Math 2 Quiz: Rational Exponents In Context

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Rational Exponents In ContextQuestion 1 of 19

A wind turbine's power output is modeled by $P = kv^3$ , where $v$ is wind speed and $k$ is a constant. Due to seasonal weather patterns, the average wind speed changes according to $v(t) = 12t^{1/4}$ mph, where $t$ is measured in months since winter began. What is the growth rate of power output with respect to time?

Power grows proportionally to

t^{1/4}

Power grows proportionally to

t^{3/4}

Power grows proportionally to

t^{12/4}

Power grows proportionally to

t^{4/3}

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Math 2 Quiz

Math 2 Quiz: Rational Exponents In Context

Practice Rational Exponents In Context in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rational Exponents In Context, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Power grows proportionally to $t^{1/4}$
Power grows proportionally to $t^{3/4}$ (correct answer)
Power grows proportionally to $t^{12/4}$
Power grows proportionally to $t^{4/3}$

Explanation: Substituting

v(t) = 12t^{1/4}

into

P = kv^3

P(t) = k(12t^{1/4})^3 = k \cdot 12^3 \cdot t^{3/4}

. Therefore, power grows proportionally to

t^{3/4}

. Choice A uses the exponent from the wind speed function directly. Choice C multiplies the coefficient 12 with the exponent. Choice D incorrectly inverts the fractional exponent.

Question 2

The intensity of radiation from a point source follows $I = \frac{P}{4\pi r^2}$ , where $P$ is power and $r$ is distance. A detector measures intensity as $I(t) = I_0 \cdot t^{-3/2}$ where $t$ represents time in seconds as the source moves away. How does the distance from source to detector change with time?

$r \propto t^{1/4}$
$r \propto t^{3/4}$ (correct answer)
$r \propto t^{2/3}$
$r \propto t^{4/3}$

Explanation: From

I = \frac{P}{4\pi r^2}

, we have

I \propto r^{-2}

, so

r \propto I^{-1/2}

. Given

I(t) = I_0 t^{-3/2}

, we get

r \propto (t^{-3/2})^{-1/2} = t^{3/4}

. Choice A uses exponent

(-3/2) \times (-1/4)

incorrectly. Choice C uses

-(-3/2) \times (-1/3)

. Choice D uses

(-3/2) \times (-2/3)

with wrong sign.

Question 3

An investment grows according to $A(t) = A_0(1 + r)^t$ , where $r$ is the annual interest rate. If the investment triples in value over a period where $t^{2/3} = 4$ , what is the annual growth rate $r$ ?

$3^{1/8} - 1$ (correct answer)
$3^{1/6} - 1$
$3^{3/8} - 1$
$3^{2/3} - 1$

Explanation: From

t^{2/3} = 4

, we get

t = 4^{3/2} = 8

. The investment triples, so

3A_0 = A_0(1+r)^8

, giving

(1+r)^8 = 3

. Therefore

1+r = 3^{1/8}

, so

r = 3^{1/8} - 1

. Choice B uses

t = 6

instead of

t = 8

. Choice C uses

3/8

by incorrectly combining

3

and

1/8

. Choice D uses the exponent

2/3

from the original constraint.

Question 4

The Doppler shift for electromagnetic waves is approximated by $\frac{\Delta f}{f} \approx \frac{v}{c}$ for $v << c$ , where $v$ is relative velocity and $c$ is the speed of light. A satellite's velocity changes as $v(t) = v_0 t^{-1/3}$ due to atmospheric drag. If the initial frequency shift is 0.001%, after what time will the frequency shift be 0.0001%?

$t = 10^{3/2}$ time units
$t = 10^2$ time units
$t = 10^3$ time units (correct answer)
$t = 10^{1/3}$ time units

Explanation: When you encounter Doppler shift problems with changing velocities, you need to track how the frequency shift varies with time through the velocity relationship. Given that

\frac{\Delta f}{f} \approx \frac{v}{c}

and

v(t) = v_0 t^{-1/3}

, the frequency shift becomes

\frac{\Delta f}{f} = \frac{v_0 t^{-1/3}}{c}

. Initially (at

t = 1

), the frequency shift is 0.001%, so:

