UNITS & MEASUREMENTS MCQs for NEET UG PHYSICS — Easy, Page 3

61 Easy

61 of 128

Application +4 | -1

Three pieces of silver have masses 2.3 kg, 41.15 g and 30.19 g. The total mass of correct significant figures is

View answer and explanation

Correct option: D

2.4 kg

Option D (2.4 kg) is correct because when adding numbers with different decimal places, the result should be rounded to the least number of decimal places.

Solution approach

Correct Answer + Core Reason: 2.4 kg. The sum is rounded to the least number of decimal places.

Step-by-Step Solution:

Step 1: Convert all masses to the same unit (kg): 2.3 kg, 41.15 g = 0.04115 kg, 30.19 g = 0.03019 kg.

Step 2: Add the masses: 2.3 kg + 0.04115 kg + 0.03019 kg = 2.37134 kg.

Step 3: Determine the least number of decimal places: 2.3 kg has only one decimal place.

Step 4: Round the sum to one decimal place: 2.37134 kg rounds to 2.4 kg.

Why other options are wrong:
- Option A: 2.37032 kg: This is the direct sum without rounding to the correct number of decimal places.
- Option B: 2.370 kg: This is rounded to three decimal places, which is not correct based on the least number of decimal places in the given values.
- Option C: 2.37 kg: This is rounded to two decimal places, which is not correct based on the least number of decimal places in the given values.

Key Concepts:
- Significant figures
- Rounding rules
- Unit conversion

Memory Aids:
- When adding or subtracting, the result should have the same number of decimal places as the quantity with the fewest decimal places.

Quick Tips:
- Always convert all values to the same unit before performing calculations.

62 Easy

62 of 128

Application +4 | -1

{N}

division on main scale of a vernier callipers coincide with

(N+1)

division of the vernier scale if each division on main scale is of "

a

" units, least count of instrument is

View answer and explanation

Correct option: B

\frac{a}{N+1}

Option B (

\frac{a}{N+1}

) is correct because the least count is calculated by finding the difference between one main scale division and one vernier scale division.

Solution approach

Correct Answer + Core Reason: $\frac{a}{N+1}$ . The least count is the difference between one main scale division and one vernier scale division.

Step-by-Step Solution:

Step 1: Define the given variables.
$N$ divisions on the main scale coincide with $(N+1)$ divisions of the vernier scale.\nEach division on the main scale is of length $a$ .

Step 2: Find the length of one vernier scale division.\nLength of $(N+1)$ vernier scale divisions = Length of $N$ main scale divisions = $Na$ \nLength of 1 vernier scale division = $\frac{Na}{N+1}$

Step 3: Calculate the least count.\nLeast count = 1 main scale division - 1 vernier scale division
$LC = a - \frac{Na}{N+1} = \frac{a(N+1) - Na}{N+1} = \frac{aN + a - Na}{N+1} = \frac{a}{N+1}$

Why other options are wrong:
- Option A: $\frac{N+1}{a}$ is incorrect because it's the inverse of the correct expression and doesn't represent the difference between main scale and vernier scale divisions.
- Option C: $\frac{N-1}{a}$ is incorrect because it's not derived from the given relationship between the main scale and vernier scale divisions.
- Option D: $\frac{a}{N-1}$ is incorrect because it's not derived from the given relationship between the main scale and vernier scale divisions.

Key Concepts:
- Vernier callipers
- Least count

Memory Aids:
- Least Count = 1 MSD - 1 VSD

Quick Tips:
- Understand the relationship between the main scale and vernier scale divisions to correctly calculate the least count.

63 Easy

63 of 128

Understanding +4 | -1

Assertion: In the measurement of physical quantities direct and indirect methods are used.

Reason : The accuracy and precision of measuring instruments along with errors in measurements should be taken into account, while expressing the result.

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

Option A is correct because both the assertion and reason are correct, and the reason correctly explains why direct and indirect methods are used in measurement.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion. The accuracy and precision of measuring instruments and errors in measurements are important considerations when choosing between direct and indirect measurement methods.

Step-by-Step Solution:

Step 1: Analyze the assertion: Direct and indirect methods are indeed used in the measurement of physical quantities.

Step 2: Analyze the reason: The accuracy and precision of measuring instruments, along with errors in measurements, are crucial factors to consider when expressing the result of a measurement.

Step 3: Determine if the reason explains the assertion: The reason explains why we use both direct and indirect methods. We choose the method that provides the best accuracy and precision, considering the limitations of the instruments and potential errors.

Why other options are wrong:
- Option B: The reason *is* a correct explanation of the assertion.
- Option C: The reason is correct, so this option is incorrect.
- Option D: Both the assertion and reason are correct, so this option is incorrect.

Key Concepts:
- Measurement methods
- Accuracy
- Precision
- Errors in measurement

Memory Aids:
- Consider accuracy and precision when choosing a measurement method.

Quick Tips:
- Understand the difference between accuracy and precision.

64 Easy

64 of 128

Understanding +4 | -1

Which of the following statement(s) is/are correct?
I. A screw gauge having a smaller value of pitch has greater accuracy.
II. The least count of screw gauge is directly proportional to the number of divisions on circular scale.

View answer and explanation

Correct option: A

Only I

Option A is correct because a smaller pitch in a screw gauge means the screw advances less per rotation, leading to higher accuracy.

Solution approach

Correct Answer + Core Reason: Only I. A smaller pitch leads to greater accuracy in a screw gauge.

Step-by-Step Solution:

Step 1: Define pitch: Pitch is the distance moved by the screw per rotation.

Step 2: Define least count: Least count (LC) is the smallest measurement that can be accurately measured by the instrument.

Step 3: The formula for least count is: $\mathrm{LC} = \frac{{\text{{Pitch}}}}{{\text{{Number of divisions on circular scale}}}}$

Step 4: Statement I: A screw gauge having a smaller value of pitch has greater accuracy. This is correct because a smaller pitch results in a smaller least count, hence greater accuracy.

Step 5: Statement II: The least count of screw gauge is directly proportional to the number of divisions on circular scale. This is incorrect. The least count is *inversely* proportional to the number of divisions on the circular scale.

