \frac{\frac{4}{3} π R_{A}^{3}}{\frac{4}{3} π R_{N}^{3}}$
$\begin{aligned} = {(\frac{R_{A}}{R_{N}})}^{3} \\ = {(\frac{10^{- 10}}{10^{- 15}})}^{3} \\ = 10^{15} \end{aligned}$

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 \times 10^{23}

carbon atoms

(_{6} C^{12}) = 12 g m

Thus, the mass of one

_{6} C^{12}

atom

= \frac{12}{6.023} \times 10^{- 23} g m

Now , 1 amu

= \frac{1}{12} \times \frac{12}{6.023} \times 10^{- 23} g m

= 1.66 \times 10^{- 24} g m

Thus, 1 amu

= 1.66 \times 10^{- 27} k g

Question:23

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

Answer:

(i)

θ

is a dimensionless physical quantity since it is represented by an angle viz. equal to arc/radius.
(ii) As

θ

is dimensionless, all its powers will also be dimensionless. Here the first term is 1 viz. also dimensionless.
Hence, the L.H.S.

f (θ)

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 ${(\frac{1}{2})}^{o}$ 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 the sun's to Earth's diameters.

Answer:

(a) Here we know that,

θ =

arc/radius

= \frac{R_{e}}{60 R_{e}}

= \frac{1}{60} r a d

Now, the angle from the moon to the diameter of the earth will be

2 θ = 2 (\frac{1}{60}) (\frac{180}{π})

= \frac{6^{o}}{π}

≅ \frac{6}{3.14} ≅ 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 r_em = x m be the distance between the earth and the moon
Now, the distance between the sun and the earth (r_se) = 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 earth are equal.
Thus,

θ_{m} = θ_{s}

......... ( the angle from the earth to the moon is

θ {_{m}}^{o}

& angle from sun to earth is

θ {_{s}}^{o}

Now since

θ_{m} = θ_{s}

\frac{D_{m}}{r_{e m}} = \frac{D_{s}}{r_{x}}

\frac{D_{m}}{x} = \frac{D_{s}}{400 x}

\frac{D_{s}}{D_{M}} = 400

Thus,

4 D_{m} = D_{e} o r D_{m} = \frac{D_{e}}{4}

Thus,

\frac{D_{s}}{\frac{D_{e}}{4}} = 400

\frac{4 D_{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 a 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: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 the galaxy to earth =

10^{25} m .

We know that speed of light =

3 \times 10^{8} m / s

Thus, the required time = distance / speed
=

\frac{10^{25}}{3} \times 10^{- 8} s e c

= \frac{1}{3} \times 10^{17} s e c

= 3.3 \bar{3} \times 10^{16} s e c

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 the parts on the Vernier scale, n = 50
Now, the no. Of parts of M.S. coinciding with n parts of the 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,

Ω_{s} = Ω_{m}

Thus,

\frac{π (\frac{D_{s}}{2})^{2}}{r {_{s}}^{2}} = \frac{π (\frac{D_{m}}{2})^{2}}{r {_{m}}^{2}}

i.e.,

\frac{D_{s}^{2}}{4 r_{s}^{2}} = \frac{D_{m}^{2}}{4 r_{m}^{2}}

Thus,

\frac{4 R_{s}^{2}}{4 r_{s}^{2}} = \frac{4 R_{m}^{2}}{4 r_{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 \times 10 \times 100^{- 2} = 100

\frac{10 M}{10000} = 100

Thus,

M = 10^{5} k g, L = 10^{1} m, F = 10^{2} N a n d T = 10^{2} s e c o n d s

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} - k g

k = 9 \times 10^{9} N - m^{2} - C^{- 2} (V a c u u m)

(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 $\frac{π}{6}$ subtends an angle of at the centre.

Answer:

Given :radius = 31 cm
Angle =

\frac{π}{6}

Length of arc = ?
Now, we know that
Angle = length of arc/radius of arc

\frac{π}{6} = \frac{x}{31}

x = 31 \times \frac{π}{6}

= 31 \times \frac{3.14}{6}

x = 16.22 c m

Question:33

Calculate the solid angle subtended by the periphery of an area of $1 c m^{2}$ at a point situated symmetrically at a distance of 5 cm from the area.

