Class 11 Physics Worksheet Solutions: Units & Measurements (Chapter 1)

Class XI Physics

Chapter - 1: Units & Measurement

Worksheet Solutions

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Q1. Calculate the length of the arc of a circle of radius 31 cm which subtends an angle of π/6 at the centre.

Formula:

The relation between angle (θ in radians), arc length (l), and radius (r) is given by:

θ = l / r

Given:

• Radius, r = 31 cm
• Angle, θ = π / 6 radians

Solution:

Rearranging the formula to find arc length:

l = r ⋅ θ

l = 31 cm × π / 6

Putting the approximate value of π ≈ 3.14:

l ≈ 31 × 3.14 / 6

l ≈ 31 × 0.5233

l ≈ 16.22 cm

Answer: The length of the arc is approximately 16.22 cm.

Q2. Calculate the solid angle subtended by the periphery of an area of 1 cm² at a point situated symmetrically at a distance of 5 cm from the area.

Formula:

The solid angle (Ω) is defined as:

Ω = Area (A) / Distance squared (r²)

Given:

• Area, A = 1 cm²
• Distance, r = 5 cm

Solution:

Ω = 1 cm² / (5 cm)²

Ω = 1 / 25

Ω = 0.04 steradians (sr)

Answer: The solid angle subtended is 0.04 sr.

Q3. 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?

Dimensional Formula of Force:

[F] = [M¹ L¹ T⁻²]

Given in new system:

• Unit of Force (F) = 100 N
• Unit of Length (L) = 10 m
• Unit of Time (T) = 100 s

Solution:

Let the unit of mass in the new system be M'. We can write the dimensional formula as a relationship of magnitudes:

F = (M × L) / T²

So, for the standard system: 1 N = (1 kg) × (1 m) / (1 s)².

Rearranging the relationship to solve for mass M in terms of other units:

M = (F × T²) / L

Now, substitute the values from the new system of units into this relationship:

M' = [ (100 N) × (100 s)² ] / (10 m)

Convert the N unit to basic units for clearer calculation: 1 N = 1 kg⋅m⋅s⁻².

M' = [ (100 kg⋅m⋅s⁻²) × (10000 s²) ] / (10 m)

Notice the s⁻² and s² cancel out, as do the m in numerator and denominator, leaving only kg.

M' = (100 × 10000) / 10 kg

M' = 1,000,000 / 10 kg

M' = 100,000 kg = 10⁵ kg

Answer: The unit of mass in this system is 10⁵ kg or 100,000 kg.

Q4. 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.

Principle of Dimensional Homogeneity:

The argument of a trigonometric function like sine must be dimensionless. This means both terms, (ωt) and (kx), are dimensionless.

Standard dimensions are:

• Displacement (y or x): [L]
• Time (t): [T]
• Amplitude (A): [L]

(i) For ω (angular frequency):

Since (ωt) is dimensionless:

[ωt] = [M⁰ L⁰ T⁰]

[ω] [t] = [1]

[ω] [T] = [1]

[ω] = [T⁻¹]

Fully expressed: [M⁰ L⁰ T⁻¹].

(ii) For k (angular wavenumber):

Since (kx) is dimensionless:

[kx] = [M⁰ L⁰ T⁰]

[k] [x] = [1]

[k] [L] = [1]

[k] = [L⁻¹]

Fully expressed: [M⁰ L⁻¹ T⁰].

Answer: (i) Dimensional formula of ω is [T⁻¹]. (ii) Dimensional formula of k is [L⁻¹].

Q5. In the expression P = E L² m⁻⁵ G⁻²; E, m, L and G denote energy, mass, angular momentum and gravitational constant, respectively. Show that P is a dimensionless quantity.

Step 1: Write dimensional formulas for each variable.

• Energy (E): From Work = Force × distance, [E] = [M¹ L¹ T⁻²] × [L] = [M¹ L² T⁻²].
• Mass (m): [m] = [M¹ L⁰ T⁰].
• Angular Momentum (L): From Momentum (p) × distance (r) = (mass × velocity) × distance,
  [L] = ([M¹] × [L¹ T⁻¹]) × [L¹] = [M¹ L² T⁻¹].
• Gravitational Constant (G): From F = G m₁m₂ / r² ⇒ G = F r² / (m₁m₂),
  [G] = ([M¹ L¹ T⁻²] × [L]²) / [M]² = [M⁻¹ L³ T⁻²].