Why other options are wrong:
- Option B: Statement II is incorrect.
- Option C: Statement II is incorrect.
- Option D: Statement I is correct.

Key Concepts:
- Screw gauge
- Pitch
- Least count
- Accuracy

Memory Aids:
- Smaller pitch = smaller LC = greater accuracy

Quick Tips:
- Understand the relationship between pitch, least count, and accuracy.

65 Easy

65 of 128

Remembering +4 | -1

SI unit of electric intensity is

View answer and explanation

Correct option: D

Newton/ coulomb

Option D (Newton/ coulomb) is correct because electric intensity is defined as force per unit charge, hence Newton/Coulomb.

Solution approach

Correct Answer + Core Reason: Newton/ coulomb is the correct SI unit of electric intensity because electric intensity is defined as force per unit charge.

Step-by-Step Solution:

Step 1: Define electric intensity (E).

Step 2: Electric intensity is the force (F) experienced by a unit positive charge (q). Therefore, $E = \frac{F}{q}$ .

Step 3: The SI unit of force (F) is Newton (N), and the SI unit of charge (q) is Coulomb (C).

Step 4: Therefore, the SI unit of electric intensity (E) is Newton/Coulomb (N/C).

Why other options are wrong:
- Option A: Coulomb is the unit of electric charge.
- Option B: $Coulomb/{m^2}$ is not the unit of electric intensity.
- Option C: Newton is the unit of force.

Key Concepts:
- Electric intensity
- SI units

Memory Aids:
- Remember the formula: $E = \frac{F}{q}$ .

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

66 Easy

66 of 128

Understanding +4 | -1

\frac{{watt}}{{kelvin}}

is the unit of

View answer and explanation

Correct option: D

Thermal conductance

Option D (Thermal conductance) is correct because thermal conductance is defined as the rate of heat flow per unit temperature difference, which corresponds to Watt/Kelvin.

Solution approach

Correct Answer + Core Reason: Thermal conductance. Thermal conductance is heat flow per unit temperature.

Step-by-Step Solution:

Step 1: Recall the definition of thermal conductance (K).

Step 2: Thermal conductance is the measure of a material's ability to conduct heat.

Step 3: It is defined as the rate of heat flow (Q/t) per unit temperature difference ( $\Delta$ T): $K = \frac{{Q/t}}{{\Delta T}}$ .

Step 4: The unit of heat flow (Q/t) is watt (W) and the unit of temperature difference ( $\Delta$ T) is kelvin (K).

Step 5: Therefore, the unit of thermal conductance is watt/kelvin (W/K).

Why other options are wrong:
- Option A: Stefan’s constant has units of $W m^{-2} K^{-4}$ .
- Option B: Wien’s constant has units of $mK$ .
- Option C: Cooling’s constant is not a standard term with a specific unit of W/K.

Key Concepts:
- Thermal conductance
- Heat flow
- Temperature difference

Memory Aids:
- Thermal conductance is related to heat flow and temperature difference.

Quick Tips:
- Remember the definition of thermal conductance.

67 Easy

67 of 128

Remembering +4 | -1

Dimension of electric current is

View answer and explanation

Correct option: A

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

Option A is correct because electric current is a fundamental quantity and its dimension is represented by [Q] or [A].

Solution approach

Correct Answer + Core Reason: $\left[ {{M^0}{L^0}{T^{ - 1}}Q} \right]$ . Electric current is a fundamental quantity, and its dimension is [Q] or [A].

Step-by-Step Solution:

Step 1: Define electric current. Electric current (I) is defined as the rate of flow of electric charge (Q) with respect to time (T), i.e., $I = \frac{Q}{T}$ .

Step 2: Determine the dimensions. The dimension of charge is [Q], and the dimension of time is [T]. Therefore, the dimension of electric current is $\frac{{\left[ Q \right]}}{{\left[ T \right]}} = \left[ {Q{T^{ - 1}}} \right]$ .

Step 3: Express in terms of M, L, and T. Since electric current does not depend on mass (M) or length (L), the dimensions of M and L are 0. Thus, the dimension of electric current is $\left[ {{M^0}{L^0}{T^{ - 1}}Q} \right]$ .

Why other options are wrong:
- Option B: This option is incorrect because it includes mass and length dimensions, which are not part of the dimension of electric current.
- Option C: This option is incorrect because it includes mass and length dimensions, which are not part of the dimension of electric current.
- Option D: This option is incorrect because it includes mass and length dimensions, which are not part of the dimension of electric current.

Key Concepts:
- Dimensions of physical quantities
- Electric current as a fundamental quantity

Memory Aids:
- Remember that electric current is charge per unit time.

Quick Tips:
- Electric current is often treated as a fundamental dimension [A] or [I], or [Q]/[T].

68 Easy

68 of 128

Understanding +4 | -1

If h is plank’s constant and

\lambda

is the wave length,

\frac{h}{\lambda }

has the dimensions of

View answer and explanation

Correct option: B

momentum

Option B (momentum) is correct because h/λ has the dimensions of momentum, derived from de Broglie's wavelength equation.

Solution approach

Correct Answer + Core Reason: momentum. $\frac{h}{\lambda}$ has dimensions of momentum.

Step-by-Step Solution:

Step 1: Recall the de Broglie wavelength equation: $\lambda = \frac{h}{p}$ , where $\lambda$ is the wavelength, h is Planck's constant, and p is momentum.

Step 2: Rearrange the equation to solve for $\frac{h}{\lambda}$ : $\frac{h}{\lambda} = p$ .

Step 3: Determine the dimensions of momentum (p): Momentum is given by $p = mv$ , where m is mass and v is velocity. The dimensions of mass are [M] and the dimensions of velocity are [LT⁻¹]. Therefore, the dimensions of momentum are [MLT⁻¹].

Step 4: Since $\frac{h}{\lambda} = p$ , the dimensions of $\frac{h}{\lambda}$ are the same as the dimensions of momentum, which are [MLT⁻¹].

