UNITS & MEASUREMENTS MCQs for NEET UG PHYSICS — Easy, Page 4

91 Easy

91 of 128

Remembering +4 | -1

The unit of Wien’s constant b is

View answer and explanation

Correct option: D

MK

Option D (MK) is correct because Wien's constant is the product of wavelength and temperature at which blackbody radiation is maximum, hence meter-Kelvin.

Solution approach

Correct Answer + Core Reason: MK. Wien's constant is the product of wavelength and temperature.

Step-by-Step Solution:

Step 1: Recall Wien's displacement law: $\lambda_{max} T = b$ , where $\lambda_{max}$ is the wavelength at which the blackbody radiation is maximum, $T$ is the temperature, and $b$ is Wien's constant.

Step 2: Determine the units of $\lambda_{max}$ : The unit of wavelength is meter (m).

Step 3: Determine the units of $T$ : The unit of temperature is Kelvin (K).

Step 4: Determine the units of $b$ : Since $b = \lambda_{max} T$ , the unit of $b$ is meter-Kelvin (mK) or MK.

Why other options are wrong:
- Option A: $\mathrm{W}{\mathrm{m}^{ - 2}}{\mathrm{K}^{ - 4}}$ is the unit for Stefan-Boltzmann constant, not Wien's constant.
- Option B: ${\mathrm{m}^{ - 1}}{\mathrm{K}^{ - 1}}$ is not related to Wien's constant.
- Option C: $\mathrm{W}{\mathrm{m}^2}$ is the unit of power per unit solid angle, not Wien's constant.

Key Concepts:
- Wien's displacement law
- Blackbody radiation
- Units of measurement

Memory Aids:
- Remember Wien's displacement law: $\lambda_{max} T = b$ .

Quick Tips:
- Always remember the formula for Wien's displacement law.
- Pay attention to the units of each quantity.

92 Easy

92 of 128

Remembering +4 | -1

Which relation is wrong

View answer and explanation

Correct option: D

1\;newton\; = \;{10^{ - 5}}dyne

Option D (

1\mathrm{\;}newton{\;}=\;{10^{ - 5}}dyne

) is wrong because the correct relation is

1\mathrm{\;}newton{\;}=\;{10^{5}}dyne

.

Solution approach

Correct Answer + Core Reason: $1\mathrm{\;}newton{\;}=\;{10^{ - 5}}dyne$ is incorrect because 1 Newton is equal to $10^5$ dynes.

Step-by-Step Solution:

Step 1: Recall the relationship between Newton and dyne.

Step 2: 1 Newton (N) is the SI unit of force, and 1 dyne is the CGS unit of force.

Step 3: The conversion factor between Newton and dyne is: 1 N = $10^5$ dynes.

Step 4: Therefore, the given relation $1\mathrm{\;}newton{\;}=\;{10^{ - 5}}dyne$ is wrong.

Why other options are wrong:
- Option A: $1\mathrm{\;}calorie{\;}=\;4.18\mathrm{\;}joule$ is correct.
- Option B: $1\mathrm{\;}{\mathrm{A}^0} = {10^{ - 10}}m$ is correct.
- Option C: $1\mathrm{\;}MeV = \;1.6\; \times {10^{ - 13}}joule$ is correct.

Key Concepts:
- Units and dimensions
- Conversion of units

Memory Aids:
- Remember that 1 N = $10^5$ dynes.

Quick Tips:
- Always check the units and their conversions carefully.

93 Easy

93 of 128

Application +4 | -1

The de-Broglie wavelength associated with a particle of mass

\mathrm{m}

and energy

\mathrm{E}

is

\mathrm{h} / \sqrt{\mathrm{2 m E}}

. The dimensional formula for Planck's constant is :

View answer and explanation

Correct option: B

\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]

Option B (

\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]

) is correct because the de-Broglie wavelength is given, and we can solve for Planck's constant (h) dimensionally.

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$ . Dimensional analysis of Planck's constant.

Step-by-Step Solution:

Step 1: Given: $\lambda = \frac{h}{\sqrt{2mE}}$

Step 2: Rearrange to solve for h: $h = \lambda \sqrt{2mE}$

Step 3: Dimensions of $\lambda$ (wavelength) are [L].

Step 4: Dimensions of m (mass) are [M].

Step 5: Dimensions of E (energy) are [ML²T^-2].

Step 6: Substitute the dimensions: $h = [L] \sqrt{2[M][ML^2T^{-2}]} = [L] \sqrt{[M^2L^2T^{-2}]} = [L][MLT^{-1}] = [ML^2T^{-1}]$

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

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

Memory Aids:
- Remember the de Broglie wavelength formula.

Quick Tips:
- Use the given formula to derive the dimensions of Planck's constant.

94 Easy

94 of 128

Application +4 | -1

Data for four screwgauge's given below, arrange them increasing order of accuracy.
(a) Pitch

=1 \mathrm{~mm}

, no. of division on C.S.

=100

(b) Pitch

=0.5 \mathrm{~mm}

, no. of division on C.S.

=100

(c) Pitch

=3 / 4 \mathrm{~mm}

, no. of division on C.S.

=60

(d) Pitch

=1 / 2 \mathrm{~mm}

, no. of division on C.S.

=50

View answer and explanation

Correct option: C

c

<

d

=

a

<

b

Option C (c < d = a < b) is correct because the accuracy is inversely proportional to the least count, and the least count is calculated as Pitch / Number of divisions on circular scale.

Solution approach

Correct Answer + Core Reason: c < d = a < b. Accuracy increases as the least count decreases. Least count = Pitch / Number of divisions on circular scale.

