Class 11 Physics (Physics Part-I) Chapter 1 Units And Measurement

Key Concept: Base Units

c) Ampere

[Solution Description] The SI base unit for measuring electric current is the ampere, often symbolized as A. This is one of the seven fundamental units in the International System of Units.

Key Concept: Complex Rounding

c) 3.5

[Solution Description] The digit to be dropped is 6 (in 3.465), which is more than 5, so we increase the preceding digit (4 to 5). Hence, rounding off to two significant figures gives us 3.5.

Key Concept: Basic Definition

b) Steradian

[Solution Description] The unit of solid angle in the SI system is called a steradian, which is denoted by “sr”. This is the official unit used worldwide for measuring solid angles.

Key Concept: Simple Calculations

c) 15 J

[Solution Description] Using the formula for work $W = F \cdot d$, where F is the force and d is the displacement:

$W = 5 \, \text{N} \times 3 \, \text{m} = 15 \, \text{J}$

Thus, the work done is 15 Joules.

Key Concept: Unit Symbol Identification

b) m

[Solution Description] The symbol for the metre, which is the SI unit of length, is ‘m’.

Key Concept: Basic Definition

a) Radian

[Solution Description] The unit for measuring plane angle in the SI system is the radian, which is defined as the ratio of the length of an arc to its radius.

Key Concept: System Comparison

a) CGS

[Solution Description] The CGS system is a historical system of units that uses centimetre, gram, and second as its base units for length, mass, and time respectively.

Key Concept: Definition Recall

c) Planck constant

[Solution Description] The kilogram is defined by the Planck constant $$h$$, with a value of $$6.62607015 \times 10^{-34}$$ J s.

Key Concept: Dimensional Formula

c) $$[M L^2 T^{-2}]$$

[Solution Description] The dimensional formula for energy can be derived from its relation with work, which is defined as force applied over a distance. Force has the dimension $$[M L T^{-2}]$$ and distance has the dimension $$[L]$$. Thus, the dimensional formula for energy is the product of these two:

$$[M L T^{-2}] \times [L] = [M L^2 T^{-2}]$$

Key Concept: Dimensional Homogeneity

b) $[M^{-1} L^3 T^{-2}]$

[Solution Description] The dimensions of force F are $[M L T^{-2}]$, mass $m_1$ and $m_2$ both have dimensions [M], and distance r has dimensions [L]. Substitute these into the equation:

$[G] = \frac{[F][r]^2}{[m_1][m_2]} = \frac{[M L T^{-2}][L]^2}{[M][M]} = \frac{[M L^3 T^{-2}]}{[M^2]}$

Simplifying gives:

$[G] = [M^{-1} L^3 T^{-2}]$

Key Concept: Basic Addition Rule

b) No

[Solution Description]

The principle of homogeneity states that physical quantities can only be added or subtracted if they have the same dimensions. Energy has the dimension $$[ML^2T^{-2}]$$, while force has the dimension $$[MLT^{-2}]$$. Since their dimensions are not the same, energy cannot be added to force.

Key Concept: Dimensional Formula

a) $[M L^2 T^{-2}]$

[Solution Description]

Energy can be expressed as work done, and the formula for work is $W = F \times d$ where F is force and d is distance. The dimensional formula for force is $[M L T^{-2}]$. Therefore,

$\text{Dimensional formula for energy} = [M L T^{-2}] \times [L] = [M L^2 T^{-2}].$

Key Concept: Advanced Consistency Check, Higher Order Reasoning

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

To verify the assertion, we need to ensure that both sides of the equation $E = \frac{1}{2}mv^2 + mgh$ have the same dimensions.

– The left side of the equation is energy $E$, having a dimensional formula $[M L^2 T^{-2}]$.

– The first term on the right is kinetic energy $\frac{1}{2}mv^2$. Its dimensional formula:

$\text{Kinetic Energy} = \frac{1}{2} m v^2 = \text{mass} \times (\text{velocity})^2$

$= [M] \times ([L T^{-1}])^2 = [M L^2 T^{-2}]$

– The second term on the right is potential energy $mgh$. Its dimensional formula:

$\text{Potential Energy} = mgh = \text{mass} \times \text{acceleration due to gravity} \times \text{height}$

$= [M] \times [L T^{-2}] \times [L] = [M L^2 T^{-2}]$

Since both terms ($\frac{1}{2}mv^2$ and $mgh$) and the left side have the same dimensional formula $[M L^2 T^{-2}]$, the equation is dimensionally consistent, making Assertion true.

