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NCERT Exemplar Class 11 Physics Solutions Chapter 2 Units and Measurement

NCERT Exemplar Class 11 Physics Solutions Chapter 2 Units and Measurement

Edited By Vishal kumar | Updated on Jul 16, 2025 01:02 AM IST

Have you ever thought how scientists determine the extremely accurate distance between the Earth and the Moon, or how engineers construct machines in the range of one millimetre precision? All this becomes achievable due to the precise units and measurements, and they are considered the foundation of any scientific observations. NCERT Exemplar Class 11 Physics Chapter 2 explores the methodologies of making precise measurements of the physical quantities using standard practices. You will learn about concepts such as SI units, dimensional analysis, significant figures, error types and the rules of checking the physical equations with the help of dimension. Such knowledge is critical to the accurate and clear solution of numerical problems in physics.

New: JEE Main/NEET 2027 - Physics Important Formulas for Class 10

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This Story also Contains
  1. NCERT Exemplar Class 11 Physics Solutions Chapter 2: MCQ I
  2. NCERT Exemplar Class 11 Physics Solutions Chapter 2: MCQ II
  3. NCERT Exemplar Class 11 Physics Solutions Chapter 2: Very Short Answer
  4. NCERT Exemplar Class 11 Physics Solutions Chapter 2: Short Answer
  5. NCERT Exemplar Class 11 Physics Solutions Chapter 2: Long Answer
  6. Class 11 Physics NCERT Exemplar Solutions Chapter 2 Topics:
  7. NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
NCERT Exemplar Class 11 Physics Solutions Chapter 2 Units and Measurement
NCERT Exemplar Class 11 Physics Solutions Chapter 2 Units and Measurement

NCERT Exemplar Class 11 Physics Solutions Chapter 2 answer is in form of steps with a number of questions of different types being covered such as multiple choice questions (MCQs) and the short answer type, the long answer and the very long answer. These solutions are made by subject professionals which make even the difficult subjects easy to understand. With these guideline solutions, the students will be able to improve their problem-solving abilities, reduce the number of calculation mistakes, and enhance their background in terms of CBSE exams, JEE and NEET, and other 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 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. The sum of the numbers can be calculated as 663.821 arithmetically. The number with least decimal places is 227.2 is correct to only one decimal place. The final result should, therefore be rounded off to one decimal place i.e., 663.8.

Question:3

The mass and volume of a body are 4.237g and 2.5cm3 , respectively. The density of the material of the body in correct significant figures is:
a) 1.6048g/cm3
b) 1.69g/cm3
c) 1.7g/cm3
d) 1.695g/cm3

Answer:

The answer is the option (c)1.7gcm3
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

As rounding off the number upto 2 significant figures, we get density =1.7gcm3

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)2.74and2.74
Explanation:

(i)2.7452.74
Here the fourth digit is five and is preceded by, viz., an even no. Hence, it becomes 2.74.
(ii)2.7352.74
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 16.2cm and 10.1cm, respectively. The area of the sheet in appropriate significant figures and error is:
a) 164±3cm2
b) 163.62±2.6cm2
c) 163.6±2.6cm2
d) 163.62±3cm2

Answer:

The answer is the option (i) 164±3cm2
Explanation: Given :
l=16.2cm,andΔl=0.1
b=10.1cm,Δb=0.1
l=16.1±0.1andb=10.1±0.1
Now, Area (A) =I×b
A=16.2×10.1
A=163.62cm2
A=164cm2
ΔAA=Δll+Δbb
ΔAA=0.116.2+0.110.1
ΔA164=(10.1×0.1+16.2×0.116.2×10.1)
Thus, ΔA=2.63cm2
We get ΔA=3cm2 by rounding off ΔlandΔb
Thus,ΔA=(164±3)cm2

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 option (c): Tension & surface tension
Explanation: We know that,
Work = force × displacement =[ML2T2]
& Torque = force × distance =[ML2T2]
Hence, their dimensions are the same.
(b) Dimensions of-
Angular momentum =[ML2T1]
& Planck’s constant =[ML2T1]
Hence, their dimensions are the same.
(c) Dimensions of-
Tension =[MLT2]
Surface tension=[ML0T2]
Their dimensions are not the same, hence opt (c).
(d) Dimensions of-
Impulse =[MLT1]
Momentum =[MLT1]
Thus, their dimensions are also the same.

