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.