All practice questions

Find the real value of $x$ that satisfies the logarithmic equation $\log_2(x^2 + 3x - 22) - \log_2(x - 2) = 3$ .

Show answer & explanation

Answer: 6

Answer

The real value of

x

that satisfies the equation is

6

.

Applying the logarithmic quotient rule transforms

\log_2(x^2 + 3x - 22) - \log_2(x - 2) = 3

into

\log_2\left(\frac{x^2 + 3x - 22}{x - 2}\right) = 3

. Expressing this in exponential form yields

\frac{x^2 + 3x - 22}{x - 2} = 8

, which simplifies to

x^2 - 5x - 6 = 0

. Factoring gives

(x - 6)(x + 1) = 0

. Since the argument of a logarithm must be strictly positive (

x - 2 > 0 \implies x > 2

),

x = -1

is extraneous and

x = 6

is the only valid solution.

Step-by-Step Solution

1

Combine the logarithmic terms using the quotient law of logarithms.

\log_2\left(\frac{x^2 + 3x - 22}{x - 2}\right) = 3

The difference of two logarithms with the same base equals the logarithm of their quotient:

\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right)

.

2

Convert the logarithmic equation into its equivalent exponential form.

\frac{x^2 + 3x - 22}{x - 2} = 2^3 = 8

If

\log_b Y = c

, then

Y = b^c

.

3

Clear the denominator and simplify to form a quadratic equation.

x^2 - 5x - 6 = 0

Multiplying both sides by

(x - 2)

yields

x^2 + 3x - 22 = 8x - 16

, which rearranges to

x^2 - 5x - 6 = 0

.

4

Solve the quadratic equation by factoring.

(x - 6)(x + 1) = 0 \implies x = 6 \text{ or } x = -1

The factors of

-6

that sum to

-5

are

-6

and

1

.

5

Test roots against domain restrictions to eliminate extraneous solutions.

x = 6

Logarithmic arguments must be strictly positive. For

\log_2(x - 2)

to be defined,

x > 2

. Thus,

x = -1

is extraneous, leaving

x = 6

as the unique valid solution.

Key Concept

Logarithmic Equations and Domain Restrictions

Question 7802Question →

A faulty liquid-in-glass thermometer registers a reading of $5^\circ\text{C}$ at the melting ice point and $95^\circ\text{C}$ at the steam point of pure water under standard atmospheric pressure. What is the actual temperature in degrees Celsius when this thermometer registers a reading of $41^\circ\text{C}$ ?

36^\circ\text{C}

40^\circ\text{C}

41^\circ\text{C}

45.6^\circ\text{C}

Show answer & explanation

Answer: $40^\circ\text{C}$

Answer

The actual temperature is

40^\circ\text{C}

.

The correct answer is obtained by setting up the linear interpolation formula for thermometric property values:

\theta = \frac{X_\theta - X_0}{X_{100} - X_0} \times 100

. Substituting

X_0 = 5

,

X_{100} = 95

, and

X_\theta = 41

gives

\theta = \frac{36}{90} \times 100 = 40^\circ\text{C}

.

Step-by-Step Solution

1

Determine the fundamental interval of the faulty thermometer.

Fundamental interval

= 95^\circ\text{C} - 5^\circ\text{C} = 90\,\text{divisions}

.

The total interval between the lower fixed point and upper fixed point represents

100^\circ\text{C}

on the standard Celsius scale.

2

Calculate the measured change from the ice point.

Measured difference

= 41^\circ\text{C} - 5^\circ\text{C} = 36\,\text{divisions}

.

The zero error of

+5^\circ\text{C}

must be subtracted from the observed reading.

3

Apply the linear scale interpolation formula to find the actual temperature

\theta

.

\theta = \frac{41 - 5}{95 - 5} \times 100^\circ\text{C} = \frac{36}{90} \times 100^\circ\text{C} = 40^\circ\text{C}

.

The ratio of the measured interval to the total fundamental interval equals the true fraction of

100^\circ\text{C}

.

Key Concept

Linear interpolation on non-standard or faulty thermometer scales using fixed points.

Question 7803Question →

What is the $x$ -coordinate of the maximum stationary point of the curve $y = \sin(2x) - x$ in the interval $0 \le x \le \pi$ ?

