Physics Ncert Exemplar 2019 Solutions for Class 11 Science Physics Chapter 2 - Units And Measurements

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Page No 5:

Question 2.1:

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

Answer:

According to the rules to calculate the number of significant figures in a given number ,

1.) All non zero numbers will be significant.
2.) A zero will be significant if it appears between two non zero numbers.
3.) Leading zeros cannot be significant.
4.) Trailing zeros will be significant.

In 0.06900 there are 2 trailing zeros + 2 non zero numbers hence, the total number of significant figures are 4.

Hence the correct answer is option (b).

Page No 5:

Question 2.2:

The sum 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:

In the addition of the numbers 436.32, 227.2 and 0.301
The result will have same number of decimal places as the decimal number present in the number having least number of decimal places. That means the result should have only one decimal place like in 227.2

436.32 + 227.2 + 0.301 = ( 663.821) = 664 after rounding off.

Hence the correct answer is option (b).

Page No 5:

Question 2.3:

The mass and volume of a body are 4.237 g and 2.5 cm³, 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:

We know that ,

$D e n s i t y = \frac{m a s s}{V o l u m e}$

The result will have same number of decimal places as the decimal number present in the number having least number of decimal places. That means the result should have only one decimal place like in 2.5.

$D e n s i t y = \frac{4.237}{2.5} = 1.6948 {gcm}^{- 3} = 1.7 {gcm}^{- 3} (after round off)$

Hence the correct answer is option (c).

Page No 6:

Question 2.4:

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:

Rules for rounding off digits:

1.) The last digit should be kept unchanged if the figure dropped is less than 5.
2.) The last digit should be increased by 1 if the figure dropped is greater than 5.
3.) If the figure dropped is 5 and if all the figure following 5 are zero or there is no figure after 5 then the last figure should be kept unchanged if that last figure is even.
4.) If the figure dropped is 5 and if all the figure following 5 are zero or there is no figure after 5 then the last figure should be increased by 1 if that last figure is odd.
5.) If the first figure dropped is 5 and there are any figures following the 5 that are not zero then the last figure should be increased by 1.

In case of 2.745 , the digit dropped will be 5 , the preceding digit will be unchanged as it is even,
Hence 2.745 will be round off to 2.74

In case of 2.735 , the digit dropped again will be 5 , the preceding digit will be increased by 1 as it is odd,
Hence 2.735 will be round off to 2.74

Hence the correct answer is option (d).

Page No 6:

Question 2.5:

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 ± 3 cm²
(b) 163.62 ± 2.6 cm²
(c) 163.6 ± 2.6 cm²
(d) 163.62 ± 3 cm²

Answer:

We have,
Length, l = 16.2 cm
Breadth, b = 10.1 cm
Area, $A = l \times b = 16.2 \times 10.1 = 163.2 {cm}^{2}$

Error in the quantity which is in the multiplication is depicted by :

$A = l \times b \frac{∆ A}{A} = \frac{∆ l}{l} + \frac{∆ b}{b}$ .... eq(1)

According to the problem,
$l = (16.2 \pm 0.1) cm b = (10.2 \pm 0.1) cm$

According to equation (1):

$\frac{∆ A}{A} = \frac{0.1}{16.2} + \frac{0.1}{10.1} \Rightarrow ∆ A = A (\frac{0.1}{16.2} + \frac{0.1}{10.1}) {cm}^{2} \Rightarrow ∆ A = 163.62 \times \frac{2.63}{163.62} {cm}^{2} \Rightarrow ∆ A = 2.63 c m^{2} = 3 {cm}^{2} (after round off) Thus, A = A \pm ∆ A = (164 \pm 3) {cm}^{2}$

Hence the correct answer is option (a).

Page No 6:

Question 2.6:

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:

Lets take up the option one by one :
Option (a)
Work = F.d
$W = [M L^{1} T^{- 2}] [L] W = [M L^{2} T^{- 2}] Torque = r \times F Torque = [M L^{2} T^{- 2}]$

Option(b)

$Angular Momentum = m v r L = [M] [L T^{- 1}] [L] L = [M L^{2} T^{- 1}] Planks constant (h) = \frac{E}{v} [h] = \frac{[M L^{2} T^{- 2}]}{[T^{- 1}]} [h] = [M L^{2} T^{- 1}]$

Option(c)

$T e n s i o n = F o r c e = m a = [M] [L T^{- 2}] T = [M L T^{- 2}] S u r f a c e T e n s i o n = \frac{F}{l} [S] = \frac{[M L T^{- 2}]}{[L]} [S] = [M T^{- 2}] Not same$

Option(d)
$Impulse (J) = F \times t [J] = [M L T^{- 2}] . [T] [J] = [M L T^{- 1}] Momentum (p) = m v [p] = [M] [L T^{- 1}] [p] = [M L T^{- 1}] same$