Step-by-Step Solution:

Step 1: Calculate the least count for each screw gauge.
(a) Pitch = 1 mm, No. of divisions = 100
$LC_a = \frac{1}{100} = 0.01 \mathrm{~mm}$
(b) Pitch = 0.5 mm, No. of divisions = 100
$LC_b = \frac{0.5}{100} = 0.005 \mathrm{~mm}$
(c) Pitch = 3/4 mm, No. of divisions = 60
$LC_c = \frac{3/4}{60} = \frac{3}{240} = \frac{1}{80} = 0.0125 \mathrm{~mm}$
(d) Pitch = 1/2 mm, No. of divisions = 50
$LC_d = \frac{1/2}{50} = \frac{1}{100} = 0.01 \mathrm{~mm}$

Step 2: Arrange the least counts in increasing order.
$LC_b < LC_a = LC_d < LC_c$

Step 3: Arrange the screw gauges in increasing order of accuracy (decreasing order of least count).\nc < d = a < b

Why other options are wrong:
- Option A: b < d = a < c is incorrect because the order of accuracy is not correctly determined based on the least counts.
- Option B: a = d < b < c is incorrect because the order of accuracy is not correctly determined based on the least counts.
- Option D: a = d < c < b is incorrect because the order of accuracy is not correctly determined based on the least counts.

Key Concepts:
- Screw gauge
- Least count
- Accuracy

Memory Aids:
- Accuracy is inversely proportional to the least count.

Quick Tips:
- Calculate the least count accurately for each instrument before comparing their accuracies.

95 Easy

95 of 128

Understanding +4 | -1

Statement I: The order of accuracy of measurement depends on the least count of the measuring instrument.

Statement II: The smaller the least count, the greater is the number of significant figures in the measured value.

View answer and explanation

Correct option: B

Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I

Option B (Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I) is correct because both statements are true, but the least count doesn't directly explain the order of accuracy.

Solution approach

Correct Answer + Core Reason: Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I. Both statements are correct, but the second doesn't explain the first.

Step-by-Step Solution:

Step 1: Analyze Statement I: The order of accuracy of measurement depends on the least count of the measuring instrument.
- This is true. A smaller least count allows for more precise measurements, thus increasing the order of accuracy.

Step 2: Analyze Statement II: The smaller the least count, the greater is the number of significant figures in the measured value.
- This is also true. A smaller least count allows for more digits to be reliably recorded, increasing the number of significant figures.

Step 3: Determine if Statement II explains Statement I.
- While both are true, the number of significant figures doesn't directly explain why the order of accuracy depends on the least count. The least count directly determines the precision of the measurement, which affects accuracy.

Why other options are wrong:
- Option A (Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I): Statement II doesn't directly explain Statement I.
- Option C (Statement I is true; Statement II is false): Statement II is also true.
- Option D (Statement I is false; Statement II is true): Statement I is also true.

Key Concepts:
- Least count
- Accuracy
- Significant figures

Memory Aids:
- Smaller least count = higher precision = higher potential accuracy.

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

96 Easy

96 of 128

Application +4 | -1

The force F on the sphere of radius ‘a’ moving in a medium with velocity '

v

' is given by

F = 6\pi \eta a\upsilon

. The dimensions of η are

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it represents the dimensions of viscosity (η) derived from the given formula using dimensional analysis.

Solution approach

Correct Answer + Core Reason: $M{L^{ - 1}}{T^{ - 1}}$ . This is the dimension of viscosity.

Step-by-Step Solution:

Step 1: Given the formula $F = 6\pi \eta a v$ , where $F$ is force, $a$ is radius, and $v$ is velocity.

Step 2: We need to find the dimensions of $\eta$ .

Step 3: Rearrange the formula to solve for $\eta$ : $\eta = \frac{F}{6\pi a v}$

Step 4: Write the dimensions of each quantity:
- $[F] = MLT^{-2}$
- $[a] = L$
- $[v] = LT^{-1}$
- $[6\pi] = 1$ (dimensionless)

Step 5: Substitute the dimensions into the formula for $\eta$ :
$\eta = \frac{MLT^{-2}}{L \cdot LT^{-1}} = \frac{MLT^{-2}}{L^2T^{-1}} = ML^{-1}T^{-1}$

Step 6: Final Answer: $ML^{-1}T^{-1}$

Why other options are wrong:
- Option B: $MT^{-1}$ is not the correct dimension for viscosity.
- Option C: $MLT^{-2}$ is the dimension of force, not viscosity.
- Option D: $ML^{-3}$ is the dimension of density, not viscosity.

Key Concepts:
- Dimensional Analysis
- Viscosity

Memory Aids:
- Remember the formula $F = 6\pi \eta a v$ for viscous force.

Quick Tips:
- Always write the dimensions of each quantity before substituting into the formula.

97 Easy

97 of 128

Application +4 | -1

If a current of 2.33 amp is passes through a resistance of 10.485 ohm, the potential is 24.43005 V its value is proper significant figures would be

View answer and explanation

Correct option: B

24.4 V

Option B (24.4 V) is correct because the result should have the same number of significant figures as the value with the fewest significant figures.

Solution approach

Correct Answer + Core Reason: 24.4 V. The result is rounded to three significant figures.

Step-by-Step Solution:

Step 1: Calculate the potential: Potential (V) = Current (I) x Resistance (R) = 2.33 amp x 10.485 ohm = 24.43005 V.

Step 2: Identify the number of significant figures in the given values: 2.33 has 3 significant figures, and 10.485 has 5 significant figures.

Step 3: Round the calculated potential to the least number of significant figures, which is 3. So, 24.43005 is rounded to 24.4.

Why other options are wrong:
- Option A: 24.23 V: This is incorrect and doesn't follow the rules of significant figures.
- Option C: 24.430 V: This has 5 significant figures, which is more than the least number of significant figures in the given values.
- Option D: 24.43005 V: This has 7 significant figures, which is more than the least number of significant figures in the given values.

Key Concepts:
- Significant figures
- Rounding rules
- Ohm's Law (V = IR)

Memory Aids:
- When multiplying or dividing, the result should have the same number of significant figures as the quantity with the fewest significant figures.

Quick Tips:
- Always identify the least number of significant figures in the given values before performing calculations.