The reason states that the dimensional formula of $$mgh$$ is the same as that of energy, which is indeed correct. Therefore, Reason is also true, and provides an explanation for why the assertion is true.

Key Concept: Dimensionless Quantity

a) Angle

[Solution Description]

Dimensionless quantities are those which have no physical dimension and are just numbers. These often occur in ratios or as arguments for trigonometric, logarithmic, or exponential functions. Among the given options, angle is a dimensionless quantity because it is expressed as the ratio of two lengths (arc length/radius), thus having no dimension.

Key Concept: Complex Dimensional Analysis, Distinguishing Equations with Same Dimensions

a) $$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$$

[Solution Description] Frequency $$f$$ has the dimension $$[T^{-1}]$$. Check each option:

– For a), $$\left[\sqrt{\frac{k}{m}}\right] = \left[\sqrt{\frac{MLT^{-2}}{M}}\right] = [T^{-1}]$$, which is dimensionally consistent.

– For b), $$\left[\frac{k}{2\pi m}\right] = \left[\frac{MLT^{-2}}{M}\right] = [LT^{-2}]$$, which does not match.

– For c), $$\left[\sqrt{\frac{km}{2\pi}}\right] = \left[\sqrt{(MLT^{-2})M}\right] = [M^{1/2}L T^{-1}]\neq[T^{-1}]$$, which does not match.

– For d), $$\left[2\pi \sqrt{\frac{m}{k}}\right] = \left[\sqrt{\frac{M}{MLT^{-2}}}\right] = [L^{1/2}T]\neq[T^{-1}]$$, which does not match.

Therefore, option a) $$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$$ is dimensionally consistent with frequency.

Key Concept: Derived Quantity Analysis

c) $[M L T^{-1}]$

[Solution Description]

Momentum (p) is calculated using the formula $p = mv$, where m is mass and v is velocity. The dimensional formula for mass is [M] and for velocity is $[M^0 L T^{-1}]$. Thus, the dimensional formula for momentum is:

$[M][M^0 L T^{-1}] = [M L T^{-1}]$

Therefore, the correct dimensional formula for momentum is $[M L T^{-1}]$.

Key Concept: Constant Value

d) $$1.380649\times10^{-23}$$

[Solution Description] The fixed numerical value of the Boltzmann constant $$k$$ specified in the definition of kelvin is $$1.380649 \times 10^{-23} \, \text{J} \, \text{K}^{-1}$$.

Key Concept: Unit Identification

a) Calorie

[Solution Description] The unit “calorie” is a non-SI unit commonly used to measure energy, particularly in food and nutrition contexts. It is retained for general use alongside joules, which are the SI unit for energy.

Key Concept: Definition Recall

c) Second

[Solution Description] The SI unit of time is second, which is symbolized by $$s$$. It is defined based on the caesium frequency.

Key Concept: Homogeneity Principle

c) $v = u + at$

[Solution Description]

An equation is dimensionally homogeneous if each term has the same dimensions. Check each option:

a) $x = v_0 t + \frac{1}{2} a t^2$: Dimensions are consistent as all terms have [L].

b) $x = v_0 + a t$: Dimensions are inconsistent ($[L]$ vs $[L/T]$).

c) $v = u + at$: Dimensions are consistent as both sides have $[L T^{-1}]$.

d) $F = ma + x$: Dimensions are inconsistent ($[M L T^{-2}]$$ vs $$[L]$).

Thus, options a) and c) are dimensionally homogeneous, but only one can be chosen in this context, so:

Key Concept: Zero Dimension Concept

a) Mass Density

[Solution Description] A quantity with zero dimension in length and time will have all exponents for these parameters as zero. From the options given, mass density has the dimensional formula:

$$[M L^{-3} T^0]$$

This formula indicates that it has dimensions in mass but not in length or time.

Key Concept: Multi-step Calculation

c) 19.139 g

[Solution Description]

Perform the addition

12.345 g+6.789 g+0.00456 g=19.13856 g

Determine the number of significant figures

12.345g has 5 significant figures.

6.789g has 4 significant figures.