Question:7

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

Answer:

The answer is the option (a) (0.25±0.08)m
Explanation: By using the given values of A & B,
X=AB=2.5×0.10=0.25m
ΔXX=ΔAA+ΔBB
ΔXX=0.52.5+0.010.10
ΔXX=0.0750.25
Thus, ΔX=0.0750.08
Thus, AB=(0.25±0.08)m

Question:8

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

Answer:

The answer is the option (d) 1.4±0.2m
There are two significant figures in 1.0 & 2.0
AB=1.0×2.0=2=1.414m
After rounding up 1.0 &2.0, we got two significant features, i.e.
X=AB=1.4
Now, ΔXX=12[ΔAA+ΔBB]


ΔX1.4=0.1[2.0+1.01.0×2.0]


ΔX=0.1(32)((1.4)=0.21 m after rounding up

Question:9

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

Answer:

The answer is option (a), 5.00mm
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:10

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

Answer:

The answer is option (a), 4.9cm
Explanation :

|Δa1|=|54.9|=0.1cm,|Δa2|=|54.805|=0.195cm
|Δa3|=|55.25|=0.25cm,|Δa4|=|55.4|=0.4cm
Thus,|Δa1| it is the most acurate.

Question:11

Young’s modulus of steel is 1.9×1011N/m2. When expressed in CGS units of dynes/cm2, it will be equal to:
a) 1.9×1010
b) 1.9×1011
c) 1.9×1012
d) 1.9×1013

Answer:

The answer is option (c) 1.9×1012
Explanation: Young's modulus (Y) =1.9×1011N/m
Converting into dyne/cm2-
Y=1.9×1011×105dynes104cm2


Y=1.9×1011+54


Y=1.9×1012dyne/cm2

Question:12

If momentum (p), area (A), and time (T) are taken to be fundamental quantities, then energy has the dimensional formula
a)(p1A1T1)
b) (p2A1T1)
c) (p1A1/2T1)
d) (p1A1/2T1)

Answer:

The answer is the option d) (P1A1/2T1)

Given, fundamental quantities are momentum (p), area (A) and time (T). We can write energy E as

EpaAbTcE=kpaAATc

where k is dimensionless constant of proportionality.

Dimensions of

E=[E]=[ML2 T2] and [p]=[MLT1][A]=[L2][T]=[T][E]=[K][p]a[ A]b[ T]c


Putting all the dimensions, we get

ML2 T2=[MLT1]a[ L2]b[ T]c=Ma L2b+a Ta+c

By principle of homogeneity of dimensions,

a=1,2b+a=22b+1=2b=1/2a+c=2c=2+a=2+1=1 Hence, E=pA1/2T1

NCERT Exemplar Class 11 Physics Solutions Chapter 2: MCQ II

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=asin2πtT
b) y=asinvt
c) y=aTsin(ta)
d) y=a2[sin(2πtt)cos(2πtT)]

Answer:

The answer are the options (b) y=asinvt and (c) y=aTsin(ta)

Now, by principle of homogeneity of dimensions LHS and RHS of (a) and (d) will be same and is L.
For (c)

[LHS]=L[RHS]=L T=LT1[LHS][RHS]


Hence, (c) is not correct option.
In option (b) dimension of angle is [vt]i.e., L

 RHS =L.L=L2 and LHS=L LHS  RHS .


So, option (b) is also not correct.

Question:14

If P, Q, R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity?
a)(PQ)R
b)PQR
c)PQR
d)(PRQ2)R
e)(R+Q)P

Answer:

The answer are the options a)(PQ)R and e)(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 option (d), dimensions of PR & Q2 may be equal.
(iii) In option (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 option (c).