Step 2: Substitute these dimensional formulas into the equation for P.

[P] = [E L² m⁻⁵ G⁻²]

[P] = [M¹ L² T⁻²]¹ × [M¹ L² T⁻¹]² × [M¹]⁻⁵ × [M⁻¹ L³ T⁻²]⁻²]

Step 3: Combine powers for each fundamental dimension.

• For M: (1) + 2(1) + (−5)(1) + (−2)(−1) = 1 + 2 − 5 + 2 = 0
• For L: (2) + 2(2) + (−5)(0) + (−2)(3) = 2 + 4 + 0 − 6 = 0
• For T: (−2) + 2(−1) + (−5)(0) + (−2)(−2) = −2 − 2 + 0 + 4 = 0

Step 4: Combine the results.

[P] = [M⁰ L⁰ T⁰]

Since all powers are zero, P is a dimensionless quantity. **Hence proved.**

Q6. Taking velocity, time and force as the fundamental quantities, find the dimensions of mass.

This requires expressing mass in terms of force, velocity, and time.

Step 1: Relate given quantities.

We know Newton's second law: Force (F) = mass (M) × acceleration (a).

Acceleration can be expressed as change in velocity over time: a ≈ Velocity (V) / Time (T).

Substitute this into the force equation: F = M × V / T.

Step 2: Rearrange to solve for mass M.

M = F × T / V

Expressing this dimensionally in terms of the required fundamental quantities (F, V, T):

[M] = [F] [T] / [V]

[M] = [F¹ V⁻¹ T¹]

Answer: The dimensions of mass in terms of velocity, time, and force are [F V⁻¹ T].

Q7. The value of G in CGS system is 6.67 × 10⁻⁸ dyne cm² g⁻². Calculate the value in SI units.

Conversion from CGS to SI units. We'll use the dimensional formula of G from Q5: [G] = [M⁻¹ L³ T⁻²].

Conversion parameters:

• Mass (M): 1 g = 10⁻³ kg
• Length (L): 1 cm = 10⁻² m
• Time (T): 1 s = 1 s (does not change)

Solution using direct substitution into units:

G_CGS = 6.67 × 10⁻⁸ dyne cm² / g²

Recall conversion for dyne: 1 dyne = 10⁻⁵ N (from 1 N = 1 kg⋅m/s² = 1000g ⋅ 100cm/s² = 10⁵ dyne).

Substitute conversion factors:

G_SI = 6.67 × 10⁻⁸ × [ (10⁻⁵ N) × (10⁻² m)² ] / [ (10⁻³ kg)² ]

G_SI = 6.67 × 10⁻⁸ × [ (10⁻⁵ N × 10⁻⁴ m²) / (10⁻⁶ kg²) ]

Combine numeric powers of 10 and keep units separate:

G_SI = 6.67 × 10⁻⁸ × (10^{−5 − 4 + 6}) N⋅m²/kg²

G_SI = 6.67 × 10⁻⁸ × (10^{−9 + 6}) N⋅m²/kg²

G_SI = 6.67 × 10⁻⁸ × 10⁻³ N⋅m²/kg²

G_SI = 6.67 × 10⁻¹¹ N⋅m²/kg²

Answer: The value of G in SI units is 6.67 × 10⁻¹¹ N⋅m²/kg².

Q8. Find the value of 60 J per min on a system that has 100 g, 100 cm and 1 min as the base units.

This is conversion of power to a new unit system.

Step 1: Identify quantity and dimensions.

Quantity is energy per time, which is Power.

Dimensional formula of power: [P] = Work / Time = [M¹ L² T⁻²] / [T] = [M¹ L² T⁻³].

Exponents are a=1, b=2, c=−3.

Step 2: Define conversion parameters. We use the formula n₂ = n₁ [u₁/u₂].

Standard System (SI):

• Numeric value, n₁ = 60 J / min = 60 J / 60 s = 1 J/s = 1 Watt.
• Base units: M₁ = 1 kg, L₁ = 1 m, T₁ = 1 s.

