In the relation $x = \cos \left( {{\rm{\omega }}t + kx} \right),$ the dimensions of ω are

$\left[ {{{\rm{M}}^0}{{\rm{L}}^0}{{\rm{T}}^{ - 1}}} \right]$

$\left[ {{{\rm{M}}^0}{\rm{LT}}} \right]$

$\left[ {{{\rm{M}}^0}{{\rm{L}}^{ - 1}}{{\rm{T}}^0}} \right]$

$\left[ {{{\rm{M}}^0}{\rm{L}}{{\rm{T}}^{ - 1}}} \right]$

UNITS & MEASUREMENTS MCQs for NEET UG PHYSICS — Easy

1 Easy

1 of 128

Understanding +4 | -1

Statement-I: Random errors can be positive or negative.
Statement-II: Cause of random errors are uncertain.

View answer and explanation

Correct option: A

Statement-I is true and statement-II is true

Option A (Statement-I is true and statement-II is true) is correct because random errors can indeed be positive or negative, and their causes are uncertain.

Solution approach

Correct Answer + Core Reason: Statement-I is true and statement-II is true. Random errors are by definition both positive and negative, and their causes are unknown.

Step-by-Step Solution:

Step 1: Define random errors. Random errors are unpredictable fluctuations in experimental measurements.

Step 2: Analyze Statement-I: Random errors can be positive or negative. This is true because the fluctuations can cause the measured value to be either higher or lower than the true value.

Step 3: Analyze Statement-II: Cause of random errors are uncertain. This is also true because if the cause were known, the error could be corrected, and it would no longer be a random error but a systematic error.

Step 4: Conclusion: Both statements are true.

Why other options are wrong:
- Option B (Statement-I is true and statement-II is false): Statement II is also true.
- Option C (Statement-I is false and statement-II is true): Statement I is also true.
- Option D (Statement-I is false and statement-II is false): Both statements are true.

Key Concepts:
- Random errors
- Systematic errors

Memory Aids:
- Random errors are unpredictable and affect precision.

Quick Tips:
- Differentiate between random and systematic errors.

2 Easy

2 of 128

Understanding +4 | -1

2.34 is obtained by rounding of the number

View answer and explanation

Correct option: C

2.335

Option C (2.335) is correct because when rounding 2.335 to two decimal places, the 5 causes the preceding 3 to round up to 4, resulting in 2.34.

Solution approach

Correct Answer + Core Reason: 2.335. Rounding 2.335 to two decimal places results in 2.34.

Step-by-Step Solution:

Step 1: Identify the number to be rounded: 2.335

Step 2: Determine the desired number of decimal places: 2

Step 3: Look at the digit in the third decimal place (the digit after the second decimal place), which is 5.

Step 4: Since the digit is 5, round up the second decimal place.

Step 5: The second decimal place is 3, so it becomes 4.

Step 6: The rounded number is 2.34.

Why other options are wrong:
- Option A (2.346): Rounding 2.346 to two decimal places results in 2.35, not 2.34.
- Option B (2.355): Rounding 2.355 to two decimal places results in 2.36, not 2.34.
- Option D (2.334): Rounding 2.334 to two decimal places results in 2.33, not 2.34.

Key Concepts:
- Rounding rules
- Significant figures

Memory Aids:
- 5 or more, round up; less than 5, round down.

Quick Tips:
- Always look at the digit immediately following the desired number of decimal places to determine whether to round up or down.

3 Easy

3 of 128

Application +4 | -1

The radius of the proton is about

{10^{ - 15}}m

. The radius of the observable universe is

{10^{26}}

m. identify the distance which is half-way between these two extremes on a logarithmic scale.

View answer and explanation

Correct option: B

{10^6}m

Option B (

{10^6}m

) is correct because it represents the geometric mean of the two extreme distances on a logarithmic scale.

Solution approach

Correct Answer + Core Reason: ${10^6}m$ . The logarithmic midpoint is the geometric mean of the two distances.

Step-by-Step Solution:

Step 1: Define Logarithmic Scale: A logarithmic scale represents numbers using their logarithms.

Step 2: Find the Logarithms: The logarithm (base 10) of ${10^{-15}}$ is -15, and the logarithm of ${10^{26}}$ is 26.

Step 3: Calculate the Midpoint: The midpoint between -15 and 26 is $\frac{{-15 + 26}}{{2}} = \frac{{11}}{{2}} = 5.5$ .

Step 4: Convert Back to Linear Scale: The distance corresponding to the midpoint is ${10^{5.5}} \approx {10^6}$ m (since 5.5 is close to 6).

Why other options are wrong:
- Option A: ${10^{21}}m$ : This is not the logarithmic midpoint.
- Option C: ${10^{ - 6}}m$ : This is not the logarithmic midpoint.
- Option D: ${10^0}m$ : This is not the logarithmic midpoint.

Key Concepts:
- Logarithmic Scale
- Geometric Mean

Memory Aids:
- Logarithmic midpoint is the geometric mean.

Quick Tips:
- When dealing with extremes on a large scale, use logarithms to find the midpoint.

4 Easy

4 of 128

Understanding +4 | -1

Statement-I: In the measurement of

g

using simple pendulum generally we take central position (mean position) of the oscillation as reference position for measuring time of oscillation.
Statement-II: This reduces the human error in measurement of time.

View answer and explanation

Correct option: A

Statement-I is true and statement-II is true

Option A (Statement-I is true and statement-II is true) is correct because taking the mean position as a reference reduces human error in timing oscillations.

Solution approach

Correct Answer + Core Reason: Statement-I is true and statement-II is true. Using the mean position reduces human error in timing.

Step-by-Step Solution:

Step 1: Analyze Statement-I: In a simple pendulum experiment, the central (mean) position is indeed generally taken as the reference point for measuring the time of oscillation.

Step 2: Analyze Statement-II: Using the mean position as a reference point helps reduce human error because the pendulum's velocity is highest at this point, making it easier to accurately mark the time.

Step 3: Combine the analyses: Both statements are true, and Statement-II correctly explains why Statement-I is true.

Why other options are wrong:
- Option B: Statement-II is true, so this option is wrong.
- Option C: Statement-I is true, so this option is wrong.
- Option D: Both statements are true, so this option is wrong.

Key Concepts:
- Simple pendulum
- Measurement of time period
- Error reduction in experiments

Memory Aids:
- Mean position = highest velocity = easier timing.

Quick Tips:
- Always choose a reference point where the object's motion is most distinct to minimize timing errors.

5 Easy

5 of 128

Understanding +4 | -1

The dimensions of

M{L^2}{T^{ - 2}}

may correspond to

View answer and explanation

Correct option: D

Energy

Option D (Energy) is correct because the dimensions of energy (or work) are Mass x (Length)² / (Time)², which matches the given dimensions.

Solution approach

Correct Answer + Core Reason: Energy. The dimensions match the formula for energy.

Step-by-Step Solution:

Step 1: Recall the dimensions of various physical quantities.

