CBSE Class 11 Physics Chapter 2 Motion in a Straight Line Exercise 2.1 to 2.7 Solutions

-  Dr.Sanjaykumar Pawar  

CBSE Class 11 Physics Chapter 2 Motion in a Straight Line NCERT Exercise 2.1 to 2.7 Solved Answers.

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NCERT Class 11 Physics
Chapter 2: Motion in a Straight Line
Exercise Solutions

Question 2.1

In which of the following examples of motion can the body be considered approximately a point object?

1. A railway carriage moving without jerks between two stations.
2. A monkey sitting on top of a man cycling smoothly on a circular track.
3. A spinning cricket ball that turns sharply on hitting the ground.
4. A tumbling beaker that has slipped off the edge of a table.

A body can be treated as a point object if its size is negligible compared to the distance travelled.

• (a) Yes. The distance between stations is much greater than the size of the carriage.
• (b) Yes. The radius of the circular track is much larger than the dimensions of the monkey-man system.
• (c) No. Rotation and spin are important.
• (d) No. The tumbling motion depends on the size and shape of the beaker.

Final Answer: (a) and (b)

Question 2.2

The position-time (x-t) graphs for two children A and B returning from school to their homes are shown in Fig. 2.9. Choose the correct entries in the brackets.

(a) (A/B) lives closer to the school than (B/A).

Since OP is less than OQ, A lives closer to the school.

Answer: A lives closer to the school than B.

(b) (A/B) starts from the school earlier than (B/A).

A starts at t = 0 while B starts later.

Answer: A starts earlier than B.

(c) (A/B) walks faster than (B/A).

The slope of an x-t graph represents speed. B has a steeper graph.

Answer: B walks faster than A.

(d) A and B reach home at the (same/different) time.

Both graphs terminate at the same value of time.

Answer: Same time.

(e) (A/B) overtakes (B/A) on the road (once/twice).

The graphs intersect only once and after that B remains ahead.

Answer: B overtakes A once.

Question 2.3

A woman starts from her home at 9.00 am, walks with a speed of 5 km h⁻¹ on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h⁻¹. Plot the x-t graph of her motion.

Given:

• Distance = 2.5 km
• Walking speed = 5 km h⁻¹
• Auto speed = 25 km h⁻¹

Time taken to reach office:

t = Distance / Speed = 2.5 / 5 = 0.5 hour = 30 minutes

She reaches the office at 9:30 am.

Time taken to return:

t = 2.5 / 25 = 0.1 hour = 6 minutes

She reaches home at 5:06 pm.

Time	Position (km)
9:00 am	0
9:30 am	2.5
5:00 pm	2.5
5:06 pm	0

The x-t graph consists of an increasing straight line, a horizontal line, and a steep decreasing straight line.

Question 2.4

A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward repeatedly. Each step is 1 m long and takes 1 s. Determine how long he takes to fall in a pit 13 m away.

Net displacement in one cycle = 5 − 3 = 2 m

Time taken in one cycle = 8 s

After 4 cycles:

Displacement = 4 × 2 = 8 m

Time = 4 × 8 = 32 s

Then he moves forward:

Time (s)	Position (m)
33	9
34	10
35	11
36	12
37	13

Therefore, the drunkard falls into the pit after 37 seconds.

Question 2.5

A car moving along a straight highway with speed 126 km h⁻¹ is brought to a stop within a distance of 200 m. What is the retardation and how long does it take to stop?

Initial speed, u = 126 km h⁻¹ = 35 m s⁻¹

Final speed, v = 0

Distance, s = 200 m

Using:

v² = u² + 2as

0 = (35)² + 2(a)(200)

a = −3.06 m s⁻²

Retardation = 3.06 m s⁻²

Using:

v = u + at

0 = 35 − 3.06t

t = 11.4 s

Retardation = 3.06 m s⁻²
Time taken = 11.4 s

Question 2.6

A player throws a ball upwards with an initial speed of 29.4 m s⁻¹.

(a) What is the direction of acceleration during upward motion?

Acceleration is always vertically downward due to gravity.