98 Easy

98 of 128

Application +4 | -1

If the unit of force is

\mathrm{kN}

, the length is

1 \mathrm{~km}

and time

100 \mathrm{~s}

, then what will be the unit of mass?

View answer and explanation

Correct option: C

10000 \mathrm{~kg}

Option C (

10000 \mathrm{~kg}

) is correct because force = mass * acceleration, so mass = force / acceleration. Converting the units, we get mass = (1000 N) / (1000 m / (100 s)²) = 10000 kg.

Solution approach

Correct Answer + Core Reason: $10000 \mathrm{~kg}$ . Mass is force divided by acceleration.

Step-by-Step Solution:

Step 1: Recall the formula: $F = ma$ , where F is force, m is mass, and a is acceleration.

Step 2: Solve for mass: $m = \frac{F}{a}$

Step 3: Determine the unit of force: The unit of force is kN, so $F = 1 \text{ kN} = 1000 \text{ N}$

Step 4: Determine the unit of length: The unit of length is km, so $L = 1 \text{ km} = 1000 \text{ m}$

Step 5: Determine the unit of time: The unit of time is 100 s, so $t = 100 \text{ s}$

Step 6: Calculate the acceleration: $a = \frac{L}{t^2} = \frac{1000 \text{ m}}{(100 \text{ s})^2} = \frac{1000}{10000} \text{ m/s}^2 = 0.1 \text{ m/s}^2$

Step 7: Calculate the mass: $m = \frac{F}{a} = \frac{1000 \text{ N}}{0.1 \text{ m/s}^2} = 10000 \text{ kg}$

Why other options are wrong:
- Option A: $1000 \mathrm{~kg}$ is incorrect.
- Option B: $1 \mathrm{~kg}$ is incorrect.
- Option D: $100 \mathrm{~kg}$ is incorrect.

Key Concepts:
- Newton's second law of motion
- Units and dimensions

Memory Aids:
- $F = ma$

Quick Tips:
- Always convert units to SI units before performing calculations.

99 Easy

99 of 128

Understanding +4 | -1

Which of the following is not equal to watt

View answer and explanation

Correct option: D

ampere/volt

Option D (ampere/volt) is correct because watt is a unit of power and can be expressed as joule/second, ampere × volt, or (ampere)² × ohm, but not ampere/volt.

Solution approach

Correct Answer + Core Reason: ampere/volt. Watt is a unit of power, and ampere/volt is not a valid expression for it.

Step-by-Step Solution:

Step 1: Recall the definition of watt (W) as a unit of power.

Step 2: Recall the formulas for power: $P = VI = I^2R = \frac{V^2}{R}$ , where V is voltage, I is current, and R is resistance.

Step 3: Analyze each option:
- A. joule/second: This is the fundamental definition of power (1 watt = 1 joule/second).
- B. ampere $\times$ volt: This is equivalent to $VI$ , which is power.
- C. ${(ampere)^2} \times ohm$ : This is equivalent to $I^2R$ , which is power.
- D. ampere/volt: This is equivalent to $I/V = 1/R$ , which is the inverse of resistance (conductance), not power.

Why other options are wrong:
- Option A: joule/second is incorrect because it is equal to watt.
- Option B: ampere $\times$ volt is incorrect because it is equal to watt.
- Option C: ${(ampere)^2} \times ohm$ is incorrect because it is equal to watt.
- Option D: ampere/volt is the reciprocal of resistance and is not equal to watt.

Key Concepts:
- Power (Watt)
- Ohm's Law
- Electrical Units

Memory Aids:
- Remember the power formulas: $P = VI = I^2R = \frac{V^2}{R}$

Quick Tips:
- Know the relationships between power, voltage, current, and resistance.

100 Easy

100 of 128

Analyzing +4 | -1

Assertion: The percentage change in time period is

1.5 \%

, if the length of simple pendulum increases by

3 \%

.
Reason: Time period is directly proportional to length of pendulum.

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 reason is false; the time period is proportional to the square root of the length, not the length itself.

Solution approach

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

Step-by-Step Solution:

Step 1: Recall the formula for the time period of a simple pendulum: $T = 2\pi \sqrt{\frac{l}{g}}$ .

Step 2: Calculate the percentage change in time period: $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l}$ .

Step 3: Substitute the given percentage change in length: $\frac{\Delta l}{l} = 3\% = 0.03$ .

Step 4: Calculate the percentage change in time period: $\frac{\Delta T}{T} = \frac{1}{2} \times 0.03 = 0.015 = 1.5\%$ . Therefore, the assertion is true.

Step 5: Analyze the reason: The time period is proportional to the square root of the length, not the length itself. Therefore, the reason is false.

Why other options are wrong:
- Option A: If both Assertion and Reason are true and Reason is the correct explanation of Assertion is incorrect because the reason is false.
- Option B: If both Assertion and Reason are true, but Reason is not the correct explanation of Assertion is incorrect because the reason is false.
- Option D: If both Assertion and Reason are false is incorrect because the assertion is true.

Key Concepts:
- Simple pendulum
- Time period
- Percentage error

Memory Aids:
- Remember the formula for the time period of a simple pendulum: $T = 2\pi \sqrt{\frac{l}{g}}$ .

Quick Tips:
- The time period is proportional to the square root of the length.

101 Easy

101 of 128

Application +4 | -1

The area of a rectangular field (in

{m^2}

) of length 55.3 m and breadth 25 m after rounding off the value for correct significant digits is :

View answer and explanation

Correct option: A

14 \times {10^2}

Option A (

14 \times {10^2}

) is correct because the area should be rounded to the least number of significant figures present in the given values, which is two.

Solution approach

Correct Answer + Core Reason: $14 \times {10^2}$ . The area is rounded to two significant figures.