Hence the correct answer is option (c)

Page No 6:

Question 2.7:

Measure of two quantities along with the precision of respective measuring instrument is
A = 2.5 m s^–1 ± 0.5 m s^–1
B = 0.10 s ± 0.01 s
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:

Let, Z = AB
We have,
A = 2.5 m s^–1± 0.5 m s^–1
B = 0.10 s ± 0.01 s

$Z = A . B Z = (2.5) \times (0.1) Z = 0.25$

According to the formula:
$\frac{∆ Z}{Z} = \frac{∆ A}{A} + \frac{∆ B}{B} ∆ Z = Z (\frac{∆ A}{A} + \frac{∆ B}{B}) ∆ Z = 0.25 (\frac{0.5}{2.5} + \frac{0.01}{0.1}) ∆ Z = 0.25 \times 0.3 ∆ Z = 0.075 ∆ Z = 0.08 (after rounding off)$

$Thus, Z = 0.25 \pm 0.08$ m

Hence the correct answer is option (a)

Page No 6:

Question 2.8:

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

Answer:

We have,
A = 1.0 m ± 0.2 m
B = 2.0 m ± 0.2 m

Let,
$P = \sqrt{A . B}$

$P = \sqrt{(1.0) \times (2.0)} p = 1.414 m$

According to the formula,

$\frac{∆ p}{p} = \frac{1}{2} . \frac{∆ A}{A} + \frac{1}{2} . \frac{∆ B}{B} ∆ p = \frac{p}{2} (\frac{∆ A}{A} + \frac{∆ B}{B}) ∆ p = \frac{1.414}{2} (\frac{0.2}{1} + \frac{0.2}{2}) ∆ p = \frac{1.414}{2} (0.2 + 0.1) ∆ p = 0.707 \times 0.3 = 0.212 ∆ p = 0.2 (after rounding off) Thus, p = (1.41 \pm 0.2) m$

Hence the correct answer is option (d).

Page No 7:

Question 2.9:

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:

Precision is the degree to which several measurement are very close to each other. Lesser the deviation more precise it'll be. In all the given options the magnitude is same and are correct up to two decimal places but their units are different. Now in the option (a), which is 5.00 mm is the one which has smallest unit. Absolute error in the option (a) will be 0.01 mm which is least one, hence it is more precise.

Hence the correct answer is option (a).

Page No 7:

Question 2.10:

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:

A value is said to be accurate if it is very close to the true value or actual value, out of these 4.9 cm is the once which is very close to the true value (5 cm) .

Hence the correct answer is option (a).

Page No 7:

Question 2.11:

Young’s modulus of steel is 1.9 × 10¹¹ N/m². When expressed in CGS units of dynes/cm², it will be equal to (1N = 10⁵ dyne, 1m² = 10⁴ cm²)
(a) 1.9 × 10¹⁰
(b) 1.9 × 10¹¹
(c) 1.9 × 10¹²
(d) 1.9 × 10¹³

Answer:

We have ,
Y= 1.9 × 10¹¹ N/m2
We know that,
1 N= 10⁵dyne and 1 m² = 10⁴cm²

Now,
$Y = 1.9 \times 10^{11} \times 10^{5} \times 10^{- 4} Y = 1.9 \times 10^{12} dyne / {cm}^{2}$

Hence the correct answer is option (c).

Page No 7:

Question 2.12:

If momentum (P), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula
(a) (P¹ A^–1 T¹)
(b) (P² A¹ T¹)
(c) (P¹ A^–1/2 T¹)
(d) (P¹ A^1/2 T^–1)

Answer:

Let P, A and T are the fundamental quantities , then Energy can be represented as:
$E \propto {[P]}^{a} {[A]}^{b} {[T]}^{c} E \propto {[M L T^{- 1}]}^{a} {[L^{2}]}^{b} {[T]}^{c} [M L^{2} T^{- 2}] = k {[M]}^{a} {[L]}^{a + 2 b} {[T]}^{c - a} Compare R . H . S and L . H . S, we get, a = 1, a + 2 b = 2, c - a = - 2 Solving we get : b = \frac{1}{2} a n d c = - 1$
Hence,
$E \propto [P] {[A]}^{\frac{1}{2}} {[T]}^{- 1}$

Hence the correct answer is option (d).

Page No 7:

Question 2.13:

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 2πt /T

(b) y = a sin vt.

