UNITS & MEASUREMENTS MCQs for NEET UG PHYSICS — Easy, Page 5

121 Easy

121 of 128

Remembering +4 | -1

The number 0.004 has significant figures equal to

View answer and explanation

Correct option: A

1

Option A (1) is correct because leading zeros are not significant figures.

Solution approach

Correct Answer + Core Reason: 1. Leading zeros are not significant.

Step-by-Step Solution:

Step 1: Identify the number: 0.004

Step 2: Apply the rules for significant figures: Leading zeros are not significant. Only the non-zero digit 4 is significant.

Step 3: Count the significant figures: There is only one significant figure.

Step 4: Final Answer: 1

Why other options are wrong:
- Option B: Incorrect because it counts some of the leading zeros as significant.
- Option C: Incorrect because it counts some of the leading zeros as significant.
- Option D: Incorrect because it counts all the leading zeros as significant.

Key Concepts:
- Significant figures rules

Memory Aids:
- Leading zeros never count.

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

122 Easy

122 of 128

Application +4 | -1

A physical quantity

Q

is calculated according to the expression

Q=\frac{A^3 B^3}{C \sqrt{D}}

If percentage errors in

A, B, C, D

are

2 \%, 1 \%, 3 \%

and

4 \%

, respectively. What is the percentage error in

Q

?

View answer and explanation

Correct option: C

\pm 14 \%

Option C (

\pm 14 \\%

) is correct because the percentage error in Q is calculated by summing the percentage errors in A, B, C, and D, each multiplied by its respective power.

Solution approach

Correct Answer + Core Reason: $\pm 14 \\%$ . The percentage error in Q is calculated by summing the percentage errors in A, B, C, and D, each multiplied by its respective power.

Step-by-Step Solution:

Step 1: Given expression: $Q=\frac{A^3 B^3}{C \sqrt{D}}$

Step 2: Percentage errors: $\Delta A = 2 \\%, \Delta B = 1 \\%, \Delta C = 3 \\%, \Delta D = 4 \\%$

Step 3: Formula for percentage error in Q:
$\frac{\Delta Q}{Q} \times 100 = 3 \frac{\Delta A}{A} \times 100 + 3 \frac{\Delta B}{B} \times 100 + \frac{\Delta C}{C} \times 100 + \frac{1}{2} \frac{\Delta D}{D} \times 100$

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

Step 5: Calculate:
$\frac{\Delta Q}{Q} \times 100 = 6 \\% + 3 \\% + 3 \\% + 2 \\% = 14 \\%$

Step 6: Final Answer: $\pm 14 \\%$

Why other options are wrong:
- **Option A ( $\pm 8 \\%$ ): This is incorrect as it doesn't account for the correct propagation of errors.
- Option B ( $\pm 10 \\%$ ): This is incorrect as it doesn't account for the correct propagation of errors.
- Option D ( $\pm 12 \\%$ ): This is incorrect as it doesn't account for the correct propagation of errors.

Key Concepts:
- Error Propagation
- Percentage Error

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

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

123 Easy

123 of 128

Understanding +4 | -1

Dimensions of frequency are

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because frequency is inverse of time period, hence only time dimension is involved.

Solution approach

Correct Answer + Core Reason: ${M^0}{L^0}{T^{ - 1}}$ is the dimension of frequency because frequency is the inverse of the time period.

Step-by-Step Solution:

Step 1: Define frequency (f).

Step 2: Frequency is the number of oscillations per unit time. It is the inverse of the time period (T). Therefore, $f = \frac{1}{T}$ .

Step 3: The dimension of time period (T) is T.

Step 4: Therefore, the dimension of frequency (f) is $T^{-1}$ , which can also be written as ${M^0}{L^0}{T^{ - 1}}$ .

Why other options are wrong:
- Option A: ${M^0}{L^{ - 1}}{T^0}$ is incorrect.
- Option C: ${M^0}{L^0}T$ is the dimension of time period.
- Option D: $M{T^{ - 2}}$ is incorrect.

Key Concepts:
- Dimensions of physical quantities
- Frequency

Memory Aids:
- Remember that frequency is the inverse of the time period.

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

124 Easy

124 of 128

Application +4 | -1

In a simple pendulum experiment for determination of acceleration due to gravity

(\mathrm{g})

, time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be

30 \mathrm{~s}

. The length of pendulum is measured by using a meter scale of least count

1 \mathrm{~mm}

and the value obtained is

55.0 \mathrm{~cm}

. The percentage

(\%)

error in the determination of

g

is close to

View answer and explanation

Correct option: A

6.8

Option A (6.8) is correct because the percentage error in g is calculated using the formula involving the errors in time period and length.