(b) What are the velocity and acceleration at the highest point?

Velocity = 0

Acceleration = 9.8 m s⁻² downward

v = 0, a = 9.8 m s⁻² downward

(c) Sign convention:

• During upward motion: Position positive, Velocity negative, Acceleration positive.
• During downward motion: Position positive, Velocity positive, Acceleration positive.

(d) Maximum height and total time of flight

Using:

v² = u² − 2gh

0 = (29.4)² − 2(9.8)h

h = 44.1 m

Time to rise:

t = u/g = 29.4/9.8 = 3 s

Total time:

T = 2 × 3 = 6 s

Maximum height = 44.1 m
Total time = 6 s

Question 2.7

State whether the following statements are true or false with reasons.

(a) A particle with zero speed may have non-zero acceleration.

True. Example: Ball at highest point of vertical motion.

(b) A particle with zero speed may have non-zero velocity.

False. Velocity is zero when speed is zero.

(c) A particle moving with constant speed must have zero acceleration.

False. Uniform circular motion has non-zero acceleration.

(d) A particle with positive acceleration must be speeding up.

False. Positive acceleration can reduce speed if velocity is negative.

Units and Measurements Class 11 CBSE Physics Exemplar Question & Answer Book with Solutions

 

NCERT Class 11 Physics Exemplar solutions helping students master concepts with clear problem-solving practice.

Dr.Sanjaykumar Pawar 

 Units And Measurements 

 

Units and Measurements - MCQs with Step-by-Step Solutions

Q.1 The number of significant figures in 0.06900 is

(a) 5
(b) 4
(c) 2
(d) 3

Thinking Process

If the number is less than 1, the zeros before the first non-zero digit are not significant. Trailing zeros after a decimal point are significant.

Number = 0.06900

• 0 → Not significant
• 0 → Not significant
• 6 → Significant
• 9 → Significant
• 0 → Significant
• 0 → Significant

Total significant figures = 4

Answer: (b) 4

Q.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

Step 1: Add the Numbers

436.32 + 227.2 + 0.301 = 663.821

Step 2: Apply Significant Figure Rule

For addition and subtraction, the result should have the same number of decimal places as the quantity with the least decimal places.

• 436.32 → 2 decimal places
• 227.2 → 1 decimal place
• 0.301 → 3 decimal places

Least decimal places = 1

663.821 ≈ 663.8 ≈ 664

Answer: (b) 664

Note: In addition and subtraction, focus on decimal places, not significant figures.

Q.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

Formula

Density = Mass / Volume

Density = 4.237 / 2.5

Density = 1.6948 g cm⁻³

Apply Significant Figure Rule

• 4.237 has 4 significant figures
• 2.5 has 2 significant figures

Final answer should contain only 2 significant figures.

1.6948 ≈ 1.7

Answer: (c) 1.7 g cm⁻³

Q.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

For 2.745

The digit to be dropped is 5. The preceding digit is 4 (even). Therefore, 4 remains unchanged.

2.745 → 2.74

For 2.735

The digit to be dropped is 5. The preceding digit is 3 (odd). Therefore, increase 3 by 1.

2.735 → 2.74

Answer: (d) 2.74 and 2.74

Q.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

Given

Length = (16.2 ± 0.1) cm

Breadth = (10.1 ± 0.1) cm

Step 1: Area

Area = Length × Breadth

A = 16.2 × 10.1 = 163.62 cm²

Rounded to 3 significant figures:

A = 164 cm²

Step 2: Relative Error

ΔA/A = Δl/l + Δb/b

ΔA/A = (0.1/16.2) + (0.1/10.1)

ΔA/A = 0.01607

Step 3: Absolute Error

ΔA = 163.62 × 0.01607

ΔA = 2.63 cm²

Rounded to one significant figure:

ΔA = 3 cm²

Area = (164 ± 3) cm²

Answer: (a) (164 ± 3) cm²

Q.6 Which of the following pairs of physical quantities does not have the same dimensional formula?