Step 5: Final Answer: momentum

Why other options are wrong:
- Option A: energy: Energy has dimensions of [ML²T⁻²].
- Option C: moment of Inertia: Moment of inertia has dimensions of [ML²].
- Option D: frequency: Frequency has dimensions of [T⁻¹].

Key Concepts:
- de Broglie Wavelength
- Momentum
- Planck's Constant
- Dimensional Analysis

Memory Aids:
- $\lambda = \frac{h}{p}$ (de Broglie's equation).

Quick Tips:
- Remember the de Broglie wavelength equation and its components.

69 Easy

69 of 128

Application +4 | -1

If the displacement

y

of a particle is

y=A

\sin ({pt}+{qx})

, then the dimensions of '

{p} / {q}

' is

View answer and explanation

Correct option: C

M^{\circ} L T^{-1}

Option C (

M^{\circ} L T^{-1}

) is correct because pt and qx must be dimensionless, so p has dimensions of T^-1 and q has dimensions of L^-1.

Solution approach

Correct Answer + Core Reason: $M^{\circ} L T^{-1}$ . Dimensional analysis of the given equation.

Step-by-Step Solution:

Step 1: Since $y = A \sin ({pt}+{qx})$ , the arguments of the sine function must be dimensionless.

Step 2: Therefore, pt is dimensionless, which means the dimensions of p are T^-1.

Step 3: Similarly, qx is dimensionless, which means the dimensions of q are L^-1.

Step 4: Find the dimensions of p/q: (T^-1) / (L^-1) = L T^-1

Step 5: Express in terms of M, L, and T: $M^{\circ} L T^{-1}$

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

Key Concepts:
- Dimensional Analysis
- Trigonometric Functions

Memory Aids:
- Remember that the arguments of trigonometric functions must be dimensionless.

Quick Tips:
- The argument of sine, cosine, etc., must be dimensionless.

70 Easy

70 of 128

Remembering +4 | -1

Unit of stress is

View answer and explanation

Correct option: C

N/{m^2}

Option C (

N/{m^2}

) is correct because stress is defined as force per unit area, hence N/m².

Solution approach

Correct Answer + Core Reason: $N/{m^2}$ is the unit of stress because stress is defined as force per unit area.

Step-by-Step Solution:

Step 1: Define stress.

Step 2: Stress is the force (F) acting per unit area (A). Therefore, $Stress = \frac{F}{A}$ .

Step 3: The SI unit of force (F) is Newton (N), and the SI unit of area (A) is square meter ( $m^2$ ).

Step 4: Therefore, the SI unit of stress is Newton per square meter ( $N/{m^2}$ ), which is also known as Pascal (Pa).

Why other options are wrong:
- Option A: N/m is not the unit of stress.
- Option B: N-m is the unit of torque or moment.
- Option D: $N - {m^2}$ is dimensionally incorrect.

Key Concepts:
- Stress
- SI units

Memory Aids:
- Remember the formula: $Stress = \frac{F}{A}$ .

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

71 Easy

71 of 128

Understanding +4 | -1

Henry/ohm can be expressed in

View answer and explanation

Correct option: A

Second

Option A (Second) is correct because Henry/ohm represents the time constant (L/R) in an RL circuit, which has units of seconds.

Solution approach

Correct Answer + Core Reason: Second. Henry/ohm represents the time constant in an RL circuit.

Step-by-Step Solution:

Step 1: Recall that the time constant ( $\tau$ ) in an RL circuit is given by $\tau = \frac{L}{R}$ , where L is inductance (in Henry) and R is resistance (in ohms).

Step 2: The time constant has units of time, which is seconds.

Why other options are wrong:
- Option B: Coulomb is incorrect; it's a unit of electric charge.
- Option C: Mho is incorrect; it's a unit of conductance (inverse of resistance).
- Option D: Metre is incorrect; it's a unit of length.

Key Concepts:
- RL circuit
- Time constant
- Inductance
- Resistance

Memory Aids:
- Remember the formula for the time constant in an RL circuit: $\tau = \frac{L}{R}$

Quick Tips:
- Know the units of inductance (Henry) and resistance (ohm).

72 Easy

72 of 128

Application +4 | -1

N

divisions on the main scale of a vernier calliper coincide with

(N+1)

divisions of the vernier scale. If each division of main scale is '

a

' units, then the least count of the instrument is

View answer and explanation

Correct option: D

\frac{a}{N+1}

Option D (

\frac{a}{N+1}

) is correct because the least count is calculated by finding the difference between one main scale division and one vernier scale division.

Solution approach

Correct Answer + Core Reason: $\frac{a}{N+1}$ . The least count is the difference between one main scale division and one vernier scale division.

Step-by-Step Solution:

Step 1: Define the given variables.
$N$ divisions on the main scale coincide with $(N+1)$ divisions of the vernier scale.\nEach division on the main scale is of length $a$ .

Step 2: Find the length of one vernier scale division.\nLength of $(N+1)$ vernier scale divisions = Length of $N$ main scale divisions = $Na$ \nLength of 1 vernier scale division = $\frac{Na}{N+1}$

Step 3: Calculate the least count.\nLeast count = 1 main scale division - 1 vernier scale division
$LC = a - \frac{Na}{N+1} = \frac{a(N+1) - Na}{N+1} = \frac{aN + a - Na}{N+1} = \frac{a}{N+1}$

Why other options are wrong:
- Option A: $a$ is incorrect because it represents the length of one main scale division, not the least count.
- Option B: $\frac{a}{N}$ is incorrect because it's a miscalculation of the least count.
- Option C: $\frac{N}{N+1} \times a$ is incorrect because it represents the length of one vernier scale division, not the least count.

Key Concepts:
- Vernier callipers
- Least count

Memory Aids:
- Least Count = 1 MSD - 1 VSD

Quick Tips:
- Understand the relationship between the main scale and vernier scale divisions to correctly calculate the least count.

73 Easy

73 of 128

Remembering +4 | -1

Dimensional formula for linear momentum is

View answer and explanation

Correct option: C

ML{T^{ - 1}}

Option C (

ML{T^{ - 1}}

) is correct because linear momentum is mass times velocity, so its dimensions are

MLT^{-1}

.