Answer:

We know that,
Solid angle

(Ω)

=area/ (distance)²
=

\frac{1 c m^{2}}{(5 c m)^{2}}

= \frac{1}{25}

= 4 \times 10^{- 2} s t e r a d i a n

Question:34

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

Answer:

According to the principle of homogeneity, the dimensional formula on L.H.S. & R.H.S. is equal.
Thus, the dimension of

A \sin (ω t - k x)

= Dimension of y

ω t - k x

has no dimensions because it is an angle of sin.

\frac{2 π}{T} t = K x

[M^{0} L^{0} T^{0}] = k [L]

Thus,

ω

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

| Δ t_{1} | = | \bar{t} - t_{1} | = | 39.7 - 39.6 | = | 0.1 | = 0.1 s

| Δ t_{2} | = | \bar{t} - t_{2} | = | 39.7 - 39.9 | = | 0.2 | = 0.2 s

| Δ t_{3} | = | \bar{t} - t_{3} | = | 39.7 - 39.5 | = | 0.2 | = 0.2 s

Now, Mean absolute error

= \frac{0.1 + 0.2 + 0.2}{3}

= \frac{0.5}{3} ≅ 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 $α$ kg, unit of length $β$ 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

= [M L^{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 k g

M_{2} = α k g

L_{1} = 1 m

L_{2} = β m

T_{1} = 1 s e c o n d

T_{2} = γ s e c

Thus,

n_{2} = 5 [\frac{1 k g}{α k g}]^{1} [\frac{1 m}{β m}]^{2} [\frac{1 s}{γ s}]^{- 2}

= 5 [α^{- 1} β^{- 2} γ^{2}]

Thus, by the new system,

[α^{- 1} β^{- 2} γ^{2}]

Question:37

The volume of a liquid flowing out per second of a pipe of length l and radius r is written by a student as
$v = \frac{π}{8} \frac{P r^{4}}{η l}$
where P is the pressure difference between the two ends of the pipe and η is the coefficient of viscosity of the liquid having dimensional formula $M L^{- 1} T^{- 1}$ . Check whether the equation is dimensionally correct.

Answer:

Let us write dimensions of -
Volume per second

(V) = \frac{V}{T}

= [L^{3} T^{- 1}]

P = \frac{F}{A} = [M L^{- 1} T^{- 2}]

γ = [L]

η = [M L^{- 1} T^{- 1}]

l = [L]

Now, Dimensions of L.H.S.

= [M^{0} L^{3} T^{- 1}]

& dimensions of R.H.S.

= [M^{0} L^{3} T^{- 1}]

The equation is dimensionally correct since the dimensions of both sides are equal.

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, the percentage error in

a = (\frac{Δ a}{a}) (100) = 1^{o} /_{o}

As per the information given in the question, the Percentage error in

b = (\frac{Δ b}{b}) (100) = 2^{o} /_{o}

As per the information given in the question, the Percentage error in

c = (\frac{Δ c}{c}) (100) = 3^{o} /_{o}

As per the information given in the question, the Percentage error in

d = (\frac{Δ d}{d}) (100) = 4^{o} /_{o}

Percentage error in

X = (\frac{Δ x}{x}) (100)

\frac{Δ X}{X} \times 100 = \pm (2 \times 1 + 3 \times 2 + 2.5 \times 3 + 2 \times 4) \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 $P = E I^{2} m^{- 5} G^{- 2}, E, m, I a n d G$ denote energy, mass, angular momentum and gravitational constant, respectively. Show that P is a dimensionless quantity.

Answer:

The dimensional formula of –

E = [M L^{2} T^{- 2}]

I = [M L^{2} T^{- 1}]

G = [M^{- 1} L^{3} T^{- 2}]

Now let us find out the dimension of P

P = E I^{2} m^{- 5} G^{- 2}

[P] = \frac{[E] [I^{2}]}{[m^{5}] [G^{2}]}

= \frac{[M L^{2} T^{- 2}] [M L^{2} T^{- 1}]^{2}}{[M]^{5} [M^{- 1} L 3 T^{- 2}]^{2}}

= \frac{M^{3} L^{6} T^{- 4}}{M^{3} L^{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{[M L^{2} T^{- 2}]}{[T^{- 1}]}

= [M L^{2} T^{- 1}]

C = [L T^{- 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 [M L^{2} T^{- 1}]^{a} [L T^{- 1}]^{b} [M^{- 1} L^{3} T^{- 1}]^{c}