Question:15

Photon is quantum of radiation with energy E = hν where ν 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 are the options (b) Angular Impulse and (d) Angular momentum
Explanation: It is given that E = hν
Hence, h (Planck’s constant)=Eν=[ML2T2][T1]=[ML2T1]
Now, Linear Impulse = F.t=dpdt.dt=dp=mv=[MLT1]
& Angular Impulse =τ.dt=dLdt.dt=dL=mvr
dL=[M][LT1][L]=[ML2T1]
Now, Linear momentum = mv=[MLT1]
& Angular momentum (L) = mvr= [ML2T1]

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 are the optsion (a) Mass of electron (me) and (b) Universal gravitational constant (G) and (d) Mass of proton (mp)

 We know that dimension of h=[h]=[ML2 T1][c]=[LT1],[me]=M[G]=[M1 L3 T2][e]=[AT],[mp]=[M][hcG]=[ML2 T1][LT1][M1 L3 T2]=[M2]M=hcG

Similarly,

hc=[ML2 T1][LT1]=[ML]

 As, L=hcM=hcGhc=Ghc3/2c=LT1[T]=[L][c]=Ghc3/2c=Ghc5/2


Hence, (a), (b) or (d) any can be used to express L,M and T in terms of three chosen fundamental quantities.

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 (b) Energy/Volume
Explanation:

Let us start with calculating the dimension of pressure,
We know that Pressure(P) =FA=[MLT2][L2]
This also verifies the option (a).
Now let us compare it with other options-
(b)EV=[ML2T2][L3]=[ML1T2]= Dimesions of P
Thus, option (b) is verified as well.
(c)EA=[ML2T2][L3]=[ML0T2]

This is not equal to the Dimensions of P
(d)FV=[MLT2][L3]=[ML2T2]

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

Question:18

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

Answer:

The answer are the options (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 1A 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 =1A=1010 m
Radius of nucleus =1 fermi =1015 m
Volume of atom =VA=43πRA3
Volume of nucleus VN=43πRN3

VAVN=43πRA343πRN3


VAVN=(RARN)3=(10101015)3=1015

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=112 the mass of one C12 atom
Mass of 6.023×1023 carbon atoms (6C12)=12gm
Thus, the mass of one 6C12 atom =126.023×1023gm
Now , 1 amu =112×126.023×1023gm
1amu=1.66×1024gm
Thus, 1 amu =1.66×1027kg

Question:23

A function f(θ) is defined as
f(θ)=1θ+θ22!θ33!+θ44!+.........
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 (12)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 =Re60Re=160rad
Angular Diameter
Now, the angle from the moon to the diameter of the earth will be
2θ=2(160)(180π)=6oπ63.142o
(b) If the moon is seen from the earth diametrically, the angle will be 12o
& if the earth is seen from the moon, the angle will be 2o
Thus, the size of the moon/ size of the earth = 12o2o=14
Therefore, as compared to the earth, the size of the moon is 14 the size(diameter) of the earth.
(c)Let rem = x be the distance between the earth and the moon
Now, the distance between the sun and the earth (rse) = 400x
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, θm=θs......... ( the angle from the earth to the moon is θmo & angle from sun to earth is θso
Now since θm=θs
Dmrem=Dsrx

Dmx=Ds400x

DsDM=400

Thus, 4Dm=De  orDm=De4

Thus, DsDe4=400

4DsDe=400

DsDe=4004=100

Therefore, Ds=100De

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 = 110 sec

Digital watch = 1100 sec &

Atomic clock = 11013 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 = 1025m.
We know that speed of light = 3×108m/s
Thus, the required time, t = distance / speed
t=10253×108sec
t=13×1017sec
t=3.33¯×1016sec

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:

Consider the diagram given above
Rme= Distance of moon from earth
Rse = Distance of sun from earth
Let angle made by sun and moon is θ, we can write

θ=AsunRse2=AmoonRme2
Here,
Asun = Area of the sun
Amonn = Area of the moon

θ=πRs2Rse2=πRm2Rme2(RsRse)2=(RmRme)2RsRse=RmRmeRsRm=RseRme

(Here, radius of sun and moon represents their sizes respectively)

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 = [M1L1T2]=100 N ..........(i)
Length = [L1]= 10 m ...........(ii)
Time = [T1]= 100 sec ...........(iii)
Now let us substitute the values of eq. (ii) & (iii) in eq. (i)
We get,
M×10×1002=100

10M10000=100

Thus, M=105kg,L=101m,F=102NandT=102seconds

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=6.67×1011Nm2kg
k=9×109Nm2C2(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 π6 subtends an angle of at the centre.