Step 2: Analyze the given dimensions: $M{L^2}{T^{ - 2}}$ .

Step 3: Consider the dimensions of energy (E). Energy has the same dimensions as work, which is Force x Distance. Force = Mass x Acceleration, so Energy = Mass x Acceleration x Distance = [M] x [L][T]^-2 x [L] = [M][L]²[T]^-2.

Step 4: Consider the dimensions of pressure (P). Pressure = Force / Area = [M][L][T]^-2 / [L]² = [M][L]^-1[T]^-2.

Step 5: Consider the dimensions of density ( $\rho$ ). Density = Mass / Volume = [M] / [L]³ = [M][L]^-3.

Step 6: Consider the dimensions of area (A). Area = [L]².

Step 7: Comparing the given dimensions with the dimensions of the options, we find that the dimensions of energy match the given dimensions.

Why other options are wrong:
- Option A: Pressure has dimensions of $M{L^{ - 1}}{T^{ - 2}}$ .
- Option B: Density has dimensions of $M{L^{ - 3}}$ .
- Option C: Area has dimensions of ${L^2}$ .

Key Concepts:
- Energy
- Pressure
- Density
- Area
- Dimensional analysis

Memory Aids:
- Energy = Force x Distance
- Force = Mass x Acceleration

Quick Tips:
- Remember the formulas for the given quantities and derive their dimensions.

6 Easy

6 of 128

Analyzing +4 | -1

Statement-I: The quantity

\frac{e^2}{\epsilon_0 c h}

is dimensionless
Statement II:

\frac{1}{\sqrt{\mu_0 \epsilon_0 }}

has the dimensions of velocity and is numerically equal of velocity of light.

View answer and explanation

Correct option: A

Statement-I is true and statement-II is true

Statement-I is true and statement-II is true because

\frac{e^2}{\epsilon_0 c h}

is a dimensionless constant (fine-structure constant) and

\frac{1}{\sqrt{\mu_0 \epsilon_0 }}

represents the speed of light.

Solution approach

Correct Answer + Core Reason: Statement-I is true and statement-II is true. Both statements are factually correct regarding dimensionless quantities and the speed of light.

Step-by-Step Solution:

Step 1: Analyze Statement-I: The quantity $\frac{e^2}{\epsilon_0 c h}$ is the fine-structure constant, which is a dimensionless quantity.
- e is the elementary charge (dimension: AT)
- $\epsilon_0$ is the permittivity of free space (dimension: $M^{-1}L^{-3}T^4A^2$ )
- c is the speed of light (dimension: $LT^{-1}$ )
- h is Planck's constant (dimension: $ML^2T^{-1}$ )
Therefore, the dimensions of $\frac{e^2}{\epsilon_0 c h}$ are $\frac{(AT)^2}{(M^{-1}L^{-3}T^4A^2)(LT^{-1})(ML^2T^{-1})} = \frac{A^2T^2}{A^2T^2} = 1$ (dimensionless).

Step 2: Analyze Statement-II: $\frac{1}{\sqrt{\mu_0 \epsilon_0 }}$ has the dimensions of velocity and is numerically equal to the velocity of light.
- $\mu_0$ is the permeability of free space (dimension: $MLT^{-2}A^{-2}$ )
- $\epsilon_0$ is the permittivity of free space (dimension: $M^{-1}L^{-3}T^4A^2$ )
Therefore, the dimensions of $\frac{1}{\sqrt{\mu_0 \epsilon_0 }}$ are $\frac{1}{\sqrt{(MLT^{-2}A^{-2})(M^{-1}L^{-3}T^4A^2)}} = \frac{1}{\sqrt{L^{-2}T^2}} = \frac{1}{L^{-1}T} = LT^{-1}$ (velocity).
Numerically, $\frac{1}{\sqrt{\mu_0 \epsilon_0 }} = c$ (speed of light).

Step 3: Conclusion: Both statements are true.

Why other options are wrong:
- Option B: Statement-I is true and statement-II is false: Statement-II is also true.
- Option C: Statement-I is false and statement-II is true: Statement-I is also true.
- Option D: Statement-I is false and statement-II is false: Both statements are true.

Key Concepts:
- Dimensional analysis
- Fine-structure constant
- Permittivity and permeability of free space
- Speed of light

Memory Aids:
- Remember the formula for the speed of light in terms of $\mu_0$ and $\epsilon_0$ .

Quick Tips:
- Know the dimensions of fundamental constants.

7 Easy

7 of 128

Analyzing +4 | -1

Assertion The error in the measurement of radius of the sphere is

0.3 \%

. The permissible error in its surface area is

0.6 \%

.

Reason The permissible error is calculated by the formula

\frac{\Delta A}{A}=4 \frac{\Delta r}{r}

View answer and explanation

Correct option: C

If Assertion is true, but Reason is false.

Option C (If Assertion is true, but Reason is false) is correct because the assertion is true, but the correct formula is

\frac{\Delta A}{A}=2 \frac{\Delta r}{r}

.

Solution approach

Correct Answer + Core Reason: If Assertion is true, but Reason is false. The assertion is correct, but the formula provided in the reason is incorrect.

Step-by-Step Solution:

Step 1: Analyze the assertion.\nThe error in the radius is 0.3%, so $\frac{\Delta r}{r} = 0.003$ . The surface area is $A = 4\pi r^2$ . The percentage error in the surface area is $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r} = 2 \times 0.003 = 0.006$ , which is 0.6%. Thus, the assertion is true.

Step 2: Analyze the reason.\nThe correct formula for the percentage error in surface area is $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$ , not $\frac{\Delta A}{A} = 4 \frac{\Delta r}{r}$ . Therefore, the reason is false.

Why other options are wrong:
- Option A: The reason is not correct, so this option is incorrect.
- Option B: The reason is not correct, so this option is incorrect.
- Option D: The assertion is true, so this option is incorrect.

Key Concepts:
- Error propagation
- Surface area of a sphere

Memory Aids:
- Remember the power rule for error propagation: if $y = x^n$ , then $\frac{\Delta y}{y} = n \frac{\Delta x}{x}$ .

Quick Tips:
- Always check the formulas provided in the reason before evaluating the assertion.

8 Easy

8 of 128

Remembering +4 | -1

The dimensional formula for Reynold's number is

View answer and explanation

Correct option: A

\left[\mathrm{L}^0 \mathrm{M}^0 \mathrm{~T}^0\right]

Option A ([L⁰ M⁰ T⁰]) is correct because Reynold's number is a dimensionless quantity, representing the ratio of inertial forces to viscous forces.

Solution approach

Correct Answer + Core Reason: [L⁰ M⁰ T⁰]. Reynold's number is dimensionless.

Step-by-Step Solution:

Step 1: Define Reynold's number.\nReynold's number ( $Re$ ) is given by the formula: $Re = \frac{{\rho v L}}{{\mu}}$ , where:
\ $\rho$ is the density of the fluid,\nv is the flow speed,\nL is a characteristic linear dimension,
\ $\mu$ is the dynamic viscosity of the fluid.