[M^{1} L^{0} T^{0}] = k [M^{a - c} L^{2 a + b + 3 c} T^{- a - b - 2 c}]

Comparing the powers on R.H.S. & L.H.S.

a - c = 1

Thus,

a = c + 1

2 a + b + 3 c = 0 . . . . . (i)

- a - b - 2 c = 0 . . . . . (i i)

Substituting the value of a in eq (i) & (ii)

2 (c + 1) + b + 3 c = 0

2 c + 2 + b + 3 c = 0

b + 5 c = - 2 . . . . . (i i i)

- (c + 1) - b - 2 c = 0

- b - 3 c = 1 . . . . . . (i v)

By adding (iii) & (iv), we get,

c = \frac{1}{2}

n o w, 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 = k h^{\frac{1}{2}} C^{\frac{1}{2}} G^{\frac{- 1}{2}}

M = k \sqrt{\frac{h c}{G}}

Now, let

L α C^{a} h^{b} G^{c}

Thus,

[L^{1}] = k [L T^{- 1}]^{a} [M L^{2} T^{- 1}]^{b} [M^{- 1} L^{3} T^{2}]^{c}

= k [M^{b - c} L^{a + 2 b + 3 c} T^{- a - b - 2 c}]

Equating the powers on both sides

b - c = 0

Thus,

b = c

a + 2 b + 3 c = 1 . . . . . . (i)

- a - b - 2 c = 0 . . . . (i i)

Now, since

b = c

a + 2 b + 3 b = 1

becomes

a + 5 b = 1 . . . . . (i i i)

& eq. (ii)

a + b + 2 c = 0

becomes

a + 3 b = 0 . . . . . . (i v)

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{h G}{c^{3}}}

Let us consider

T α G^{a} h^{b} C^{c}

Thus,

[M^{0} L^{0} T^{- 1}] = k [M^{- 1} L^{3} T^{- 2}]^{a} [M L^{2} T^{- 1}] [L T^{- 1}]^{c}

= k [M^{- a + b} L^{3 a + 2 b + c} T^{- 2 a - b - c}]

- a + b = 0 . . . (i)

3 a + 2 b + c = 0 . . . . . . . (i i)

- 2 a - b - c = 1 . . . . (i i i)

On adding eq (ii) & (iii) we get,

a + b = 1 . . . . . (i v)

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{h G}{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 $T = \frac{k}{R} \sqrt{\frac{r 3}{g}}$ 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} α r^{3}

T α r^{\frac{3}{2}}

Now, R & g also affects T
Thus,

T α g^{a} R^{b} r^{\frac{3}{2}}

T = K [L T^{- 2}]^{a} [L^{b}] [L]^{\frac{3}{2}}

[T^{1}] = K [L^{a + b + \frac{3}{2}} T^{- 2 a}]

Now, comparing M, L & T according to the principle of homogeneity

a + b + \frac{3}{2} = 0 . . . . . (i)

- 2 a = 1 . . . . . (i i)

a = \frac{- 1}{2}

Thus,

\frac{- 1}{2} + b + \frac{3}{2} = 0

b + 1 = 0

B = - 1

- 2 a = 1

T = K g^{\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, the 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 the 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} m l

= \frac{1}{400} m l

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 \times 10^{5} A . U .

(b)Angle of the sun’s diameter

{(\frac{1}{2})}^{o}

is subtended by 1 A.U. since the distance from the sun increases angle subtended in the same ratio.
Now,

2 \times 10^{5} A . U .

will form an angle of

θ = {(\frac{1}{4 \times 10^{5}})}^{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 $1 M e V = 1.6 \times 10^{- 13} J$ , 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 $1 u = 931.5 M e V$ . The teacher points out that the relation is dimensionally incorrect. Write the correct relation.