Answer:

Given :radius = 31 cm
Angle = π6
Length of arc = ?
Now, we know that
Angle = length of arc/radius of arc
π6=x31
x=31×π6
x=31×3.146
x=16.22cm

Question:33

Calculate the solid angle subtended by the periphery of an area of 1cm2 at a point situated symmetrically at a distance of 5 cm from the area.

Answer:

We know that solid angle
Ω= Area ( Distance )2=1 cm2(5 cm)2=125=4×102 steradian 

Question:34

The displacement of a progressive wave is represented by y=Asin(wtkx), 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 Asin(ωtkx) = Dimension of y
ωtkx has no dimensions because it is an angle of sin.
2πTt=Kx
[M0L0T0]=k[L]
Thus, ω has no dimension & dimension of k=[M0L1T0].

Question:35

Time for 20 oscillations of a pendulum is measured as t1=39.6s;t2=39.9s;t3=39.5s. What is the precision in the measurements? What is the accuracy of the measurement?

Answer:

Given data :
t1=39.6s
t2=39.9s
and t3=39.5s
L.C. of the instrument is 0.1 sec, hence, Precision (LC) = 0.1 sec.
Now,
For 20 oscillations, the mean value will be
=39.6+39.9+39.53
=1192
=39.7s
Now the absolute errors in measurement
|Δt1|=|t¯t1|=|39.739.6|=|0.1|=0.1s
|Δt2|=|t¯t2|=|39.739.9|=|0.2|=0.2s
|Δt3|=|t¯t3|=|39.739.5|=|0.2|=0.2s

Now, Mean absolute error =0.1+0.2+0.23=0.530.2s
& Accuracy of measurement =±0.2s

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 =[ML2T2]
Let n1 be the SI system of unit & n2 be the new system of unit
Thus, n2u2=n1u1
n2=n1(u1u2)=n1[M1M2]1[L1L2]2[T1T2]2
Now,
M1=1kg & M2=αkg
L1=1m & L2=βm
T1=1second & T2=γsec

Thus, n2=5[1kgαkg]1[1mβm]2[1sγs]2=5[α1β2γ2]
Thus, by the new system, [α1β2γ2].

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=π8Pr4ηl
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 ML1T1. Check whether the equation is dimensionally correct.

Answer:

Let us write dimensions of -
Volume per second (V)=VT=[L3T1]
P=FA=[ML1T2]
γ=[L]
η=[ML1T1]
l=[L]
Now, Dimensions of L.H.S.=[M0L3T1]
& dimensions of R.H.S. =[M0L3T1]
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=a2b3c52d2. 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=(Δaa)(100)=1
As per the information given in the question, the Percentage error in b=(Δbb)(100)=2
As per the information given in the question, the Percentage error in c=(Δcc)(100)=3
As per the information given in the question, the Percentage error in d=(Δdd)(100)=4

Percentage error in X=(ΔXX)(100)

ΔXX×100=±(2×1+3×2+2.5×3+2×4)±(2+6+7.5+8)=±23.5o/o

Now, let us calculate the mean absolute error =±23.5100=±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=EI2m5G2,E,m,IandG denote energy, mass, angular momentum and gravitational constant, respectively. Show that P is a dimensionless quantity.

Answer:

The dimensional formula of –
E=[ML2T2]
I=[ML2T1]
G=[M1L3T2]
Now let us find out the dimension of P
P=EI2m5G2
[P]=[E][I2][m5][G2]=[ML2T2][ML2T1]2[M]5[M1L3T2]2

[P]=M3L6T4M3L6T4

[P]=[M0L0T0]
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=Ev=[ML2T2][T1]=[ML2T1]
C=[LT1]
G=[M1L3T2]
(i) Let us consider that Mass m is directly proportional to [h]a[C]b[G]c
[M1L0T0]=k[ML2T1]a[LT1]b[M1L3T1]c