Solution approach

Correct Answer + Core Reason: 6.8. The percentage error in g is calculated using the formula derived from the pendulum equation.

Step-by-Step Solution:

Step 1: The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$ , where $T$ is the time period, $L$ is the length, and $g$ is the acceleration due to gravity.

Step 2: Rearrange the formula to solve for $g$ : $g = 4\pi^2 \frac{L}{T^2}$ .

Step 3: The percentage error in $g$ is given by: $\frac{\Delta g}{g} \times 100 = \left( \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} \right) \times 100$ .

Step 4: Calculate $\Delta L = 1 \text{{ mm}} = 0.1 \text{{ cm}}$ , so $\frac{\Delta L}{L} = \frac{0.1}{55.0} = 0.001818$ .

Step 5: Calculate $\Delta T$ . The least count of the watch is 1 second, and the time is measured for 20 oscillations. Therefore, the error in the time period $T$ is $\frac{1}{20} = 0.05 \text{{ s}}$ . The mean value of time taken for 20 oscillations is 30 s, so the time period $T = \frac{30}{20} = 1.5 \text{{ s}}$ . Thus, $\frac{\Delta T}{T} = \frac{0.05}{1.5} = 0.0333$ .

Step 6: Substitute the values into the percentage error formula: $\frac{\Delta g}{g} \times 100 = \left( 0.001818 + 2 \times 0.0333 \right) \times 100 = (0.001818 + 0.0666) \times 100 = 0.068418 \times 100 = 6.8418 \approx 6.8 \\%$ .

Why other options are wrong:
- Option B (0): This is incorrect as the errors in length and time period contribute to the error in g.
- Option C (0): This is incorrect as the errors in length and time period contribute to the error in g.
- Option D (0): This is incorrect as the errors in length and time period contribute to the error in g.

Key Concepts:
- Simple pendulum
- Error propagation
- Percentage error

Memory Aids:
- Remember the formula for the time period of a simple pendulum and how to calculate percentage error.

Quick Tips:
- Pay attention to the units and convert them appropriately before calculations.

125 Easy

125 of 128

Understanding +4 | -1

Energy per unit volume represents

View answer and explanation

Correct option: A

Pressure

Option A (Pressure) is correct because pressure is defined as force per unit area, and energy per unit volume has the same dimensions as pressure.

Solution approach

Correct Answer + Core Reason: Pressure. Energy per unit volume has the same dimensions as pressure.

Step-by-Step Solution:

Step 1: Determine the dimensions of energy per unit volume.\nEnergy = Force x Distance = MLT^-2 x L = ML²T^-2\nVolume = L³\nEnergy/Volume = (ML²T^-2) / L³ = ML^-1T^-2

Step 2: Determine the dimensions of pressure.\nPressure = Force/Area = MLT^-2 / L² = ML^-1T^-2

Step 3: Compare the dimensions. Energy/Volume and Pressure have the same dimensions.

Why other options are wrong:
- Option B: Force has dimensions MLT^-2, which is different.
- Option C: Thrust is a type of force and has dimensions MLT^-2, which is different.
- Option D: Work has dimensions ML²T^-2, which is different.

Key Concepts:
- Dimensions of Physical Quantities
- Pressure
- Energy
- Volume

Memory Aids:
- Remember the formula for pressure (Force/Area).
- Remember the formula for energy per unit volume.

Quick Tips:
- Knowing the dimensional formulas of common quantities can help in quickly solving such problems.

126 Easy

126 of 128

Application +4 | -1

A physical quantity

X

is given by

X=\frac{2 k^3 l^2}{m \sqrt{n}}

The percentage error in the measurements of

k, l, m

and

n

are

1 \%, 2 \%, 3 \%

and

4 \%

, respectively. The value of

X

is uncertain by

View answer and explanation

Correct option: C

12 \%

Option C (12%) is correct because the percentage error in X is calculated by summing the percentage errors in k, l, m, and n, each multiplied by their respective powers in the formula.

Solution approach

Correct Answer + Core Reason: $12 \\%$ . The percentage error is calculated by considering the powers of each variable.

