UNITS & MEASUREMENTS MCQs for NEET UG PHYSICS — Easy, Page 2

31 Easy

31 of 128

Analyzing +4 | -1

If

A, B

and

C

are physical quantities, having different dimensions, which of the following combination can never be meaningful quantity?

View answer and explanation

Correct option: A

(A-B) /C

Option A ((A-B)/C) is correct because for (A-B) to be meaningful, A and B must have the same dimensions, but the question states that A, B, and C have different dimensions.

Solution approach

Correct Answer + Core Reason: $(A-B) /C$ ; For subtraction to be meaningful, A and B must have the same dimensions, which contradicts the problem statement.

Step-by-Step Solution:

Step 1: Analyze the condition for meaningful physical quantities.
- For addition or subtraction, physical quantities must have the same dimensions.
- For multiplication or division, physical quantities can have different dimensions.

Step 2: Evaluate each option.
- A. $(A-B) /C$ : For (A-B) to be meaningful, A and B must have the same dimensions. Since A, B, and C have different dimensions, this combination cannot be meaningful.
- B. $A B-C$ : This can be meaningful if AB and C have the same dimensions.
- C. $A B / C$ : This can be meaningful as A, B, and C can have different dimensions.
- D. $A B-C^2$ : This can be meaningful if AB and C $^2$ have the same dimensions.

Why other options are wrong:
- Option B: $A B-C$ : This can be meaningful if AB and C have the same dimensions.
- Option C: $A B / C$ : This can be meaningful as A, B, and C can have different dimensions.
- Option D: $A B-C^2$ : This can be meaningful if AB and C $^2$ have the same dimensions.

Key Concepts:
- Dimensional analysis
- Principle of homogeneity of dimensions

Memory Aids:
- Remember that only quantities with the same dimensions can be added or subtracted.

Quick Tips:
- Always check if the quantities being added or subtracted have the same dimensions.

32 Easy

32 of 128

Analyzing +4 | -1

Which of the following pairs does not have similar dimensions

View answer and explanation

Correct option: C

Tension and surface tension

Option C (Tension and surface tension) is correct because tension is a force and surface tension is force per unit length, hence they have different dimensions.

Solution approach

Correct Answer + Core Reason: Tension and surface tension. Tension is a force, while surface tension is force per unit length, so their dimensions differ.

Step-by-Step Solution:

Step 1: Write down the dimensions of each quantity:\nStress = $[ML^{-1}T^{-2}]$ \nPressure = $[ML^{-1}T^{-2}]$ \nAngle = $[M^0L^0T^0]$ \nStrain = $[M^0L^0T^0]$ \nTension = $[MLT^{-2}]$ \nSurface tension = $[MT^{-2}]$ \nPlanck's constant = $[ML^2T^{-1}]$ \nAngular momentum = $[ML^2T^{-1}]$

Step 2: Compare the dimensions of each pair:\nOption A: Stress and pressure have the same dimensions.\nOption B: Angle and strain have the same dimensions (dimensionless).\nOption C: Tension and surface tension have different dimensions.\nOption D: Planck's constant and angular momentum have the same dimensions.

Why other options are wrong:
- Option A: Stress and pressure both have dimensions of force per unit area, hence the same dimensions.
- Option B: Angle and strain are both dimensionless quantities.
- Option D: Planck's constant and angular momentum both have dimensions of $[ML^2T^{-1}]$ .

Key Concepts:
- Dimensional analysis
- Stress, pressure, angle, strain, tension, surface tension, Planck's constant, and angular momentum

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.

33 Easy

33 of 128

Application +4 | -1

If the measurement of volume of solid sphere using the formula

V = \frac{4}{3}\pi {r^3}

if the error committed in the measurement of the radius r is 1.5%, the percentage error in the volume measurement is

View answer and explanation

Correct option: C

0.045

Option C (0.045) is correct because the percentage error in volume is three times the percentage error in the radius.

Solution approach

Correct Answer + Core Reason: 0.045. The percentage error in volume is three times the percentage error in the radius.

Step-by-Step Solution:

Step 1: Volume Formula: $V = \frac{4}{3} \pi r^3$

Step 2: Percentage Error Formula: $\frac{{\Delta V}}{V} \times 100 = 3 \times \frac{{\Delta r}}{r} \times 100$

Step 3: Substitute the Given Value: $\frac{{\Delta V}}{V} \times 100 = 3 \times 1.5\% = 4.5\%$

Step 4: Convert Percentage to Decimal: 4.5% = 0.045

Why other options are wrong:
- Option A: 0.03: This is incorrect; it's likely the error in radius was not multiplied by 3.
- Option B: 0.015: This is incorrect; it's the error in the radius, not the volume.
- Option D: 0.0075: This is incorrect; it's likely the error in radius was divided by 2 instead of multiplied by 3.

Key Concepts:
- Percentage Error Propagation
- Volume of a Sphere

Memory Aids:
- If $V = r^n$ , then $\frac{{\Delta V}}{V} = n \frac{{\Delta r}}{r}$

Quick Tips:
- When a quantity is raised to a power, the percentage error is multiplied by that power.

34 Easy

34 of 128

Understanding +4 | -1

Dimensions of impulse are same as that of

View answer and explanation

Correct option: B

Momentum

The correct answer is Momentum because impulse is defined as the change in momentum.

Solution approach

Correct Answer + Core Reason: Momentum. Impulse is defined as the change in momentum.

Step-by-Step Solution:

Step 1: Define impulse. Impulse (J) is defined as the change in momentum ( $\Delta p$ ), i.e., $J = \Delta p$ .

Step 2: Consider other options. Force is the rate of change of momentum, energy is the capacity to do work, and acceleration is the rate of change of velocity.

Why other options are wrong:
- Option A: Force is the rate of change of momentum, not the same as impulse.
- Option C: Energy is the capacity to do work and has different dimensions than impulse.
- Option D: Acceleration is the rate of change of velocity and has different dimensions than impulse.

Key Concepts:
- Impulse and momentum

Memory Aids:
- Remember that Impulse = Change in Momentum.

Quick Tips:
- Know the definitions of impulse, momentum, force, energy, and acceleration.

35 Easy

35 of 128

Application +4 | -1

The vernier constant of Vernier callipers is

0.1 \mathrm{~mm}

and it has zero error of

(-0.05) \mathrm{cm}

. While measuring diameter of a sphere, the main scale reading is

1.7 \mathrm{~cm}

and coinciding vernier division is 5 . The corrected diameter will be

\times 10^{-2} \mathrm{~cm}

.