Solution approach

Correct Answer + Core Reason: $ML{T^{ - 1}}$ . Linear momentum is the product of mass and velocity.

Step-by-Step Solution:

Step 1: Define linear momentum: Linear momentum ( $p$ ) is given by $p = m \times v$ , where $m$ is mass and $v$ is velocity.

Step 2: Determine the dimensions of mass: The dimension of mass ( $m$ ) is $M$ .

Step 3: Determine the dimensions of velocity: The dimension of velocity ( $v$ ) is $LT^{-1}$ .

Step 4: Calculate the dimensions of linear momentum: The dimension of linear momentum is $M \times LT^{-1} = MLT^{-1}$ .

Why other options are wrong:
- Option A: $M{L^0}{T^{ - 1}}$ is incorrect. It represents mass divided by time, which is not momentum.
- Option B: ${M^0}L{T^{ - 1}}$ is incorrect. It represents velocity only, without considering mass.
- Option C: $ML{T^{ - 1}}$ is the correct dimensional formula for linear momentum.
- Option D: $M{L^{ - 1}}T$ is incorrect. It does not represent the correct combination of mass, length, and time for momentum.

Key Concepts:
- Linear momentum
- Dimensional analysis

Memory Aids:
- Remember that momentum is mass times velocity.

Quick Tips:
- Always start with the formula and then substitute the dimensions.

74 Easy

74 of 128

Application +4 | -1

Applying the principle of homogeneity of dimensions, determine which one is correct. where

\mathrm{T}

is time period,

\mathrm{G}

is gravitational constant,

\mathrm{M}

is mass,

\mathrm{r}

is radius of orbit.

View answer and explanation

Correct option: C

\mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}^3}{\mathrm{GM}}

Option C (

\mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}^3}{\mathrm{GM}}

) is correct because it is the only equation that is dimensionally consistent, matching the dimensions of time squared.

Solution approach

Correct Answer + Core Reason: $\mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}^3}{\mathrm{GM}}$ . This equation is dimensionally correct.

Step-by-Step Solution:

Step 1: Write the dimensions of each quantity:
- $[T] = T$
- $[G] = M^{-1}L^3T^{-2}$
- $[M] = M$
- $[r] = L$

Step 2: Check the dimensions of each option:
- Option A: $T^2 = \frac{L}{M^{-1}L^3T^{-2}M^2} = \frac{L}{ML^3T^{-2}} = M^{-1}L^{-2}T^2$ (Incorrect)
- Option B: $T^2 = L^3$ (Incorrect)
- Option C: $T^2 = \frac{L^3}{M^{-1}L^3T^{-2}M} = \frac{L^3}{L^3T^{-2}} = T^2$ (Correct)
- Option D: $T^2 = \frac{L^2}{M^{-1}L^3T^{-2}M} = \frac{L^2}{L^3T^{-2}} = L^{-1}T^2$ (Incorrect)

Step 3: Final Answer: $\mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}^3}{\mathrm{GM}}$

Why other options are wrong:
- Option A: The dimensions on both sides of the equation do not match.
- Option B: The dimensions on both sides of the equation do not match.
- Option D: The dimensions on both sides of the equation do not match.

Key Concepts:
- Dimensional Analysis
- Principle of Homogeneity of Dimensions

Memory Aids:
- Remember the dimensions of G, M, r, and T.

Quick Tips:
- Always check the dimensions of each term in the equation.

75 Easy

75 of 128

Application +4 | -1

The density of a material in CGS system is

2 \mathrm{~g} / \mathrm{cm}^3

. In a system of units in which unit of length is

2 \mathrm{~cm}

and unit of mass is

4 \mathrm{~g}

, what is the numerical value of the density of the material?

View answer and explanation

Correct option: A

4

Option A (4) is correct because it involves converting the density from CGS to the new system of units by adjusting for the changes in length and mass.

Solution approach

Correct Answer + Core Reason: 4. The density is converted to the new system of units by adjusting for the changes in length and mass.

Step-by-Step Solution:

Step 1: Given density in CGS: $\rho = 2 \frac{g}{cm^3}$

Step 2: New unit of length: $L' = 2 cm$ \nNew unit of mass: $M' = 4 g$

Step 3: We want to find the density in the new system, $\rho'$ such that $\rho = x \frac{M'}{L'^3}$ where x is the numerical value we want to find.

Step 4: Substitute the new units: $2 \frac{g}{cm^3} = x \frac{4g}{(2cm)^3} = x \frac{4g}{8cm^3}$

Step 5: Solve for x: $x = 2 \frac{g}{cm^3} \times \frac{8cm^3}{4g} = 2 \times 2 = 4$

Step 6: Final Answer: 4

Why other options are wrong:
- Option B (0): This is incorrect as it doesn't follow from the correct unit conversion.
- Option C (0): This is incorrect as it doesn't follow from the correct unit conversion.
- Option D (0): This is incorrect as it doesn't follow from the correct unit conversion.

Key Concepts:
- Unit Conversion
- Density

Memory Aids:
- Density = Mass/Volume

Quick Tips:
- Always pay attention to the units and how they change in different systems.
- Use the conversion factors correctly to avoid errors.

76 Easy

76 of 128

Application +4 | -1

If the unit of length is 4 m and that of velocity is

6\,m{s^{ - 1}},

the unit of time is (in second)

View answer and explanation

Correct option: A

2/3

Option A (2/3) is correct because time = length / velocity, so the unit of time is 4 m / 6 m/s = 2/3 s.

Solution approach

Correct Answer + Core Reason: The unit of time is 2/3 seconds.

Step-by-Step Solution:

Step 1: Relationship between length, velocity, and time: $v = \frac{l}{t}$ .

Step 2: Rearrange for time: $t = \frac{l}{v}$ .

Step 3: Given length unit: $l = 4 \text{ m}$ .

Step 4: Given velocity unit: $v = 6 \text{ m/s}$ .

Step 5: Calculate the unit of time: $t = \frac{4 \text{ m}}{6 \text{ m/s}} = \frac{2}{3} \text{ s}$ .