\frac{\pi}{6}

\frac{5\pi}{6}

\frac{\pi}{3}

\pi

Show answer & explanation

Answer: $\frac{\pi}{6}$

Answer

The

x

-coordinate of the maximum stationary point is

\frac{\pi}{6}

.

Differentiating

y = \sin(2x) - x

gives

\frac{dy}{dx} = 2\cos(2x) - 1

. Setting this derivative to zero yields

\cos(2x) = \frac{1}{2}

, giving

x = \frac{\pi}{6}

and

x = \frac{5\pi}{6}

in the given interval. Checking the second derivative

\frac{d^2y}{dx^2} = -4\sin(2x)

at

x = \frac{\pi}{6}

gives

-2\sqrt{3} < 0

, which confirms that the maximum stationary point occurs at

\frac{\pi}{6}

.

Step-by-Step Solution

1

Find the first derivative of

y = \sin(2x) - x

using the chain rule.

\frac{dy}{dx} = 2\cos(2x) - 1

Stationary points occur where the rate of change

\frac{dy}{dx} = 0

.

2

Set

\frac{dy}{dx} = 0

and solve for

x

within

0 \le x \le \pi

.

2\cos(2x) - 1 = 0 \implies \cos(2x) = \frac{1}{2} \implies 2x = \frac{\pi}{3}

or

2x = \frac{5\pi}{3} \implies x = \frac{\pi}{6}

or

x = \frac{5\pi}{6}

The trigonometric equation

\cos(\theta) = \frac{1}{2}

has solutions

\frac{\pi}{3}

and

\frac{5\pi}{3}

in

[0, 2\pi]

.

3

Evaluate the second derivative

\frac{d^2y}{dx^2}

to determine the nature of the stationary points.

\frac{d^2y}{dx^2} = -4\sin(2x)

. At

x = \frac{\pi}{6}

,

\frac{d^2y}{dx^2} = -4\sin\left(\frac{\pi}{3}\right) = -2\sqrt{3} < 0

.

A negative second derivative (

\frac{d^2y}{dx^2} < 0

) indicates a local maximum.

Key Concept

Determining maximum stationary points of trigonometric functions using the first and second derivative tests.

Question 7804Question →

A straight line $L$ passes through the point $(2, -3)$ and is perpendicular to the line $4x + 5y - 20 = 0$ . Calculate the $x$ -intercept of line $L$ .

Show answer & explanation

Answer: 4.4

Answer

The

x

-intercept of line

L

is

4.4

(or

\frac{22}{5}

).

The given line

4x + 5y - 20 = 0

has a slope of

-\frac{4}{5}

. The line perpendicular to it must have a slope equal to the negative reciprocal, which is

\frac{5}{4}

. Using the point-slope formula with point

(2, -3)

, the equation of line

L

is

y + 3 = \frac{5}{4}(x - 2)

, simplifying to

y = \frac{5}{4}x - \frac{11}{2}

. Setting

y = 0

yields

\frac{5}{4}x = \frac{11}{2}

, giving an

x

-intercept of

x = \frac{22}{5} = 4.4

.

Step-by-Step Solution

1

Determine the gradient of the given line

4x + 5y - 20 = 0

Gradient

m_1 = -\frac{4}{5}

Converting to slope-intercept form

y = -\frac{4}{5}x + 4

reveals the slope.

2

Calculate the perpendicular gradient for line

L

Gradient

m = \frac{5}{4}

Perpendicular lines have negative reciprocal gradients (

m_1 \cdot m_2 = -1

).

3

Derive the equation of line

L

using point

(2, -3)

y = \frac{5}{4}x - \frac{11}{2}

Substitute the point

(2, -3)

and gradient

m = \frac{5}{4}

into point-slope form.

4

Solve for the

x

-intercept by setting

y = 0

x = 4.4

The

x

-intercept is defined as the point where the line crosses the

x

-axis (

y = 0

).

Key Concept

Perpendicular line slope relationships and x-intercept calculations

Estimated Time:1m 30s

Question 7805Question →

A mechanical wave propagating along a string is defined by the displacement equation $y = 0.05 \sin\left(160\pi t - 8\pi x + \frac{\pi}{3}\right)$ , where $x$ and $y$ are in meters and $t$ is in seconds. The wave transitions into a different section of string where its propagation speed drops by $25\%$ . Calculate the minimum distance (in meters) between two points in this second section that have a phase difference of $\frac{2\pi}{3}\text{ rad}$ .