Step-by-Step Solution:

Step 1: Calculate the area: Area = length x breadth = 55.3 m x 25 m = 1382.5 ${m^2}$

Step 2: Identify the number of significant figures in the given values: 55.3 has 3 significant figures, and 25 has 2 significant figures.

Step 3: Round the calculated area to the least number of significant figures, which is 2. So, 1382.5 is rounded to 1400.

Step 4: Express the rounded area in scientific notation: 1400 = $14 \times {10^2}$

Why other options are wrong:
- Option B: $138 \times {10^1}$ : This has 3 significant figures, which is more than the least number of significant figures in the given values.
- Option C: 1382: This has 4 significant figures, which is more than the least number of significant figures in the given values.
- Option D: 1382.5: This has 5 significant figures, which is more than the least number of significant figures in the given values.

Key Concepts:
- Significant figures
- Rounding rules
- Area calculation

Memory Aids:
- When multiplying or dividing, the result should have the same number of significant figures as the quantity with the fewest significant figures.

Quick Tips:
- Always identify the least number of significant figures in the given values before performing calculations.

102 Easy

102 of 128

Application +4 | -1

The dimensions of K in the equation

W = \frac{1}{2}\;K{x^2}

is

View answer and explanation

Correct option: A

{M^1}{L^0}{T^{ - 2}}

Option A (M¹L⁰T⁻²) is correct because K represents the spring constant, and its dimensional formula is derived from the work done equation.

Solution approach

Correct Answer + Core Reason: The dimensions of K in the equation $W = \frac{1}{2}Kx^2$ is ${M^1}{L^0}{T^{-2}}$ .

Step-by-Step Solution:

Step 1: Given equation: $W = \frac{1}{2}Kx^2$ .

Step 2: Rearrange for K: $K = \frac{2W}{x^2}$ .

Step 3: Dimensional formula for Work (W): $[W] = [ML^2T^{-2}]$ .

Step 4: Dimensional formula for Displacement (x): $[x] = [L]$ .

Step 5: Dimensional formula for K: $[K] = \frac{[ML^2T^{-2}]}{[L^2]} = [MT^{-2}] = [M^1L^0T^{-2}]$ .

Why other options are wrong:
- Option B: ${M^0}{L^1}{T^{-1}}$ : Incorrect dimensions.
- Option C: ${M^1}{L^1}{T^{-2}}$ : Incorrect dimensions.
- Option D: ${M^1}{L^0}{T^{-1}}$ : Incorrect dimensions.

Key Concepts:
- Dimensional analysis
- Work done by a spring

Memory Aids:
- Work = (1/2)Kx²

Quick Tips:
- Remember the formula for work done by a spring.

103 Easy

103 of 128

Analyzing +4 | -1

Choose the incorrect options among the following

View answer and explanation

Correct option: D

Both (2) and (3)

Option D (Both (2) and (3)) is correct because a physical quantity with dimensions must have a unit, and a physical quantity having a constant value can have dimensions.

Solution approach

Correct Answer + Core Reason: Both (2) and (3). A physical quantity with dimensions must have a unit, and a physical quantity having a constant value can have dimensions.

Step-by-Step Solution:

Step 1: Analyze each statement:\nStatement 1: A physical quantity with units may not have dimensions. This is incorrect. If a quantity has units, it must have dimensions.\nStatement 2: A physical quantity with dimensions may not have a unit. This is incorrect. If a quantity has dimensions, it must have a unit.\nStatement 3: A physical quantity having constant value is dimensionless. This is incorrect. A constant value can have dimensions (e.g., gravitational constant).

Step 2: Identify the incorrect options.\nOptions 2 and 3 are incorrect.

Why other options are wrong:
- Option A: Statement 1 is not incorrect.
- Option B: Statement 2 is incorrect, but statement 3 is also incorrect.
- Option C: Statement 3 is incorrect, but statement 2 is also incorrect.

Key Concepts:
- Dimensional analysis
- Units and dimensions

Memory Aids:
- Remember the relationship between units and dimensions.

Quick Tips:
- Understand the definitions of units and dimensions.
- Consider examples to verify the statements.

104 Easy

104 of 128

Understanding +4 | -1

Which of the following physical quantities has neither dimensions nor unit?

View answer and explanation

Correct option: C

Coefficient of friction

Option C (Coefficient of friction) is correct because it is a dimensionless ratio of two forces and therefore has no unit.

Solution approach

Correct Answer + Core Reason: Coefficient of friction. It is a dimensionless ratio.

Step-by-Step Solution:

Step 1: Analyze the dimensions and units of each quantity.

Step 2: Angle is defined as the ratio of arc length to radius, both having dimensions of length (L), so it is dimensionless but has a unit (radian or degree).

Step 3: Luminous intensity has a unit (candela) and dimension.

Step 4: Current has a unit (ampere) and dimension.

Step 5: Coefficient of friction is the ratio of frictional force to normal force, both having the same dimensions (MLT^-2), making it dimensionless and unitless.

Why other options are wrong:
- Option A: Angle has a unit (radian or degree), even though it is dimensionless.
- Option B: Luminous intensity has both dimensions and a unit (candela).
- Option D: Current has both dimensions and a unit (ampere).

Key Concepts:
- Dimensions and Units
- Dimensionless Quantities
- Coefficient of Friction

Memory Aids:
- Remember that the coefficient of friction is a ratio of two forces.

Quick Tips:
- Check if the quantity is a ratio of similar quantities; if so, it's likely dimensionless.

105 Easy

105 of 128

Analyzing +4 | -1

The dimensions of

(\text { velocity })^2 \div

radius are the same as that of

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the dimensions of (velocity)² / radius do not match any of the given options.

Solution approach

Correct Answer + Core Reason: None of these; The dimensions of (velocity) $^2$ / radius do not match any of the given options.