(a) Work and Torque
(b) Angular Momentum and Planck’s Constant
(c) Tension and Surface Tension
(d) Impulse and Linear Momentum

(a) Work and Torque

Work = Force × Distance

= [MLT⁻²][L]

= [ML²T⁻²]

Torque = Force × Distance = [ML²T⁻²]

Same dimensions ✔

(b) Angular Momentum and Planck's Constant

Angular Momentum = mvr

= [M][LT⁻¹][L]

= [ML²T⁻¹]

Planck's Constant = Energy/Frequency

= [ML²T⁻²]/[T⁻¹]

= [ML²T⁻¹]

Same dimensions ✔

(c) Tension and Surface Tension

Tension = Force

= [MLT⁻²]

Surface Tension = Force/Length

= [MLT⁻²]/[L]

= [MT⁻²]

Different dimensions ✘

(d) Impulse and Momentum

Impulse = Force × Time

= [MLT⁻²][T]

= [MLT⁻¹]

Momentum = Mass × Velocity

= [M][LT⁻¹]

= [MLT⁻¹]

Same dimensions ✔

Answer: (c) Tension and Surface Tension

Important Exam Note: Tension and Surface Tension sound similar but have different dimensional formulae.

Physics - Units, Measurements and Errors

Q.7

Measure of two quantities along with the precision of respective measuring instruments are:

A = (2.5 ± 0.5) m
B = (0.10 ± 0.01) s

The value of AB will be:

1. (0.25 ± 0.08)
2. (0.25 ± 0.5)
3. (0.25 ± 0.05)
4. (0.25 ± 0.135)

Answer: (a)

Solution:

AB = 2.5 × 0.10 = 0.25

Relative error:

Δ(AB)/AB = ΔA/A + ΔB/B

= 0.5/2.5 + 0.01/0.10

= 0.20 + 0.10 = 0.30

Δ(AB) = 0.25 × 0.30 = 0.075

Rounded off: Δ(AB) ≈ 0.08

AB = (0.25 ± 0.08)

Q.8

You measure two quantities as:

A = (1.0 ± 0.2) m
B = (2.0 ± 0.2) m

The correct value of √AB is:

1. 1.4 ± 0.4 m
2. 1.41 ± 0.15 m
3. 1.4 ± 0.3 m
4. 1.4 ± 0.2 m

Answer: (d)

Solution:

Y = √AB

Y = √(1.0 × 2.0)

Y = √2 = 1.414

Rounded value: Y = 1.4 m

Relative error:

ΔY/Y = ½[(ΔA/A) + (ΔB/B)]

= ½[(0.2/1.0) + (0.2/2.0)]

= ½(0.2 + 0.1)

= 0.15

ΔY = 1.414 × 0.15

= 0.212

Rounded: ΔY ≈ 0.2 m

Y = (1.4 ± 0.2) m

Q.9

Which of the following measurements is most precise?

1. 5.00 mm
2. 5.00 cm
3. 5.00 m
4. 5.00 km

Answer: (a)

Reason:

All measurements have two decimal places. Precision depends on the smallest unit.

mm < cm < m < km

5.00 mm is the most precise measurement.

Q.10

The mean length of an object is 5 cm. Which of the following measurements is most accurate?

1. 4.9 cm
2. 4.805 cm
3. 5.25 cm
4. 5.4 cm

Answer: (a)

Solution:

True value = 5 cm

Error in (a): |5 − 4.9| = 0.1 cm

Error in (b): |5 − 4.805| = 0.195 cm

Error in (c): |5 − 5.25| = 0.25 cm

Error in (d): |5 − 5.4| = 0.4 cm

Least error = 0.1 cm

Therefore, 4.9 cm is the most accurate measurement.

Q.11

Young's modulus of steel is 1.9 × 10¹¹ N/m². Express it in dyne/cm².