0.001\% = \frac{v_0}{c}

This means

\frac{v_0}{c} = 10^{-5}

(converting percentage to decimal). At time

t

, when the frequency shift is 0.0001% =

10^{-6}

10^{-6} = \frac{v_0 t^{-1/3}}{c} = 10^{-5} \cdot t^{-1/3}

Solving for

t

10^{-6} = 10^{-5} \cdot t^{-1/3}

10^{-1} = t^{-1/3}

t^{1/3} = 10

t = 10^3

Answer C (

t = 10^3

) is correct. Answer A (

t = 10^{3/2}

) results from incorrectly taking the square root instead of cubing. Answer B (

t = 10^2

) comes from forgetting to account for the

t^{-1/3}

exponent properly. Answer D (

t = 10^{1/3}

) inverts the relationship, treating the exponent as positive rather than negative. Strategy tip: In problems involving power-law time dependencies, always set up the ratio of initial to final conditions first, then solve the resulting equation step-by-step. Pay careful attention to negative exponents—they often reverse intuitive expectations about how quantities change over time.

Question 5

The wavelength of light emitted by a heated object follows Wien's displacement law: $\lambda_{max} = \frac{b}{T}$ , where $b$ is a constant and $T$ is temperature. If temperature varies as $T(t) = T_0 \cdot (2t)^{4/3}$ over time $t$ , how does the peak wavelength change?

$\lambda_{max} \propto t^{-4/3}$
$\lambda_{max} \propto (2t)^{-4/3}$
$\lambda_{max} \propto 2^{4/3}t^{-4/3}$
$\lambda_{max} \propto 2^{-4/3}t^{-4/3}$ (correct answer)

Explanation: Since

\lambda_{max} = \frac{b}{T}

and

T(t) = T_0(2t)^{4/3}

, we have

\lambda_{max}(t) = \frac{b}{T_0(2t)^{4/3}} = \frac{b}{T_0} \cdot (2t)^{-4/3} = \frac{b}{T_0} \cdot 2^{-4/3} \cdot t^{-4/3}

. Therefore

\lambda_{max} \propto 2^{-4/3}t^{-4/3}

. Choice A ignores the factor of 2. Choice B doesn't separate the constant factor. Choice C has wrong sign on the power of 2.

Question 6

A manufacturing process scales production cost according to $C = k \cdot n^{3/4}$ , where $n$ is the number of units and $k$ is a constant. If the company wants to reduce per-unit cost by 20%, by what factor must production volume increase?

$(1.25)^{4/3}$
$(1.25)^3$
$(0.8)^{-4}$
$(1.25)^4$ (correct answer)

Explanation: Per-unit cost is

\frac{C}{n} = \frac{kn^{3/4}}{n} = kn^{-1/4}

. For 20% reduction, new per-unit cost is

0.8

times original. So

kn_{new}^{-1/4} = 0.8 \cdot kn_{old}^{-1/4}

, giving

\left(\frac{n_{new}}{n_{old}}\right)^{-1/4} = 0.8

. Therefore

\frac{n_{new}}{n_{old}} = (0.8)^{-4} = \left(\frac{4}{5}\right)^{-4} = \left(\frac{5}{4}\right)^4 = (1.25)^4

. Choice A uses wrong exponent. Choice B uses exponent 3 instead of 4. Choice C doesn't invert properly.

Question 7

The half-life of a radioactive isotope can be expressed as $t_{1/2} = \frac{\ln(2)}{\lambda}$ , where $\lambda$ is the decay constant. In a particular experiment, the decay constant follows $\lambda(T) = \lambda_0 T^{2/3}$ , where $T$ is temperature. If temperature is reduced by a factor of 8, how does the half-life change?

Half-life increases by a factor of 2
Half-life increases by a factor of 4 (correct answer)
Half-life increases by a factor of 8
Half-life decreases by a factor of 4

Explanation: When temperature reduces by factor 8:

T_{new} = \frac{T_{old}}{8}

. Then

\lambda_{new} = \lambda_0 \left(\frac{T_{old}}{8}\right)^{2/3} = \lambda_0 T_{old}^{2/3} \cdot 8^{-2/3} = \lambda_{old} \cdot (2^3)^{-2/3} = \lambda_{old} \cdot 2^{-2} = \frac{\lambda_{old}}{4}

. Since

t_{1/2} \propto \frac{1}{\lambda}

, the new half-life is

\frac{t_{1/2,old}}{\lambda_{old}/4} = 4t_{1/2,old}

. Choice A uses factor 2 instead of 4. Choice C uses the temperature reduction factor directly. Choice D inverts the relationship.