Why other options are wrong:
- Option B: 4/3: Incorrect calculation.
- Option C: 3/2: Incorrect calculation.
- Option D: 24: Incorrect calculation.

Key Concepts:
- Relationship between length, velocity, and time

Memory Aids:
- Velocity = Length / Time

Quick Tips:
- Remember the relationship between length, velocity, and time.

77 Easy

77 of 128

Understanding +4 | -1

Identify the pair whose dimensions are equal.

View answer and explanation

Correct option: A

Torque and work

Option A (Torque and work) is correct because both torque and work are defined as force times distance, resulting in the same dimensions.

Solution approach

Correct Answer + Core Reason: Torque and work. Both are force times distance.

Step-by-Step Solution:

Step 1: Recall the definition of torque ( $\tau$ ). Torque = Force x Distance.

Step 2: Recall the definition of work (W). Work = Force x Distance.

Step 3: Since both torque and work are defined as Force x Distance, they have the same dimensions.

Step 4: Stress is defined as Force / Area, and energy is related to work.

Why other options are wrong:
- Option B: Stress and energy have different dimensions. Stress is Force/Area, Energy is Force x Distance.
- Option C: Force and stress have different dimensions. Stress is Force/Area.
- Option D: Force and work have different dimensions. Work is Force x Distance.

Key Concepts:
- Torque
- Work
- Stress
- Force
- Energy

Memory Aids:
- Torque and Work are both Force x Distance.

Quick Tips:
- Remember the definitions of the quantities.

78 Easy

78 of 128

Remembering +4 | -1

The dimensions of power are

View answer and explanation

Correct option: A

{M^1}{L^2}{T^{ - 3}}

Option A (

{M^1}{L^2}{T^{ - 3}}

) is correct because power is work done per unit time, and its dimensional formula is derived from that.

Solution approach

Correct Answer + Core Reason: ${M^1}{L^2}{T^{ - 3}}$ . Power is work done per unit time.

Step-by-Step Solution:

Step 1: Define power. Power (P) is defined as the rate at which work is done, i.e., $P = \frac{W}{t}$ , where W is work and t is time.

Step 2: Determine the dimensions of work. Work (W) is defined as force (F) times distance (d), i.e., $W = F \times d$ . The dimension of force is $\left[ {ML{T^{ - 2}}} \right]$ , and the dimension of distance is [L]. Therefore, the dimension of work is $\left[ {ML{T^{ - 2}}} \right] \times \left[ L \right] = \left[ {M{L^2}{T^{ - 2}}} \right]$ .

Step 3: Determine the dimensions of power. Since $P = \frac{W}{t}$ , the dimension of power is $\frac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ T \right]}} = \left[ {M{L^2}{T^{ - 3}}} \right]$ .

Why other options are wrong:
- Option B: This option is incorrect because the dimensions of length and time are incorrect for power.
- Option C: This option is incorrect because the dimension of time is incorrect for power.
- Option D: This option is incorrect because the dimensions of length and time are incorrect for power.

Key Concepts:
- Dimensions of physical quantities
- Power, work, and energy

Memory Aids:
- Remember that Power = Work / Time.

Quick Tips:
- Start with the basic definition of the quantity and derive its dimensions from there.

79 Easy

79 of 128

Application +4 | -1

If

x = a - b

, then the maximum percentage error in the measurement of x will be

View answer and explanation

Correct option: A

\left( {\frac{{\Delta a + \Delta b}}{{a - b}}} \right) \times 100\%

Option A is correct because the maximum percentage error in subtraction is the sum of the absolute errors divided by the difference, expressed as a percentage.

Solution approach

Correct Answer + Core Reason: $\left( {\frac{{\Delta a + \Delta b}}{{a - b}}} \right) \times 100\%$ . The maximum error in subtraction is the sum of absolute errors divided by the result.

Step-by-Step Solution:

Step 1: Identify the given information: $x = a - b$ , and we want to find the maximum percentage error in x.

Step 2: Recall the rule for error propagation in subtraction: If $x = a - b$ , then $\Delta x = \Delta a + \Delta b$ .

Step 3: Calculate the fractional error: $\frac{{\Delta x}}{x} = \frac{{\Delta a + \Delta b}}{{a - b}}$ .

Step 4: Convert to percentage error: Percentage error = $\left( {\frac{{\Delta a + \Delta b}}{{a - b}}} \right) \times 100\%$ .

Why other options are wrong:
- Option B: $\left( {\frac{{\Delta a}}{a} - \frac{{\Delta b}}{b}} \right) \times 100\%$ : This is incorrect because it represents the difference of fractional errors, which is not how errors propagate in subtraction.
- Option C: $\left( {\frac{{\Delta a}}{{a - a}} + \frac{{\Delta b}}{{a - b}}} \right) \times 100\%$ : This option has a typo in the denominator of the first term (a-a), and it's not the correct way to calculate percentage error in subtraction.
- Option D: $\left( {\frac{{\Delta a}}{{a - a}} - \frac{{\Delta b}}{{a - b}}} \right) \times 100\%$ : This option also has a typo in the denominator of the first term (a-a), and it's not the correct way to calculate percentage error in subtraction.

Key Concepts:
- Error propagation in subtraction
- Percentage error calculation

Memory Aids:
- In subtraction, absolute errors add up.

Quick Tips:
- Always add absolute errors when dealing with subtraction or addition.

80 Easy

80 of 128

Application +4 | -1

What are the units of

K = 1/4\pi {\varepsilon _0}

View answer and explanation

Correct option: B

N{m^2}{C^{ - 2}}

Option B (

Nm^2C^{-2}

) is correct because K is the electrostatic constant in Coulomb's law, and its unit is derived from the formula F = k(q1q2/r²).

Solution approach

Correct Answer + Core Reason: The units of $K = 1/4\pi \epsilon_0$ are $Nm^2C^{-2}$ .

Step-by-Step Solution:

Step 1: Recall Coulomb's Law: $F = K \frac{q_1 q_2}{r^2}$ .

Step 2: Rearrange for K: $K = \frac{Fr^2}{q_1 q_2}$ .

Step 3: Units of Force (F): Newtons (N).