Given: 1 N = 10⁵ dyne, 1 m = 100 cm

1. 1.9 × 10¹⁰
2. 1.9 × 10¹¹
3. 1.9 × 10¹²
4. 1.9 × 10¹³

Answer: (c)

Solution:

Y = 1.9 × 10¹¹ N/m²

1 N = 10⁵ dyne

Y = 1.9 × 10¹¹ × 10⁵

= 1.9 × 10¹⁶ dyne/m²

1 m² = (100 cm)² = 10⁴ cm²

Y = (1.9 × 10¹⁶) / 10⁴

Y = 1.9 × 10¹² dyne/cm²

Q.12

If momentum (p), area (A) and time (T) are taken as fundamental quantities, then the dimensional formula of energy is:

1. [pA^1/2T^-1]
2. [p²AT]
3. [pA^-1T]
4. [pA^1/2T^-1]

Answer: (d)

Solution:

Let

E = kp^aA^bT^c

Dimensions:

[E] = [ML²T⁻²]

[p] = [MLT⁻¹]

[A] = [L²]

[T] = [T]

Substituting:

ML²T⁻² = (MLT⁻¹)^a(L²)^b(T)^c

Comparing powers:

a = 1

a + 2b = 2

1 + 2b = 2

b = 1/2

−a + c = −2

−1 + c = −2

c = −1

E = pA^1/2T^-1

Dimensions and Units - Multiple Correct MCQs

Q.13 On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing Simple Harmonic Motion (SHM) is not correct?

(a) y = a sin(2πt/T)
(b) y = a sin(vt)
(c) y = a sin(v)
(d) y = a sin(2πt/T) − a cos(2πt/T)

Thinking Process

For trigonometric functions (sin, cos), the angle must be dimensionless. Also, displacement y and amplitude a both have dimensions of length (L).

Answer: (b) and (c)

Option (a)

Angle = t/T = T/T = 1 (dimensionless)

Therefore, RHS has dimension L and matches LHS. Correct.

Option (b)

[vt] = [LT⁻¹ × T] = [L]

Angle has dimension L, not dimensionless. Incorrect.

Option (c)

[v] = [LT⁻¹]

Angle is not dimensionless. Incorrect.

Option (d)

[t/T] = dimensionless

Both terms have dimension L. Correct.

Q.14 If P, Q and 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) PQR
(d) (PR)/Q − R
(e) (R + Q)/P

Thinking Process

Addition or subtraction is possible only between quantities having the same dimensions.

Answer: (a) and (e)

Option (a)

P − Q

P and Q have different dimensions, so subtraction is impossible.

Option (e)

R + Q

R and Q have different dimensions, so addition is impossible. Options (b), (c), and (d) may or may not be meaningful depending on dimensions.

Q.15 Photon is a 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: (b) and (d)

Step 1

E = hν

[h] = [E]/[ν]

[E] = ML²T⁻²

[ν] = T⁻¹

[h] = ML²T⁻¹

Linear Impulse

MLT⁻¹

Not equal.

Angular Impulse

ML²T⁻¹

Equal.

Linear Momentum

MLT⁻¹

Not equal.

Angular Momentum

ML²T⁻¹

Equal.

Q.16 If Planck's constant (h) and speed of light (c) are taken as two fundamental quantities, which one of the following can additionally be taken to express length, mass and time?

(a) Mass of electron (me)
(b) Universal Gravitational Constant (G)
(c) Charge of electron (e)
(d) Mass of proton (mp)

Answer: (a), (b) and (d)

Dimensions

[h] = ML²T⁻¹

[c] = LT⁻¹

[me] = M

[G] = M⁻¹L³T⁻²

[e] = AT

[mp] = M

Electron mass, proton mass, and gravitational constant help express M, L and T. Charge introduces current dimension A, so it cannot be used.

Q.17 Which of the following ratios express pressure?

(a) Force / Area
(b) Energy / Volume
(c) Energy / Area
(d) Force / Volume

Answer: (a) and (b)

Pressure Formula

Pressure = Force / Area

Therefore option (a) is correct.

For Option (b)

Energy = Force × Distance

Volume = Area × Distance

Energy / Volume = (Force × Distance)/(Area × Distance) = Force/Area

Hence option (b) also represents pressure. Options (c) and (d) do not have dimensions of pressure.