Question 8

The resonant frequency of a vibrating string is given by $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$ , where $L$ is length, $T$ is tension, and $\mu$ is mass per unit length. If the tension increases according to $T(t) = T_0 \cdot 4^t$ and length decreases as $L(t) = L_0 \cdot 2^{-t/2}$ , how does frequency change with time?

$f \propto 4^{t/2}$
$f \propto 2^{t/2}$
$f \propto 2^{3t/2}$ (correct answer)
$f \propto 2^{2t}$

Explanation: When you encounter physics formulas with multiple variables changing over time, you need to substitute the time-dependent expressions and use exponent rules to simplify. Starting with the resonant frequency formula

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

, substitute the given time-dependent functions:

T(t) = T_0 \cdot 4^t

and

L(t) = L_0 \cdot 2^{-t/2}

. Since

\mu

remains constant, we get:

f(t) = \frac{1}{2L_0 \cdot 2^{-t/2}}\sqrt{\frac{T_0 \cdot 4^t}{\mu}}

Simplifying the length term:

\frac{1}{2^{-t/2}} = 2^{t/2}

For the square root term, since

4^t = (2^2)^t = 2^{2t}

\sqrt{4^t} = \sqrt{2^{2t}} = 2^t

Combining everything:

f(t) = \frac{2^{t/2}}{2L_0} \cdot 2^t \cdot \sqrt{\frac{T_0}{\mu}} = \text{constant} \cdot 2^{t/2 + t} = \text{constant} \cdot 2^{3t/2}

Therefore,

f \propto 2^{3t/2}

, making C correct. Choice A (

4^{t/2}

) incorrectly keeps the base 4 instead of converting to base 2. Choice B (

2^{t/2}

) only accounts for the length change, ignoring the tension effect. Choice D (

2^{2t}

) incorrectly handles the square root of the tension term. Key strategy: When dealing with exponential functions in physics, convert everything to the same base (usually 2 or e) before combining exponents. This prevents calculation errors and makes the algebra much cleaner.

Question 9

The efficiency of a solar panel is modeled by $E = 0.2T^{-1/2}$ , where $T$ is temperature in Kelvin above 300K. If the temperature increases from 50K above baseline to 200K above baseline, what happens to the efficiency?

It decreases by a factor of $2$ (correct answer)
It decreases by a factor of $\sqrt{2}$
It decreases by a factor of $4$
It increases by a factor of $2$

Explanation: Initial efficiency:

E_1 = 0.2(50)^{-1/2} = \frac{0.2}{\sqrt{50}}

. Final efficiency:

E_2 = 0.2(200)^{-1/2} = \frac{0.2}{\sqrt{200}}

. The ratio is

\frac{E_1}{E_2} = \frac{\sqrt{200}}{\sqrt{50}} = \sqrt{\frac{200}{50}} = \sqrt{4} = 2

. Since

E_2 = \frac{E_1}{2}

, efficiency decreases by factor 2. Choice B confuses the square root in the calculation. Choice C squares the factor. Choice D ignores the negative exponent.

Question 10

The surface area of a spherical balloon is modeled by $S = 4\pi r^2$ , where $r$ is the radius. If the balloon is inflated so that its volume increases by a factor of 8, by what factor does its surface area increase?

2
4 (correct answer)
6
8

Explanation: Volume of a sphere is

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

. If volume increases by factor 8, then

\frac{4}{3}\pi r_{new}^3 = 8 \cdot \frac{4}{3}\pi r_{old}^3

, so

r_{new}^3 = 8r_{old}^3

, giving

r_{new} = 2r_{old}

. Surface area scales as

r^2

, so the new surface area is

4\pi(2r_{old})^2 = 4 \cdot 4\pi r_{old}^2 = 4

times the original. Choice A confuses radius scaling with surface area scaling. Choice C adds radius factor and surface area factor. Choice D uses the volume scaling factor directly.

Question 11

The gravitational force between two masses follows $F = \frac{Gm_1m_2}{r^2}$ . In a binary star system, the separation distance changes as $r(t) = r_0 t^{3/4}$ due to orbital decay. If one star's mass grows as $m_1(t) = m_0 t^{1/2}$ while the other remains constant at $m_2$ , how does the gravitational force change over time?