(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:

Lets take up the options one by one :
Option (a) : an angle be is a dimensionless quantity hence whatever you are putting in sin , cos , tan etc should be dimensionless.
therefore,
$\frac{2 πt}{T} = dimensionless and [y] = [L], [a] = [L] In this case L . H . S = R . H . S$

Option (b):
$[v t] = [L T^{- 1}] [T] = [L] angle cannot have dimensions, hence this equation is not possible .$

Option (c):

$[\frac{t}{a}] = [L^{- 1}] [T] = [L^{- 1} T] angle cannot have dimensions, hence this equation is not possible .$

Option (d):

$\frac{2 πt}{T} = dimensionless and y = [L], a = [L] In this case L . H . S = R . H . S$

Hence the correct answers are option (b) and (c).

Page No 7:

Question 2.14:

If P, Q, R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity?
(a) (P – Q)/R
(b) PQ – R
(c) PQ/R
(d) (PR – Q²)/R
(e) (R + Q)/P

Answer:

According to the principle of homogeneity same physical quantities can be added or subtracted. but division and multiplication can be done with different physical quantities.
In the given options, option (b), (c) and (d) will be having some meaningful quantities.
and option (a) and (e) can never be meaningful quantities.

Hence the correct answers are option (a) and (e).

Page No 8:

Question 2.15:

Photon is quantum of radiation with energy E = $h ν$ 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:

$Planks constant (h) = \frac{E}{ν} \Rightarrow [h] = \frac{[M L^{2} T^{- 2}]}{[T^{- 1}]} \Rightarrow [h] = [M L^{2} T^{- 1}]$

Lets take up all the options one by one :
option (a):
$Linear Impulse (J) = m (v - u) [J] = [M] [L T^{- 1}] [J] = [M L T^{- 1}] option (b) : Angular impulse (J_{R}) = m (v - u) . R [J_{R}] = [M] [L T^{- 1}] [L] [J_{R}] = [M L^{2} T^{- 1}] option (c) : Linear momentum (P) = m v [P] = [M] [L T^{- 1}] [P] = [M L T^{- 1}] option (d) : Angular momentum (L) = m v r [L] = [M] [L T^{- 1}] [L] [L] = [M L^{2} T^{- 1}]$

Dimensions of the plank's constant is matching with angular impulse nad angular momentum.

Hence the correct answers are option (b) and (d).

Page No 8:

Question 2.16:

If Planck’s constant (h) and speed of light in vacuum (c) are taken as two fundamental quantities, which one 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:

We know the dimensions of plank's constant , light speed and gravitational constant :
$h = \frac{E}{v} [h] = \frac{[M L^{2} T^{- 2}]}{[T^{- 1}]} = [M L^{2} T^{- 1}]$

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

$[G] = [M^{- 1} L^{3} T^{- 2}]$
Let mass relates with h, c and G as:
$m \propto {(h)}^{a} {(c)}^{b} {(G)}^{c}$
$[M L^{0} T^{0}] = {[M L^{2} T^{- 1}]}^{a} {[L T^{- 1}]}^{b} {[M^{- 1} L^{3} T^{- 2}]}^{c} comparing both sides and we get, a - c = 1 2 a + b + 3 c = 0 - a - b - 2 c = 0 Solving above equations we get : a = \frac{1}{2}, b = \frac{1}{2} and c = - \frac{1}{2} So, m \propto \sqrt{\frac{c h}{G}} S imilarly for length, L = \sqrt{\frac{h G}{c^{3}}} and for Time T = \sqrt{\frac{h G}{c^{5}}}$
Mass can be expressed in the terms of m_eand m_p , hence option (a), (b) and (d) can be used to express L, M , T fundamental physical quantities.
Hence the correct answer are options (a), (b) and (d).

Page No 8:

Question 2.17:

Which of the following ratios express pressure?
(a) Force/ Area
(b) Energy/ Volume
(c) Energy/ Area
(d) Force/ Volume

Answer:

Lets express the options one by one :

(a) $[Pressure] = \frac{Force}{Area} = \frac{[M L T^{- 2}]}{[L^{2}]} = [M L^{- 1} T^{- 2}]$

$(b) [\frac{Energy}{Volume}] = \frac{[M L^{2} T^{- 2}]}{[L^{3}]} = [M L^{- 1} T^{- 2}] Which is the dimension of pressure . (c) [\frac{E n e r g y}{A r e a}] = \frac{[M L^{2} T^{- 2}]}{[L^{2}]} = [M L^{0} T^{- 2}] Which is different from dimension of pressure .$

$(d) [\frac{F o r c e}{V o l u m e}] = \frac{[M L T^{- 2}]}{[L^{3}]} = [M L^{- 2} T^{- 2}] Which is different from dimension of pressure .$

Hence the correct answers are option (a) and (b)

Page No 8:

Question 2.18:

Which of the following are not a unit of time?
(a) Second
(b) Parsec
(c) year
(d) Light year

Answer:

Parsec and light year are the units through which we measure the distance between the celestial bodies. Second and year are the unit of time.