Quick Revision Table

Question	Correct Answer
Q.13	(b), (c)
Q.14	(a), (e)
Q.15	(b), (d)
Q.16	(a), (b), (d)
Q.17	(a), (b)

CBSE Class 11 Physics
Units and Measurements
Questions 18–30

Q.18 Which of the following are not a unit of time?

(a) Second
(b) Parsec
(c) Year
(d) Light year

Answer: (b) Parsec and (d) Light year

Explanation:

Second and year are units of time.

1 Light Year = Distance travelled by light in one year. Therefore it is a unit of distance.

1 Parsec = 3.08 × 10¹⁶ m. It is also a unit of distance used in astronomy.

Q.19 Why do we have different units for the same physical quantity?

The value of a physical quantity may vary over a very large range. Therefore different units are used for convenience.

Examples:

• Length of a pen → centimetre (cm)
• Height of a tree → metre (m)
• Distance between cities → kilometre (km)
• Distance between stars → light year (ly)

Q.20 The radius of an atom is of the order of 1 Å and radius of nucleus is of the order of 1 fermi. How many times greater is the volume of atom than the volume of nucleus?

Given:

Radius of atom = 1 Å = 10⁻¹⁰ m

Radius of nucleus = 1 fermi = 10⁻¹⁵ m

Volume of sphere ∝ R³

Vatom / Vnucleus = (Ratom / Rnucleus)³

= (10⁻¹⁰ / 10⁻¹⁵)³

= (10⁵)³

= 10¹⁵

Answer: Volume of atom is 10¹⁵ times larger than the volume of nucleus.

Q.21 Name the device used for measuring the mass of atoms and molecules.

Answer: Mass Spectrograph

Q.22 Express unified atomic mass unit (u) in kg.

One atomic mass unit is one-twelfth of the mass of a Carbon-12 atom.

1 u = 1.66 × 10⁻²⁷ kg

Q.23 A function f(θ) = 1 + θ + θ²/2! + θ³/3! + ... is defined. Why is it necessary for θ to be dimensionless?

All terms are added together.

According to the principle of homogeneity, quantities can be added only if they have the same dimensions.

Since 1 is dimensionless, θ must also be dimensionless.

Q.24 Why are length, mass and time chosen as base quantities in mechanics?

Reason 1: Length, mass and time are independent quantities.

Reason 2: All mechanical quantities can be expressed in terms of length, mass and time.

Q.25 (a) The earth-moon distance is about 60 earth radii. What will be the diameter of the earth (approximately in degrees) as seen from the moon?

θ = l/r

θ = RE / 60RE = 1/60 rad

θ ≈ 1°

Diameter corresponds to 2θ.

Diameter = 2°

Answer: 2°

Q.25 (b) Moon is seen to be of 0.5° diameter from the earth. What must be the relative size compared to the earth?

Earth diameter = 2°

Moon diameter = 0.5°

2 / 0.5 = 4

Answer: Diameter of Earth = 4 × Diameter of Moon

Q.25 (c) The sun is about 400 times farther away than the moon. Estimate the ratio of Sun-Earth diameters.

Dsun = 400 Dmoon

Dearth = 4 Dmoon

Dsun = 100 Dearth

Answer: Diameter of Sun = 100 × Diameter of Earth

Q.26 Which of the following time measuring devices is most precise?

(a) Wall clock
(b) Stop watch
(c) Digital watch
(d) Atomic clock

Answer: Atomic Clock

An atomic clock has an accuracy of about 1 second in 10¹³ seconds. Hence it is the most precise.

Q.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.

Distance = 10²⁵ m

Speed of light = 3 × 10⁸ m/s

t = Distance / Speed

t = 10²⁵ / (3 × 10⁸)

t = 3.3 × 10¹⁶ s

Answer: Order of magnitude = 10¹⁶ s

Q.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.

50 VSD = 49 MSD

1 VSD = 49/50 MSD

Least Count = 1 MSD − 1 VSD

= (1/50) MSD

= 0.5/50

= 0.01 mm

Answer: 0.01 mm

Q.29 During a total solar eclipse the moon almost entirely covers the sphere of the sun. Write the relation between distances and sizes of the sun and moon.