$F \propto t^{-2}$
$F \propto t^{-3/2}$
$F \propto t^{-1}$ (correct answer)
$F \propto t^{1/2}$

Explanation: When you encounter problems involving functions that change over time, the key is to substitute all the time-dependent expressions into the original equation and carefully track how the exponents combine. Starting with the gravitational force equation

F = \frac{Gm_1m_2}{r^2}

, you need to substitute the given time dependencies:

m_1(t) = m_0 t^{1/2}

m_2

(constant), and

r(t) = r_0 t^{3/4}

. Substituting these into the force equation:

F(t) = \frac{G \cdot m_0 t^{1/2} \cdot m_2}{(r_0 t^{3/4})^2} = \frac{Gm_0 m_2 t^{1/2}}{r_0^2 t^{3/2}}

Simplifying the exponents:

F(t) = \frac{Gm_0 m_2}{r_0^2} \cdot t^{1/2 - 3/2} = \frac{Gm_0 m_2}{r_0^2} \cdot t^{-1}

Therefore,

F \propto t^{-1}

, making C correct. Choice A (

t^{-2}

) would result if you only considered the changing distance and ignored the mass growth entirely. Choice B (

t^{-3/2}

) represents the isolated effect of distance change alone, since

(t^{3/4})^2 = t^{3/2}

in the denominator. Choice D (

t^{1/2}

) incorrectly treats the growing mass as if it were in the numerator without accounting for the distance effect. Study tip: In physics problems with multiple time-dependent variables, write out each substitution explicitly and combine exponents algebraically. Don't try to reason about the "net effect" mentally—the math will guide you to the right answer.

Question 12

A pharmaceutical company models the bioavailability of a drug using the function $B(t) = 100 \cdot t^{2/3}$ , where $B(t)$ represents the percentage of drug absorbed after $t$ hours. If the bioavailability increases from 400% to 500%, what is the ratio of the corresponding time values?

$\left(\frac{5}{4}\right)^{3/2}$ (correct answer)
$\left(\frac{5}{4}\right)^{2/3}$
$\left(\frac{5}{4}\right)^{1/2}$
$\left(\frac{5}{4}\right)^{3/4}$

Explanation: Setting up equations:

400 = 100t_1^{2/3}

gives

t_1^{2/3} = 4

, so

t_1 = 4^{3/2} = 8

. Similarly,

500 = 100t_2^{2/3}

gives

t_2^{2/3} = 5

, so

t_2 = 5^{3/2}

. The ratio is

\frac{t_2}{t_1} = \frac{5^{3/2}}{4^{3/2}} = \left(\frac{5}{4}\right)^{3/2}

. Choice B uses the original exponent 2/3. Choice C uses 1/2 instead of 3/2. Choice D incorrectly combines exponents as 3/4.

Question 13

A spherical balloon's volume increases according to $V = 10t^{3/2}$ cubic inches, where $t$ is time in minutes. Since volume of a sphere is $V = \frac{4}{3}\pi r^3$ , which expression correctly represents how the radius changes with time?

$r = \left(\frac{30t^{3/2}}{4\pi}\right)^{1/3} = \left(\frac{30}{4\pi}\right)^{1/3} \cdot t^{1/2}$ (correct answer)
$r = \left(\frac{40t^{3/2}}{3\pi}\right)^{1/3} = \left(\frac{40}{3\pi}\right)^{1/3} \cdot t^{1/2}$
$r = \left(\frac{30t^{3/2}}{4\pi}\right)^{1/3} = \left(\frac{30}{4\pi}\right)^{1/3} \cdot t^{3/2}$
$r = \left(\frac{10t^{3/2}}{\pi}\right)^{1/3} = \left(\frac{10}{\pi}\right)^{1/3} \cdot t^{1/2}$

Explanation: Setting

10t^{3/2} = \frac{4}{3}\pi r^3

and solving for

r

r^3 = \frac{30t^{3/2}}{4\pi}

, so

r = \left(\frac{30t^{3/2}}{4\pi}\right)^{1/3}

. Using exponent rules:

r = \left(\frac{30}{4\pi}\right)^{1/3} \cdot (t^{3/2})^{1/3} = \left(\frac{30}{4\pi}\right)^{1/3} \cdot t^{1/2}

. Choice B has wrong coefficient (40 instead of 30). Choice C has wrong final exponent on

t

. Choice D omits the factor of 4/3 from the sphere volume formula.