[M1L0T0]=k[MacL2a+b+3cTab2c]

Comparing the powers on R.H.S. & L.H.S.
ac=1
Thus, a=c+1
2a+b+3c=0.....(i)
ab2c=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)b2c=0
b3c=1......(iv)
By adding (iii) & (iv), we get,
c=12
now,a=c+1
Thus, a=12+1=12
Substituting these values in eq. (iii)
b+5(1/2)=2
b=2+25
b=12
Thus, m=kh12C12G12
M=khcG
Now, let LαCahbGc
Thus, [L1]=k[LT1]a[ML2T1]b[M1L3T2]c
=k[MbcLa+2b+3cTab2c]
Equating the powers on both sides
bc=0
Thus,b=c
a+2b+3c=1......(i)
ab2c=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=12
Thus, c=12
And eq (iv) will be
a+3(12)=0
a=32
Therefore, I=kC32h12G12
L=khGc3
Let us consider TαGahbCc
Thus, [M0L0T1]=k[M1L3T2]a[ML2T1][LT1]c
=k[Ma+bL3a+2b+cT2abc]
a+b=0...(i)
3a+2b+c=0.......(ii)
2abc=1....(iii)
On adding eq (ii) & (iii) we get,
a+b=1.....(iv)
From eq (i) &(iv)
b=12
a=12
Thus, c=52
T=kG12h12C52
Thus, T=khGc5

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=kRr3g where k is a dimensionless constant and g is acceleration due to gravity.

Answer:

According to Kepler’s 3rd law of planetary motion,
T2αr3
Tαr32
Now, R & g also affects T
Thus,
TαgaRbr32
T=K[LT2]a[Lb][L]32
[T1]=K[La+b+32T2a]
Now, comparing M, L & T according to the principle of homogeneity
a+b+32=0.....(i)
2a=1.....(ii)
a=12
Thus, 12+b+32=0

b+1=0
B=1
2a=1
T=Kg12R1r32
i.e., T=KRr3g

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) Oleic acid does not dissolve in water hence, it is dissolved in alcohol.
(b) Lycopodium powder spreads over the entire surface of water when it is sprinkled evenly. When a drop of prepared solution is dropped on water, oleic acid does not dissolve in water. Instead it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls. We can therefore, measure the area over which oleic acid spreads.
(c) In each mL of solution prepared volume of oleic acid =120 mL×120=1400 mL
(d) Volume of n drops of this solution of oleic acid can be calculated by means of a burette and measuring cylinder and measuring the number of drops.
(e) If n drops of the solution make 1 mL , the volume of oleic acid in one drop will be 1(400)n mL.

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 = 1A.U.1 parsec

Thus, 1 parsec = 1A.U.1 arc sec

Thus, 1 parsec= 630(3600)11=206182.8=2×105A.U.

(b)Angle of the sun’s diameter (12)o is subtended by 1 A.U. since the distance from the sun increases angle subtended in the same ratio.
Now,
2×105A.U. will form an angle of θ=(14×105)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 =(1.25×106)o
Thus, angle =75×106 min
When it is seen with a telescope that has a magnification of 100, the angle formed will be 7.5×103 min, viz., less than a minute.
Hence, it can’t be observed by a telescope.

(c)

Given that Dmars Dearth =12....(i)
where D represents diameter.
From answer 25(e)
we know that,

Dearth Dsun =1100Dmars Dsun =12×1100[fromEq.(i)]

 At 1AU sun's diameter =(12) mar's diameter =12×1200=1400 At 12AU, mar's diameter =1400×2=(1200) With 100 magnification, Mar's diameter =1200×100=(12)=30


This is larger than resolution limit due to atmospheric fluctuations. Hence, it looks magnified.

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 1MeV=1.6×1013J, 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 1u=931.5MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.

Answer:

(a) Given :
m=1
u=1.67×1027kg
c=3×108m/s
By the formula E=mc2
E=1.67×1027×3×108×3×108

E=1.67×1027+16×9J

E=(1.67)(9)(1011)(1.6)(1013)MeV

E=939.4MeV931.5MeV

(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=mc21uc2931.5MeV, will be dimensionally correct.

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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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