Since the Sun and Moon appear to have the same angular diameter:

Rs / Rm = Ds / Dm

where:

• Rs = Radius of Sun
• Rm = Radius of Moon
• Ds = Distance of Sun from Earth
• Dm = Distance of Moon from Earth

Q.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?

Given:

F = MLT⁻²

Force unit = 100 N

Length unit = 10 m

Time unit = 100 s

M = FT²/L

M = 100 × (100)² / 10

M = 100000 kg

M = 10⁵ kg

Answer: 10⁵ kg

Quick Revision Table

Question	Answer
Q.18	(b), (d)
Q.19	Different units are used for convenience.
Q.20	10¹⁵
Q.21	Mass Spectrograph
Q.22	1.66 × 10⁻²⁷ kg
Q.23	θ must be dimensionless
Q.24	L, M and T are independent quantities
Q.25(a)	2°
Q.25(b)	Earth = 4 × Moon
Q.25(c)	Sun = 100 × Earth
Q.26	Atomic Clock
Q.27	10¹⁶ s
Q.28	0.01 mm
Q.29	Rs/Rm = Ds/Dm
Q.30	10⁵ kg

CBSE Class 11 Physics
Units and Measurements
Solutions (Q.31 - Q.37)

Q.31 Give an example of:

(a) A physical quantity which has a unit but no dimensions.

Answer: Plane Angle (θ)
Plane angle is defined as:

θ = Arc Length / Radius

Since both numerator and denominator have dimension L:

θ = L/L = 1

Therefore, it has no dimensions but its SI unit is radian (rad).

(b) A physical quantity which has neither unit nor dimensions.

Answer: Strain

Strain = Change in Length / Original Length

= L/L = 1

Hence strain has neither unit nor dimensions.

(c) A constant which has a unit.

Answer: Gravitational Constant (G)

G = 6.67 × 10⁻¹¹ N m² kg⁻²

Therefore it is a constant having units.

(d) A constant which has no unit.

Answer: Reynolds Number
It is dimensionless and unitless.

Q.32 Calculate the length of the arc of a circle of radius 31.0 cm which subtends an angle of π/6 at the centre.

Given:
Radius, r = 31 cm
Angle, θ = π/6

θ = l/r

Therefore,

l = rθ

Substituting values:

l = 31 × (π/6)

l = 16.24 cm

Rounded to three significant figures:

l = 16.2 cm

Answer: 16.2 cm

Q.33 Calculate the solid angle subtended by an area of 12 cm² at a point situated symmetrically at a distance of 5 cm from the area.

Given:
Area = 12 cm²
Distance = 5 cm

Ω = Area / Distance²

Ω = 12 / (5²)

Ω = 12 / 25

Ω = 0.48 steradian

Answer: 0.48 steradian

Q.34 The displacement of a progressive wave is represented by y = A sin(ωt − kx), where x is distance and t is time. Write the dimensional formula of (i) ω and (ii) k.

By the principle of homogeneity, the quantity inside the sine function must be dimensionless.

(i) Dimension of ω

ωt = dimensionless

[ω][T] = 1

[ω] = T⁻¹

Answer: [ω] = T⁻¹

(ii) Dimension of k

kx = dimensionless

[k][L] = 1

[k] = L⁻¹

Answer: [k] = L⁻¹

Q.35 Time for 20 oscillations of a pendulum is measured as t₁ = 39.6 s, t₂ = 39.9 s and t₃ = 39.5 s. What is the precision and accuracy of the measurement?

Step 1: Precision
Since readings are measured up to one decimal place:

Least Count = 0.1 s

Therefore,

Precision = 0.1 s

Step 2: Mean Value

t̄ = (39.6 + 39.9 + 39.5)/3

t̄ = 39.7 s

Step 3: Absolute Errors

Δt₁ = |39.7 - 39.6| = 0.1 s

Δt₂ = |39.7 - 39.9| = 0.2 s

Δt₃ = |39.7 - 39.5| = 0.2 s

Step 4: Mean Absolute Error

Δt = (0.1 + 0.2 + 0.2)/3

Δt = 0.17 s ≈ 0.2 s

Answer:
Precision = 0.1 s
Accuracy = ±0.2 s

Q.36 A new system of units is proposed in which unit of mass is a kg, unit of length is b m and unit of time is g s. How much will 5 J measure in this new system?