Hence the correct answers are (b) and (d).

Page No 8:

Question 2.19:

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

Answer:

Magnitude of a given physical unit may vary over a wide range , therefore different units of the same physical quantities is required:

for example mass ranges from $10^{- 30}$ (mass of electron ) to $10^{53}$ (celestial bodies) . We need different units like Kg, g and mg to measure mass.

similarly in case of length, the length of a pencil can be measured in cm, height of a building can be measured in meters, distance between two cities can be measured in km or distance between two celestial bodies can be measured in light years.

Page No 8:

Question 2.20:

The radius of atom is of the order of 1 Å 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:

We have ,
$Radius of an atom : (r_{a}) = 10^{- 10} m Volume of the atom : V_{a} = \frac{4}{3} {π r_{a}}^{3} Radius of the nuclues : (r_{n}) = 10^{- 15} m Volume of the nucleu s : V_{n} = \frac{4}{3} {π r_{n}}^{3} \frac{V_{a}}{V_{n}} = {(\frac{r_{a}}{r_{n}})}^{3} = {(\frac{10^{- 10}}{10^{- 15}})}^{3} = 10^{15}$
Hence the volume of atom is 10¹⁵ times more than the volume of nucleus of that atom.

Page No 8:

Question 2.21:

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

Answer:

Mass Sprectrograph is a device that is used to measure the mass of atoms and molecules.

Page No 9:

Question 2.22:

Express unified atomic mass unit in kg.

Answer:

Unified atomic mass unit (amu) or 1 u = $\frac{1}{12}$ of â€‹ the mass of one carbon (C₁₂ ) atom.
Mass of one mole of $C_{12}$ atom = 12 g
Number of atoms in one mole = $6.022 \times 10^{23}$

Mass of a single $C_{12}$ atom = $\frac{12}{6.022 \times 10^{23}}$ g
Unified atomic mass unit (amu) or
â€‹ $1 u = \frac{1}{12} \times \frac{12}{6.022 \times 10^{23}} g = 1.67 \times 10^{- 24} g = 1.67 \times 10^{- 27} kg$

Page No 9:

Question 2.23:

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 f (θ) to be a dimensionless quantity?

Answer:

According to the principle of homogeneity same dimensions terms can be added or subtracted only. If we look at R.H.S , As 1 and $θ$ is dimensionless hence each term is dimensionless in the equation, therefore term on L.H.S has be of same dimensions.
Hence, f (θ) must have to be a dimensionless quantity.

Page No 9:

Question 2.24:

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

Answer:

Generally each physical quantity has unit and sometimes a physical quantity has more than one units. It has been found that in mechanics, if we choose length , mass and time as a base unit then we can derive the all other physical quantities used in mechanics with the the help of these these three units, so length , mass , time are chosen to be base units in mechanics.

Page No 9:

Question 2.25:

(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 (½)° 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) As the distance between the earth and moon is too large as compared to the radius of the earth , hence we can treat the radius of earth as an arc.
$l = R_{E} = Length of arc Distance betwwn moon and earth = 60 R_{E} So, angle subtended at distance r due to an arc of length l is : θ = \frac{l}{r} = \frac{R_{E}}{60 R_{E}} = \frac{1}{60} rad = (\frac{1}{60} \times \frac{180}{π}) = (\frac{3}{3.14}) \approx 1^{°} Hence, angle substended by diameter of the earth is 2 θ = 2^{°} .$

(b)
$According to the problem, moon is seen as {(\frac{1}{2})}^{°} and diameter from earth and earth is seen as 2^{°} diameter from moon . As θ is propotional to diameter, \Rightarrow \frac{Diameter of earth}{Diameter of moon} = \frac{2}{\frac{1}{2}} = 4 Hence, diameter of earth is 4 times the diameter of the moon .$

(c)
$From parallax measurement : Sun is at a distance of about 400 times the earth moon distance, \Rightarrow \frac{r_{s u n}}{r_{m o o n}} = 400 Sun and moon both appear to be of the same angular diameter as seen from earth . \frac{D_{s u n}}{r_{s u n}} = \frac{D_{m o o n}}{r_{m o o n}} But, \frac{D_{e a r t h}}{D_{m o o n}} = 4 \Rightarrow \frac{D_{s u n}}{D_{e a r t h}} = 100 (Here r stands for the distance and D for the diameter)$

Page No 9:

Question 2.26:

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

Answer:

A clock can measure time up to one second. A stop watch can measure time up to fractions of second and a digital watch can measure time up to fraction of a second whereas an atomic clock is the most accurate timekeeper and is based on the characteristics frequency of radiation emitted by certain atoms. An atomic clock is the most precise time measuring device because atomic oscillations are repeated with a precision of 1s in 10¹³ s.