New System:

• Numeric value, n₂ = ? (to be found).
• Base units: M₂ = 100 g = 0.1 kg, L₂ = 100 cm = 1 m, T₂ = 1 min = 60 s.

Step 3: Apply the conversion formula.

n₂ = n₁ (M₁/M₂)^a (L₁/L₂)^b (T₁/T₂)^c

n₂ = 1 (1 kg / 0.1 kg)¹ (1 m / 1 m)² (1 s / 60 s)⁻³

n₂ = 1 × (10)¹ × (1)² × (1 / 60)⁻³

Recalling x⁻ⁿ = 1/xⁿ, so (1/60)⁻³ = 60³.

n₂ = 10 × (60 × 60 × 60)

n₂ = 10 × 216,000

n₂ = 2,160,000

Answer: The value in the new system is 2,160,000 new units.

Q9. The density of mercury is 13.6 g⋅cm⁻³ in CGS system. Find its values in SI units.

Conversion from CGS to SI.

Step 1: Identify quantity and dimensions.

Quantity is density (ρ).

Dimensional formula of density: [ρ] = Mass / Volume = [M] / [L]³ = [ML⁻³].

Exponents are a=1, b=−3, c=0.

Step 2: Define conversion parameters.

CGS System (System 1): n₁ = 13.6, M₁ = 1g, L₁ = 1cm, T₁ = 1s.

SI System (System 2): n₂ = ?, M₂ = 1kg = 1000g, L₂ = 1m = 100cm, T₂ = 1s.

Step 3: Apply conversion formula.

n₂ = n₁ (M₁/M₂)^a (L₁/L₂)^b (T₁/T₂)^c

n₂ = 13.6 (1g / 1000g)¹ (1cm / 100cm)⁻³ (1s / 1s)⁰

n₂ = 13.6 × (10⁻³) × (10⁻²)⁻³ × (1)

n₂ = 13.6 × 10⁻³ × 10^{(−2 × −3)}

n₂ = 13.6 × 10⁻³ × 10⁶

n₂ = 13.6 × 10³ = 13,600

Answer: The value of density in SI units is 13,600 kg⋅m⁻³.

Q10. Find the dimensions of a/b in the equation: F = a√x + bt², where F is force, x is distance and t is time.

Use the Principle of Dimensional Homogeneity.

Step 1: State principle and find basic dimensions.

All terms on both sides must have same dimensions. So [F] = [a√x] = [bt²].

• Force (F): [M¹ L¹ T⁻²]
• Distance (x): [L¹]. So, √x = x^1/2 ⇒ [√x] = [L^1/2]
• Time (t): [T¹]

Step 2: Find dimensions of a.

From [F] = [a√x] ⇒ [a] = [F] / [√x].

[a] = [M¹ L¹ T⁻²] / [L^1/2] = [M¹ L^{(1 − 1/2)} T⁻²] = [M¹ L^1/2 T⁻²].

Step 3: Find dimensions of b.

From [F] = [bt²] ⇒ [b] = [F] / [t²].

[b] = [M¹ L¹ T⁻²] / [T]² = [M¹ L¹ T^{−2 − 2}] = [M¹ L¹ T⁻⁴].

Step 4: Find dimensions of a/b.

[a/b] = [a] / [b] = [M¹ L^1/2 T⁻²] / [M¹ L¹ T⁻⁴].

M cancels out. [a/b] = [L^{(1/2 − 1)} T^{(−2 − (−4))}].

[a/b] = [L^−1/2 T²].

Answer: The dimensional formula of a/b is [L^−1/2 T²].

Q11. The Van der Waal's equation for a gas is (P + a/V²)(V − b) = RT, determine the dimension of a and b. Hence write the SI units of a and b.

Use the Principle of Dimensional Homogeneity (can only add/subtract same dimensions).

Step 1: Identify dimensions of standard quantities.

• Pressure (P): Force / Area = [MLT⁻²] / [L²] = [ML⁻¹T⁻²].
• Volume (V): [L]³ = [L³].

Step 2: Find dimensions of b.

In (V − b), b is subtracted from V, so dimensions are equal.

[b] = [V] = [L³].

Step 3: Find dimensions of a.