0.00456g has 3 significant figures.

When adding numbers, we round to the least number of decimal places. The smallest number of decimal places here is 3 decimal places (from 0.00456).

Step 3: Round the result to 3 decimal places

19.13856g≈19.139g

Final Answer:

The final mass, rounded to the correct significant figures, is 19.139 g.

Key Concept: Trailing Zeros With Decimal

c) 5

[Solution Description] Trailing zeros that come after a decimal point are considered significant. In the number 12.300, the digits 1, 2, 3, 0, and 0 are all significant, making it a total of 5 significant figures.

Key Concept: Dimensional Check

c) $v = v_0 + at$

[Solution Description] To check for dimensional consistency, we ensure that each term in the equation has the same dimensions. Consider the equation $v = v_0 + at$. The dimensions are:

$[v] = L T^{-1}, \quad [v_0] = L T^{-1}, \quad [at] = [a][t] = (L T^{-2})(T) = L T^{-1}$

Since all terms have the same dimensions $L T^{-1}$, the equation is dimensionally consistent.

Key Concept: Dimensional Consistency

a) Yes

[Solution Description]

The left side of the equation is distance (s), with dimensions [L]. The right side consists of two terms:

– First term: ut, where u (velocity) has dimensions $[LT^{-1}]$ and t has dimensions [T]. Thus, ut has dimensions [L].

– Second term: $\frac{1}{2}at^2$, where a (acceleration) has dimensions $[LT^{-2}]$ and $t^2$ has dimensions $[T^2]$. Thus, $at^2$ also has dimensions [L].

Both sides have dimensions [L], so the equation is dimensionally consistent.

Key Concept: Dimensional Formula Derivation

c) $[L T^{-2}]$

[Solution Description] Acceleration is defined as the rate of change of velocity with respect to time. The dimensional formula for velocity is $[L T^{-1}]$. Therefore, the dimensional formula of acceleration, being velocity per unit time, would be:

$\text{Acceleration} = \frac{\text{Velocity}}{\text{Time}} = \frac{[L T^{-1}]}{[T]} = [L T^{-2}]$

Thus, the dimensional formula of acceleration is $[L T^{-2}]$.

Key Concept: Complex Unit Conversion, Arithmetic Operations

b) 2

[Solution Description]

To find the area, multiply the length and width to get:

$$\text{Area} = 3.752 \times 4.6 = 17.2492 \text{ m}^2$$

Convert square meters to square millimeters:

$$1 \text{ m}^2 = 1,000,000 \text{ mm}^2$$

So,

$$17.2492 \text{ m}^2 = 17.2492 \times 1,000,000 = 17,249,200 \text{ mm}^2$$

The limiting factor for significant figures is the measurement with the least number of significant figures, which is $$4.6$$ (2 significant figures). Thus, round the result to two significant figures:

$$17,249,200 \approx 17,000,000 \text{ mm}^2$$

Therefore, the final answer should have two significant figures.

Key Concept: Base Units Identification

c) second

[Solution Description] The SI base unit for time is defined as the second, which is one of the seven base units in the International System of Units (SI).

Key Concept: Luminous Efficacy Value

b) 683

[Solution Description] The candela is defined by the fixed numerical value of the luminous efficacy of monochromatic radiation of frequency $$540 \times 10^{12}$$ Hz, and this value is 683 when expressed in the unit lm W$$^{-1}$$.

Key Concept: SI Units

c) Kelvin

[Solution Description] The SI unit for measuring temperature is the Kelvin (K). It is one of the seven base units in the International System of Units (SI).

Key Concept: Basic Definition

c) Mole

[Solution Description]

The SI unit for the amount of substance is the mole, abbreviated as “mol”. This unit measures the number of specified elementary entities such as atoms, molecules, or ions.

Key Concept: Dimensional Analysis

a) $F = m \cdot a$

[Solution Description]

To check dimensional consistency, match dimensions on both sides of each equation.

Option a: $F = m \cdot a$, is Force = Mass $\times$ Acceleration. Dimensions are $[M L T^{-2}] = [M][L T^{-2}]$. Consistent.

Option b: $v = a \cdot t^2$, Velocity = Acceleration $\times$ Time squared. Dimensions are $[L T^{-1}] = [L T^{-2}][T^2] \Rightarrow [L] \neq [L T^{-1}]$. Inconsistent.