View answer and explanation

Correct option: A

180

Option A (180) is correct because the corrected diameter is calculated by adding the main scale reading and vernier scale reading, then correcting for the zero error.

Solution approach

Correct Answer + Core Reason: 180. The corrected diameter is calculated by adding the main scale reading and vernier scale reading, then correcting for the zero error.

Step-by-Step Solution:

Step 1: Identify the given information.\nVernier constant (VC) = 0.1 mm = 0.01 cm\nZero error = -0.05 cm\nMain scale reading (MSR) = 1.7 cm\nVernier coinciding division (VCD) = 5

Step 2: Calculate the vernier scale reading.\nVernier scale reading = VCD * VC = 5 * 0.01 cm = 0.05 cm

Step 3: Calculate the observed diameter.\nObserved diameter = MSR + Vernier scale reading = 1.7 cm + 0.05 cm = 1.75 cm

Step 4: Correct for zero error.\nCorrected diameter = Observed diameter - Zero error = 1.75 cm - (-0.05 cm) = 1.75 cm + 0.05 cm = 1.80 cm

Step 5: Express the answer in the required format.\nCorrected diameter = 1.80 cm = 180 × 10^-2 cm

Why other options are wrong:
- Option B: 0 is incorrect because it doesn't account for the main scale reading, vernier scale reading, and zero error.
- Option C: 0 is incorrect because it doesn't account for the main scale reading, vernier scale reading, and zero error.
- Option D: 0 is incorrect because it doesn't account for the main scale reading, vernier scale reading, and zero error.

Key Concepts:
- Vernier callipers
- Vernier constant
- Zero error

Memory Aids:
- Corrected reading = Observed reading - Zero error

Quick Tips:
- Always pay attention to the sign of the zero error.
- Ensure all units are consistent before performing calculations.

36 Easy

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Application +4 | -1

A torque meter is calibrated to reference standards of mass, length and time each with

5 \%

accuracy. After calibration, the measured torque with this torque meter will have net accuracy of

View answer and explanation

Correct option: B

25\%

Option B (

25\%

) is correct because the accuracy of torque is related to the product of mass, length, and time, and their errors add up.

Solution approach

Correct Answer + Core Reason: $25\%$ . The errors in mass, length, and time contribute to the overall error in torque measurement.

Step-by-Step Solution:

Step 1: Torque is related to mass, length, and time. The exact relationship isn't given, but we know torque has dimensions of $[ML^2T^{-2}]$ .

Step 2: Assume the torque is proportional to $M^a L^b T^c$ . The percentage error in torque will be $a(\% \text{{ error in M}}) + b(\% \text{{ error in L}}) + |c|(\% \text{{ error in T}})$ .

Step 3: From the dimensions, we have $a=1$ , $b=2$ , and $c=-2$ .

Step 4: Calculate the percentage error in torque: $1(5\%) + 2(5\%) + |-2|(5\%) = 5\% + 10\% + 10\% = 25\%$ .

Why other options are wrong:
- Option A: This is incorrect because it only considers the sum of the individual errors without considering the exponents in the dimensional formula.
- Option C: This is incorrect as it is an arbitrary value and doesn't relate to the propagation of errors.
- Option D: This is incorrect because it only considers the error in one quantity and ignores the others.

Key Concepts:
- Error propagation
- Dimensional analysis

Memory Aids:
- Remember that errors in multiplication and division add up.

Quick Tips:
- Use the dimensional formula to determine how the errors propagate.

37 Easy

37 of 128

Understanding +4 | -1

Dimensional formula for the plank’s constant is

View answer and explanation

Correct option: A

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

Option A (

ML^2T^{-1}

) is correct because Planck's constant relates energy and frequency (E=hf), and its dimensional formula reflects this relationship.

Solution approach

Correct Answer + Core Reason: $M{L^2}{T^{ - 1}}$ . Planck's constant relates energy and frequency.

Step-by-Step Solution:

Step 1: Recall the equation relating energy and frequency: $E = hf$ , where E is energy, h is Planck's constant, and f is frequency.

Step 2: Determine the dimensions of Energy (E): Energy has dimensions of [ML²T⁻²].

Step 3: Determine the dimensions of Frequency (f): Frequency has dimensions of [T⁻¹].

Step 4: Solve for the dimensions of Planck's constant (h): $h = \frac{E}{f}$ , so the dimensions of h are $\frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML²T⁻¹]$ .

Step 5: Final Answer: $M{L^2}{T^{ - 1}}$

Why other options are wrong:
- Option B: ${M^{ - 1}}{L^2}{T^{ - 3}}$ : Incorrect dimensions.
- Option C: ${M^{ - 1}}L{T^{ - 2}}$ : Incorrect dimensions.
- Option D: ${M^{ - 2}}{L^3}{T^{ - 1}}$ : Incorrect dimensions.

Key Concepts:
- Planck's Constant
- Energy-Frequency Relationship
- Dimensional Analysis

Memory Aids:
- E = hf (Energy equals Planck's constant times frequency).

Quick Tips:
- Remember the relationship between energy and frequency involving Planck's constant.

38 Easy

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Application +4 | -1

The dimension of

\left(\mu_0 \varepsilon_0\right)^{-1 / 2}

are

View answer and explanation

Correct option: C

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

Option C

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

is correct because

\left(\mu_0 \varepsilon_0\right)^{-1 / 2}

represents the speed of light, which has dimensions of length divided by time.

Solution approach

Correct Answer + Core Reason: $\left[L T^{-1}\right]$ . The expression represents the speed of light.

Step-by-Step Solution:

Step 1: Recognize that $\left(\mu_0 \varepsilon_0\right)^{-1 / 2}$ represents the speed of light ( $c$ ).

Step 2: Recall the dimensions of speed (or velocity). Speed is distance/time.

Step 3: Write the dimensions of speed as $\frac{[L]}{[T]} = [L T^{-1}]$ .

Step 4: Therefore, the dimensions of $\left(\mu_0 \varepsilon_0\right)^{-1 / 2}$ are $[L T^{-1}]$ .