Hence the correct answer is option (d).

Page No 9:

Question 2.27:

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:

Distance of the galaxy = $10^{25} m$
Speed of the light = $3 \times 10^{8} m / s$
Time taken to reach the light,
$t = \frac{Distance}{speed} = \frac{10^{25}}{3 \times 10^{8}} = 0.33 \times 10^{17} s \Rightarrow t = 3.33 \times 10^{16} s$

Page No 9:

Question 2.28:

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:

According to the problem , 50 divisions of Vernier scale which coincide with 49 main scale division.

$50 VSD = 49 MSD 1 MSD = \frac{50}{49} VSD o r, 1 VSD = \frac{49}{50} MSD S o, 1 MSD - 1 VSD = 1 MSD - \frac{49}{50} MSD = \frac{1}{50} MSD it is given that 1 MSD = 0.5 mm Hence, minimum inaccuracy = \frac{1}{50} \times 0.5 mm = \frac{1}{100} = 0.01 mm$

Page No 9:

Question 2.29:

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:

A small object subtends the same solid angle as the angle made by far object, Diagram shows that moon is much smaller than sun , it is also much closer to earth.

$R_{e m} = Distance of moon from earth R_{e s} = Distance of sun from earth Let the solid angle made by sun and moon is θ, we can right, θ = \frac{A_{s u n}}{_{{R^{2}}_{e s}}} = \frac{A_{m o o n}}{_{{R^{2}}_{e m}}} A_{s u n} = Area of the sun = {π R^{2}}_{s} A_{m o o n} = Area of the moon = {π R^{2}}_{m} θ = \frac{{π R^{2}}_{s}}{_{R^{2} e s}} = \frac{{π R^{2}}_{m}}{_{{R^{2}}_{e m}}} \Rightarrow \frac{R_{s}}{_{R_{e s}}} = \frac{R_{m}}{_{R_{e m}}} \Rightarrow \frac{R_{s}}{_{R_{m}}} = \frac{R_{e s}}{_{R_{e m}}}$
This is the required relation between the distances and sizes of the sun and moon.

Page No 9:

Question 2.30:

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:

Force = $[M L T^{- 2}] = 100 N . . . eq (1)$
Length= $[L] = 10 m$
Time= $[T] = 100 s$
Substituting the values of L and T in the eq(1) we get,
$M \times 10 \times 100^{- 2} = 100 M = 10^{5} kg$
This is the unit of mass in this system of units.

Page No 10:

Question 2.31:

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) Solid angle, $Ω = \frac{Area}{r^{2}}$ and the plane angle $θ = \frac{length}{r}$ both are having units (steradian and radian) but none of them having dimensions.

(b) Ratio of the same physical units will not be having units as well as dimensions like relative density = $\frac{density of medium}{density of water}$
(c) Gravitational Constant (G)= $6.67 \times 10^{- 11} N - m^{2} / {Kg}^{2}$

(d) Reynold's number

Page No 10:

Question 2.32:

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

Answer:

$Angle, θ = \frac{length of arc}{radius} \Rightarrow \frac{π}{6} = \frac{l}{0.31} \Rightarrow l = \frac{3.14 \times 0.31}{6} \Rightarrow l = 0.162 m \Rightarrow l = 16.2 cm$

Page No 10:

Question 2.33:

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:

Area = 1 cm²
radius = 5 cm
Solid angle,
$Ω = \frac{Area}{r^{2}} = \frac{1}{5^{2}} = \frac{1}{25} = 4 \times 10^{- 2} steradian$

Page No 10:

Question 2.34:

The displacement of a progressive wave is represented by y = A sin(ω t – kx), 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 , physical quantities having same dimensions can be added or subtracted.
According to the given equation ,
$y = A \sin (ω t - k x) (ω t - k x) = Dimensionless \Rightarrow [ω t] = [M^{0} L^{0} T^{0}] \Rightarrow [ω] = \frac{[M^{0} L^{0} T^{0}]}{[T]} = [M^{0} L^{0} T^{- 1}] Similarly, \Rightarrow [k x] = [M^{0} L^{0} T^{0}] \Rightarrow [k] = \frac{[M^{0} L^{0} T^{0}]}{[L]} = [M^{0} L^{- 1} T^{0}]$

Page No 10:

Question 2.35:

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

Answer:

t₁ = 39.6 s
t₂ = 39.9 s
t₃ = 39.5 s
we can observe from the given readings that the least count of the watch is 0.1 s.
Precision in the measurement = Least count of the device.
Precision in 20 oscillations = 0.1 s
Therefore, precision in 1 oscillation =
$\frac{0.1}{20} = 0.005 Average time, t_{a v g} = \frac{t_{1} + t_{2} + t_{3}}{3} t_{a v g} = (\frac{39.6 + 39.9 + 39.5}{3}) = 39.6 s Time period = (\frac{39.6}{20}) = 1.98 s Measured time period of the second pendulum = (2 - 0.005) = 1.995 s Accuracy in the measurement = (1.995 - 1.980) = 0.015 s$

Page No 10:

Question 2.36:

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:

For a physical Quantity be:
$Q = n_{1} u_{1} = n_{2} u_{2} Let M_{1}, L_{1}, T_{1} and M_{2}, L_{2}, T_{2} are the units of mass, length and time in given two systems . So, n_{2} = n_{1} {[\frac{M_{1}}{M_{2}}]}^{a} \times {[\frac{L_{1}}{L_{2}}]}^{b} \times {[\frac{T_{1}}{T_{2}}]}^{c} We know that dimnsion of energy [U] = [M L^{2} T^{- 2}] According to the problem, M_{1} = 1 Kg, L_{1} = 1 m, T_{1} = 1 s M_{2} = α Kg, L_{2} = β m, T_{2} = γ s Substituting the values, we get n_{2} = 5 \times [\frac{M_{1}}{M_{2}}] \times {[\frac{L_{1}}{L_{2}}]}^{2} \times {[\frac{T_{1}}{T_{2}}]}^{- 2} = 5 [\frac{1}{α}] \times {[\frac{1}{β}]}^{2} \times {[\frac{1}{γ}]}^{- 2} = \frac{5 γ^{2}}{α β^{2}} This will be the value of energy in the new system of units .$

Page No 10:

Question 2.37:

The volume of a liquid flowing out per second of a pipe of length l and radius r is written by a student as

$ν = \frac{π}{8} \frac{P r^{4}}{η l}$

where P is the pressure difference between the two ends of the pipe and η is coefficent of viscosity of the liquid having dimensional formula ML^–1 T^–1.
Check whether the equation is dimensionally correct.

Answer:

If the dimensions of the L.H.S is equal to the dimensions of the R.H.S, then we can say that the given formula is correct.
Dimensions of L.H.S=
$[V] = \frac{dimensions of volume}{dimensions of time} = \frac{[L^{3}]}{[T]} = [L^{3} T^{- 1}], [P] = [M L^{- 1} T^{- 2}], [η] = [M L^{- 1} T^{- 1}], [l] = [L], [r] = [L] L . H . S = [V] = \frac{[L^{3}]}{[T]} = [L^{3} T^{- 1}] R . H . S = [\frac{π}{8} \frac{P r^{4}}{η l}] = \frac{[M L^{- 1} T^{- 2}] \times [L^{4}]}{[M L^{- 1} T^{- 1}] \times [L]} = [L^{3} T^{- 1}] Dimensionally, L . H . S = R . H . S Hence the given formula is dimensionally correct$

Page No 10:

Question 2.38:

A physical quantity X is related to four measurable quantities a, b, c and d as follows:
X = a²b³c^5/2d^–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:

The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4%, respectively.
$Percentage error in quantity X is given by, \frac{∆ x}{x} \times 100 According to the problem, physical quantity is X = a^{2} b^{3} c^{\frac{5}{2}} d^{2} Percentage error in a = (\frac{∆ a}{a} \times 100) = 1 % Percentage error in b = (\frac{∆ b}{b} \times 100) = 2 % Percentage error in c = (\frac{∆ c}{c} \times 100) = 3 % Percentage error in d = (\frac{∆ d}{d} \times 100) = 4 % Maximum percentage error in X is : \frac{∆ x}{x} \times 100 = \pm [2 (\frac{∆ a}{a} \times 100) + 3 (\frac{∆ b}{b} \times 100) + \frac{5}{2} (\frac{∆ c}{c} \times 100) + 2 (\frac{∆ d}{d} \times 100)] = \pm [2 (1) + 3 (2) + \frac{5}{2} \times (3) + 2 (4)] % = \pm 23.5 % Percentage error in the quantity X = \pm 23.5 % Mean absolute error in X = \pm 0.235 = \pm 0.24 (rounding off upto two significant digits)$
Since the error is in first decimal, hence the result should be rounded off as 2.8.

Page No 11:

Question 2.39:

In the expression P = El²m^–5G^–2, E, m, l and G denote energy, mass, angular momentum and gravitational constant, respectively. Show that P is a dimensionless quantity.