Step-by-Step Solution:

Step 1: Determine the dimensions of (velocity) $^2$ / radius.
- Velocity (v) = $LT^{-1}$
- (Velocity) $^2$ = $(LT^{-1})^2 = L^2T^{-2}$
- Radius (r) = $L$
- (Velocity) $^2$ / Radius = $L^2T^{-2} / L = LT^{-2}$

Step 2: Determine the dimensions of the given options.
- Planck's constant (h): $ML^2T^{-1}$
- Gravitational constant (G): $M^{-1}L^3T^{-2}$
- Dielectric constant ( $\epsilon_0$ ): $M^{-1}L^{-3}T^4A^2$

Step 3: Compare the dimensions.
- $LT^{-2}$ does not match any of the options.

Why other options are wrong:
- Option A: Planck's constant: Incorrect dimensions.
- Option B: Gravitational constant: Incorrect dimensions.
- Option C: Dielectric constant: Incorrect dimensions.

Key Concepts:
- Dimensional analysis
- Dimensions of velocity, radius, Planck's constant, gravitational constant, and dielectric constant

Memory Aids:
- Remember the basic dimensions of mass (M), length (L), and time (T).

Quick Tips:
- Always write down the dimensions of each quantity before comparing them.

106 Easy

106 of 128

Understanding +4 | -1

Which of the following system of units is not based on units of mass, length and time alone

View answer and explanation

Correct option: A

SI

Option A (SI) is correct because the SI system includes units beyond mass, length, and time, such as electric current (ampere) and temperature (kelvin).

Solution approach

Correct Answer + Core Reason: SI. The SI system includes units beyond mass, length, and time.

Step-by-Step Solution:

Step 1: Identify the base units for each system.

Step 2: MKS (meter, kilogram, second) is based on length, mass, and time.

Step 3: FPS (foot, pound, second) is based on length, mass, and time.

Step 4: CGS (centimeter, gram, second) is based on length, mass, and time.

Step 5: SI (International System of Units) includes base units like ampere (current), kelvin (temperature), and candela (luminous intensity) in addition to meter, kilogram, and second.

Why other options are wrong:
- Option B: MKS is based only on mass, length, and time.
- Option C: FPS is based only on mass, length, and time.
- Option D: CGS is based only on mass, length, and time.

Key Concepts:
- Systems of Units (SI, MKS, CGS, FPS)
- Base Units

Memory Aids:
- Remember the base units of each system.

Quick Tips:
- Know the fundamental units of each system to quickly identify the correct answer.

107 Easy

107 of 128

Understanding +4 | -1

A physical quantity is measured and its value is found to be nu where where n= numerical value and u= unit. Then which of the following relations is true

View answer and explanation

Correct option: D

n \propto \frac{1}{u}

Option D (

n \propto \frac{1}{u}

) is correct because the product of numerical value and unit is constant for a given physical quantity.

Solution approach

Correct Answer + Core Reason: $n \propto \frac{1}{u}$ is correct because the product of the numerical value (n) and the unit (u) of a physical quantity is constant.

Step-by-Step Solution:

Step 1: Understand the relationship between numerical value and unit.

Step 2: For a given physical quantity, the product of its numerical value (n) and its unit (u) is constant. Therefore, $nu = constant$ .

Step 3: This implies that $n \propto \frac{1}{u}$ .

Why other options are wrong:
- Option A: $n \propto {u^2}$ is incorrect.
- Option B: $n\; \propto u$ is incorrect.
- Option C: $n \propto \sqrt u$ is incorrect.

Key Concepts:
- Units and dimensions

Memory Aids:
- Remember that the product of numerical value and unit is constant.

Quick Tips:
- If the unit increases, the numerical value decreases proportionally, and vice versa.

108 Easy

108 of 128

Analyzing +4 | -1

Which of the following sets of quantities have same dimensional formula?

View answer and explanation

Correct option: D

Work, energy and torque

Option D (Work, energy and torque) is correct because all three quantities have the same dimensional formula of [ML²T⁻²].

Solution approach

Correct Answer + Core Reason: Work, energy and torque. All three have the same dimensional formula.

Step-by-Step Solution:

Step 1: Write down the dimensions of each quantity:\nFrequency = $[T^{-1}]$ \nAngular frequency = $[T^{-1}]$ \nAngular momentum = $[ML^2T^{-1}]$ \nSurface tension = $[MT^{-2}]$ \nStress = $[ML^{-1}T^{-2}]$ \nSpring constant = $[MT^{-2}]$ \nAcceleration = $[LT^{-2}]$ \nMomentum = $[MLT^{-1}]$ \nRetardation = $[LT^{-2}]$ \nWork = $[ML^2T^{-2}]$ \nEnergy = $[ML^2T^{-2}]$ \nTorque = $[ML^2T^{-2}]$

Step 2: Compare the dimensions of each set:\nOption A: Frequency and angular frequency have the same dimensions, but angular momentum has different dimensions.\nOption B: Surface tension and spring constant have the same dimensions, but stress has different dimensions.\nOption C: Acceleration and retardation have the same dimensions, but momentum has different dimensions.\nOption D: Work, energy, and torque have the same dimensions.

Why other options are wrong:
- Option A: Angular momentum has different dimensions than frequency and angular frequency.
- Option B: Stress has different dimensions than surface tension and spring constant.
- Option C: Momentum has different dimensions than acceleration and retardation.

Key Concepts:
- Dimensional analysis
- Frequency, angular frequency, angular momentum, surface tension, stress, spring constant, acceleration, momentum, retardation, work, energy, and torque

Memory Aids:
- Remember the definitions of each quantity.

Quick Tips:
- Always write down the dimensions of each quantity before comparing.
- Pay attention to the definitions of the quantities.

109 Easy

109 of 128

Understanding +4 | -1

The unit of L/R is (where L=inductance and R= Resistance)

View answer and explanation

Correct option: A

Sec

Option A (Sec) is correct because the ratio L/R has the dimensions of time.

Solution approach

Correct Answer + Core Reason: Sec. The ratio L/R has the dimensions of time.