Why other options are wrong:
- Option A: $\left[L^{1 / 2} T^{-1 / 2}\right]$ : This is incorrect as it does not represent the dimensions of speed.
- Option B: $\left[L^{-1} T\right]$ : This is incorrect as it is the inverse of speed.
- Option D: $\left[L^{1 / 2} T^{1 / 2}\right]$ : This is incorrect as it does not represent the dimensions of speed.

Key Concepts:
- Speed of light and its relation to permeability and permittivity of free space.
- Dimensional analysis.

Memory Aids:
- Remember that the speed of light $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ .

Quick Tips:
- Knowing the formula for the speed of light in terms of $\mu_0$ and $\varepsilon_0$ makes this problem straightforward.

39 Easy

39 of 128

Understanding +4 | -1

Which of the following represents a volt

View answer and explanation

Correct option: B

Watt/ampere

Option B (Watt/ampere) is correct because Volt = Power/Current, and Watt is the unit of power, and Ampere is the unit of current.

Solution approach

Correct Answer + Core Reason: Watt/ampere. Volt = Power/Current.

Step-by-Step Solution:

Step 1: Define Volt. Volt (V) is the unit of electric potential difference.

Step 2: Relate Volt to Power and Current. Electric power (P) is given by $P = VI$ , where V is voltage and I is current. Therefore, $V = \frac{P}{I}$ .

Step 3: Determine the units of Power and Current. Power (P) is measured in Watts (W), and current (I) is measured in Amperes (A).

Step 4: Express Volt in terms of Watt and Ampere. Since $V = \frac{P}{I}$ , the unit of Volt is Watt/Ampere.

Why other options are wrong:
- Option A: Joule/second is the unit of power (Watt), not Volt.
- Option C: Watt/coulomb does not represent Volt. Volt is energy per unit charge (Joule/Coulomb).
- Option D: Coulomb/joule is the inverse of Volt.

Key Concepts:
- Electric potential difference (Voltage)
- Electric power
- Units of physical quantities

Memory Aids:
- Remember that Power = Voltage x Current.

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

40 Easy

40 of 128

Remembering +4 | -1

The dimensions of calorie are

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because calorie is a unit of energy, and energy has dimensions of

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

.

Solution approach

Correct Answer + Core Reason: $M{L^2}{T^{ - 2}}$ is the correct dimension of calorie because calorie is a unit of energy, and energy has these dimensions.

Step-by-Step Solution:

Step 1: Recognize that a calorie is a unit of energy (heat).

Step 2: Recall the formula for kinetic energy: $KE = \frac{1}{2}mv^2$ , where m is mass and v is velocity.

Step 3: The dimensions of mass (m) are M.

Step 4: The dimensions of velocity (v) are $LT^{-1}$ . Therefore, the dimensions of $v^2$ are $(LT^{-1})^2 = L^2T^{-2}$ .

Step 5: The dimensions of kinetic energy (KE) are $M L^2 T^{-2}$ . Since calorie is a unit of energy, its dimensions are also $M L^2 T^{-2}$ .

Why other options are wrong:
- Option B: $ML{T^{ - 2}}$ is the dimension of force.
- Option C: $M{L^2}{T^{ - 1}}$ is not the dimension of energy.
- Option D: $M{L^2}{T^{ - 3}}$ is the dimension of power.

Key Concepts:
- Dimensions of physical quantities
- Energy

Memory Aids:
- Remember the dimensions of kinetic energy: $M L^2 T^{-2}$ .

Quick Tips:
- Relate the quantity to a known formula to derive its dimensions.

41 Easy

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Application +4 | -1

What is the dimensional formula of

m{c^2}

, where the letters have their usual meanings?

View answer and explanation

Correct option: C

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

Option C ([ML²T⁻²]) is correct because mc² represents energy, and its dimensional formula is the same as that of kinetic energy or work done.

Solution approach

Correct Answer + Core Reason: The dimensional formula of $mc^2$ is $[ML^2T^{-2}]$ .

Step-by-Step Solution:

Step 1: Identify the quantities: $m$ is mass and $c$ is the speed of light.

Step 2: Dimensional formula for Mass: $[m] = [M]$ .

Step 3: Dimensional formula for Speed of Light: $[c] = [LT^{-1}]$ .

Step 4: Dimensional formula for $mc^2$ : $[mc^2] = [M][LT^{-1}]^2 = [M][L^2T^{-2}] = [ML^2T^{-2}]$ .

Why other options are wrong:
- Option A: $[MLT^{-1}]$ : This is the dimension of momentum.
- Option B: $[M L^0 T^0]$ : This is the dimension of a dimensionless quantity multiplied by mass.
- Option D: $[M^{-1}L^3T^6]$ : Incorrect dimensions.

Key Concepts:
- Dimensional analysis
- Energy-mass equivalence

Memory Aids:
- E = mc²

Quick Tips:
- Remember the dimensions of mass and speed of light.

42 Easy

42 of 128

Analyzing +4 | -1

Which of the following pair of quantities do not have the same dimensions :

View answer and explanation

Correct option: C

Light year, time period

Option C (Light year, time period) is correct because light year is a unit of distance, while time period is a unit of time.

Solution approach

Correct Answer + Core Reason: Light year, time period. Light year is a unit of distance, while time period is a unit of time, so they have different dimensions.

Step-by-Step Solution:

Step 1: Determine the dimensions of each pair of quantities.

Step 2: Potential gradient and electric field:
- Potential gradient is the rate of change of electric potential with distance: $\frac{dV}{dr}$ . Its dimensions are $[V/L] = [ML^2T^{-3}A^{-1}]/[L] = [MLT^{-3}A^{-1}]$ .
- Electric field (E) is defined as force per unit charge: $\frac{F}{q}$ . Its dimensions are $[MLT^{-2}]/[AT] = [MLT^{-3}A^{-1}]$ .
They have the same dimensions.

Step 3: Torque and kinetic energy:
- Torque ( $\tau$ ) is defined as $rF\sin\theta$ . Its dimensions are $[L][MLT^{-2}] = [ML^2T^{-2}]$ .
- Kinetic energy (KE) is defined as $\frac{1}{2}mv^2$ . Its dimensions are $[M][LT^{-1}]^2 = [ML^2T^{-2}]$ .
They have the same dimensions.

Step 4: Light year and time period:
- Light year is a unit of distance. Its dimensions are [L].
- Time period (T) is a unit of time. Its dimensions are [T].
They do not have the same dimensions.