Step 2: Determine the dimensions of each quantity.\nDensity (\ $\rho$ ): [M L⁻³]\nVelocity (v): [L T⁻¹]\nLength (L): [L]\nDynamic Viscosity (\ $\mu$ ): [M L⁻¹ T⁻¹]

Step 3: Substitute the dimensions into the formula.
$Re = \frac{{[M L^{-3}] [L T^{-1}] [L]}}{{[M L^{-1} T^{-1}]}}$

Step 4: Simplify the expression.
$Re = \frac{{[M L^{-1} T^{-1}]}}{{[M L^{-1} T^{-1}]}} = [L^0 M^0 T^0]$

Step 5: Final Answer
[L⁰ M⁰ T⁰]

Why other options are wrong:
- Option B: [LMT] This is incorrect because Reynold's number is dimensionless, meaning it has no physical units.
- Option C: [L⁻¹ MT] This is incorrect because Reynold's number is dimensionless, meaning it has no physical units.
- Option D: [LMT⁻¹] This is incorrect because Reynold's number is dimensionless, meaning it has no physical units.

Key Concepts:
- Dimensional Analysis
- Reynold's Number

Memory Aids:
- Remember that dimensionless quantities have no physical units.

Quick Tips:
- Always write down the formula for the quantity you are trying to find the dimensions of.
- Substitute the dimensions of each quantity into the formula.
- Simplify the expression to find the final dimensions.

9 Easy

9 of 128

Application +4 | -1

In the relation

x = \cos \left( {\mathrm{\omega }t + kx} \right),

the dimensions of ω are

View answer and explanation

Correct option: C

\left[ {{\mathrm{M}^0}{\mathrm{L}^0}{\mathrm{T}^{ - 1}}} \right]

Option C

\left[ {{\mathrm{M}^0}{\mathrm{L}^0}{\mathrm{T}^{ - 1}}} \right]

is correct because the argument of a trigonometric function must be dimensionless, so ωt must be dimensionless, making ω have dimensions of inverse time.

Solution approach

Correct Answer + Core Reason: $\left[ {{\mathrm{M}^0}{\mathrm{L}^0}{\mathrm{T}^{ - 1}}} \right]$ . The argument of a trigonometric function must be dimensionless.

Step-by-Step Solution:

Step 1: Analyze the Argument: In the expression $\cos(\omega t + kx)$ , the term $(\omega t + kx)$ must be dimensionless.

Step 2: Determine Dimensions of ωt: Since $\omega t$ is dimensionless, the dimensions of $\omega$ must be the inverse of the dimensions of $t$ .

Step 3: Dimensions of Time: The dimension of time, $t$ , is [T].

Step 4: Dimensions of ω: Therefore, the dimension of $\omega$ is $[T^{-1}]$ , which can be written as $\left[ {{\mathrm{M}^0}{\mathrm{L}^0}{\mathrm{T}^{ - 1}}} \right]$ .

Why other options are wrong:
- Option A: $\left[ {{\mathrm{M}^0}\mathrm{LT}} \right]$ : This is incorrect because it includes dimensions of length and time, which is not consistent with the dimensionless argument of the cosine function.
- Option B: $\left[ {{\mathrm{M}^0}{\mathrm{L}^{ - 1}}{\mathrm{T}^0}} \right]$ : This is incorrect because it includes dimensions of inverse length, which is not consistent with the dimensionless argument of the cosine function.
- Option D: $\left[ {{\mathrm{M}^0}\mathrm{L}{\mathrm{T}^{ - 1}}} \right]$ : This is incorrect because it includes dimensions of length and inverse time, which is not consistent with the dimensionless argument of the cosine function.

Key Concepts:
- Dimensional Analysis
- Trigonometric Functions

Memory Aids:
- Arguments of trigonometric functions are dimensionless.

Quick Tips:
- Remember that the arguments of trigonometric functions, exponential functions, and logarithms must be dimensionless.

10 Easy

10 of 128

Application +4 | -1

A physical quantity

Q

depends on other physical quantities

a, b

and

c

as

Q=\frac{a^4 b^3}{c^2}

If maximum percentage error in measurement of

a

,

b

and

c

are

3 \%, 4 \%

and

5 \%

respectively, then find maximum percentage error in measurement of

Q

.

View answer and explanation

Correct option: A

34

Option A (34) is correct because the maximum percentage error in Q is calculated by summing the percentage errors in a, b, and c, each multiplied by its respective power.

Solution approach

Correct Answer + Core Reason: 34. The maximum percentage error in Q is calculated by summing the percentage errors in a, b, and c, each multiplied by its respective power.

Step-by-Step Solution:

Step 1: Given expression: $Q=\frac{a^4 b^3}{c^2}$

Step 2: Percentage errors: $\Delta a = 3 \\%, \Delta b = 4 \\%, \Delta c = 5 \\%$

Step 3: Formula for percentage error in Q:
$\frac{\Delta Q}{Q} \times 100 = 4 \frac{\Delta a}{a} \times 100 + 3 \frac{\Delta b}{b} \times 100 + 2 \frac{\Delta c}{c} \times 100$

Step 4: Substitute the given values:
$\frac{\Delta Q}{Q} \times 100 = 4(3 \\%) + 3(4 \\%) + 2(5 \\%)$

Step 5: Calculate:
$\frac{\Delta Q}{Q} \times 100 = 12 \\% + 12 \\% + 10 \\% = 34 \\%$

Step 6: Final Answer: 34

Why other options are wrong:
- Option B (44): This is incorrect as it doesn't account for the correct propagation of errors.
- Option C (54): This is incorrect as it doesn't account for the correct propagation of errors.
- Option D (64): This is incorrect as it doesn't account for the correct propagation of errors.

Key Concepts:
- Error Propagation
- Percentage Error

Memory Aids:
- When quantities are multiplied or divided, their percentage errors are added.
- When a quantity is raised to a power, the percentage error is multiplied by the power.

Quick Tips:
- Always remember to multiply the percentage error by the power of the corresponding quantity.
- Be careful with the signs when adding errors.

11 Easy

11 of 128

Understanding +4 | -1

Select the pair whose dimensions are same

View answer and explanation

Correct option: A

Pressure and stress

Option A (Pressure and stress) is correct because both pressure and stress are defined as force per unit area, hence having the same dimensions.

Solution approach

Correct Answer + Core Reason: Pressure and stress have the same dimensions as both are force per unit area.

Step-by-Step Solution:

Step 1: Define Pressure: Pressure is defined as force per unit area, $P = \frac{F}{A}$ .

Step 2: Define Stress: Stress is also defined as force per unit area, $\sigma = \frac{F}{A}$ .

Step 3: Dimensional formula for Force: $[F] = [MLT^{-2}]$ .

Step 4: Dimensional formula for Area: $[A] = [L^2]$ .

Step 5: Dimensional formula for Pressure: $[P] = \frac{[F]}{[A]} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$ .

Step 6: Dimensional formula for Stress: $[\sigma] = \frac{[F]}{[A]} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$ .