Option c: $v^2 = 2as$, Velocity squared = 2 $\times$ Acceleration $\times$ Distance. Dimensions are $[L^2 T^{-2}] = [L T^{-2}][L] \Rightarrow [L^2 T^{-2}] = [L^2 T^{-2}]$. Consistent.

Option d: $F = m \cdot v$, is Force = Mass $\times$ Velocity. Dimensions are $[M L T^{-2}] \neq [M][L T^{-1}]$. Inconsistent.

Key Concept: Base Units

c) Second

[Solution Description] The International System of Units (SI) defines a set of base units for seven fundamental physical quantities. Among these, the base unit for time is the second.

Key Concept: Derived Units

c) newton

[Solution Description] A derived unit is formed by combining two or more base units. From the given options, newton is a combination of kilogram, meter, and second: $$\text{newton} = \text{kg m s}^{-2}$$. Therefore, it is a derived unit.

Key Concept: Base Units Identification

d) Ampere

[Solution Description] The SI base unit for electric current is the ampere (A). This unit is one of the seven fundamental units in the International System of Units (SI).

Key Concept: Basic Concept

c) Mass

[Solution Description] In the International System of Units (SI), the kilogram is the base unit for measuring mass.

Key Concept: Dimensional Validity

b) $$[M L T^{-2}]$$

[Solution Description] The dimensional formula for force is $$[M L T^{-2}]$$. We need to check which option has this dimension.

$$\text{Option a: } [M L^2 T^{-3}] = \text{incorrect dimensions}$$

$$\text{Option b: } [M L T^{-2}] = \text{correct dimensions}$$

$$\text{Option c: } [M^2 L T^{-1}] = \text{incorrect dimensions}$$

$$\text{Option d: } [M L^{-1} T^{2}] = \text{incorrect dimensions}$$

Therefore, the correct answer is option b).

Key Concept: Basic Application

c) 4 s

[Solution Description] Using the formula $$T = 2\pi \sqrt{\frac{l}{g}}$$, we substitute $$l = 4 \, \text{m}$$ and $$g = 9.8 \, \text{m/s}^2$$. Therefore,

$$T = 2\pi \sqrt{\frac{4}{9.8}} \approx 2\pi \times 0.637 \approx 4 \, \text{s}.$$

Key Concept: Historical Systems

a) CGS

[Solution Description]

Historical systems of units before SI included CGS (Centimetre-Gram-Second), FPS (Foot-Pound-Second), and MKS (Metre-Kilogram-Second).

Key Concept: Dimensionless Constants

b) Dimensional analysis cannot determine dimensionless constants.

[Solution Description] Dimensional analysis cannot determine dimensionless constants. It is used for checking the dimensional validity of equations but does not provide values or relationships involving these constants.

Key Concept: Basic Definition

c) Kilogram

[Solution Description] The SI system includes seven base units which are metre, kilogram, second, ampere, kelvin, mole, and candela. From these options, kilogram is a base unit in the SI system.

Key Concept: Error Propagation

b) $14.40 \pm 0.08 \text{ cm}^2$

[Solution Description]

The area A of the rectangle is calculated using $A = l \times w$, where $l = 4.50$ cm and $w = 3.20$ cm.

$A = 4.50 \times 3.20 = 14.40 \text{ cm}^2$

Error in multiplication: $\left(\frac{\Delta A}{A}\right) = \sqrt{\left(\frac{\Delta l}{l}\right)^2 + \left(\frac{\Delta w}{w}\right)^2}$, where $\Delta l = 0.02$ cm and $\Delta w = 0.01$ cm.

$\frac{\Delta A}{14.40} = \sqrt{\left(\frac{0.02}{4.50}\right)^2 + \left(\frac{0.01}{3.20}\right)^2}$

$\frac{\Delta A}{14.40} = \sqrt{(0.00444)^2 + (0.003125)^2} \approx 0.00536$

$\Delta A \approx 0.00536 \times 14.40 \approx 0.077 \text{ cm}^2$

Thus, the area with uncertainty is $14.40 \pm 0.08 \text{ cm}^2$.