Step 5: Impedance and reactance:
- Impedance (Z) and reactance (X) both have dimensions of resistance, which is $[ML^2T^{-3}A^{-2}]$ .
They have the same dimensions.

Why other options are wrong:
- Option A: Potential gradient, electric field: They have the same dimensions.
- Option B: Torque, kinetic energy: They have the same dimensions.
- Option D: Impedance, reactance: They have the same dimensions.

Key Concepts:
- Dimensions of physical quantities
- Potential gradient
- Electric field
- Torque
- Kinetic energy
- Light year
- Time period
- Impedance
- Reactance

Memory Aids:
- Remember the definitions of each quantity.

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

43 Easy

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Analyzing +4 | -1

Assertion : The number of significant figures depends on the least count of measuring instrument.

Reason : Significant figures define the accuracy of measuring instrument.

View answer and explanation

Correct option: B

If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

Option B (If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion) is correct because both statements are true, but the reason doesn't explain the assertion.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion. Both statements are individually correct, but the reason doesn't explain the assertion.

Step-by-Step Solution:

Step 1: Analyze the assertion.\nThe number of significant figures is related to the precision of the measuring instrument, which is determined by its least count. A smaller least count allows for more precise measurements and thus more significant figures. So, the assertion is correct.

Step 2: Analyze the reason.\nSignificant figures do define the accuracy of a measuring instrument. More significant figures indicate a more accurate measurement. So, the reason is correct.

Step 3: Determine if the reason explains the assertion.\nThe reason does not explain why the number of significant figures depends on the least count. The least count determines the precision, which then affects the number of significant figures. The reason simply states that significant figures define accuracy, which is true but doesn't explain the assertion.

Why other options are wrong:
- Option A: The reason does not explain the assertion.
- Option C: The reason is correct.
- Option D: Both the assertion and reason are correct.

Key Concepts:
- Significant figures
- Least count of a measuring instrument
- Accuracy and precision

Memory Aids:
- Remember that significant figures reflect the precision of a measurement.

Quick Tips:
- Distinguish between accuracy and precision. Accuracy refers to how close a measurement is to the true value, while precision refers to the repeatability of a measurement.

44 Easy

44 of 128

Understanding +4 | -1

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : In Vernier calliper if positive zero error exists, then while taking measurements, the reading taken will be more than the actual reading.

Reason (R) : The zero error in Vernier Calliper might have happened due to manufacturing defect or due to rough handling.
In the light of the above statements, choose the
correct answer from the options given below:

View answer and explanation

Correct option: C

Both (A) and (R) are correct and (R) is the
correct explanation of (A)

Option C (Both (A) and (R) are correct and (R) is the correct explanation of (A)) is correct because a positive zero error in Vernier callipers leads to overestimation, and manufacturing defects or rough handling can cause this error.

Solution approach

Correct Answer + Core Reason: Both (A) and (R) are correct and (R) is the correct explanation of (A). Positive zero error leads to overestimation, and manufacturing defects/rough handling can cause this.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): In Vernier calliper if positive zero error exists, then while taking measurements, the reading taken will be more than the actual reading. This is correct. A positive zero error means the instrument reads a value even when it should read zero, thus overestimating measurements.

Step 2: Analyze Reason (R): The zero error in Vernier Calliper might have happened due to manufacturing defect or due to rough handling. This is also correct. These are common causes of zero errors in measuring instruments.

Step 3: Determine the relationship between (A) and (R). The reason (R) explains why the assertion (A) is true. Manufacturing defects or rough handling can cause the positive zero error mentioned in the assertion.

Why other options are wrong:
- Option A: (A) is true but (R) is false is incorrect because both A and R are true.
- Option B: Both (A) and (R) are correct but (R) is not the correct explanation of (A) is incorrect because R is the correct explanation of A.
- Option D: (A) is false but (R) is true is incorrect because A is true.

Key Concepts:
- Vernier calliper
- Zero error (positive and negative)
- Causes of zero error

Memory Aids:
- Positive zero error: Instrument reads a value even when it should read zero (overestimation).
- Negative zero error: Instrument reads zero before it should (underestimation).

Quick Tips:
- Always check for zero error before taking measurements with a Vernier calliper or screw gauge.

45 Easy

45 of 128

Application +4 | -1

If V denotes the potential difference across the plates of a capacitor of capacitance C, the dimensions of

C{V^2}

are

View answer and explanation

Correct option: D

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

Option D (

ML^2T^{-2}

) is correct because CV² represents energy stored in a capacitor, which has the same dimensions as work or kinetic energy.

Solution approach

Correct Answer + Core Reason: The dimensions of $CV^2$ are $ML^2T^{-2}$ .

Step-by-Step Solution:

Step 1: Identify the expression: $CV^2$ represents energy stored in a capacitor.

Step 2: Energy stored in a capacitor: $E = \frac{1}{2}CV^2$ .

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

Step 4: Therefore, the dimensions of $CV^2$ are the same as energy, which is $[ML^2T^{-2}]$ .

Why other options are wrong:
- Option A: Not expressible in MLT: It is expressible in MLT.
- Option B: $MLT^{-2}$ : This is the dimension of force.
- Option C: $M^2LT^{-1}$ : Incorrect dimensions.

Key Concepts:
- Energy stored in a capacitor
- Dimensional analysis

Memory Aids:
- E = (1/2)CV²

Quick Tips:
- Remember the formula for energy stored in a capacitor.

46 Easy

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Remembering +4 | -1

Velocity of light is equal to

View answer and explanation

Correct option: A

\sqrt{\frac{1}{\varepsilon_0 \mu_0}}

Option A (

\sqrt{\frac{1}{\varepsilon_0 \mu_0}}

) is correct because the speed of light is related to the permittivity and permeability of free space by this formula.

Solution approach

Correct Answer + Core Reason: $\sqrt{\frac{1}{\varepsilon_0 \mu_0}}$ . The velocity of light is related to permittivity and permeability of free space.

Step-by-Step Solution:

Step 1: Recall the formula for the speed of light. The speed of light (c) in vacuum is given by the formula $c = \frac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }}$ , where ${\varepsilon _0}$ is the permittivity of free space and ${\mu _0}$ is the permeability of free space.

Step 2: Rearrange the formula. The formula can be rewritten as $c = \sqrt {\frac{1}{{{\varepsilon _0}{\mu _0}}}}$ .