In (P + a/V²), a/V² is added to P, so dimensions are equal.

[a/V²] = [P] &rArr [a] = [P] [V]².

[a] = [ML⁻¹T⁻²] × [L³]² = [ML⁻¹T⁻²] × [L⁶] = [ML⁽⁻¹⁺⁶⁾ T⁻²] = [ML⁵T⁻²].

Step 4: Write SI units.

• For b (dimension [L³]): Unit is meter³, or m³.
• For a (dimension [ML⁵T⁻²]): Unit is kg⋅m⁵/s², or kg⋅m⁵⋅s⁻². Alternatively, since [a] = [P][V]², unit is Pascal⋅m⁶, which is equivalent to N⋅m⁻²⋅m⁶ = N⋅m⁴. Both are correct; kg⋅m⁵⋅s⁻² uses base SI units.

Answer: Dimensions of b: [L³], SI unit: m³. Dimensions of a: [ML⁵T⁻²], SI unit: kg⋅m⁵⋅s⁻².

Q12. A small spherical ball of radius r falls with velocity v through a liquid having coefficient of viscosity η. Obtain an expression for viscous drag force F on the ball assuming it depends on η, r and v. (Take k = 6π)

Derivation using the method of dimensions.

Step 1: Express relationship. Let F ∝ η^a r^b v^c. Introduce constant k.

F = k η^a r^b v^c -- (Equation 1).

Step 2: Write standard dimensional formulas.

• Force (F): [MLT⁻²].
• Coefficient of viscosity (η): From F = η A dv/dx, [η] = [F] / [A] [dv/dx] = [MLT⁻²] / [L²] [LT⁻¹/L] = [ML⁻¹T⁻¹].
• Radius (r): [L].
• Velocity (v): [LT⁻¹].
• k is dimensionless.

Step 3: Put dimensions in Equation 1.

[MLT⁻²] = [ML⁻¹T⁻¹]^a [L]^b [LT⁻¹]^c.</

Combine terms on right side: [MLT⁻²] = [M^a L^{(−a + b + c)} T^{(−a − c)}].

Step 4: Equate powers.

• For M: a = 1.
• For T: −a − c = −2. Substitute a=1 ⇒ −1 − c = −2 ⇒ c = 1.
• For L: −a + b + c = 1. Substitute a=1, c=1 ⇒ −1 + b + 1 = 1 ⇒ b = 1.

Step 5: Substitute a, b, c back into Equation 1.

F = k η¹ r¹ v¹.

Given k = 6π. The final formula is: F = 6πηrv. (Stokes' Law).

Q13. Write the number of significant figures in the following:

Rules applied: non-zero digits are significant; leading zeros are not; trailing zeros with a decimal point are; trailing zeros in a whole number are usually not significant placeholders (unless specified, which we assume here not).

• (i) 0.005 m² : Only one non-zero digit. Answer: 1.
• (ii) 2.63 × 10²⁸ kg : The digits before the power are significant. Answer: 3.
• (iii) 0.2560 gcm⁻³ : Trailing zero after decimal counts. Answer: 4.
• (iv) 6.320 N : Trailing zero counts. Answer: 4.
• (v) 5.003 J : Middle zeros count. Answer: 4.
• (vi) 0.0006032 : Leading zeros don't count. Answer: 4.
• (vii) 12500 : Trailing zeros are assumed placeholders. Answer: 3.

Q14. If dimensions of length are expressed as G^x c^y h^z, where G, c and h are the universal gravitational constant, speed of light and Planck's constant respectively, what are the value of x, y and z?

Step 1: Write dimensional formulas.

• Length (L): [M⁰L¹T⁰].
• c (Velocity): [LT⁻¹].
• G (from Q5): [M⁻¹L³T⁻²].
• h: From E = hν ⇒ h = E/ν. Frequency ν is 1/Time, dim [T⁻¹]. E (Energy, Q5) dim [ML²T⁻²].
  [h] = [ML²T⁻²] / [T⁻¹] = [ML²T⁻¹].

Step 2: Set up the dimensional equation.

[M⁰L¹T⁰] = [M⁻¹L³T⁻²]^x [LT⁻¹]^y [ML²T⁻¹]^z.