Key Concept: Dimensional Consistency

c) $ML^2T^{-2}$

[Solution Description] To check dimensional consistency, we compare each option’s dimensions against $$[M L^2 T^{-2}]$ for energy:

a) $ML^3T^{-2}$ $\rightarrow$ This indicates force times length, not energy.

b) $M^2L^2T^{-2}$ $\rightarrow$ Incorrect due to extra mass dimension.

c) $ML^2T^{-2}$ $\rightarrow$ Matches energy dimensions perfectly.

d) $MLT^{-1}$ $\rightarrow$ Represents momentum, not energy.

Key Concept: Dimensional Consistency

b) No, it is not consistent

[Solution Description]

The dimensional formula for work is derived by considering that it involves force $$F$$ and displacement s.

Given:

– Dimensional formula of force F: $[M L T^{-2}]$

– Dimensional formula of displacement s: [L]

Thus, according to the given equation $W = F \cdot s^2$, we have:

$\text{Dimensions of } W = [M L T^{-2}] \times [L]^2 = [M L^3 T^{-2}]$

The correct dimensional formula for work should be $[M L^2 T^{-2}]$. Therefore, the given equation is not dimensionally consistent.

Key Concept: Basic Definition

c) The ratio of arc length to radius

[Solution Description] A plane angle is defined as the ratio of the length of an arc to its radius.

Key Concept: Dimensional Equations

a) $V = W/q$

[Solution Description]

Electric potential (V) has the unit volt (V), which can also be expressed as joules per coulomb (J/C). Dimensional analysis involves ensuring units are consistent on both sides of an equation.

For option A, $V = W/q$, dimensions are:

$\left[\dfrac{\text{M L}^2 \text{T}^{-2}}{\text{A T}}\right]$ = $\left[\dfrac{\text{M L}^2}{\text{A T}^3}\right]$

This matches the expected dimensional formula for volts: $\left[\dfrac{\text{M L}^2}{\text{A T}^3}\right]$.

Checking other options:

– Option B: $V = q/W$

Dimensions: $\left[\dfrac{\text{A T}}{\text{M L}^2 \text{T}^{-2}}\right]$ = $\left[\dfrac{\text{A T}^3}{\text{M L}^2}\right]$

– Option C: $V = W^2/q$

Dimensions: $\left[\dfrac{\text{M}^2 \text{L}^4 \text{T}^{-4}}{\text{A T}}\right]$ = $\left[\dfrac{\text{M}^2 \text{L}^4}{\text{A T}^5}\right]$

– Option D: $V = q^2/W$

Dimensions: $\left[\dfrac{(\text{A T})^2}{\text{M L}^2 \text{T}^{-2}}\right]$ = $\left[\dfrac{\text{A}^2 \text{T}^4}{\text{M L}^2}\right]$

Only option A has consistent dimensions with electric potential.

Key Concept: Unit Conversion Basics

b) 1 C/s

[Solution Description] By definition, $$1$$ ampere is equal to a flow of $$1$$ coulomb of charge per second. Therefore, $$1 \text{ A} = 1 \text{ C/s}$$.

Key Concept: Dimensional Consistency

c) $v = u + at$

[Solution Description] To check dimensional consistency, verify that dimensions on both sides are equal.

For equation $s = ut + \frac{1}{2} a t^3$, where:

– s has the dimension [L]

– ut has the dimension $[L T^{-1}] \cdot [T] = [L]$

– $\frac{1}{2} a t^3$ has the dimension $[L T^{-2}] \cdot [T^3] = [L T]$

Thus, $[L] \neq [L] + [L T]$. The equation is not consistent.

Repeat for other options similarly, and identify the correct one as follows:

Equation $$v = u + at$$, where:

– v and u both have the dimension $[L T^{-1}]$

– at has the dimension $[L T^{-2}] \cdot [T] = [L T^{-1}]$

Hence, $[L T^{-1}] = [L T^{-1}] + [L T^{-1}]$ confirms dimensional consistency.

Key Concept: Dimensional Equation Construction

c) $$[M L^{-1} T^{-2}]$$

[Solution Description]

Pressure is defined as force per unit area. The dimensional formula for force is $$[M L T^{-2}]$$ and for area is $$[L^2]$$. Therefore, the dimensional formula for pressure is calculated as:

$$[P] = \frac{[F]}{[A]} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}]$$

Thus, the dimensional equation for pressure is $$[M L^{-1} T^{-2}]$$.