Why other options are wrong:
- Option B: This option is incorrect because it has the permittivity and permeability in the wrong positions within the square root.
- Option C: This option is incorrect because it does not have the square root and the positions of permittivity and permeability are incorrect.
- Option D: This option is incorrect because it does not represent the correct relationship between the speed of light, permittivity, and permeability.

Key Concepts:
- Speed of light
- Permittivity and permeability of free space

Memory Aids:
- Remember the formula $c = \frac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }}$

Quick Tips:
- Know the fundamental constants and their relationships.

47 Easy

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Application +4 | -1

A physical quantity A is related to four observables a,b,c and d as follows

\;A = \frac{{{a^2}{b^3}}}{{c\sqrt d }}

The percentage errors of measurement in a,b,c and d are 1%, 3%, 2% and 2% respectively. What is the percentage error in the quantity A?

View answer and explanation

Correct option: D

14\%

Option D (14%) is correct because percentage errors are added after multiplying by the power of each variable in the formula.

Solution approach

Correct Answer + Core Reason: $14\%$ . Percentage errors are added after multiplying by the powers.

Step-by-Step Solution:

Step 1: Identify the given formula: $A = \frac{{{{a^2}{b^3}}}}{{c\sqrt d }}$

Step 2: Identify the percentage errors: $\frac{{\Delta a}}{a} \times 100\% = 1\%$ , $\frac{{\Delta b}}{b} \times 100\% = 3\%$ , $\frac{{\Delta c}}{c} \times 100\% = 2\%$ , $\frac{{\Delta d}}{d} \times 100\% = 2\%$ .

Step 3: Write the error equation: $\frac{{\Delta A}}{A} \times 100\% = 2 \times (\frac{{\Delta a}}{a} \times 100\%) + 3 \times (\frac{{\Delta b}}{b} \times 100\%) + 1 \times (\frac{{\Delta c}}{c} \times 100\%) + \frac{1}{2} \times (\frac{{\Delta d}}{d} \times 100\%)$

Step 4: Substitute the given values: $\frac{{\Delta A}}{A} \times 100\% = 2(1\%) + 3(3\%) + 1(2\%) + \frac{1}{2}(2\%)$

Step 5: Calculate the percentage error: $\frac{{\Delta A}}{A} \times 100\% = 2\% + 9\% + 2\% + 1\% = 14\%$

Why other options are wrong:
- Option A: $12\%$ : This is incorrect because it might be calculated without considering the power of d or with a mistake in adding the errors.
- Option B: $7\%$ : This is incorrect because it might be calculated by simply adding the percentage errors without multiplying by the powers.
- Option C: $5\%$ : This is incorrect as it is a significantly lower value than the correct percentage error, likely due to an incorrect calculation of error propagation.

Key Concepts:
- Error propagation in multiplication and division
- Percentage error calculation

Memory Aids:
- Remember to multiply the percentage error by the power of the variable.

Quick Tips:
- When a quantity is raised to a power, multiply the percentage error by that power.

48 Easy

48 of 128

Understanding +4 | -1

Unit of impulse is

View answer and explanation

Correct option: C

kg-m/s

Option C (kg-m/s) is correct because impulse is change in momentum, which has units of kg-m/s.

Solution approach

Correct Answer + Core Reason: kg-m/s is the unit of impulse because impulse is equal to the change in momentum.

Step-by-Step Solution:

Step 1: Define impulse (J).

Step 2: Impulse is the change in momentum ( $\Delta p$ ). Therefore, $J = \Delta p = m(v_f - v_i)$ , where m is mass, $v_f$ is final velocity, and $v_i$ is initial velocity.

Step 3: The SI unit of mass (m) is kilogram (kg), and the SI unit of velocity (v) is meter per second (m/s).

Step 4: Therefore, the SI unit of impulse (J) is kg-m/s.

Why other options are wrong:
- Option A: Newton is the unit of force.
- Option B: kg-m is not the unit of impulse.
- Option D: Joule is the unit of energy.

Key Concepts:
- Impulse
- Momentum
- SI units

Memory Aids:
- Remember that impulse is equal to the change in momentum.

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

49 Easy

49 of 128

Analyzing +4 | -1

A, B, C

and

D

are four different physical quantities having different dimensions. None of them is dimensionless. But we know that the equation

\mathrm{AD}=\mathrm{C} \ln (\mathrm{BD})

holds true. Then which of the combination is not a meaningful quantity?

View answer and explanation

Correct option: D

\frac{(\mathrm{A}-\mathrm{C})}{\mathrm{D}}

Option D ((A-C)/D) is correct because for (A-C) to be meaningful, A and C must have the same dimensions, but the problem states they have different dimensions.

Solution approach

Correct Answer + Core Reason: $\frac{(\mathrm{A}-\mathrm{C})}{\mathrm{D}}$ ; For subtraction to be meaningful, A and C must have the same dimensions, which contradicts the problem statement.

Step-by-Step Solution:

Step 1: Analyze the given equation: $\mathrm{AD}=\mathrm{C} \ln (\mathrm{BD})$ .
- Since the argument of a logarithm must be dimensionless, BD is dimensionless. Therefore, the dimensions of B are the inverse of the dimensions of D.
- AD and C must have the same dimensions.

Step 2: Evaluate each option.
- A. $\frac{\mathrm{C}}{\mathrm{BD}}-\frac{\mathrm{AD}^2}{\mathrm{C}}$ : Since BD is dimensionless, this simplifies to C - (AD $^2$ /C). Since AD and C have the same dimensions, this expression can be meaningful if AD $^2$ /C has the same dimensions as C.
- B. $\mathrm{A}^2-\mathrm{B}^2 \mathrm{C}^2$ : This can be meaningful if A $^2$ and B $^2$ C $^2$ have the same dimensions.
- C. $\frac{\mathrm{A}}{\mathrm{B}}-\mathrm{C}$ : This can be meaningful if A/B and C have the same dimensions.
- D. $\frac{(\mathrm{A}-\mathrm{C})}{\mathrm{D}}$ : For (A-C) to be meaningful, A and C must have the same dimensions. However, the problem states that A, B, C, and D have different dimensions. Therefore, this combination cannot be meaningful.