Question 14

The intensity of light decreases with distance according to $I = I_0 \cdot d^{-2/3}$ , where $I_0$ is the initial intensity and $d$ is the distance. If the intensity at distance 8 units is 64 lumens, and the intensity at distance 27 units is $k$ lumens, what is the value of $k$ ?

$k = 64 \cdot \left(\frac{8}{27}\right)^{2/3} = 64 \cdot \frac{4}{9} = \frac{256}{9}$ (correct answer)
$k = 64 \cdot \left(\frac{27}{8}\right)^{2/3} = 64 \cdot \frac{9}{4} = 144$
$k = 64 \cdot \left(\frac{8}{27}\right)^{3/2} = 64 \cdot \frac{8}{81} = \frac{64}{9}$
$k = 64 \cdot \left(\frac{27}{8}\right)^{3/2} = 64 \cdot \frac{27}{16} = 108$

Explanation: Using the ratio of intensities:

\frac{I_{27}}{I_8} = \frac{d_8^{2/3}}{d_{27}^{2/3}} = \left(\frac{8}{27}\right)^{2/3}

. Since

8 = 2^3

and

27 = 3^3

, we have

\left(\frac{8}{27}\right)^{2/3} = \left(\frac{2^3}{3^3}\right)^{2/3} = \frac{2^2}{3^2} = \frac{4}{9}

. Therefore

k = 64 \cdot \frac{4}{9} = \frac{256}{9}

. Choice B inverts the ratio. Choices C and D use the wrong exponent (3/2 instead of 2/3).

Question 15

A chemical reaction follows the rate law $R = k[A]^{3/4}[B]^{1/2}$ , where $[A]$ and $[B]$ are concentrations. If $[A]$ is doubled while $[B]$ is reduced to one-fourth its original value, by what factor does the reaction rate change?

The rate increases by a factor of $2^{3/4} \cdot \left(\frac{1}{4}\right)^{1/2} = \frac{2^{3/4}}{2} = 2^{-1/4} = \frac{1}{\sqrt[4]{2}}$
The rate increases by a factor of $2^{3/4} \cdot \left(\frac{1}{4}\right)^{1/2} = \frac{\sqrt[4]{8}}{2} = \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$ (correct answer)
The rate decreases by a factor of $2^{4/3} \cdot \left(\frac{1}{4}\right)^{2} = \frac{2^{4/3}}{16} = \frac{\sqrt[3]{16}}{16}$
The rate increases by a factor of $2^{3/4} \cdot 4^{1/2} = \sqrt[4]{8} \cdot 2 = 4\sqrt[4]{2}$

Explanation: The rate changes by the factor

(2)^{3/4} \cdot \left(\frac{1}{4}\right)^{1/2}

. Since

\left(\frac{1}{4}\right)^{1/2} = \frac{1}{2}

and

2^{3/4} = (2^3)^{1/4} = \sqrt[4]{8}

, the factor is

\frac{\sqrt[4]{8}}{2}

. Since

\sqrt[4]{8} = \sqrt[4]{4 \cdot 2} = \sqrt{2}

, we get

\frac{\sqrt{2}}{2}

. Choice A has an error in simplification. Choice C uses wrong exponents (4/3 and 2 instead of 3/4 and 1/2). Choice D incorrectly uses

4^{1/2}

instead of

(1/4)^{1/2}

Question 16

An investment grows according to $A = P(1 + r)^t$ , where $P$ is principal, $r$ is the annual rate, and $t$ is time in years. If an account triples every 4 years, what is the equivalent continuous compounding rate $k$ in the formula $A = Pe^{kt}$ ?