Dimension of energy:

[E] = ML²T⁻²

Let:

M₂ = a kg

L₂ = b m

T₂ = g s

Using:

n₁u₁ = n₂u₂

New unit of energy:

u₂ = ab²g⁻² u₁

Therefore:

5u₁ = n₂(ab²/g²)u₁

n₂ = 5g²/(ab²)

Answer:

5J = 5g²/(ab²)

in the new system.

Q.37 The volume of a liquid flowing out per second of a pipe of length l and radius r is given by:

V = (πpr⁴)/(8ηl)

where p is pressure difference and η is coefficient of viscosity having dimensional formula [ML⁻¹T⁻¹]. Check whether the equation is dimensionally correct.

Dimension of LHS:

[V] = Volume / Time

[V] = L³T⁻¹

Dimension of RHS:

[p] = ML⁻¹T⁻²

[r] = L

[l] = L

[η] = ML⁻¹T⁻¹

Substituting:

RHS = (ML⁻¹T⁻² × L⁴)/(ML⁻¹T⁻¹ × L)

RHS = L³T⁻¹

Comparing:

LHS = RHS = L³T⁻¹

Therefore the equation is dimensionally correct. Answer: The given equation is dimensionally correct.

CBSE Class 11 Physics
Units and Measurements (Q38–Q44)

Q.38

A physical quantity X is related to four measurable quantities a, b, c and d as follows:

X = a² b³ c^(5/2) d²

The percentage error in a, b, c and d are 1%, 2%, 3% and 4% respectively. Find the percentage error in X. If X = 2.763, round off the result appropriately.

Solution:

Step 1: Error formula

% error in X = |2|×1% + |3|×2% + |5/2|×3% + |2|×4%

Step 2: Calculation

= 2 + 6 + 7.5 + 8 = 23.5%

Step 3: Absolute error

ΔX = (23.5/100) × 2.763 ≈ 0.65

Step 4: Final value

X ≈ 2.8

Final Answer: 23.5%, X = 2.8

Q.39

Show that P is dimensionless where

P = (E L²) / (m² G)

Solution:

Step 1: Dimensions

E = ML²T⁻² m = M L = ML²T⁻¹ G = M⁻¹L³T⁻²

Step 2: Substitute

P = (ML²T⁻² × M²L²T⁻²) / (M² × M⁻¹L³T⁻²)

Step 3: Simplify

P = 1

Final Answer: P is dimensionless

Q.40

Express mass, length and time in terms of c, h and G.

Solution:

c = LT⁻¹ h = ML²T⁻¹ G = M⁻¹L³T⁻²

Final Results:

M = √(hc/G)

L = Gh / c³

T = Gh / c⁵

Q.41

Show that T = k √(r³ / Rg)

Solution:

Step 1: Assume T = k rᵃ Rᵇ gᶜ

T = L^(a+b+c) T^(-2c)

Step 2: Compare powers

-2c = 1 → c = -1/2 a + b + c = 0

Final result:

T = k √(r³ / Rg)

Q.42

Oleic acid experiment (conceptual)

(a) Dissolved in alcohol because oleic acid is insoluble in water.

(b) Lycopodium powder helps visualize spreading area.

(c) Volume per mL = 1/400 mL

(d) Volume of n drops = n × (volume of one drop)

(e) Volume in one drop = 1/(400n) mL

Q.43

Astronomy and units

(a) 1 parsec = 2 × 10⁵ AU

(b) Angular size ≈ 1 arc minute (due to atmospheric limit)

(c) Mars apparent size ≈ 30 arc minutes

Q.44

Einstein mass-energy relation

(a)

E = mc² 1u = 931.5 MeV

(b) Correct relation:

1u c² = 931.5 MeV

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