Step 4: Units of Distance (r): meters (m).

Step 5: Units of Charge (q): Coulombs (C).

Step 6: Units of K: $K = \frac{Nm^2}{C^2} = Nm^2C^{-2}$ .

Why other options are wrong:
- Option A: $C^2N^{-1}m^{-2}$ : This is the inverse of the correct unit.
- Option C: $Nm^2C^2$ : Incorrect unit.
- Option D: Unitless: K has units.

Key Concepts:
- Coulomb's Law
- Electrostatic constant

Memory Aids:
- Remember Coulomb's Law formula.

Quick Tips:
- Remember Coulomb's Law and rearrange it to find the units of K.

81 Easy

81 of 128

Remembering +4 | -1

Ampere-hour is a unit of

View answer and explanation

Correct option: A

Quantity of electricity

Option A (Quantity of electricity) is correct because Ampere-hour is the unit of charge, which represents the quantity of electricity.

Solution approach

Correct Answer + Core Reason: Quantity of electricity. Ampere-hour is a unit of charge.

Step-by-Step Solution:

Step 1: Define Ampere-hour:\nAmpere-hour (Ah) is a unit of electric charge.
1 Ampere-hour = 1 Ampere × 1 hour

Step 2: Relate Ampere-hour to charge:\nCharge (Q) = Current (I) × Time (t)\nQ = I × t\nUnit of charge = Ampere × hour = Ampere-hour

Step 3: Therefore, Ampere-hour is a unit of quantity of electricity (charge).

Why other options are wrong:
- Option B: Strength of electric current: Ampere is the unit of electric current, not Ampere-hour.
- Option C: Power: Watt is the unit of power, not Ampere-hour.
- Option D: Energy: Joule is the unit of energy, not Ampere-hour.

Key Concepts:
- Electric charge
- Electric current
- Units of measurement

Memory Aids:
- Remember that charge is the product of current and time.

Quick Tips:
- Know the definitions and units of basic electrical quantities.

82 Easy

82 of 128

Analyzing +4 | -1

A physical quantity

\mathrm{T}=\frac{\sqrt{\mathrm{abc}^2}}{\mathrm{~d}^3 \mathrm{e}^{1 / 6}}

is determined by measuring a, b, c, d, e separately with the percentage error of

1 \%, 2 \%, 3 \%, 4 \%, 3 \%

respectively. Minimum amount of error is contributed by the measurement of :

View answer and explanation

Correct option: C

a, e

Option C (a, e) is correct because the percentage errors of a and e contribute the least to the overall error in the calculated quantity T.

Solution approach

Correct Answer + Core Reason: a, e. The minimum error is contributed by the measurements with the smallest coefficients in the error propagation formula.

Step-by-Step Solution:

Step 1: Write the expression for T: $\mathrm{T}=\frac{\sqrt{\mathrm{abc}^2}}{\mathrm{~d}^3 \mathrm{e}^{1 / 6}}$

Step 2: Write the expression for the percentage error in T: $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta a}{a} + \frac{1}{2} \frac{\Delta b}{b} + \frac{2}{2} \frac{\Delta c}{c} + 3 \frac{\Delta d}{d} + \frac{1}{6} \frac{\Delta e}{e}$

Step 3: Substitute the given percentage errors:
$\frac{\Delta T}{T} = \frac{1}{2}(1\%) + \frac{1}{2}(2\%) + 1(3\%) + 3(4\%) + \frac{1}{6}(3\%)$

Step 4: Calculate the contribution of each variable:\na: $\frac{1}{2}(1\%) = 0.5\%$ \nb: $\frac{1}{2}(2\%) = 1\%$ \nc: $1(3\%) = 3\%$ \nd: $3(4\%) = 12\%$ ≠: $\frac{1}{6}(3\%) = 0.5\%$

Step 5: Identify the minimum contributions: a and e have the smallest contributions (0.5% each).

Step 6: Conclusion: The minimum amount of error is contributed by the measurement of a and e.

Why other options are wrong:
- Option A: a alone is not the complete answer, as 'e' also contributes the minimum error.
- Option B: c contributes a 3% error, which is not minimal.
- Option D: c and d contribute 3% and 12% errors respectively, which are not minimal.

Key Concepts:
- Error propagation
- Percentage error
- Identifying the contribution of each variable to the total error

Memory Aids:
- Remember to multiply the percentage error of each variable by its corresponding power in the formula.

Quick Tips:
- When calculating the overall error, remember to take the absolute values of the errors and add them.

83 Easy

83 of 128

Understanding +4 | -1

Which pair has the same dimensions

View answer and explanation

Correct option: C

Momentum and impulse

Option C (Momentum and impulse) is correct because impulse is the change in momentum, so they have the same dimensions [MLT⁻¹].

Solution approach

Correct Answer + Core Reason: Momentum and impulse. Impulse is the change in momentum.

Step-by-Step Solution:

Step 1: Determine the dimensions of Momentum (p): Momentum is given by $p = mv$ , where m is mass and v is velocity. The dimensions of mass are [M] and the dimensions of velocity are [LT⁻¹]. Therefore, the dimensions of momentum are [MLT⁻¹].

Step 2: Determine the dimensions of Impulse (J): Impulse is given by $J = F\Delta t$ , where F is force and $\Delta t$ is the change in time. The dimensions of force are [MLT⁻²] and the dimensions of time are [T]. Therefore, the dimensions of impulse are [MLT⁻²][T] = [MLT⁻¹].

Step 3: Compare the dimensions: Both momentum and impulse have dimensions of [MLT⁻¹].

Step 4: Final Answer: Momentum and impulse

Why other options are wrong:
- Option A: Work and power: Work has dimensions of [ML²T⁻²], and power has dimensions of [ML²T⁻³].
- Option B: Density and relative density: Density has dimensions of [ML⁻³], while relative density is dimensionless.
- Option D: Stress and strain: Stress has dimensions of [ML⁻¹T⁻²], while strain is dimensionless.

Key Concepts:
- Momentum
- Impulse
- Dimensional Analysis

Memory Aids:
- Impulse is the change in momentum.

Quick Tips:
- Remember the definitions of momentum and impulse.