$k = \frac{\ln(3^{1/4})}{1} = \frac{\ln(3)}{4}$ per year, but this gives the wrong equivalence for the 4-year period
$k = \frac{4}{\ln(3)}$ , since setting $3P = Pe^{4k}$ gives $\ln(3) = 4k$ and $k = \frac{4}{\ln(3)}$
$k = \ln(3) \cdot 4 = 4\ln(3)$ , since the growth factor per year is $3^{1/4}$ compounded continuously
$k = \frac{\ln(3)}{4}$ , since setting $3P = Pe^{4k}$ gives $3 = e^{4k}$ and $k = \frac{\ln(3)}{4}$ (correct answer)

Explanation: When you encounter problems involving equivalent interest rates between discrete and continuous compounding, the key is setting the two formulas equal and solving for the unknown rate. Since the account triples every 4 years, after 4 years we have

A = 3P

. To find the equivalent continuous compounding rate, we set the two formulas equal at

t = 4

3P = Pe^{4k}

Dividing both sides by

P

3 = e^{4k}

Taking the natural logarithm of both sides:

\ln(3) = 4k

Solving for

k

k = \frac{\ln(3)}{4}

This confirms answer D is correct. Let's examine why the other options are wrong: Option A correctly calculates

k = \frac{\ln(3)}{4}

but then claims this gives "wrong equivalence for the 4-year period," which is false—this is exactly the right equivalence. Option B makes an algebraic error, incorrectly solving

\ln(3) = 4k

to get

k = \frac{4}{\ln(3)}

instead of

k = \frac{\ln(3)}{4}

. Option C misunderstands the relationship entirely, multiplying

\ln(3) \times 4

instead of dividing. While it's true that the annual growth factor is

3^{1/4}

, this doesn't translate to

k = 4\ln(3)

in continuous compounding. Study tip: For equivalent rate problems, always set the two expressions equal at a specific time point where you know the relationship, then solve algebraically. Don't try to reason through growth factors—stick to the direct algebraic approach.

Question 17

The surface area of a cube with side length $s$ is $6s^2$ . If the volume of a cube increases by a factor of 27, by what factor does its surface area increase?

The surface area increases by a factor of 27, since both volume and surface area scale proportionally with the overall size change of the cube
The surface area increases by a factor of 18, since if volume increases by 27, the total change in surface area is $27^{2/3} = 9$ times 2 faces per dimension
The surface area increases by a factor of $27^{2/3} = 9$ , but this assumes linear scaling which doesn't apply to three-dimensional surface relationships
The surface area increases by a factor of 9, since if volume increases by 27, each side length increases by $27^{1/3} = 3$ , and surface area scales as $s^2$ , giving $3^2 = 9$ (correct answer)

Explanation: When you encounter problems about scaling geometric figures, the key is understanding how different measurements scale at different rates. Linear dimensions scale by one factor, areas by the square of that factor, and volumes by the cube of that factor. Let's work through this step-by-step. A cube's volume is

s^3

, so if the volume increases by a factor of 27, we have: new volume =

27s^3

. This means the new side length must satisfy

(s_{new})^3 = 27s^3

. Taking the cube root of both sides:

s_{new} = 27^{1/3} \cdot s = 3s

. So each side length increases by a factor of 3. Since surface area scales as the square of linear dimensions, and each side length increased by a factor of 3, the surface area increases by a factor of

3^2 = 9

. Answer A incorrectly assumes volume and surface area scale proportionally—they don't. Volume scales as length cubed while surface area scales as length squared. Answer B makes an error in both calculation and reasoning, incorrectly applying

27^{2/3}

and then multiplying by 2. Answer C correctly calculates

27^{2/3} = 9

but then dismisses this correct result with confused reasoning about "linear scaling not applying to three-dimensional relationships." Answer D correctly identifies the logical sequence: volume factor of 27 → side length factor of

27^{1/3} = 3

→ surface area factor of

3^2 = 9

. Study tip: Remember the scaling hierarchy: linear dimensions scale by factor

k

, areas by

k^2

, and volumes by

k^3

. Always find the linear scaling factor first, then apply the appropriate power.

Question 18

The brightness of a star as observed from Earth follows $B = L \cdot d^{-2}$ , where $L$ is intrinsic luminosity and $d$ is distance. Two stars have the same intrinsic luminosity, but Star A appears 16 times brighter than Star B. If Star B is 100 light-years away, how far is Star A?

