📝 Logarithmic scales pH decibel Richter (13 MCQs)

📖 From Calculus • 1. Basics before calculus • 13 questions available

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What is Logarithmic scales pH decibel Richter?

Definition:
Logarithmic scales, such as pH ( $\mathrm{pH} = -\log_{10}[\mathrm{H}^+]$ ), decibels ( $\mathrm{dB} = 10 \log_{10}(I/I_0)$ ), and Richter magnitude ( $M = \log_{10}(A/A_0)$ ), express wide-ranging quantities in compact, human-readable numbers.

Example:
A solution with $[\mathrm{H}^+] = 10^{-5}$ has pH = 5; a sound with intensity 1000 times reference has $\mathrm{dB} = 10 \log_{10}(1000) = 30$ .

Reason:
Log scales compress exponential ranges (e.g., acidity, loudness, earthquake energy) into linear intervals, making comparisons and interpretations practical.

Easy

Medium

Hard

📝 All Logarithmic scales pH decibel Richter MCQs

Q1. If the sound intensity is increased by a factor of 10, how many decibels are added to the sound level $\beta$ ?

A.5 dB

B.10 dB ✅

C.15 dB

D.20 dB

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Using the definition $\beta = 10\log_{10}(I/I_0)$ , multiplying $I$ by 10 adds $10\log_{10}10 = 10$ dB to $\beta$ . Therefore a ten‑fold increase corresponds to a 10‑dB rise, making option B the correct choice.

Q2. A sound has level $\beta_1 = 80$ dB while another has $\beta_2 = 90$ dB. What is the ratio $I_1/I_2$ of their intensities?

A.0.1 ✅

B.0.316

C.1

D.3.16

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The intensity ratio follows $I_1/I_2 = 10^{(\beta_1-\beta_2)/10} = 10^{-10/10}=10^{-1}=0.1$ . Hence the first sound is one‑tenth as intense as the second, which matches option A.

Q3. Raising a sound level from 60 dB to 100 dB corresponds to what multiplicative factor in intensity?

A.100

B.1000

C.10000 ✅

D.100000

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The intensity factor is $10^{(\Delta\beta)/10}=10^{(100-60)/10}=10^{4}=10,000$ . This large increase reflects the logarithmic nature of the decibel scale, so option C is correct.

Q4. Which statement best explains why the decibel scale is preferred over a linear intensity scale for human hearing measurements?

A.It expands the range of values

B.It compresses the range making differences more perceptible

C.It eliminates the need for a reference intensity

D.It converts multiplicative changes into additive ones ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: The decibel scale uses logarithms, turning multiplicative changes in intensity into additive changes in $\beta$ . This property simplifies comparison of sounds that differ by many orders of magnitude, which is why option D correctly describes the advantage.

Q5. A 30 dB increase in sound level represents what factor increase in intensity?

A.10

B.30

C.100

D.1000 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: An increase of $\Delta\beta =30$ dB gives an intensity factor of $10^{\Delta\beta/10}=10^{3}=1,000$ . Thus the intensity grows a thousand‑fold, corresponding to option D.

Q6. Regarding the pH scale and the decibel scale, which of the following statements is true?

A.Both use base‑10 logarithms; pH is a negative log of concentration while dB is a positive log of an intensity ratio ✅

B.Both are defined as

-\log_{10}

of a ratio

C.pH uses natural log while dB uses base‑10

D.Both increase linearly with the underlying physical quantity

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The pH value is defined as $\text{pH} = -\log_{10}[H^+]$ , whereas the decibel level is $\beta = 10\log_{10}(I/I_0)$ . Both rely on base‑10 logarithms, but the signs differ, making option A the accurate comparison.

Q7. With the reference intensity $I_0 =10^{-12}$  W/m $^2$ , what intensity corresponds to a sound level of 0 dB?

10^{-12}

 W/m

^2

✅

10^{-10}

 W/m

^2

C.1 W/m

^2

D.0 W/m

^2

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Setting $\beta =0$ gives $0 =10\log_{10}(I/I_0)$ → $I/I_0 =10^{0}=1$ . Therefore $I = I_0 =10^{-12}$  W/m $^2$ , which matches option A.

Q8. A sensor outputs voltage $V =k\beta$ . If the sound intensity doubles, how does the voltage change approximately (assume the original level is 50 dB)?

A.3 % increase

B.6 % increase ✅

C.10 % increase

D.20 % increase

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Doubling intensity adds $10\log_{10}2 \approx 3.01$  dB to the level, so the new level is $53.01$  dB. The voltage changes from $k·50$ to $k·53.01$ , a relative increase of $3.01/50 ≈0.0602$ or about 6 %, which corresponds to option B.

Q9. Two earthquakes have Richter magnitudes 5.0 and 7.0. How many times greater is the energy release of the 7.0 quake compared to the 5.0 quake?

A.10

B.100 ✅

C.1000

D.10000

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The Richter magnitude $M = \log_{10}(E/E_0)$ . A difference of $\Delta M =2$ implies $E_{7}/E_{5}=10^{2}=100$ . Thus the larger quake releases one hundred times more energy, matching option B.

Q10. In acoustic measurements, what does the abbreviation “dB” stand for?

A.decibel ✅

B.deca‑bel

C.derivative bel

D.digital bel

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The unit dB is short for decibel, a logarithmic unit named after Alexander Graham Bell. It quantifies sound level relative to a reference intensity, making option A the correct definition.

Q11. If the distance from a point source is doubled, how does the sound level in decibels change?

A.Increase by 6 dB

B.Decrease by 3 dB

C.Decrease by 6 dB ✅

D.No change

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Intensity follows the inverse‑square law: $I \propto 1/d^{2}$ . Doubling distance reduces intensity by a factor of 4, giving a decibel change of $10\log_{10}(1/4) ≈ -6.02$  dB. Hence the level decreases by about 6 dB, which is option C.

Q12. System A uses a 20 dB gain amplifier, while System B uses a voltage gain of 2×. Which system provides the larger increase in acoustic intensity?

A.System A ✅

B.System B

C.Both equal

D.Cannot determine

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A 20 dB gain corresponds to an intensity factor of $10^{20/10}=100$ . A voltage gain of 2 multiplies intensity by $2^{2}=4$ . Since 100 > 4, System A yields the greater intensity increase, so option A is correct.

Q13. A tone at 0.5 kHz has a sound level of 70 dB but is perceived as 65 phons. What does this indicate about the relationship between $\beta$ and perceived loudness across frequencies?

A.Loudness equals sound level at all frequencies

B.Loudness is always lower than sound level

C.Loudness differs from sound level at frequencies away from 1 kHz ✅

D.Loudness is unrelated to sound level

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Phons are defined to match decibel levels only near 1 kHz. The discrepancy at 0.5 kHz shows that perceived loudness does not follow the same numerical value as $\beta$ when frequency deviates from 1 kHz, supporting option C.

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