Step 7: Conclusion: Since both pressure and stress have the same dimensional formula $[ML^{-1}T^{-2}]$ , they have the same dimensions.

Why other options are wrong:
- Option B: Stress and strain: Strain is dimensionless, while stress has dimensions of $[ML^{-1}T^{-2}]$ .
- Option C: Pressure and force: Pressure is force per unit area, so it has dimensions $[ML^{-1}T^{-2}]$ , while force has dimensions $[MLT^{-2}]$ .
- Option D: Power and force: Power is work per unit time, so it has dimensions $[ML^2T^{-3}]$ , while force has dimensions $[MLT^{-2}]$ .

Key Concepts:
- Pressure definition
- Stress definition
- Dimensional analysis

Memory Aids:
- Pressure = Force/Area, Stress = Force/Area

Quick Tips:
- Remember the definitions of pressure and stress. Both are force per unit area.

12 Easy

12 of 128

Application +4 | -1

If

\varepsilon_0

is the permittivity of free space and

E

is the electric field, then

\varepsilon_0 E^2

has the dimensions:

View answer and explanation

Correct option: B

\left[{M} {L}^{-1}{T}^{-2}\right]

Option B (

\left[{M} {L}^{-1}{T}^{-2}\right]

) is correct because

\epsilon_0 E^2

represents energy density, which has dimensions of pressure (Force/Area).

Solution approach

Correct Answer + Core Reason: $\left[{M} {L}^{-1}{T}^{-2}\right]$ . Dimensional analysis of the given expression.

Step-by-Step Solution:

Step 1: Recognize that $\varepsilon_0 E^2$ has the dimensions of energy density or pressure.

Step 2: Dimensions of electric field E are [MLT^-3A^-1].

Step 3: Dimensions of $\varepsilon_0$ are [M^-1L^-3T⁴A²].

Step 4: Dimensions of $\varepsilon_0 E^2$ are [M^-1L^-3T⁴A²] * [MLT^-3A^-1]² = [M^-1L^-3T⁴A²] * [M²L²T^-6A^-2] = [ML^-1T^-2]

Why other options are wrong:
- Option A: Incorrect dimensions.
- Option C: Incorrect dimensions.
- Option D: Incorrect dimensions.

Key Concepts:
- Dimensional Analysis
- Electric Field
- Permittivity
- Energy Density

Memory Aids:
- Remember the dimensions of electric field and permittivity.

Quick Tips:
- Recall the dimensions of electric field and permittivity.

13 Easy

13 of 128

Application +4 | -1

When 57.986 is rounded off to 4 significant figures then it becomes

View answer and explanation

Correct option: D

57.99

Option D (57.99) is correct because when rounding 57.986 to 4 significant figures, the 6 causes the 8 to round up to 9.

Solution approach

Correct Answer + Core Reason: 57.99. Rounding to 4 significant figures requires looking at the 5th digit.

Step-by-Step Solution:

Step 1: Identify the number: 57.986

Step 2: Determine the desired number of significant figures: 4

Step 3: Look at the fifth significant digit, which is 6.

Step 4: Since the digit is 6 (greater than or equal to 5), round up the fourth significant digit.

Step 5: The fourth significant digit is 8, so it becomes 9. Therefore, 57.98 becomes 57.99.

Why other options are wrong:
- Option A (58): This is rounded to two significant figures.
- Option B (57): This is rounded to two significant figures.
- Option C (57.9): This is rounded to three significant figures.

Key Concepts:
- Rounding rules
- Significant figures

Memory Aids:
- 5 or more, round up; less than 5, round down.

Quick Tips:
- Always look at the digit immediately following the desired number of significant figures to determine whether to round up or down.

14 Easy

14 of 128

Understanding +4 | -1

The dimensions of universal gas constant is

View answer and explanation

Correct option: A

\left[ {M{L^2}{T^{ - 2}}{\theta ^{ - 1}}} \right]

Option A (

\left[ {M{L^2}{T^{ - 2}}{\theta ^{ - 1}}} \right]

) is correct because the ideal gas law (PV = nRT) can be used to derive the dimensions of the universal gas constant R.

Solution approach

Correct Answer + Core Reason: $\left[ {M{L^2}{T^{ - 2}}{\theta ^{ - 1}}} \right]$ . Use the ideal gas law to find the dimensions.

Step-by-Step Solution:

Step 1: Recall the ideal gas law: $PV = nRT$ , where P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is temperature.

Step 2: Rearrange the formula to solve for R: $R = \frac{{PV}}{{nT}}$ .

Step 3: Determine the dimensions of pressure (P). Pressure = Force / Area = [M][L][T]^-2 / [L]² = [M][L]^-1[T]^-2.

Step 4: Determine the dimensions of volume (V). Volume = [L]³.

Step 5: Determine the dimensions of the number of moles (n). Moles are a dimensionless quantity, so [n] = 1.

Step 6: Determine the dimensions of temperature (T). Temperature is represented by [ $\theta$ ].

Step 7: Substitute the dimensions into the equation for R: [R] = ([M][L]^-1[T]^-2 [L]³) / (1 [ $\theta$ ]) = [M][L]²[T]^-2[ $\theta$ ]^-1.

Why other options are wrong:
- Option B: $\left[ {{M^2}L{T^{ - 2}}\theta } \right]$ is incorrect. These dimensions do not correspond to the universal gas constant.
- Option C: $\left[ {M{L^3}{T^{ - 1}}{\theta ^{ - 1}}} \right]$ is incorrect. These dimensions do not correspond to the universal gas constant.
- Option D: None of these is incorrect because option A is the correct answer.

Key Concepts:
- Universal gas constant
- Ideal gas law
- Pressure
- Volume
- Temperature
- Dimensional analysis

Memory Aids:
- PV = nRT (Ideal Gas Law)

Quick Tips:
- Remember the ideal gas law and the dimensions of pressure, volume, and temperature.

15 Easy

15 of 128

Application +4 | -1

The displacement of an oscillating particle is given by

y=A \sin (B x+C t+D)

. The dimensional formula for

(A B C D)

is

View answer and explanation

Correct option: B

\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]

Option B (

\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]

) is correct because the argument of the sine function must be dimensionless, implying B has dimensions of

L^{-1}

and C has dimensions of

T^{-1}

. A has dimensions of L. Therefore, the product ABCD has dimensions of

L \times L^{-1} \times T^{-1} \times 1 = T^{-1}

.

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]$ . The argument of the sine function must be dimensionless.