Why other options are wrong:
- Option A: $\frac{\mathrm{C}}{\mathrm{BD}}-\frac{\mathrm{AD}^2}{\mathrm{C}}$ : This can be meaningful under certain conditions.
- Option B: $\mathrm{A}^2-\mathrm{B}^2 \mathrm{C}^2$ : This can be meaningful under certain conditions.
- Option C: $\frac{\mathrm{A}}{\mathrm{B}}-\mathrm{C}$ : This can be meaningful under certain conditions.

Key Concepts:
- Dimensional analysis
- Principle of homogeneity of dimensions
- Dimensionless quantities

Memory Aids:
- Remember that the argument of a logarithm must be dimensionless.

Quick Tips:
- Always check if the quantities being added or subtracted have the same dimensions.
- Remember that dimensionless quantities do not have dimensions.

50 Easy

50 of 128

Analyzing +4 | -1

Assertion :

\sqrt{\frac{\text { Modulus of elasticity }}{\text { Density }}}

has the unit

\mathrm{ms}^{-1} \text {. }

Reason: Acceleration has the dimensions of

\frac{1}{\left(\sqrt{\varepsilon_0 \mu_0}\right) t} \text {. }

View answer and explanation

Correct option: B

If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

Option B is correct because both the Assertion and Reason are correct, but the Reason does not explain the Assertion.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

Step-by-Step Solution:

Step 1: Analyze the Assertion: $\sqrt{\frac{\text { Modulus of elasticity }}{\text { Density }}}$ has the unit $\mathrm{ms}^{-1}$ .
- Modulus of elasticity (E) has dimensions of stress, which is force per unit area: $[E] = [ML^{-1}T^{-2}]$ .
- Density ( $\rho$ ) has dimensions of mass per unit volume: $[\rho] = [ML^{-3}]$ .
- Therefore, the dimensions of $\sqrt{\frac{E}{\rho}}$ are $\sqrt{\frac{ML^{-1}T^{-2}}{ML^{-3}}} = \sqrt{L^2T^{-2}} = LT^{-1}$ , which is the dimension of velocity. Thus, the unit is $\mathrm{ms}^{-1}$ . The assertion is correct.

Step 2: Analyze the Reason: Acceleration has the dimensions of $\frac{1}{\left(\sqrt{\varepsilon_0 \mu_0}\right) t}$ .
- $\frac{1}{\sqrt{\varepsilon_0 \mu_0}}$ has dimensions of velocity (as shown in a previous question), which is $LT^{-1}$ .
- Therefore, $\frac{1}{\left(\sqrt{\varepsilon_0 \mu_0}\right) t}$ has dimensions of $\frac{LT^{-1}}{T} = LT^{-2}$ , which is the dimension of acceleration. The reason is correct.

Step 3: Determine if the Reason explains the Assertion: The Reason is about the dimensions of acceleration in terms of permittivity and permeability, while the Assertion is about the dimensions of a quantity involving the modulus of elasticity and density. They are not directly related. Therefore, the Reason does not explain the Assertion.

Why other options are wrong:
- Option A: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion: The Reason does not explain the Assertion.
- Option C: If Assertion is true, but Reason is false: The Reason is also true.
- Option D: If Assertion is false, but Reason is true: The Assertion is also true.

Key Concepts:
- Dimensions of physical quantities
- Modulus of elasticity
- Density
- Permittivity and permeability of free space
- Acceleration

Memory Aids:
- Remember the dimensions of modulus of elasticity and density.

Quick Tips:
- Analyze each statement independently before determining if one explains the other.

51 Easy

51 of 128

Application +4 | -1

The dimensional formula of coefficient of viscosity is

\left( {F = \eta A\frac{{\Delta V}}{{\Delta x}}} \right)

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because the coefficient of viscosity relates force to area and velocity gradient, and its dimensions can be derived from the given formula.

Solution approach

Correct Answer + Core Reason: $M{L^{ - 1}}{T^{ - 1}}$ . The coefficient of viscosity relates force to area and velocity gradient.

Step-by-Step Solution:

Step 1: Analyze the given formula: $F = \eta A \frac{\Delta V}{\Delta x}$

Step 2: Solve for the coefficient of viscosity: $\eta = \frac{F}{A \frac{\Delta V}{\Delta x}}$

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

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

Step 5: Determine the dimensions of velocity gradient: $\left[\frac{\Delta V}{\Delta x}\right] = \frac{L T^{-1}}{L} = T^{-1}$

Step 6: Calculate the dimensions of the coefficient of viscosity: $[\eta] = \frac{[F]}{[A] \left[\frac{\Delta V}{\Delta x}\right]} = \frac{M L T^{-2}}{L^2 T^{-1}} = M L^{-1} T^{-1}$

Why other options are wrong:
- Option A: $M{L^{ - 2}}{T^{ - 2}}$ is not the dimension of viscosity.
- Option B: ${M^ \circ }{L^ \circ }{T^ \circ }$ is dimensionless and not the dimension of viscosity.
- Option C: $M{L^2}{T^{ - 2}}$ is the dimension of energy or torque, not viscosity.

Key Concepts:
- Coefficient of viscosity
- Dimensions of force, area, and velocity gradient

Memory Aids:
- Remember the formula: $F = \eta A \frac{\Delta V}{\Delta x}$

Quick Tips:
- Isolate the desired quantity and express it in terms of other known quantities with known dimensions.

52 Easy

52 of 128

Application +4 | -1

Dimensional formula for the universal gravitational constant G is

View answer and explanation

Correct option: C

\left[ {{\mathrm{M}^{ - 1}}{\mathrm{L}^3}{\mathrm{T}^{ - 2}}} \right]

Option C ([M⁻¹L³T⁻²]) is correct because it is derived from Newton's law of gravitation, F = G(m1m2/r²).

Solution approach

Correct Answer + Core Reason: The dimensional formula for the universal gravitational constant G is $[M^{-1}L^3T^{-2}]$ .

Step-by-Step Solution:

Step 1: Recall Newton's Law of Gravitation: $F = G \frac{m_1 m_2}{r^2}$ .

Step 2: Rearrange for G: $G = \frac{Fr^2}{m_1 m_2}$ .

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

Step 4: Dimensional formula for Distance (r): $[r] = [L]$ .

Step 5: Dimensional formula for Mass (m): $[m] = [M]$ .

Step 6: Dimensional formula for G: $[G] = \frac{[MLT^{-2}][L^2]}{[M][M]} = [M^{-1}L^3T^{-2}]$ .