Answer:

(a) Given :

m = 1

u = 1.67 \times 10^{- 27} k g

c = 3 \times 10^{8} m / s

By the formula

E = m c^{2}

E = 1.67 \times 10^{- 27} \times 3 \times 10^{8} \times 3 \times 10^{8}

= 1.67 \times 10^{- 27 + 16} \times 9 J

= \frac{(1.67) (9) (10^{- 11})}{(1.6) (10^{- 13})} M e V

= 939.4 M e v

≅ 931.5 M e v

(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 = m c^{2} \to 1 u c^{2} ≅ 931.5 M e v

, 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

Introduction
The International System Of Unit
Measurement Of Length
Measurement Of Mass
Measurement Of Time
Accuracy, Precision Of Instruments, And Errors Of Measurements
Significant Figures
Dimensions Of Physical Quantities
Dimensional Formulae And Dimensional Equations
Applications Of Dimensional Analysis

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

This chapter helps students understand how measurement methods have changed over time and teaches them about basic units. With the help of NCERT Exemplar Class 11 Physics Chapter 2 Solutions, students can connect these concepts to real-life measuring tools. It also shows how to use accurate measurements to solve practical problems—like finding the mass of rain clouds, speed of an aircraft using dimensional analysis, or even calculating the distance between stars.

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

NCERT Exemplar Class 11 Physics Chapter 2 Solutions help students learn about different units used around the world and the standard unit system accepted internationally. The chapter explains how to measure physical quantities like length and mass, as well as abstract ones like time.

You'll also learn about dimensional formulas and equations, which are super useful in solving real-life problems using dimensional analysis. The chapter also explains why significant figures matter in measurements.

It teaches you the difference between accuracy and precision and how to deal with errors in measurements. You'll also get to know the types of errors that can occur in instruments and how to spot and fix them easily.

NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

Chapter 2 Units and Measurement

Chapter 3 Motion in a Straight Line

Chapter 4 Motion in a Plane

Chapter 5 Laws of Motion

Chapter 6 Work, Energy, and Power

Chapter 7 System of Particles and Rotational Motion

Chapter 8 Gravitation

Chapter 9 Mechanical Properties of Solids

Chapter 10 Mechanical Properties of Fluids

Chapter 11 Thermal Properties of Matter

Chapter 12 Thermodynamics

Chapter 13 Kinetic Theory

Chapter 14 Oscillations

Chapter 15 Waves

NCERT Exemplar Class 11th Solutions

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Check Class 11 Physics Chapter-wise Solutions

NCERT solutions for Class 11 Physics Chapter 1 Units and Measurement

NCERT solutions for Class 11 Physics Chapter 2 Motion in a straight line

NCERT solutions for Class 11 Physics Chapter 3 Motion in a Plane

NCERT solutions for Class 11 Physics Chapter 4 Laws of Motion

NCERT solutions for Class 11 Physics Chapter 5 Work, Energy and Power

NCERT solutions for Class 11 Physics Chapter 6 System of Particles and Rotational Motion

NCERT solutions for Class 11 Physics Chapter 7 Gravitation

NCERT solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids

NCERT solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

NCERT solutions for Class 11 Physics Chapter 10 Thermal Properties of Matter

NCERT solutions for Class 11 Physics Chapter 11 Thermodynamics

NCERT solutions for Class 11 Physics Chapter 12 Kinetic Theory

NCERT solutions for Class 11 Physics Chapter 13 Oscillations

NCERT solutions for Class 11 Physics Chapter 14 Waves

Also, Read NCERT Solution subject-wise -

Check NCERT Notes subject-wise -

Also, Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (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.

4. Why is learning about significant figures important?

Significant figures show how accurate your measurement is. They help you write your answers correctly and avoid silly mistakes in calculations

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NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reactions

NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions For Class 11 Biology Chapter 10 - Cell Cycle and Cell Division

NCERT Solutions for Class 11 Biology Chapter 11 Photosynthesis in Higher Plants

NCERT Solutions for Class 11 Physics Chapter 15 Waves

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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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NCERT Exemplar Class 11 Physics Solutions Chapter 2 Units and Measurement

NCERT Exemplar Class 11 Physics Solutions Chapter 2: MCQ I

NCERT Exemplar Class 11 Physics Solutions Chapter 2: MCQ II

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

NCERT Exemplar Class 11 Physics Solutions Chapter 2: Short Answer

NCERT Exemplar Class 11 Physics Solutions Chapter 2: Long Answer

Class 11 Physics NCERT Exemplar Solutions Chapter 2 Topics:

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

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-Wise

NCERT Exemplar Class 11th Solutions

Check Class 11 Physics Chapter-wise Solutions

Also, Read NCERT Solution subject-wise -

Check NCERT Notes subject-wise -

Also, Check NCERT Books and NCERT Syllabus here

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