Combine powers on right side:
[M⁰L¹T⁰] = [M^(−x+z) L^(3x+y+2z) T^{(−2x−y−z)}].

Step 3: Form linear equations by equating powers.

1. For M: −x + z = 0 ⇒ x = z -- (Eq 1).
2. For T: −2x − y − z = 0 -- (Eq 2).
3. For L: 3x + y + 2z = 1 -- (Eq 3).

Step 4: Solve the equations. Substitute x=z into others.

New Eq 2: −2x − y − x = 0 ⇒ −3x = y ⇒ y = −3x -- (Eq 4).

New Eq 3: 3x + y + 2x = 1 ⇒ 5x + y = 1.

Substitute y = −3x into this: 5x + (−3x) = 1 ⇒ 2x = 1 ⇒ x = 1/2.

Find y from Eq 4: y = −3(1/2) ⇒ y = −3/2.

Find z from Eq 1: z = x ⇒ z = 1/2.

Answer: The values are x = 1/2, y = −3/2, z = 1/2.

NCERT Solutions Class 11 Physics Chapter 3: Point Object Concept

Motion in a Straight Line: Point Object Analysis

The Rule: If the distance moved by an object is much greater than its linear size, the object can be considered a "Point Object."

POINT OBJECT

(a) Railway Carriage

Carriage Length: ~15 - 30 meters

Distance (Station to Station): ~10 - 50 km

Size is negligible compared to the distance between stations.

POINT OBJECT

(b) Man's Cap on Track

Cap Diameter: ~0.2 meters

Circular Track Length: ~400 meters

The cap's dimensions are insignificant relative to the lap distance.

NOT A POINT OBJECT

(c) Spinning Cricket Ball

Ball Diameter: ~0.07 meters

Spin/Deviation: ~0.2 to 0.5 meters

The distance it "turns" is comparable to its own size; spin mechanics matter here.

NOT A POINT OBJECT

(d) Tumbling Beaker

Beaker Height: ~0.1 meters

Table Height: ~0.75 - 1.0 meters

The height of fall is only about 7-10 times the size of the object. Not negligible.

Q1. In which of the following examples of motion can the body be considered approximately a point object?

• (a) A railway carriage moving without jerks between two stations.
• (b) A cap on top of a man cycling smoothly on a circular track.
• (c) A spinning cricket ball that turns sharply on hitting the ground.
• (d) A tumbling beaker that has slipped off the edge of a table.

Answer: (a) and (b)

Reasoning: A body is considered a point object if the distance it moves is much larger than its own linear dimensions.

• Cases (a) & (b): The size of the carriage and the cap is negligible compared to the distance between stations or the length of a race track. Hence, they are point objects.
• Cases (c) & (d): The size of the ball is comparable to its bounce distance, and the beaker's size is comparable to the height of the table. Hence, they cannot be treated as point objects.

Class 11 Physics | Chapter 3: Motion in a Straight Line

NCERT Solutions & Conceptual Notes

Q1. In which of the following examples of motion can the body be considered approximately a point object?

•(a) A railway carriage moving without jerks between two stations.

Understanding the Point Object concept: When does size matter in Physics?

•(b) A cap on top of a man cycling smoothly on a circular track. •(c) A spinning cricket ball that turns sharply on hitting the ground. •(d) A tumbling beaker that has slipped off the edge of a table.

Detailed Explanation:

YES (POINT OBJECT)

Cases (a) and (b)

The distance between two stations or the length of a racing track is vastly larger than the dimensions of a carriage or a cap.

Condition: Object Size << Distance Traveled

NO (EXTENDED BODY)

Cases (c) and (d)

The size of the ball or beaker is comparable to the distance it moves (the turn or the table height). Internal motion (spinning/tumbling) matters here.

Condition: Object Size ≈ Distance Traveled

💡 Core Concept: In Physics, a "Point Object" is an idealization. We ignore the internal structure and size of a body only when its motion covers a distance many times its own lengths 

5. Internal Links

 * Prerequisite Knowledge: What is Kinematics? An Introduction to Motion

 * Next Lesson: Average Speed vs. Instantaneous Velocity - Class 11 Notes

 * Practice: MCQ Quiz: Motion in a Straight Line (Level 1)