Why other options are wrong:
- Option A: $[M^{-1}L^2T^{-2}]$ : Incorrect dimensions.
- Option B: $[M^0L^0T^0]$ : G is not dimensionless.
- Option D: $[M^{-1}L^3T^{-1}]$ : Incorrect dimensions.

Key Concepts:
- Newton's Law of Gravitation
- Universal Gravitational Constant

Memory Aids:
- Remember Newton's Law of Gravitation formula.

Quick Tips:
- Remember Newton's Law of Gravitation and rearrange it to find the dimensions of G.

53 Easy

53 of 128

Understanding +4 | -1

Statement-I: The number of significant figures in

0.001

is 1 while in

0.100

it is 3.
Statement-II: Zeros before a non-zero significant digit are not counted while zeros after a non-zero significant digit are counted.

View answer and explanation

Correct option: A

Statement-I is true and statement-II is true

Option A (Statement-I is true and statement-II is true) is correct because the number of significant figures in 0.001 is 1, and in 0.100 it is 3, and the rule about zeros is correctly stated.

Solution approach

Correct Answer + Core Reason: Statement-I is true and statement-II is true. The rules for significant figures are correctly applied and stated.

Step-by-Step Solution:

Step 1: Analyze Statement-I: The number of significant figures in 0.001 is 1 while in 0.100 it is 3.
- 0.001 has one significant figure (the 1). Leading zeros are not significant.
- 0.100 has three significant figures (the 1 and the two trailing zeros). Trailing zeros after the decimal point are significant.
- Therefore, Statement-I is true.

Step 2: Analyze Statement-II: Zeros before a non-zero significant digit are not counted while zeros after a non-zero significant digit are counted.
- This statement correctly describes the rules for determining significant figures.
- Therefore, Statement-II is true.

Step 3: Conclusion: Both statements are true.

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

Key Concepts:
- Significant figures
- Rules for counting significant figures

Memory Aids:
- Atlantic-Pacific Rule: If the decimal point is Absent, start counting from the Atlantic (right) side at the first non-zero digit. If the decimal point is Present, start counting from the Pacific (left) side at the first non-zero digit.

Quick Tips:
- Remember the rules for identifying significant figures.

54 Easy

54 of 128

Remembering +4 | -1

In which of the following systems of units, a Weber is the unit of magnetic flux?

View answer and explanation

Correct option: C

SI

Option C (SI) is correct because the Weber is defined as the SI unit of magnetic flux.

Solution approach

Correct Answer + Core Reason: SI. The Weber is the SI unit of magnetic flux.

Step-by-Step Solution:

Step 1: Recall the definition of magnetic flux and its units in different systems.

Step 2: The Weber (Wb) is defined as the unit of magnetic flux in the International System of Units (SI).

Step 3: In the CGS system, the unit of magnetic flux is the Maxwell.

Step 4: In the MKS system, the unit of magnetic flux is also the Weber, but MKS is less commonly used than SI.

Step 5: FPS is the foot-pound-second system, which does not have a standard unit for magnetic flux.

Why other options are wrong:
- Option A: CGS uses Maxwell as the unit of magnetic flux.
- Option B: MKS uses Weber, but SI is the standard system.
- Option D: FPS does not have a standard unit for magnetic flux.

Key Concepts:
- Magnetic flux
- SI units
- CGS units
- MKS units
- FPS units

Memory Aids:
- Weber = SI unit of magnetic flux

Quick Tips:
- Remember the units of magnetic flux in different systems.

55 Easy

55 of 128

Remembering +4 | -1

The value of Planck's constant in SI unit is

View answer and explanation

Correct option: D

6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}

Option D (

6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}

) is correct because it is the experimentally determined value of Planck's constant in SI units.

Solution approach

Correct Answer + Core Reason: $6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ . This is the standard value of Planck's constant.

Step-by-Step Solution:

Step 1: Recall Planck's constant: Planck's constant (h) is a fundamental constant in quantum mechanics.

Step 2: State the value: The value of Planck's constant in SI units is approximately $6.626 \times 10^{-34} \mathrm{~J \cdot s}$ . For this question, $6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ is the closest and most appropriate value.

Step 3: Final Answer: The value of Planck's constant is $6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ .

Why other options are wrong:
- Option A: $6.63 \times 10^{-31} \mathrm{~J}-\mathrm{s}$ : This value is incorrect. The exponent is wrong.
- Option B: $6.63 \times 10^{-30} \mathrm{~kg}-\mathrm{m} / \mathrm{s}$ : This value is incorrect. The exponent and units are wrong. Planck's constant is not measured in kg-m/s.
- Option C: $6.63 \times 10^{-32} \mathrm{~kg}-\mathrm{m}^2$ : This value is incorrect. The exponent and units are wrong. Planck's constant is not measured in kg-m².

Key Concepts:
- Planck's Constant
- Quantum Mechanics

Memory Aids:
- Remember the approximate value: $6.63 \times 10^{-34}$ .

Quick Tips:
- Know the fundamental constants and their values.

56 Easy

56 of 128

Application +4 | -1

The dimensional formula for Boltzmann’s constant is

View answer and explanation

Correct option: A

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

Option A ([ML²T⁻²θ⁻¹]) is correct because Boltzmann's constant relates energy to temperature, and its dimensional formula reflects this relationship.

Solution approach

Correct Answer + Core Reason: $\left[ {M{L^2}{T^{ - 2}}{\theta ^{ - 1}}} \right]$ . Boltzmann's constant relates energy and temperature.

Step-by-Step Solution:

Step 1: Define the problem: Find the dimensional formula for Boltzmann's constant (k).

Step 2: Recall the relationship: Boltzmann's constant relates energy (E) to temperature (T) through the equation $E = kT$ , where E is energy, k is Boltzmann's constant, and T is temperature.

Step 3: Express Boltzmann's constant: $k = \frac{E}{T}$ .

Step 4: Determine the dimensions of energy: Energy has the same dimensions as work, which is force times distance: $[E] = [F \times d] = [M L T^{-2} \times L] = [M L^2 T^{-2}]$ .

Step 5: Determine the dimensions of temperature: Temperature is represented by $\theta$ .

Step 6: Calculate the dimensions of Boltzmann's constant: $[k] = \frac{[E]}{[T]} = \frac{[M L^2 T^{-2}]}{[\theta]} = [M L^2 T^{-2} \theta^{-1}]$ .