Step-by-Step Solution:

Step 1: Analyze the equation: $y = A \sin(Bx + Ct + D)$

Step 2: Determine the dimensions of y: [y] = L (since y is a displacement)

Step 3: Determine the dimensions of A: [A] = [y] = L

Step 4: Determine the dimensions of Bx: Since Bx is part of the argument of the sine function, it must be dimensionless. Therefore, $[Bx] = M^0 L^0 T^0 = 1$ . Since [x] = L, then $[B] = L^{-1}$

Step 5: Determine the dimensions of Ct: Since Ct is part of the argument of the sine function, it must be dimensionless. Therefore, $[Ct] = M^0 L^0 T^0 = 1$ . Since [t] = T, then $[C] = T^{-1}$

Step 6: Determine the dimensions of D: Since D is part of the argument of the sine function, it must be dimensionless. Therefore, $[D] = M^0 L^0 T^0 = 1$

Step 7: Calculate the dimensions of ABCD: $[ABCD] = [A][B][C][D] = L \times L^{-1} \times T^{-1} \times 1 = T^{-1} = M^0 L^0 T^{-1}$

Why other options are wrong:
- Option A: $\left[\mathrm{M}^{-1} \mathrm{~L}^1 \mathrm{~T}^{-1}\right]$ is incorrect.
- Option C: $\left[\mathrm{M}^0 \mathrm{~L}^{-1} \mathrm{~T}^{-1}\right]$ is incorrect.
- Option D: $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]$ is incorrect.

Key Concepts:
- Dimensional analysis
- Trigonometric functions and their arguments

Memory Aids:
- The argument of a trigonometric function must be dimensionless.

Quick Tips:
- Remember that the argument of trigonometric functions, exponential functions, and logarithms must be dimensionless.

16 Easy

16 of 128

Remembering +4 | -1

The unit of e.m.f. is

View answer and explanation

Correct option: D

Joule/coulomb

Option D (Joule/coulomb) is correct because e.m.f. is the work done per unit charge, hence Joule/Coulomb, which is Volt.

Solution approach

Correct Answer + Core Reason: Joule/coulomb is the unit of e.m.f. because e.m.f. is defined as work done per unit charge.

Step-by-Step Solution:

Step 1: Define e.m.f. (electromotive force).

Step 2: E.m.f. is the work done (W) in moving a unit positive charge (q) from one point to another. Therefore, $e.m.f. = \frac{W}{q}$ .

Step 3: The SI unit of work done (W) is Joule (J), and the SI unit of charge (q) is Coulomb (C).

Step 4: Therefore, the SI unit of e.m.f. is Joule/Coulomb (J/C), which is also known as Volt (V).

Why other options are wrong:
- Option A: Joule is the unit of energy or work.
- Option B: Joule-coulomb is not the unit of e.m.f.
- Option C: Volt-coulomb is not the unit of e.m.f.

Key Concepts:
- Electromotive force
- SI units

Memory Aids:
- Remember the formula: $e.m.f. = \frac{W}{q}$ .

Quick Tips:
- Always remember the definition of the physical quantity to determine its unit.

17 Easy

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Analyzing +4 | -1

Which of the following quantities have the same dimensions?
a. Surface tension

\quad

b. Spring constant

\quad

c. Tension

View answer and explanation

Correct option: A

a & b

Option A (a & b) is correct because surface tension and spring constant both have the dimensions of force per unit length.

Solution approach

Correct Answer + Core Reason: a & b. Surface tension and spring constant have the same dimensions, which is force per unit length.

Step-by-Step Solution:

Step 1: Determine the dimensions of each quantity.

Step 2: Surface tension ( $\gamma$ ) is defined as force per unit length: $\gamma = \frac{F}{L}$ . Therefore, its dimensions are $[MLT^{-2}]/[L] = [MT^{-2}]$ .

Step 3: Spring constant (k) is defined by Hooke's law: $F = kx$ , where x is the displacement. Therefore, $k = \frac{F}{x}$ . Its dimensions are $[MLT^{-2}]/[L] = [MT^{-2}]$ .

Step 4: Tension (T) is a force, so its dimensions are $[MLT^{-2}]$ .

Step 5: Compare the dimensions: Surface tension and spring constant have the same dimensions $[MT^{-2}]$ .

Why other options are wrong:
- Option B: a & c: Surface tension has dimensions $[MT^{-2}]$ , while tension has dimensions $[MLT^{-2}]$ . They are not the same.
- Option C: b & c: Spring constant has dimensions $[MT^{-2}]$ , while tension has dimensions $[MLT^{-2}]$ . They are not the same.
- Option D: All: This is incorrect because surface tension, spring constant, and tension do not all have the same dimensions.

Key Concepts:
- Dimensions of physical quantities
- Surface tension
- Spring constant
- Tension

Memory Aids:
- Remember the definition of surface tension and Hooke's law.

Quick Tips:
- Always write the formula for each quantity before determining its dimensions.

18 Easy

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Understanding +4 | -1

The dimensions of time constant are

View answer and explanation

Correct option: B

\left[ {{\mathrm{M}^0}{\mathrm{L}^0}\mathrm{T}} \right]

Option B ([M⁰L⁰T]) is correct because the time constant represents a characteristic time, hence its dimension is time.

Solution approach

Correct Answer + Core Reason: $\left[ {{\mathrm{M}^0}{\mathrm{L}^0}\mathrm{T}} \right]$ . The time constant has the dimension of time.

Step-by-Step Solution:

Step 1: Define time constant. The time constant is a measure of how long it takes for a system to respond to a change.

Step 2: Consider examples. In an RC circuit, the time constant is given by $\tau = RC$ , where R is resistance and C is capacitance. In an RL circuit, the time constant is given by $\tau = L/R$ , where L is inductance and R is resistance.

Step 3: Dimensional analysis of RC. Resistance has dimensions of $\left[ {\rm{ML^2T^{-3}I^{-2}}} \right]$ and capacitance has dimensions of $\left[ {\rm{M^{-1}L^{-2}T^{4}I^{2}}} \right]$ . Multiplying these gives $\left[ \mathrm{T} \right]$ .

Step 4: Dimensional analysis of L/R. Inductance has dimensions of $\left[ {\rm{ML^2T^{-2}I^{-2}}} \right]$ . Dividing by resistance gives $\left[ \mathrm{T} \right]$ .

Step 5: Conclusion. The dimensions of the time constant are $\left[ {{\mathrm{M}^0}{\mathrm{L}^0}\mathrm{T}} \right]$ .

Why other options are wrong:
- Option A: $\left[ {{\mathrm{M}^0}{\mathrm{L}^0}{\mathrm{T}^0}} \right]$ : This represents a dimensionless quantity, which is not the case for a time constant.
- Option C: $\left[ {\mathrm{MLT}} \right]$ : This represents dimensions of mass, length, and time, which is incorrect for a time constant.
- Option D: None of these: This is incorrect as option B is correct.

Key Concepts:
- Dimensional analysis
- Time constant in RC and RL circuits

Memory Aids:
- Remember that time constant always has units of time.

Quick Tips:
- When in doubt, consider the formula for time constant in a simple circuit (RC or RL) and perform dimensional analysis.

19 Easy

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Understanding +4 | -1

No. of significant figures in 0.0241 is

View answer and explanation

Correct option: A

3

Option A (3) is correct because leading zeros are not significant, and the digits 2, 4, and 1 are significant.

Solution approach

Correct Answer + Core Reason: 3. Leading zeros are not significant.