Step-by-Step Solution:

Step 1: Write the formula for X: $X=\frac{2 k^3 l^2}{m \sqrt{n}}$

Step 2: The percentage error in X is given by:
$\frac{\Delta X}{X} \times 100 = \left( 3 \frac{\Delta k}{k} + 2 \frac{\Delta l}{l} + \frac{\Delta m}{m} + \frac{1}{2} \frac{\Delta n}{n} \right) \times 100$

Step 3: Substitute the given percentage errors:
$\frac{\Delta X}{X} \times 100 = \left( 3(1 \\%) + 2(2 \\%) + 1(3 \\%) + \frac{1}{2}(4 \\%) \right) \times 100$

Step 4: Calculate the total percentage error:
$\frac{\Delta X}{X} \times 100 = (3 \\% + 4 \\% + 3 \\% + 2 \\%) = 12 \\%$

Why other options are wrong:
- Option A (8 \\%): This is incorrect as it doesn't correctly account for the powers of each variable in the formula.
- Option B (10 \\%): This is incorrect as it doesn't correctly account for the powers of each variable in the formula.
- Option D (None of these): The correct percentage error is 12%.

Key Concepts:
- Error propagation
- Percentage error

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

Quick Tips:
- Be careful with the powers of the variables when calculating the percentage error.

127 Easy

127 of 128

Analyzing +4 | -1

Statement-I: Surface tension and spring constant have the same dimensions.
Statement-II: Both are equivalent to force per unit length

View answer and explanation

Correct option: A

Statement-I is true and statement-II is true

Statement-I is true and statement-II is true because both surface tension and spring constant have dimensions of force per unit length.

Solution approach

Correct Answer + Core Reason: Statement-I is true and statement-II is true. Surface tension and spring constant both have the same dimensions, which are equivalent to force per unit length.

Step-by-Step Solution:

Step 1: Analyze Statement-I: Surface tension and spring constant have the same dimensions.
- Surface tension ( $\gamma$ ) is defined as force per unit length: $\gamma = \frac{F}{L}$ .
- Spring constant (k) is defined by Hooke's law: $F = kx$ , so $k = \frac{F}{x}$ , where x is the displacement (length).
Therefore, both have the same dimensions: $[MLT^{-2}]/[L] = [MT^{-2}]$ .

Step 2: Analyze Statement-II: Both are equivalent to force per unit length.
As shown in Step 1, both surface tension and spring constant are defined as force per unit length.

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:
- Dimensions of physical quantities
- Surface tension
- Spring constant

Memory Aids:
- Remember the definitions of surface tension and spring constant.

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

128 Easy

128 of 128

Analyzing +4 | -1

You may not know integration. But using dimensional analysis you can check or prove results. In the integral

\int \frac{d x}{\left(2 {ax}-x^2\right)^{1 / 2}}=a^n \sin ^{-1}\left(\frac{x}{a}-1\right)

the value of

n

should be-

View answer and explanation

Correct option: A

0

Option A (0) is correct because the argument of the inverse sine function must be dimensionless, and dimensional analysis shows that n must be zero.

Solution approach

Correct Answer + Core Reason: 0. The argument of the inverse sine function must be dimensionless, and dimensional analysis shows that n must be zero.

Step-by-Step Solution:

Step 1: The argument of the $\sin^{-1}$ function must be dimensionless. Therefore, $\left(\frac{x}{a}-1\right)$ is dimensionless.

Step 2: The integral $\int \frac{d x}{\left(2 {ax}-x^2\right)^{1 / 2}}$ must have the same dimensions as $a^n \sin ^{-1}\left(\frac{x}{a}-1\right)$ .

Step 3: The dimensions of the integral are:
$\left[ \int \frac{d x}{\left(2 {ax}-x^2\right)^{1 / 2}} \right] = \frac{[L]}{[L^2]^{1/2}} = \frac{[L]}{[L]} = [M^0L^0T^0] = [1]$

Step 4: The dimensions of $a^n \sin ^{-1}\left(\frac{x}{a}-1\right)$ are:
$[a^n \sin ^{-1}\left(\frac{x}{a}-1\right)] = [L]^n [1] = [L^n]$

Step 5: Equating the dimensions:
$[L^n] = [1]$ \nTherefore, n = 0.

Why other options are wrong:
- Option B: 0 is correct, so this is not wrong.
- Option C: 0 is correct, so this is not wrong.
- Option D: 0 is correct, so this is not wrong.

Key Concepts:
- Dimensional Analysis
- Dimensionless Quantities
- Integration

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

Quick Tips:
- Always check the dimensions of the arguments of trigonometric, logarithmic, and exponential functions.

Your Progress

Units & measurements practice · Easy

Continue Your Practice

Your practice