Step-by-Step Solution:

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

Step 2: Determine the units of inductance. Inductance (L) is measured in Henries (H), which is equivalent to Volt-seconds per Ampere (V·s/A).

Step 3: Determine the units of resistance. Resistance (R) is measured in Ohms ( $\Omega$ ), which is equivalent to Volts per Ampere (V/A).

Step 4: Determine the units of L/R. Since $\tau = \frac{L}{R}$ , the unit of L/R is $\frac{{V \cdot s/A}}{{V/A}} = s$ , which is seconds.

Why other options are wrong:
- Option B: $Se{c^{ - 1}}$ is the unit of frequency, not L/R.
- Option C: Volt is the unit of electric potential, not L/R.
- Option D: Ampere is the unit of electric current, not L/R.

Key Concepts:
- Inductance and resistance
- Inductive time constant
- Units of physical quantities

Memory Aids:
- Remember that the time constant in an RL circuit is L/R.

Quick Tips:
- Know the units of inductance and resistance.

110 Easy

110 of 128

Application +4 | -1

The damping force on an oscillator is directly proportional to the velocity. The unit of the constant of proportionality is

View answer and explanation

Correct option: C

\mathrm{kgs}^{-1}

Option C (

\mathrm{kgs}^{-1}

) is correct because the damping force is proportional to velocity (F = bv), so the constant of proportionality (b) has units of force/velocity.

Solution approach

Correct Answer + Core Reason: $\mathrm{kgs}^{-1}$ . The damping force is proportional to velocity.

Step-by-Step Solution:

Step 1: Express the damping force equation: F = bv, where F is the damping force, b is the constant of proportionality, and v is the velocity.

Step 2: Solve for b: b = F/v

Step 3: Determine the units of force (F): kg m s^-2

Step 4: Determine the units of velocity (v): m s^-1

Step 5: Calculate the units of b: b = (kg m s^-2) / (m s^-1) = kg s^-1

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

Key Concepts:
- Damping Force
- Units of Measurement

Memory Aids:
- Remember the damping force equation: F = bv

Quick Tips:
- Relate the constant of proportionality to force and velocity.

111 Easy

111 of 128

Understanding +4 | -1

Zero error in an instrument is a

View answer and explanation

Correct option: B

Systematic error

Option B (Systematic error) is correct because a zero error is a consistent, repeatable error that affects all measurements in the same way.

Solution approach

Correct Answer + Core Reason: Systematic error. Zero error is a consistent error that shifts all measurements in the same direction.

Step-by-Step Solution:

Step 1: Define zero error. Zero error is a type of error where an instrument gives a non-zero reading when it should read zero.

Step 2: Classify the type of error.
- Random error: Unpredictable fluctuations.
- Systematic error: Consistent, repeatable error.
- Least count error: Error due to the resolution of the instrument.
- Personal error: Error due to the observer's bias or carelessness.

Step 3: Zero error is a consistent error that affects all measurements in the same way. Therefore, it is a systematic error.

Why other options are wrong:
- Option A (Random error): Random errors are unpredictable, while zero error is consistent.
- Option C (Least count error): Least count error is related to the instrument's resolution, not a consistent offset.
- Option D (Personal error): Personal errors are due to the observer, while zero error is a characteristic of the instrument itself.

Key Concepts:
- Systematic error
- Random error
- Zero error

Memory Aids:
- Systematic errors are predictable and can be corrected.

Quick Tips:
- Identify and correct for zero errors before taking measurements.

112 Easy

112 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 universal gas constant is related to energy per mole per unit temperature.

Solution approach

Correct Answer + Core Reason: $\left[ {M{L^2}{T^{ - 2}}{\theta ^{ - 1}}} \right]$ . The universal gas constant is related to energy per mole per unit temperature.

Step-by-Step Solution:

Step 1: Recall the ideal gas law. The ideal gas law is given by $PV = nRT$ , 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: Solve for R. $R = \frac{{PV}}{{nT}}$ .

Step 3: Determine the dimensions of pressure. Pressure is force per unit area, i.e., $P = \frac{F}{A}$ . The dimension of force is $\left[ {ML{T^{ - 2}}} \right]$ , and the dimension of area is $\left[ {{L^2}} \right]$ . Therefore, the dimension of pressure is $\frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$ .

Step 4: Determine the dimensions of volume. The dimension of volume is $\left[ {{L^3}} \right]$ .

Step 5: Determine the dimensions of the number of moles. The number of moles is dimensionless.

Step 6: Determine the dimensions of temperature. The dimension of temperature is $\left[ \theta \right]$ .

Step 7: Determine the dimensions of R. Since $R = \frac{{PV}}{{nT}}$ , the dimension of R is $\frac{{\left[ {M{L^{ - 1}}{T^{ - 2}}} \right] \times \left[ {{L^3}} \right]}}{{\left[ 1 \right] \times \left[ \theta \right]}} = \left[ {M{L^2}{T^{ - 2}}{\theta ^{ - 1}}} \right]$ .

Why other options are wrong:
- Option B: This option is incorrect because the dimensions of mass, length, and temperature are incorrect.
- Option C: This option is incorrect because the dimensions of length and time are incorrect.
- Option D: This option is incorrect because it does not provide the correct dimensions.

Key Concepts:
- Dimensions of physical quantities
- Ideal gas law

Memory Aids:
- Remember the ideal gas law: PV = nRT.

Quick Tips:
- Start with the ideal gas law and derive the dimensions from there.

113 Easy

113 of 128

Remembering +4 | -1

The unit of surface tension in SI system is

View answer and explanation

Correct option: B

Newton /m

Option B (Newton /m) is correct because surface tension is defined as force per unit length, and Newton/m is the SI unit for that.

Solution approach

Correct Answer + Core Reason: Newton /m. Surface tension is force per unit length.