Step-by-Step Solution:

Step 1: Identify the number: 0.0241

Step 2: Apply the rules for significant figures: Leading zeros are not significant. Non-zero digits are always significant.

Step 3: Count the significant figures: The digits 2, 4, and 1 are significant. The leading zeros are not.

Step 4: Final Answer: 3

Why other options are wrong:
- Option B: Incorrect because it counts an extra digit.
- Option C: Incorrect because it counts even more digits.
- Option D: Incorrect because it doesn't count all the significant digits.

Key Concepts:
- Significant Figures

Memory Aids:
- Leading zeros don't count.

Quick Tips:
- Convert to scientific notation to easily identify significant figures.

20 Easy

20 of 128

Application +4 | -1

The pitch of the screw gauge is 0.5 mm. Its circular scale contains 50 divisions. The least count of the screw gauge is

View answer and explanation

Correct option: B

0.01 mm

Option B (0.01 mm) is correct because the least count of a screw gauge is calculated by dividing the pitch by the number of divisions on the circular scale.

Solution approach

Correct Answer + Core Reason: 0.01 mm. The least count is calculated by dividing the pitch by the number of divisions on the circular scale.

Step-by-Step Solution:

Step 1: Identify the given information.\nPitch of the screw gauge = 0.5 mm\nNumber of divisions on the circular scale = 50

Step 2: Calculate the least count.\nThe least count (LC) is given by:
$LC = \frac{{\text{{Pitch}}}}{{\text{{Number of divisions on circular scale}}}}$

Step 3: Substitute the values.
$LC = \frac{{0.5 \mathrm{~mm}}}{{50}}$

Step 4: Calculate the least count.
$LC = 0.01 \mathrm{~mm}$

Why other options are wrong:
- Option A: 0.001 mm is incorrect because it's a miscalculation of the least count.
- Option C: 0.02 mm is incorrect because it's a miscalculation of the least count.
- Option D: 0.025 mm is incorrect because it's a miscalculation of the least count.

Key Concepts:
- Least count of a screw gauge

Memory Aids:
- Least count = (Pitch) / (Total number of divisions on the circular scale)

Quick Tips:
- Ensure the units are consistent before calculating the least count.

21 Easy

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Application +4 | -1

Using mass

(M)

, length

(L)

, time

(T)

and current

(A)

as fundamental quantities, the dimensions of permeability are

View answer and explanation

Correct option: C

\left[\mathrm{MLT}^{-2} \mathrm{A}^{-2}\right]

Option C

\left[\mathrm{MLT}^{-2} \mathrm{A}^{-2}\right]

is correct because the permeability can be derived from the force between two parallel current-carrying wires.

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{MLT}^{-2} \mathrm{A}^{-2}\right]$ . Permeability is related to the force between current-carrying wires.

Step-by-Step Solution:

Step 1: Consider the force between two parallel current-carrying wires: $F = \frac{\mu_0 I_1 I_2 l}{2 \pi r}$ , where $F$ is the force, $\mu_0$ is the permeability, $I_1$ and $I_2$ are the currents, $l$ is the length of the wire, and $r$ is the distance between the wires.

Step 2: Rearrange the formula to solve for $\mu_0$ : $\mu_0 = \frac{2 \pi r F}{I_1 I_2 l}$ .

Step 3: Find the dimensions of $\mu_0$ : $[\mu_0] = \frac{[r][F]}{[I_1][I_2][l]} = \frac{[L][MLT^{-2}]}{[A][A][L]} = [MLT^{-2}A^{-2}]$ .

Why other options are wrong:
- Option A: $\left[\mathrm{M}^{-1} \mathrm{LT}^{-2} \mathrm{A}\right]$ : This is incorrect as it doesn't match the derived dimensions of permeability.
- Option B: $\left[\mathrm{ML}^2 \mathrm{T}^{-2} \mathrm{A}^{-1}\right]$ : This is incorrect as it doesn't match the derived dimensions of permeability.
- Option D: $\left[\mathrm{MLT}^{-1} \mathrm{A}^{-1}\right]$ : This is incorrect as it doesn't match the derived dimensions of permeability.

Key Concepts:
- Force between two parallel current-carrying wires.
- Dimensional analysis.

Memory Aids:
- Remember the formula for the force between two parallel current-carrying wires to derive the dimensions of permeability.

Quick Tips:
- Use a relevant formula that includes the quantity you want to find the dimensions of.

22 Easy

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Application +4 | -1

Dimension of mass, in the dimensional formula of magnetic flux is

View answer and explanation

Correct option: A

1

Option A (1) is correct because the dimensional formula for magnetic flux is

ML^2T^{-2}A^{-1}

, which has a mass dimension of 1.

Solution approach

Correct Answer + Core Reason: 1. The dimension of mass in magnetic flux is 1.

Step-by-Step Solution:

Step 1: Recall the formula for magnetic flux: $\phi = B \cdot A = B A \cos \theta$ , where B is magnetic field and A is area.

Step 2: Determine the dimensions of the magnetic field (B): Force on a moving charge is $F = qvB$ , so $B = \frac{F}{qv}$ . Therefore, $[B] = \frac{[F]}{[q][v]} = \frac{M L T^{-2}}{I T L T^{-1}} = M T^{-2} I^{-1}$

Step 3: Determine the dimensions of area (A): $[A] = L^2$

Step 4: Calculate the dimensions of magnetic flux: $[\phi] = [B][A] = (M T^{-2} I^{-1}) (L^2) = M L^2 T^{-2} I^{-1}$

Step 5: Express current (I) in terms of charge (Q) and time (T): $I = \frac{Q}{T}$ , so $Q = IT$ . Therefore, we can also write the dimensions as $M L^2 T^{-2} A^{-1}$ where A represents Ampere.

Step 6: Identify the dimension of mass: The dimension of mass (M) in the formula $M L^2 T^{-2} A^{-1}$ is 1.

Why other options are wrong:
- Option B: 0 is incorrect because the dimension of mass is 1.
- Option C: 0 is incorrect because the dimension of mass is 1.
- Option D: 0 is incorrect because the dimension of mass is 1.

Key Concepts:
- Magnetic flux
- Magnetic field
- Dimensions of physical quantities

Memory Aids:
- Remember the formula for magnetic flux: $\phi = B \cdot A$
- Remember the force on a moving charge: $F = qvB$

Quick Tips:
- Break down the physical quantity into its fundamental components to determine its dimensions.

23 Easy

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Remembering +4 | -1

The dimensions of force are

View answer and explanation

Correct option: C

\left[\mathrm{MLT}^{-2}\right]

Option C ([MLT^-2]) is correct because force is mass times acceleration, and acceleration is length divided by time squared.

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{MLT}^{-2}\right]$ . Force is mass times acceleration.

Step-by-Step Solution:

Step 1: Define force: Force (F) is defined as mass (m) times acceleration (a), according to Newton's second law: $F = ma$ .