Step 7: Final Answer: $\left[ {M{L^2}{T^{ - 2}}{\theta ^{ - 1}}} \right]$

Why other options are wrong:
- Option B: $\left[ {M{L^2}{T^{ - 2}}} \right]$ is wrong because it represents energy (or work), not Boltzmann's constant.
- Option C: $\left[ {M{L^0}{T^{ - 2}}{\theta ^{ - 1}}} \right]$ is wrong because it does not have the correct dimensions for Boltzmann's constant.
- Option D: $\left[ {M{L^{ - 2}}{T^{ - 1}}{\theta ^{ - 1}}} \right]$ is wrong because it does not have the correct dimensions for Boltzmann's constant.

Key Concepts:
- Boltzmann's Constant
- Dimensions of Energy
- Dimensions of Temperature

Memory Aids:
- E = kT (Energy equals Boltzmann's constant times Temperature)

Quick Tips:
- Remember the relationship between energy and temperature to find Boltzmann's constant.

57 Easy

57 of 128

Application +4 | -1

Error in measurement of radius of sphere is 4% then maximum % error in volume is

View answer and explanation

Correct option: D

0.12

Option D (0.12) is correct because the percentage error in volume is three times the percentage error in the radius.

Solution approach

Correct Answer + Core Reason: 0.12. The percentage error in volume is three times the percentage error in the radius.

Step-by-Step Solution:

Step 1: Volume Formula: $V = \frac{4}{3} \pi r^3$

Step 2: Percentage Error Formula: $\frac{{\Delta V}}{V} \times 100 = 3 \times \frac{{\Delta r}}{r} \times 100$

Step 3: Substitute the Given Value: $\frac{{\Delta V}}{V} \times 100 = 3 \times 4\% = 12\%$

Step 4: Convert Percentage to Decimal (if needed for comparison): 12% = 0.12

Why other options are wrong:
- Option A: 0.01: Incorrect calculation.
- Option B: 0.03: Incorrect calculation.
- Option C: 0.09: Incorrect calculation.

Key Concepts:
- Percentage Error Propagation
- Volume of a Sphere

Memory Aids:
- If $V = r^n$ , then $\frac{{\Delta V}}{V} = n \frac{{\Delta r}}{r}$

Quick Tips:
- When a quantity is raised to a power, the percentage error is multiplied by that power.

58 Easy

58 of 128

Remembering +4 | -1

Dimensional formula for force is

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because force is mass times acceleration, and the dimensions of acceleration are length/time².

Solution approach

Correct Answer + Core Reason: $\left[ {{M^1}{L^1}{T^{ - 2}}} \right]$ . Force is mass times acceleration.

Step-by-Step Solution:

Step 1: Define Force: Force (F) = Mass (M) x Acceleration (a).

Step 2: Express Acceleration in terms of Length (L) and Time (T): Acceleration (a) = L/T².

Step 3: Substitute to find the dimensions of Force: F = M x (L/T²) = MLT^-2.

Step 4: Write the dimensional formula using square brackets: [F] = $\left[ {{M^1}{L^1}{T^{ - 2}}} \right]$ .

Why other options are wrong:
- Option A: Incorrect dimensions for force. The length dimension is incorrect.
- Option C: Incorrect dimensions for force. The length dimension is incorrect (negative exponent).
- Option D: Incorrect dimensions for force. The length dimension is incorrect (negative exponent).

Key Concepts:
- Dimensions of physical quantities
- Force, Mass, Acceleration

Memory Aids:
- Remember the formula: Force = Mass x Acceleration

Quick Tips:
- Break down the quantity into its fundamental components (Mass, Length, Time) and derive the dimensions.

59 Easy

59 of 128

Understanding +4 | -1

If the absolute errors in two physical quantities

A

and

B

are

a

and

b

respectively, then the absolute error in the value of

A-B

are

View answer and explanation

Correct option: C

a+b

Option C (

a+b

) is correct because the absolute error in the difference of two quantities is the sum of their absolute errors.

Solution approach

Correct Answer + Core Reason: $a+b$ . The absolute error in subtraction is the sum of individual absolute errors.

Step-by-Step Solution:

Step 1: Define the given quantities: A and B are physical quantities with absolute errors a and b, respectively.

Step 2: Express the quantities with their errors: A ± a and B ± b.

Step 3: Calculate the difference A - B: (A ± a) - (B ± b) = (A - B) ± (a + b).

Step 4: Determine the absolute error in A - B: The absolute error is the sum of the individual absolute errors, which is a + b.

Why other options are wrong:
- Option A: $b-a$ is incorrect because the absolute error is always added, not subtracted.
- Option B: $a \neq b$ is irrelevant to the calculation of absolute error in subtraction.
- Option D: $a-b$ is incorrect because the absolute error is always added, not subtracted.

Key Concepts:
- Error propagation
- Absolute error
- Error in subtraction

Memory Aids:
- Error in addition/subtraction: Add the absolute errors.

Quick Tips:
- When adding or subtracting quantities, always add their absolute errors to find the total error.

60 Easy

60 of 128

Understanding +4 | -1

Dimensions of kinetic energy are

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because kinetic energy is 1/2 * mass * velocity², and its dimensional formula is derived from that.

Solution approach

Correct Answer + Core Reason: $M{L^2}{T^{ - 2}}$ . Kinetic energy is 1/2 * mass * velocity².

Step-by-Step Solution:

Step 1: Define kinetic energy. Kinetic energy (KE) is defined as $KE = \frac{1}{2}m{v^2}$ , where m is mass and v is velocity.

Step 2: Determine the dimensions of mass. The dimension of mass is [M].

Step 3: Determine the dimensions of velocity. Velocity is defined as displacement divided by time, i.e., $v = \frac{d}{t}$ . The dimension of displacement is [L], and the dimension of time is [T]. Therefore, the dimension of velocity is $\frac{{\left[ L \right]}}{{\left[ T \right]}} = \left[ {L{T^{ - 1}}} \right]$ .

Step 4: Determine the dimensions of kinetic energy. Since $KE = \frac{1}{2}m{v^2}$ , the dimension of kinetic energy is $\left[ M \right] \times {\left[ {L{T^{ - 1}}} \right]^2} = \left[ {M{L^2}{T^{ - 2}}} \right]$ .

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

Key Concepts:
- Dimensions of physical quantities
- Kinetic energy

Memory Aids:
- Remember that Kinetic Energy = 1/2 * m * v².

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

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