Answer:

Given expression is P = El² m^–5 G^–2
$[E] = [M L^{2} T^{- 2}], [m] = [M] [l] = [M L^{2} T^{- 1}], [G] = [M^{- 1} L^{3} T^{- 2}] Substituting dimensions of each physical quantity in the given expression, [P] = [M L^{2} T^{- 2}] \times {[M L^{2} T^{- 1}]}^{2} \times [M^{- 5}] \times {[M^{- 1} L^{3} T^{- 2}]}^{- 2} [P] = [M^{1 + 2 - 5 + 2} L^{2 + 4 - 6} T^{- 2 - 2 + 4}] [P] = [M^{0} L^{0} T^{0}]$
Hence, P is dimensionless quantity.

Page No 11:

Question 2.40:

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:

Dimension of velocity of light c, Planck’s constant h and gravitational constant G are:
$[h] = [M L^{2} T^{- 1}], [c] = [L T^{- 1}], [G] = [M^{- 1} L^{3} T^{- 2}] 1 .) Let, m \propto c^{a} h^{b} G^{c} . . . eq (1) m = k c^{a} h^{b} G^{c} where, k is a dimensionless constant of propotionality, Substituting dimensions of each term in eq (1) we get, [M L^{0} T^{0}] = {[L T^{- 1}]}^{a} \times {[M L^{2} T^{- 1}]}^{b} {[M^{- 1} L^{3} T^{- 1}]}^{c} Comparing powers of same terms on both sides, we ge t b - c = 1 . . . eq (2) a + 2 b + 3 c = 0 . . . eq (3) - a - b - 2 c = 0 . . . eq (4) Solving equations (2), (3) and (4), we get : a = \frac{1}{2}, b = \frac{1}{2} and c = - \frac{1}{2} Putting the values of a, b and c in eq (1) m = k c^{\frac{1}{2}} h^{\frac{1}{2}} G^{- \frac{1}{2}} = k \sqrt{\frac{c h}{G}}$

$2 .) Let, L \propto c^{x} h^{y} G^{z} . . . . eq (5) L = k c^{a} h^{b} G^{c} where, k is a dimensionless constant of propotionality, Substituting dimensions of each term in equation (5) we get, [M^{0} L T^{0}] = {[L T^{- 1}]}^{a} \times {[M L^{2} T^{- 1}]}^{b} {[M^{- 1} L^{3} T^{- 1}]}^{c} Comparing powers of same terms on both sides, we get b - c = 0 . . . (6) a + 2 b + 3 c = 1 . . . (7) - a - b - 2 c = 0 . . . (8) Solving we get : a = \frac{- 3}{2}, b = \frac{1}{2} and c = \frac{1}{2} Putting the values of a, b and c in equation (5), we get : m = k c^{\frac{- 3}{2}} h^{\frac{1}{2}} G^{\frac{1}{2}} = k \sqrt{\frac{G h}{c^{3}}}$
â€‹
$3 .) Let, T \propto c^{a} h^{b} G^{c} . . . eq (9) T = k c^{a} h^{b} G^{c} where, k is a dimensionless constant of propotionality, Substituting dimensions of each term in eq (9) we get, [M^{0} L^{0} T^{1}] = {[L T^{- 1}]}^{a} \times {[M L^{2} T^{- 1}]}^{b} {[M^{- 1} L^{3} T^{- 1}]}^{c} Comparing powers of same terms on both sides, we get b - c = 0 . . . (10) a + 2 b + 3 c = 1 . . . (11) - a - b - 2 c = 0 . . . (12) S o l v i n g w e g e t : a = \frac{- 5}{2}, b = \frac{1}{2} and c = \frac{1}{2} Substituting the values of b and c, we get Putting the values of a, b and c in equation (9) we get : m = k c^{\frac{- 5}{2}} h^{\frac{1}{2}} G^{\frac{1}{2}} = k \sqrt{\frac{G h}{c^{5}}}$

Page No 11:

Question 2.41:

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 the Kepler's Law,
Time 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.

$T^{2} \propto r^{3} T \propto r^{\frac{3}{2}} Let, T \propto r^{\frac{3}{2}} g^{a} R^{b} T = k r^{\frac{3}{2}} g^{a} R^{b} . . . . eq (1) where, k is a dimensionless constant of propotionality . Writing the dimensions of various quantities on both the sides, we get : [M^{0} L^{0} T] = {[L]}^{\frac{3}{2}} {[L T^{- 2}]}^{a} {[L]}^{b} = [M^{0} L^{a + b + \frac{3}{2}} T^{- 2 a}] On comparing the dimensions of both the sides, we get : a = \frac{- 1}{2} and b = - 1 Substituting the values of a and b in the eq (1), we get : T = k r^{\frac{3}{2}} g^{- \frac{1}{2}} R^{- 1} T = \frac{k}{R} \sqrt{\frac{r^{3}}{g}}$

Page No 11:

Question 2.42:

In an experiment to estimate the size of a molecule of oleic acid 1 mL of oleic acid is dissolved in 19 mL of alcohol. Then 1 mL of this solution is diluted to 20 mL 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 and its diameter is measured. Knowing the volume of the drop and area of 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 acid?
(e) What will be the volume of oleic acid in one drop of this solution?