84 Easy

84 of 128

Application +4 | -1

If the length of a cylinder is measured to be 50 cm with an error of 0.25 cm. The percentage error in the measured length is

View answer and explanation

Correct option: B

0.005

Option B (0.005) is correct because the percentage error is calculated as (error / measured value) * 100.

Solution approach

Correct Answer + Core Reason: 0.005. The percentage error is calculated as (error / measured value) * 100.

Step-by-Step Solution:

Step 1: Identify the Error: The error in the length measurement is 0.25 cm.

Step 2: Identify the Measured Value: The measured length of the cylinder is 50 cm.

Step 3: Calculate the Percentage Error: Percentage error = $\frac{{\text{{Error}}}}{{\text{{Measured Value}}}} \times 100 = \frac{{0.25}}{{50}} \times 100 = 0.5\%$

Step 4: Convert Percentage to Decimal: 0.5% = 0.005

Why other options are wrong:
- Option A: 0.004: This is incorrect; it might be a calculation error.
- Option C: 0.002: This is incorrect; it might be a calculation error.
- Option D: 0.001: This is incorrect; it might be a calculation error.

Key Concepts:
- Percentage Error

Memory Aids:
- Percentage Error = (Error / Measured Value) * 100

Quick Tips:
- Always divide the error by the measured value, not the true value (if known).

85 Easy

85 of 128

Application +4 | -1

A quantity

x

is defined as

x=\frac{a^3-b^2}{\sqrt{c+d}}

. Value of

a, b, c

and

d

are reported as

{a}=3 \pm 0.001, {b}=5 \pm 0.0013, {c}=6 \pm 0.24

and

{d}=10 \pm 0.4

. Percentage error in

{x}

will be-

View answer and explanation

Correct option: A

4

Option A (4) is correct because it involves calculating the percentage error in x using the given formula and error values for a, b, c, and d.

Solution approach

Correct Answer + Core Reason: 4. The percentage error in x is calculated using error propagation.

Step-by-Step Solution:

Step 1: Write the formula for x: $x = \frac{{a^3 - b^2}}{{\sqrt{{c + d}}}}$

Step 2: Calculate the percentage error in $a^3$ : $\frac{{\Delta (a^3)}}{{a^3}} = 3 \frac{{\Delta a}}{a} = 3 \times \frac{{0.001}}{3} = 0.001$

Step 3: Calculate the percentage error in $b^2$ : $\frac{{\Delta (b^2)}}{{b^2}} = 2 \frac{{\Delta b}}{b} = 2 \times \frac{{0.0013}}{5} = 0.00052$

Step 4: Calculate the absolute error in $a^3 - b^2$ : $\Delta (a^3 - b^2) = a^3 \times 0.001 + b^2 \times 0.00052 = 3^3 \times 0.001 + 5^2 \times 0.00052 = 0.027 + 0.013 = 0.04$

Step 5: Calculate $a^3 - b^2 = 3^3 - 5^2 = 27 - 25 = 2$

Step 6: Calculate the percentage error in $a^3 - b^2$ : $\frac{{\Delta (a^3 - b^2)}}{{a^3 - b^2}} = \frac{{0.04}}{2} = 0.02$

Step 7: Calculate the percentage error in $c + d$ : $\Delta (c + d) = \Delta c + \Delta d = 0.24 + 0.4 = 0.64$

Step 8: Calculate $c + d = 6 + 10 = 16$

Step 9: Calculate the percentage error in $\sqrt{{c + d}}$ : $\frac{{\Delta \sqrt{{c + d}}}}{{\sqrt{{c + d}}}} = \frac{1}{2} \frac{{\Delta (c + d)}}{{c + d}} = \frac{1}{2} \times \frac{{0.64}}{{16}} = \frac{1}{2} \times 0.04 = 0.02$

Step 10: Calculate the percentage error in x: $\frac{{\Delta x}}{x} = \frac{{\Delta (a^3 - b^2)}}{{a^3 - b^2}} + \frac{{\Delta \sqrt{{c + d}}}}{{\sqrt{{c + d}}}} = 0.02 + 0.02 = 0.04$

Step 11: Convert to percentage: $0.04 \times 100 = 4\%$

Why other options are wrong:
- Option B: 0: Incorrect calculation.
- Option C: 0: Incorrect calculation.
- Option D: 0: Incorrect calculation.

Key Concepts:
- Error Propagation
- Percentage Error

Memory Aids:
- Remember how errors propagate through different mathematical operations.

Quick Tips:
- Break down the complex formula into smaller parts and calculate errors for each part.

86 Easy

86 of 128

Understanding +4 | -1

The dimensions of shear modulus are

View answer and explanation

Correct option: C

M{L^{ - 1}}{T^{ - 2}}

Option C (

M{L^{ - 1}}{T^{ - 2}}

) is correct because shear modulus is stress divided by strain, and strain is dimensionless.

Solution approach

Correct Answer + Core Reason: $M{L^{ - 1}}{T^{ - 2}}$ . Shear modulus is defined as stress/strain. Stress has dimensions of force/area, and strain is dimensionless.

Step-by-Step Solution:

Step 1: Recall the definition of shear modulus: $G = \frac{{\text{{shear stress}}}}{{\text{{shear strain}}}}$ .

Step 2: Shear stress is force per unit area: $\text{{shear stress}} = \frac{{F}}{{A}}$ .

Step 3: Determine the dimensions of force: $F = \left[M L T^{-2}\right]$ .

Step 4: Determine the dimensions of area: $A = \left[L^2\right]$ .

Step 5: Determine the dimensions of shear stress: $\text{{shear stress}} = \frac{{\left[M L T^{-2}\right]}}{{\left[L^2\right]}} = \left[M L^{-1} T^{-2}\right]$ .

Step 6: Shear strain is dimensionless, so its dimensions are $1$ .

Step 7: Determine the dimensions of shear modulus: $G = \frac{{\left[M L^{-1} T^{-2}\right]}}{{1}} = \left[M L^{-1} T^{-2}\right]$ .