Step 2: Determine the dimensions of mass and acceleration:
- Mass (m) has dimensions [M].
- Acceleration (a) is the rate of change of velocity, so it has dimensions of velocity (v) divided by time (t). Velocity has dimensions [LT^-1], and time has dimensions [T]. Therefore, acceleration has dimensions [LT^-2].

Step 3: Combine the dimensions:
- Force (F) = mass (m) * acceleration (a) = [M] * [LT^-2] = [MLT^-2].

Step 4: Final Answer: The dimensions of force are [MLT^-2].

Why other options are wrong:
- Option A: $\left[M L^2 T^{-1}\right]$ : This is incorrect. These dimensions do not correspond to force. It might be related to angular momentum or a similar quantity.
- Option B: $\left[\mathrm{M}^2 \mathrm{~L}^3 \mathrm{~T}^{-2}\right]$ : This is incorrect. These dimensions do not correspond to force. It is a combination of mass, length, and time but not in the correct proportions for force.
- Option D: None of these: This is incorrect because option C provides the correct dimensions of force.

Key Concepts:
- Newton's Second Law of Motion
- Dimensional Analysis

Memory Aids:
- F = ma (Force equals mass times acceleration)

Quick Tips:
- Remember the base dimensions: Mass [M], Length [L], Time [T].
- Derive dimensions from fundamental equations.

24 Easy

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Application +4 | -1

The dimensions of

\frac{R}{L}

are
[here, R=electric resistance, L=self inductance]

View answer and explanation

Correct option: B

\left[ {{\mathrm{T}^{ - 1}}} \right]

Option B (

\left[ {{\mathrm{T}^{ - 1}}} \right]

) is correct because R/L is the inverse of the time constant in an RL circuit, which has dimensions of inverse time.

Solution approach

Correct Answer + Core Reason: $\left[ {{\mathrm{T}^{ - 1}}} \right]$ . R/L is the inverse of the time constant in an RL circuit.

Step-by-Step Solution:

Step 1: Recall that $\frac{L}{R}$ represents the time constant in an RL circuit, and thus $\frac{R}{L}$ is the inverse of the time constant.

Step 2: The time constant has dimensions of time [T], so its inverse has dimensions of $[T^{-1}]$ .

Step 3: Therefore, the dimensions of $\frac{R}{L}$ are $[T^{-1}]$ .

Why other options are wrong:
- Option A: $\left[ {{\mathrm{T}^{ - 2}}} \right]$ is incorrect; it doesn't represent the dimensions of R/L.
- Option C: $\left[ {\mathrm{M}{\mathrm{L}^{ - 1}}} \right]$ is incorrect; it doesn't represent the dimensions of R/L.
- Option D: $\left[ T \right]$ is incorrect; it represents time, not the inverse of time.

Key Concepts:
- RL circuit
- Time constant
- Dimensional analysis

Memory Aids:
- Remember that $\frac{R}{L}$ is the inverse of $\frac{L}{R}$ .

Quick Tips:
- Know the relationships between inductance, resistance, and time constant.

25 Easy

25 of 128

Application +4 | -1

If L and R are respectively the inductance and resistance, then the dimensions of

\frac{L}{R}

will be

View answer and explanation

Correct option: C

M{\;^0}{L^0}T

Option C (

M{\;^0}{L^0}T

) is correct because the dimensions of L/R represent time, as seen in the context of LR circuits.

Solution approach

Correct Answer + Core Reason: $M{\;^0}{L^0}T$ . The ratio L/R has the dimensions of time.

Step-by-Step Solution:

Step 1: Recall that in an LR circuit, the time constant $\tau = \frac{L}{R}$ .

Step 2: The time constant has dimensions of time [T].

Step 3: Therefore, the dimensions of $\frac{L}{R}$ are [T] = $M^0L^0T^1$ .

Why other options are wrong:
- Option A: Incorrect dimensions. It represents inverse time.
- Option B: Incorrect dimensions. It represents length.
- Option D: Incorrect statement. The dimensions can be represented in terms of M, L, and T.

Key Concepts:
- Inductance
- Resistance
- Time constant in LR circuits

Memory Aids:
- Remember the time constant formula: $\tau = \frac{L}{R}$

Quick Tips:
- Relate the ratio to a known physical quantity with known dimensions.

26 Easy

26 of 128

Understanding +4 | -1

Dimensions of the following three quantities are the same

View answer and explanation

Correct option: D

Pressure, stress, coefficient of elasticity

Option D (Pressure, stress, coefficient of elasticity) is correct because all three quantities are defined as force per unit area, hence they share the same dimensions [ML⁻¹T⁻²].

Solution approach

Correct Answer + Core Reason: Pressure, stress, coefficient of elasticity. All are force per unit area.

Step-by-Step Solution:

Step 1: Determine the dimensions of Pressure (P): Pressure is defined as force per unit area: $P = \frac{F}{A}$ . The dimensions are [MLT⁻²]/[L²] = [ML⁻¹T⁻²].

Step 2: Determine the dimensions of Stress ( $\sigma$ ): Stress is defined as force per unit area: $\sigma = \frac{F}{A}$ . The dimensions are [MLT⁻²]/[L²] = [ML⁻¹T⁻²].

Step 3: Determine the dimensions of Coefficient of Elasticity (E): Coefficient of elasticity (Young's modulus) is defined as stress/strain. Since strain is dimensionless, the dimensions of E are the same as stress, which is [ML⁻¹T⁻²].

Step 4: Compare the dimensions: Pressure, stress, and the coefficient of elasticity all have dimensions of [ML⁻¹T⁻²].

Step 5: Final Answer: Pressure, stress, coefficient of elasticity

Why other options are wrong:
- Option A: Work, energy, force: Work and energy have dimensions of [ML²T⁻²], while force has dimensions of [MLT⁻²].
- Option B: Velocity, momentum, impulse: Velocity has dimensions of [LT⁻¹], while momentum and impulse have dimensions of [MLT⁻¹].
- Option C: Potential energy, kinetic energy, momentum: Potential and kinetic energy have dimensions of [ML²T⁻²], while momentum has dimensions of [MLT⁻¹].

Key Concepts:
- Pressure
- Stress
- Coefficient of Elasticity
- Dimensional Analysis

Memory Aids:
- Remember that pressure, stress, and Young's modulus are all related to force per unit area.

Quick Tips:
- Remember the definitions of pressure, stress, and the coefficient of elasticity.

27 Easy

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Application +4 | -1

Time (T), velocity (C) and angular momentum (h) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be :

View answer and explanation

Correct option: A

[\mathrm{M}]=\left[\mathrm{T}^{-1} \mathrm{C}^{-2} \mathrm{~h}\right]

Option A (

[\mathrm{M}]=\left[\mathrm{T}^{-1} \mathrm{C}^{-2} \mathrm{~h}\right]

) is correct because by expressing mass in terms of time, velocity, and angular momentum, we can derive the correct dimensional formula using the definitions of these quantities.

Solution approach

Correct Answer + Core Reason: $[\mathrm{M}]=\left[\mathrm{T}^{-1} \mathrm{C}^{-2} \mathrm{~h}\right]$ . Mass can be expressed in terms of time, velocity, and angular momentum.