Star A is 12.5 light-years away, since $\frac{B_A}{B_B} = \frac{d_B^2}{d_A^2} = 16$ gives $d_A^2 = \frac{100^2}{16} = 625$ and $d_A = 25$ is wrong
Star A is 50 light-years away, since $\frac{B_A}{B_B} = \frac{d_B}{d_A} = 16$ gives $\frac{100}{d_A} = 16$ and $d_A = \frac{100}{16} = 6.25$ , but this is wrong
Star A is 25 light-years away, since $\frac{B_A}{B_B} = \frac{d_B^2}{d_A^2} = 16$ gives $\frac{100^2}{d_A^2} = 16$ and $d_A = \frac{100}{4} = 25$ (correct answer)
Star A is 6.25 light-years away, since brightness is inversely proportional to distance, so $d_A = \frac{100}{16} = 6.25$ light-years from the ratio

Explanation: When you encounter inverse square law problems, remember that brightness depends on the square of the distance, not just distance itself. The formula

B = L \cdot d^{-2}

means brightness is inversely proportional to distance squared. Since both stars have the same luminosity

L

, you can set up a ratio:

\frac{B_A}{B_B} = \frac{L \cdot d_A^{-2}}{L \cdot d_B^{-2}} = \frac{d_B^2}{d_A^2}

Given that Star A appears 16 times brighter than Star B, you have:

\frac{B_A}{B_B} = 16 = \frac{d_B^2}{d_A^2}

Substituting

d_B = 100

light-years:

16 = \frac{100^2}{d_A^2} = \frac{10000}{d_A^2}

Solving:

d_A^2 = \frac{10000}{16} = 625

, so

d_A = 25

light-years. Choice A makes an arithmetic error in the final calculation, getting 12.5 instead of 25. Choice B incorrectly assumes brightness is inversely proportional to distance (not distance squared), leading to the wrong relationship

\frac{B_A}{B_B} = \frac{d_B}{d_A}

. Choice D makes the same conceptual error as B, treating this as a simple inverse proportion rather than an inverse square relationship. Study tip: Inverse square laws appear frequently in physics problems involving light, gravity, and electromagnetic fields. Always remember the "square" part—if something follows an inverse square law and one quantity changes by a factor of 4, the distance changes by a factor of 2 (since

2^2 = 4

Question 19

The pressure and volume of a gas are related by $PV^{\gamma} = k$ , where $\gamma = 1.4$ and $k$ is constant. If the volume is compressed to $\frac{1}{8}$ of its original value, by what factor does the pressure increase?

Pressure increases by factor $\left(\frac{1}{8}\right)^{1.4} = 8^{-1.4} \approx 0.089$ , so pressure actually decreases significantly
Pressure increases by factor $8^{1.4} = 8^{1.4} \approx 11.3$ , using the approximation $(1/8)^{-1.4} = 8^{1.4}$
Pressure increases by factor $8^{1.4} = 8^{7/5} = (2^3)^{7/5} = 2^{21/5} = 2^4 \cdot 2^{1/5} = 16\sqrt[5]{2} \approx 18.9$ (correct answer)
Pressure increases by factor $8 \times 1.4 = 11.2$ , since the exponent $\gamma$ multiplies the volume compression ratio directly

Explanation: When you encounter gas laws with exponential relationships, you're dealing with thermodynamic processes where small volume changes can produce dramatic pressure changes due to the exponential nature of the equation. Starting with

PV^{\gamma} = k

, if volume changes from

V

\frac{V}{8}

, you need to find how pressure changes. Since the constant

k

remains the same, you can write:

P_1 V^{1.4} = P_2 \left(\frac{V}{8}\right)^{1.4}

Solving for the pressure ratio:

\frac{P_2}{P_1} = \frac{V^{1.4}}{(V/8)^{1.4}} = \frac{V^{1.4}}{V^{1.4}/8^{1.4}} = 8^{1.4}

To evaluate

8^{1.4} = 8^{7/5}

, rewrite

8 = 2^3

(2^3)^{7/5} = 2^{21/5} = 2^4 \cdot 2^{1/5} = 16\sqrt[5]{2}

. Since

\sqrt[5]{2} \approx 1.149

, this gives approximately

18.9

. Choice A incorrectly uses

(1/8)^{1.4}

, which would apply if volume increased by factor 8 rather than decreased. Choice B makes the right substitution but uses an incorrect approximation, missing the precise calculation. Choice D completely misunderstands the relationship, treating the exponent as a simple multiplier rather than recognizing the exponential nature of the equation. Strategy tip: In exponential relationships like gas laws, always set up the ratio carefully and remember that compression (volume decreasing) means the reciprocal gets raised to the power, leading to pressure increases that can be much larger than the compression ratio itself.