Step-by-Step Solution:

Step 1: Define surface tension:\nSurface tension ( $\sigma$ ) is defined as the force (F) acting per unit length (L) perpendicular to the line drawn on the surface of the liquid.
$\sigma = \frac{F}{L}$

Step 2: Determine the SI unit of surface tension:\nSI unit of force (F) is Newton (N).\nSI unit of length (L) is meter (m).\nTherefore, the SI unit of surface tension is Newton/meter (N/m).

Why other options are wrong:
- Option A: $Dyne/c{m^2}$ : This is not the unit of surface tension. It is the unit of pressure.
- Option C: Dyne/cm: This is the CGS unit of surface tension, not the SI unit.
- Option D: $Newton/{m^2}$ : This is the unit of pressure, not surface tension.

Key Concepts:
- Surface tension
- Units of measurement

Memory Aids:
- Remember the definition of surface tension as force per unit length.

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

114 Easy

114 of 128

Application +4 | -1

The dimensional formula for magnetic flux is

View answer and explanation

Correct option: A

\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]

Option A

\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]

is correct because magnetic flux is defined as the product of magnetic field and area, and its dimensions can be derived from the Lorentz force formula.

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]$ . Magnetic flux is the product of magnetic field and area.

Step-by-Step Solution:

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

Step 2: Recall the Lorentz force formula: $F = qvB$ , where $F$ is the force, $q$ is the charge, $v$ is the velocity, and $B$ is the magnetic field.

Step 3: Find the dimensions of the magnetic field $B$ from the Lorentz force formula: $B = \frac{F}{qv}$ . Therefore, $[B] = \frac{[F]}{[q][v]} = \frac{[MLT^{-2}]}{[AT][LT^{-1}]} = [MT^{-2}A^{-1}]$ .

Step 4: Find the dimensions of the area $A$ : $[A] = [L^2]$ .

Step 5: Find the dimensions of the magnetic flux $\phi$ : $[\phi] = [B][A] = [MT^{-2}A^{-1}][L^2] = [ML^2T^{-2}A^{-1}]$ .

Why other options are wrong:
- Option B: $\left[\mathrm{ML}^3 \mathrm{T}^{-2} \mathrm{A}^{-2}\right]$ : This is incorrect as it doesn't match the derived dimensions of magnetic flux.
- Option C: $\left[\mathrm{M}^0 \mathrm{L}^{-2} \mathrm{T}^{-2} \mathrm{A}^{-2}\right]$ : This is incorrect as it doesn't match the derived dimensions of magnetic flux.
- Option D: $\left[\mathrm{ML}^2 \mathrm{T}^{-1} \mathrm{A}^2\right]$ : This is incorrect as it doesn't match the derived dimensions of magnetic flux.

Key Concepts:
- Magnetic flux formula: $\phi = B \cdot A$
- Lorentz force formula: $F = qvB$
- Dimensional analysis.

Memory Aids:
- Remember the Lorentz force formula to derive the dimensions of the magnetic field.

Quick Tips:
- Always start with a known formula to derive the dimensions of a physical quantity.

115 Easy

115 of 128

Analyzing +4 | -1

Statement-I: If a physical quantity has a unit, it must not be dimensionless.
Statement-II: No physical quantity exists which has dimension but no unit.

View answer and explanation

Correct option: C

Statement-I is false and statement-II is true

Statement-I is false and statement-II is true because dimensionless quantities can have units (e.g., radian), and there are no known physical quantities with dimensions but no unit.

Solution approach

Correct Answer + Core Reason: Statement-I is false and statement-II is true. Dimensionless quantities can have units, and no physical quantity exists which has dimension but no unit.

Step-by-Step Solution:

Step 1: Analyze Statement-I: If a physical quantity has a unit, it must not be dimensionless.
This statement is false. For example, the radian is a unit for measuring angles, which are dimensionless quantities (ratio of two lengths).

Step 2: Analyze Statement-II: No physical quantity exists which has dimension but no unit.
This statement is generally considered true. If a quantity has dimensions, it must have a unit to quantify it.

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:
- Dimensions and units
- Dimensionless quantities
- Radians

Memory Aids:
- Remember that angles are dimensionless but have units (radians or degrees).

Quick Tips:
- Consider examples when analyzing statements about dimensions and units.

116 Easy

116 of 128

Understanding +4 | -1

The pair having the same dimensions is

View answer and explanation

Correct option: B

Work, torque

Option B (Work, torque) is correct because both work and torque have the same dimensions of [ML²T⁻²].

Solution approach

Correct Answer + Core Reason: Work, torque. Both have dimensions of [ML²T⁻²].

Step-by-Step Solution:

Step 1: Determine the dimensions of angular momentum (L). Angular momentum = I $\omega$ = [M][L]² [T]^-1.

Step 2: Determine the dimensions of work (W). Work = Force x Distance = [M][L][T]^-2 [L] = [M][L]²[T]^-2.

Step 3: Determine the dimensions of torque ( $\tau$ ). Torque = Force x Distance = [M][L][T]^-2 [L] = [M][L]²[T]^-2.

Step 4: Determine the dimensions of potential energy (PE). Potential energy = mgh = [M][L][T]^-2 [L] = [M][L]²[T]^-2.

Step 5: Determine the dimensions of linear momentum (p). Linear momentum = mv = [M][L][T]^-1.

Step 6: Determine the dimensions of kinetic energy (KE). Kinetic energy = 1/2 mv² = [M][L]²[T]^-2.

Step 7: Determine the dimensions of velocity (v). Velocity = [L][T]^-1.

Step 8: Compare the dimensions of each pair.

Why other options are wrong:
- Option A: Angular momentum and work have different dimensions.
- Option C: Potential energy and linear momentum have different dimensions.
- Option D: Kinetic energy and velocity have different dimensions.

Key Concepts:
- Angular momentum
- Work
- Torque
- Potential energy
- Linear momentum
- Kinetic energy
- Velocity
- Dimensional analysis

Memory Aids:
- Work and Torque have the same dimensions.

Quick Tips:
- Remember the formulas for each quantity and derive their dimensions.