Step-by-Step Solution:

Step 1: Define the fundamental quantities: T (time), C (velocity), h (angular momentum).

Step 2: Write the dimensions of velocity: $[C] = L T^{-1}$ , so $L = CT$

Step 3: Write the dimensions of angular momentum: $[h] = M L^2 T^{-1}$ .

Step 4: Substitute $L = CT$ into the dimensions of angular momentum: $[h] = M (CT)^2 T^{-1} = M C^2 T$

Step 5: Solve for mass: $M = \frac{h}{C^2 T} = h C^{-2} T^{-1}$

Step 6: Write the dimensional formula for mass: $[M] = [T^{-1} C^{-2} h]$

Why other options are wrong:
- Option B: $[\mathrm{M}]=\left[\mathrm{T}^{-1} \mathrm{C}^2 \mathrm{~h}\right]$ is incorrect.
- Option C: $[\mathrm{M}]=\left[\mathrm{T}^{-1} \mathrm{C}^{-2} \mathrm{~h}^{-1}\right]$ is incorrect.
- Option D: $[\mathrm{M}]=\left[\mathrm{T} \mathrm{C}^{-2} \mathrm{~h}\right]$ is incorrect.

Key Concepts:
- Dimensional analysis
- Angular momentum
- Fundamental quantities

Memory Aids:
- Remember the dimensions of angular momentum: $ML^2T^{-1}$

Quick Tips:
- Express all quantities in terms of the new fundamental quantities.

28 Easy

28 of 128

Application +4 | -1

The relative error in the determination of the surface area of a sphere is

\alpha

. Then the relative error in the determination of it’s volume is

View answer and explanation

Correct option: A

\frac{3}{2}\alpha

Option A (

\frac{3}{2}\alpha

) is correct because the volume is proportional to the radius cubed, and the surface area is proportional to the radius squared, so the relative error in volume is 3/2 times the relative error in surface area.

Solution approach

Correct Answer + Core Reason: $\frac{3}{2}\alpha$ . The relative error in volume is related to the relative error in the surface area.

Step-by-Step Solution:

Step 1: Let the radius of the sphere be $r$ .

Step 2: The surface area of the sphere is $A = 4\pi r^2$ .

Step 3: The volume of the sphere is $V = \frac{4}{3}\pi r^3$ .

Step 4: The relative error in the surface area is given as $\alpha = \frac{\Delta A}{A} = \frac{2 \Delta r}{r}$ .

Step 5: The relative error in the volume is $\frac{\Delta V}{V} = \frac{3 \Delta r}{r}$ .

Step 6: From Step 4, we have $\frac{\Delta r}{r} = \frac{\alpha}{2}$ .

Step 7: Substitute this into Step 5: $\frac{\Delta V}{V} = 3 \times \frac{\alpha}{2} = \frac{3}{2}\alpha$ .

Why other options are wrong:
- **Option B ( $\alpha$ ): This is incorrect as the volume depends on r cubed, while the area depends on r squared.
- Option C ( $\frac{5}{2}\alpha$ ): This is incorrect based on the relationship between surface area and volume.
- Option D ( $\frac{2}{3}\alpha$ ): This is the inverse relationship and incorrect.

Key Concepts:
- Relative error
- Surface area of a sphere
- Volume of a sphere

Memory Aids:
- Remember the formulas for surface area and volume of a sphere.

Quick Tips:**
- Relate the errors in different quantities using the formulas.

29 Easy

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Understanding +4 | -1

The physical quantity angular momentum has the same dimensions as that of

View answer and explanation

Correct option: D

Planck’s constant

Option D (Planck’s constant) is correct because both angular momentum and Planck's constant have the dimensions of [ML²T⁻¹].

Solution approach

Correct Answer + Core Reason: Planck’s constant. Both have dimensions of [ML²T⁻¹].

Step-by-Step Solution:

Step 1: Determine the dimensions of Angular Momentum (L): Angular momentum is given by $L = I\omega$ , where I is the moment of inertia and $\omega$ is the angular velocity. The dimensions of I are [ML²] and the dimensions of $\omega$ are [T⁻¹]. Therefore, the dimensions of L are [ML²T⁻¹].

Step 2: Determine the dimensions of Planck's constant (h): Planck's constant appears in the equation $E = hf$ , where E is energy and f is frequency. The dimensions of energy are [ML²T⁻²] and the dimensions of frequency are [T⁻¹]. Therefore, the dimensions of h are $\frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML²T⁻¹]$ .

Step 3: Compare the dimensions: Both angular momentum and Planck's constant have dimensions of [ML²T⁻¹].

Step 4: Final Answer: Planck’s constant

Why other options are wrong:
- Option A: Work: Work has dimensions of [ML²T⁻²], which are different from angular momentum.
- Option B: Force: Force has dimensions of [MLT⁻²], which are different from angular momentum.
- Option C: Momentum: Momentum has dimensions of [MLT⁻¹], which are different from angular momentum.

Key Concepts:
- Angular Momentum
- Planck's Constant
- Dimensional Analysis

Memory Aids:
- Remember the formula for angular momentum and the relationship between energy and frequency involving Planck's constant.

Quick Tips:
- Remember the dimensions of energy, frequency, moment of inertia, and angular velocity.

30 Easy

30 of 128

Understanding +4 | -1

Statement-I: Pressure can be subtracted from pressure gradient
Statement-II: Only like quantities can be added or subtracted from each other

View answer and explanation

Correct option: C

Statement-I is false and statement-II is true

Option C (Statement-I is false and statement-II is true) is correct because pressure and pressure gradient have different dimensions, and only quantities with the same dimensions can be added or subtracted.

Solution approach

Correct Answer + Core Reason: Statement-I is false and statement-II is true. Only like quantities can be added or subtracted.

Step-by-Step Solution:

Step 1: Analyze Statement-I: Pressure can be subtracted from pressure gradient. Pressure has dimensions of [ML⁻¹T⁻²]. Pressure gradient is the change in pressure with respect to distance, so it has dimensions of [ML⁻²T⁻²]. Since the dimensions are different, pressure cannot be subtracted from pressure gradient. Therefore, Statement-I is false.

Step 2: Analyze Statement-II: Only like quantities can be added or subtracted from each other. This is a fundamental principle of dimensional analysis. Quantities must have the same dimensions to be added or subtracted. Therefore, Statement-II is true.

Step 3: Conclusion: Statement-I is false, and Statement-II is true.

Why other options are wrong:
- Option A: Statement-I is true and statement-II is true: Statement-I is false.
- Option B: Statement-I is true and statement-II is false: Statement-I is false and Statement-II is true.
- Option D: Statement-I is false and statement-II is false: Statement-II is true.

Key Concepts:
- Dimensional Analysis
- Pressure
- Pressure Gradient

Memory Aids:
- Only quantities with the same dimensions can be added or subtracted.

Quick Tips:
- Always check the dimensions before adding or subtracting physical quantities.

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