Answer:

(a) Since oleic acid doesn't dissolve in water hence it will be dissolved in alchol.

(b) lycopodium powder dissolves spreads on the entire surface of the water when it is sprinkled evenly. when a drop of prepared solution of oleic acid and alchol is dropped on the water, oleic acid doesn't dissolve in water. Instead it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls. thus we will be able to measure the area.

(c) Since 20 ml (1ml oleic acid +19 ml of water) contains 1 ml of oleic acid , oleic acid in each mL of the solution = $\frac{1}{20}$ mL.
As this 1 mL is diluted to 20 mL by adding alcohol.
In each mL of solution prepared, volume of oleic acid is= $\frac{1}{20} \times \frac{1}{20}$ = $\frac{1}{400}$ ml

(d) Volume of the n drops of oleic acid can be calculated by the help of burette and measuring cylinder by measuring the number of drops.

(e) 1 mL of the solution contains n number of drops , then the volume of oleic acid in one drop will be = $\frac{1}{400 n} mL$

Page No 11:

Question 2.43:

(a) How many astronomical units (A.U.) make 1 parsec?
(b) Consider a sunlike star at a distance of 2 parsecs. 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)° from the earth. Due to atmospheric fluctuations, eye can’t resolve objects smaller than 1 arc minute.
(c) Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about 1/2 A.U. from the earth. Calculate what size it will appear when seen through the same telescope.

Answer:

(a) According to the definition , 1 parsec,

$1 parsec = \frac{1 A . U .}{1 arc \sec} 1 parsec = \frac{3600 \times 180}{π} A . U . (As, 1 arc \sec = \frac{1}{3600} \times \frac{π}{180} rad) = 206265 A . U . \approx 2 \times 10^{5} A . U . (Here, A . U . = astronomical unit)$

(b) Sun's angular diameter from the earth is ${(\frac{1}{2})}^{°}$ at 1 A.U.
Angular diameter of the sun like star at a distance of 2 parsecs,
= $\frac{{(\frac{1}{2})}^{°}}{2 \times 2 \times 10^{5}} = {(\frac{1}{8} \times 10^{- 5})}^{°} = {(\frac{1}{8} \times 10^{- 5})}^{°} \times 60^{'} = 7.5 \times 10^{- 5} arc \min$

When the sun like star is seen through a telescope with magnification 100 , the angular diameter of the star
$100 \times 7.5 \times 10^{- 5} = 7.5 \times 10^{- 3} arc \min$

But eye cannot resolve smaller than 1 arcmin due to atmospheric fluctuations. So angular size of sun like star appears to be 1 arcmin.
It can’t be magnified using telescope.

(c)
$\frac{D_{m a r s}}{D_{e a r t h}} = \frac{1}{2} . . . eq (1) where D represents diameter As, \frac{D_{e a r t h}}{D_{s u n}} = \frac{1}{100} \Rightarrow \frac{D_{m a r s}}{D_{s u n}} = \frac{1}{2} \times \frac{1}{100} = \frac{1}{200} At 1 A . U ., diameter of the sun = {(\frac{1}{2})}^{°} Diameter of mars = \frac{1}{2} \times \frac{1}{200} = {(\frac{1}{400})}^{°} At \frac{1}{2} A . U . Mars' s diamter, = {(\frac{1}{400})}^{°} \times 2 = {(\frac{1}{200})}^{°} With 100 magnification Mars' s diameter = {(\frac{1}{200})}^{°} \times 100 = {(\frac{1}{2})}^{°} = \frac{60}{2} arc \min = 30 arc \min This is larger than resolution limit due to atmospheric fluctuatuation Hence, it looks magnified$

Page No 12:

Question 2.44:

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 × 10^–13J; the masses are measured in unified atomic mass unit (u) where 1u = 1.67 × 10^–27 kg.
(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) We can apply Einstien's mass energy relation in this problem, E = mc² , to calculate the energy equivalent of the given mass,

$1 amu = 1 u = 1.67 \times 10^{- 27} kg Applying, E = m c^{2} E = \frac{(1.67 \times 10^{- 27}) \times {(3 \times 10^{8})}^{2}}{1.6 \times 10^{- 19}} eV \Rightarrow E = 939.4 MeV \approx 931.5 MeV$

(b)

$As, E = m c^{2} \Rightarrow m = \frac{E}{c^{2}} According to this, 1 u = \frac{931.5 MeV}{c^{2}} Hence the dimensionally correct relation 1 amu \times c^{2} = 931.5 MeV$

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