117 Easy

117 of 128

Remembering +4 | -1

Zero error is a/an

View answer and explanation

Correct option: A

Instrumental error

Option A (Instrumental error) is correct because zero error arises from imperfections or miscalibration within the measuring instrument itself.

Solution approach

Correct Answer + Core Reason: Instrumental error. Zero error is a systematic error inherent to the measuring instrument.

Step-by-Step Solution:

Step 1: Define Zero Error: Zero error is the reading shown by a measuring instrument when the true value being measured is zero.

Step 2: Classify Error Types:
- Instrumental Error: Arises from the instrument's design, calibration, or manufacturing defects.
- Random Error: Unpredictable fluctuations in measurements.
- Personal Error: Errors due to the observer's mistakes or biases.
- Parallax Error: Error due to incorrect positioning of the observer's eye.

Step 3: Relate Zero Error to Error Types: Since zero error is a consistent offset in the instrument's readings, it is directly related to the instrument itself.

Step 4: Conclusion: Zero error is classified as an instrumental error.

Why other options are wrong:
- Option B: Random error is unpredictable and fluctuates around the true value, unlike zero error which is a consistent offset.
- Option C: Personal error is due to the observer's mistakes, not the instrument itself.
- Option D: Parallax error is due to the observer's viewing angle, not the instrument's inherent error.

Key Concepts:
- Types of Errors in Measurement (Instrumental, Random, Personal, Parallax)
- Definition of Zero Error

Memory Aids:
- Instrumental errors are inherent to the instrument.

Quick Tips:
- Always check for zero error before taking measurements.
- Correct for zero error by subtracting it from all readings.

118 Easy

118 of 128

Application +4 | -1

The position of the particle moving along

Y

-axis is given as

y=A t^2-B t^3

, where

y

is measured in metre and

t

in second. Then, the dimensions of

B

are

View answer and explanation

Correct option: C

\left[\mathrm{LT}^{-3}\right]

Option C

\left[\mathrm{LT}^{-3}\right]

is correct because in the equation

y=A t^2-B t^3

, each term must have the same dimensions as

y

, which is length [L].

Solution approach

Correct Answer + Core Reason: $\left[\mathrm{LT}^{-3}\right]$ . Each term in the equation must have the same dimensions.

Step-by-Step Solution:

Step 1: The equation is $y = A t^2 - B t^3$ .

Step 2: The dimensions of $y$ are $[L]$ (length).

Step 3: The dimensions of $A t^2$ must be $[L]$ . Therefore, $[A][T^2] = [L]$ , so $[A] = [LT^{-2}]$ .

Step 4: The dimensions of $B t^3$ must be $[L]$ . Therefore, $[B][T^3] = [L]$ , so $[B] = [LT^{-3}]$ .

Step 5: Thus, the dimensions of $B$ are $[LT^{-3}]$ .

Why other options are wrong:
- Option A: $\left[\mathrm{LT}^{-2}\right]$ : This is the dimension of A, not B.
- Option B: $\left[\mathrm{LT}^{-1}\right]$ : This is incorrect as it doesn't satisfy the dimensional homogeneity of the equation.
- Option D: $\left[\mathrm{MLT}^{-2}\right]$ : This is incorrect as it doesn't satisfy the dimensional homogeneity of the equation.

Key Concepts:
- Principle of dimensional homogeneity.
- Dimensional analysis.

Memory Aids:
- Remember that each term in a physical equation must have the same dimensions.

Quick Tips:
- Always equate the dimensions of each term in an equation to ensure dimensional correctness.

119 Easy

119 of 128

Application +4 | -1

A geometry ruler has 20 divisions between 5 cm. Its least count is

View answer and explanation

Correct option: A

0.25 cm

Option A (0.25 cm) is correct because the least count is calculated by dividing the length between the divisions by the number of divisions.

Solution approach

Correct Answer + Core Reason: 0.25 cm. The least count is calculated by dividing the length between the divisions by the number of divisions.

Step-by-Step Solution:

Step 1: Identify the given information.\nNumber of divisions = 20\nLength between divisions = 5 cm

Step 2: Calculate the length of each division.\nLength of each division = $\frac{5 \mathrm{~cm}}{20} = 0.25 \mathrm{~cm}$

Step 3: The least count is the length of each division.\nLeast count = 0.25 cm

Why other options are wrong:
- Option B: 2.5 cm is incorrect because it's a miscalculation of the least count.
- Option C: 0.25 mm is incorrect because it's a miscalculation of the least count and uses the wrong units.
- Option D: 0.025 mm is incorrect because it's a miscalculation of the least count and uses the wrong units.

Key Concepts:
- Least count

Memory Aids:
- Least count = (Total length) / (Number of divisions)

Quick Tips:
- Pay attention to the units and ensure they are consistent.

120 Easy

120 of 128

Application +4 | -1

The value of

{(17.5)^2}

is

View answer and explanation

Correct option: A

306

Option A (306) is correct because the result should be rounded to the least precise value, which has no decimal places.

Solution approach

Correct Answer + Core Reason: 306. The result should be rounded to the least precise value.

Step-by-Step Solution:

Step 1: Calculate the square: ${(17.5)^2} = 17.5 \times 17.5 = 306.25$

Step 2: Consider the significant figures. 17.5 has three significant figures and one decimal place. However, the question implies rounding to the nearest whole number.

Step 3: Round the result to the nearest whole number: 306.25 rounds to 306.

Why other options are wrong:
- Option B: 306.25: This is the direct result of the calculation, but it's not rounded to the appropriate number of significant figures or decimal places.
- Option C: 306.2: This is rounded to one decimal place, but the question implies rounding to the nearest whole number.
- Option D: 306.3: This is rounded incorrectly to one decimal place.

Key Concepts:
- Significant figures
- Rounding rules

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

Quick Tips:
- Pay attention to the context of the problem to determine the appropriate rounding rules.

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