Class XI Physics Summer Assignment
Subject Enrichment Activity – I

Q.1 Define Significant Figure? Write down one example.

Definition: Significant figures are the meaningful digits in a measured or calculated quantity which are known with certainty plus one final digit that is estimated or uncertain. They indicate the precision of a measurement.

Example: In the measurement 2.308 cm, there are 4 significant figures (2, 3, 0 and 8).

Q.2 List two commonly used practical units. How are these units related to S.I. units?

1. Light Year (ly): Used for large astronomical distances.
   Relation with SI unit: 1 light year ≈ 9.46 × 10¹⁵ m
2. Horsepower (hp): Used for power.
   Relation with SI unit: 1 hp = 746 W

Q.3 Define practical units of distance: Light Year and Parallactic Second (Parsec).

Light Year (ly): It is the distance travelled by light in vacuum in one year.

Parsec (pc): It is the distance at which an arc of length 1 Astronomical Unit (AU) subtends an angle of one second of arc.

Q.4 In a system of units, the unit of length, mass and time are 10 cm, 10 g and 0.1 s respectively. Find the unit of force.

Step 1: Dimensional formula of force:

[F] = [M L T^-2]

Step 2: Using conversion formula:

n₂ = n₁ (M₁/M₂)^a (L₁/L₂)^b (T₁/T₂)^c

M₁ = 1000 g, M₂ = 10 g
L₁ = 100 cm, L₂ = 10 cm
T₁ = 1 s, T₂ = 0.1 s

n₂ = 1 × (1000/10) × (100/10) × (1/0.1)^-2

n₂ = 100 × 10 × (10)^-2 = 10

Therefore, 1 N = 10 new units of force

Hence, the unit of force in the new system is: 0.1 N

Q.5 If velocity, time and force are chosen as basic quantities, find the dimensions of mass.

Let

M = F^a V^b T^c

Dimensions:
[F] = [M L T^-2]
[V] = [L T^-1]

Comparing powers:

• a = 1
• a + b = 0 ⇒ b = -1
• -2a - b + c = 0 ⇒ c = 1

Therefore,

[M] = [F¹ V⁻¹ T¹]

Q.6 If the time period (T) of vibration of a liquid drop depends on surface tension (S), radius (r) and density (ρ), find the expression for T.

Assume

T = k S^a r^b ρ^c

Dimensions:

• [T] = [T]
• [S] = [M T^-2]
• [r] = [L]
• [ρ] = [M L^-3]

Comparing dimensions:

• -2a = 1 ⇒ a = -1/2
• a + c = 0 ⇒ c = 1/2
• b - 3c = 0 ⇒ b = 3/2

Hence,

T = k r^3/2 ρ^1/2 / S^1/2

T = k √(ρr³/S)

Q.7 Define Astronomical Unit, Light Year and Parsec. Find relation between them.

Astronomical Unit (AU): Average distance between Earth and Sun.

Light Year: Distance travelled by light in one year.

Parsec: Distance at which 1 AU subtends an angle of 1 arc second.

Relations:

• 1 AU = 1.496 × 10¹¹ m
• 1 ly = 9.46 × 10¹⁵ m
• 1 pc = 3.086 × 10¹⁶ m

Therefore,

• 1 pc ≈ 3.26 ly
• 1 ly ≈ 63240 AU
• 1 pc ≈ 206265 AU

Increasing order: 1 AU < 1 ly < 1 pc

Q.8 In Van der Waals equation, (P + a/V²)(V − b) = RT. Find the dimension of a/b.

Since P and a/V² are added:

[a/V²] = [P]

Pressure:

[P] = [M L^-1 T^-2]

Volume:

[V] = [L³]

Therefore,

[a] = [M L⁵ T⁻²]

Since (V − b):

[b] = [L³]

Hence,

[a/b] = [M L² T⁻²]

This is the dimension of Work/Energy.

Q.9 Show dimensionally the factors affecting the time period of a simple pendulum.

Let

T = k m^a l^b g^c θ^d

θ is dimensionless.

Comparing dimensions:

• a = 0
• -2c = 1 ⇒ c = -1/2
• b + c = 0 ⇒ b = 1/2

Therefore,

T = k √(l/g)

Conclusion:

• Mass (m) has no effect.
• Angular displacement (θ) cannot be determined dimensionally.
• T ∝ √l
• T ∝ 1/√g

Q.10 If c, G and h are chosen as fundamental units, find the dimensions of mass.

Let

M = c^a G^b h^c

Using dimensions:

• [c] = [L T⁻¹]
• [G] = [M⁻¹ L³ T⁻²]
• [h] = [M L² T⁻¹]

Solving,

a = 1/2, b = -1/2, c = 1/2

Therefore,

[M] = [c^1/2 G^-1/2 h^1/2]

Q.11 Explain the Principle of Homogeneity. Show that M ∝ V⁶.

Principle of Homogeneity: A physical equation is dimensionally correct only if all terms on both sides have the same dimensions.

Let

M = k V^a ρ^b g^c

Comparing dimensions:

• b = 1
• a = -2c
• a - 3b + c = 0

Substituting b = 1:

a - 3 + c = 0

and a = -2c

Solving:

c = -3, a = 6

Hence,

M ∝ V⁶

Therefore, the mass of the largest stone moved by a river varies as the sixth power of velocity.

Q.12 Write the dimensions of a and b in the relation P = (b − x²)/(at).

Given:

P = (b − x²)/(at)

Since b and x² are subtracted:

[b] = [L²]

Rearranging:

a = (b − x²)/(Pt)

Power:

[P] = [M L² T⁻³]

Therefore,

[a] = [L²] / ([M L² T⁻³] × [T])

[a] = [M⁻¹ T²]

Answer:

• [b] = [L²]
• [a] = [M⁻¹ T²]