Why other options are wrong:
- Option A: This is not the correct dimension for shear modulus.
- Option B: This is the dimension of energy or work.
- Option D: This is not the correct dimension for shear modulus.

Key Concepts:
- Shear modulus
- Stress
- Strain
- Dimensional analysis

Memory Aids:
- Remember that shear modulus is stress over strain.

Quick Tips:
- Use the defining equation to determine the dimensions.

87 Easy

87 of 128

Application +4 | -1

The angle of

1^{\prime}

(minute of arc) in radian is nearly equal to

View answer and explanation

Correct option: A

2.91 \times 10^{-4} \mathrm{~rad}

Option A (

2.91 \times 10^{-4} \mathrm{~rad}

) is correct because it represents the conversion of 1 minute of arc to radians.

Solution approach

Correct Answer + Core Reason: $2.91 \times 10^{-4} \mathrm{~rad}$ . This is the correct conversion of 1 minute of arc to radians.

Step-by-Step Solution:

Step 1: Convert 1 minute of arc to degrees. 1 minute = 1/60 degrees.

Step 2: Convert degrees to radians. 1 degree = $\pi/180$ radians.

Step 3: Combine the conversions: 1 minute = (1/60) * ( $\pi/180$ ) radians.

Step 4: Calculate the value: (1/60) * (3.14159/180) ≈ 0.000290888 radians.

Step 5: Express in scientific notation: $2.91 \times 10^{-4}$ radians.

Why other options are wrong:
- Option B: $4.85 \times 10^{-4} \mathrm{~rad}$ is incorrect. This value is not the correct conversion of 1 minute to radians.
- Option C: $4.80 \times 10^{-6} \mathrm{~rad}$ is incorrect. This value is too small and likely results from an incorrect calculation or misunderstanding of the units.
- Option D: $1.75 \times 10^{-2} \mathrm{~rad}$ is incorrect. This value is too large and likely results from an incorrect calculation or misunderstanding of the units.

Key Concepts:
- Radian measure
- Conversion between degrees and radians
- Minute of arc

Memory Aids:
- Remember that 1 degree = 60 minutes and 1 radian = 180/ $\pi$ degrees.

Quick Tips:
- Always convert to radians before using angles in formulas.

88 Easy

88 of 128

Remembering +4 | -1

The unit of self inductance of a coil is

View answer and explanation

Correct option: B

Henry

Option B (Henry) is correct because Henry is the SI unit of self-inductance.

Solution approach

Correct Answer + Core Reason: Henry is the unit of self inductance because it is defined as the ratio of induced emf to the rate of change of current.

Step-by-Step Solution:

Step 1: Define self-inductance (L).

Step 2: Self-inductance is the property of a coil to oppose the change in current flowing through it. The induced emf ( $\varepsilon$ ) is given by $\varepsilon = -L \frac{dI}{dt}$ , where L is the self-inductance and $\frac{dI}{dt}$ is the rate of change of current.

Step 3: The unit of induced emf ( $\varepsilon$ ) is Volt (V), and the unit of rate of change of current ( $\frac{dI}{dt}$ ) is Ampere/second (A/s).

Step 4: Therefore, the unit of self-inductance (L) is Volt-second/Ampere (V.s/A), which is defined as Henry (H).

Why other options are wrong:
- Option A: Farad is the unit of capacitance.
- Option C: Weber is the unit of magnetic flux.
- Option D: Tesla is the unit of magnetic field strength.

Key Concepts:
- Self-inductance
- SI units

Memory Aids:
- Remember the formula: $\varepsilon = -L \frac{dI}{dt}$ .

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

89 Easy

89 of 128

Remembering +4 | -1

The number of significant figures in a number
"1700.00200" is

View answer and explanation

Correct option: C

9

Option C (9) is correct because all digits including trailing zeros after the decimal point are significant.

Solution approach

Correct Answer + Core Reason: 9. All digits are significant, including trailing zeros after the decimal point.

Step-by-Step Solution:

Step 1: Identify the number: 1700.00200

Step 2: Apply the rules for significant figures:
- All non-zero digits are significant.
- Zeros between non-zero digits are significant.
- Trailing zeros after the decimal point are significant.

Step 3: Count the significant figures:
- 1 and 7 are significant (2 digits)
- The two zeros between 17 and .00200 are ambiguous without the decimal, but since the decimal is present, they are significant (2 digits)
- 002 are significant (3 digits)
- The last two zeros are significant (2 digits)
- Total: 2 + 2 + 3 + 2 = 9

Step 4: Final Answer: 9

Why other options are wrong:
- Option A: Incorrect because it only counts the non-zero digits and ignores the zeros.
- Option B: Incorrect because it misses some of the significant zeros.
- Option D: Incorrect because it counts an extra digit.

Key Concepts:
- Significant figures rules

Memory Aids:
- Pacific-Atlantic rule for trailing zeros.

Quick Tips:
- When in doubt, convert to scientific notation.

90 Easy

90 of 128

Understanding +4 | -1

The dimensional formula of relative density is

View answer and explanation

Correct option: D

{M^0}{L^0}{T^0}

Option D (

{M^0}{L^0}{T^0}

) is correct because relative density is a ratio of two densities, making it dimensionless.

Solution approach

Correct Answer + Core Reason: ${M^0}{L^0}{T^0}$ . Relative density is a ratio of densities.

Step-by-Step Solution:

Step 1: Recall the definition of relative density.

Step 2: Relative density is defined as the ratio of the density of a substance to the density of a reference substance (usually water).

Step 3: Since it is a ratio of two densities, the dimensions cancel out, making it a dimensionless quantity.

Step 4: Therefore, the dimensional formula of relative density is ${M^0}{L^0}{T^0}$ .

Why other options are wrong:
- Option A: $M{L^{ - 3}}{T^0}$ represents density.
- Option B: ${M^0}L{T^{ - 1}}$ represents velocity or speed.
- Option C: $ML{T^{ - 2}}$ represents force.

Key Concepts:
- Relative density
- Density
- Dimensional analysis

Memory Aids:
- Relative density is a ratio, hence dimensionless.

Quick Tips:
- Remember that ratios of similar quantities are dimensionless.

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