Show answer & explanation

Answer: 0.0625

Answer

The minimum distance between the two points in the second section is

0.0625\text{ m}

.

The correct calculation gives

0.0625\text{ m}

. Comparing

y = 0.05 \sin(160\pi t - 8\pi x + \pi/3)

with the general form

y = A \sin(\omega t - k x + \phi_0)

identifies

\omega = 160\pi\text{ rad/s}

and

k_1 = 8\pi\text{ rad/m}

, yielding an initial speed of

v_1 = \omega / k_1 = 20\text{ m/s}

. Upon transitioning into the second string section, the speed drops by

25\%

to

v_2 = 15\text{ m/s}

. Since the angular frequency

\omega = 160\pi\text{ rad/s}

remains invariant during refraction, the wave number in the second section is

k_2 = \omega / v_2 = 160\pi / 15 = 32\pi / 3\text{ rad/m}

. Substituting

k_2

and the given phase difference

\Delta \phi = 2\pi / 3\text{ rad}

into

\Delta \phi = k_2 \Delta x

yields

\Delta x = (2\pi / 3) / (32\pi / 3) = 2/32 = 0.0625\text{ m}

.

Step-by-Step Solution

1

Extract angular frequency and wave number from the displacement equation

\omega = 160\pi\text{ rad/s}

and

k_1 = 8\pi\text{ rad/m}

The standard progressive wave equation is formatted as

y = A \sin(\omega t - k x + \phi_0)

.

2

Calculate the initial wave propagation speed

v_1

v_1 = \frac{\omega}{k_1} = \frac{160\pi}{8\pi} = 20\text{ m/s}

Wave speed is equal to the ratio of angular frequency to wave number.

3

Determine the wave speed

v_2

in the second section

v_2 = 20 \times (1 - 0.25) = 15\text{ m/s}

The wave speed decreases by

25\%

, making

v_2 = 0.75 v_1

.

4

Find the new wave number

k_2

in the second section

k_2 = \frac{\omega}{v_2} = \frac{160\pi}{15} = \frac{32\pi}{3}\text{ rad/m}

Frequency and angular frequency remain invariant when a wave passes from one medium to another.

5

Calculate the spatial separation

\Delta x

for the specified phase difference

\Delta x = \frac{\Delta \phi}{k_2} = \frac{2\pi / 3}{32\pi / 3} = \frac{2}{32} = 0.0625\text{ m}

Phase difference relates to spatial distance via

\Delta \phi = k \Delta x

.

Key Concept

Wave Equation Parameter Extraction and Invariance of Frequency in Refraction

Question 7806Question →

A metal container with an initial volume of $500\text{ cm}^3$ is completely filled with an organic liquid at $15^\circ\text{C}$ . The real cubic expansivity of the liquid is $4.0 \times 10^{-4}\text{ K}^{-1}$ . When the container and liquid are heated together to a final temperature of $65^\circ\text{C}$ , exactly $8.5\text{ cm}^3$ of the liquid overflows. What is the linear expansivity of the metal container, in units of $10^{-5}\text{ K}^{-1}$ ?

Show answer & explanation

Answer: 2

Answer

The linear expansivity of the metal container is

2.0 \times 10^{-5}\text{ K}^{-1}

, which corresponds to a numerical value of

2.0

in units of

10^{-5}\text{ K}^{-1}

.

The apparent expansivity of the liquid is derived from the overflow volume (8.5 cm^3 / (500 cm^3 * 50 K) = 3.4 x 10^-4 K^-1). Subtracting this from the liquid's real expansivity (4.0 x 10^-4 K^-1) yields the container's cubic expansivity of 0.6 x 10^-4 K^-1 (or 6.0 x 10^-5 K^-1). Dividing by 3 gives the linear expansivity of the container material as 2.0 x 10^-5 K^-1.

Step-by-Step Solution

1

Determine the temperature increase

\Delta T = 50\text{ K}

Temperature change drives the thermal expansion process.

2

Compute the apparent cubic expansivity of the liquid

\gamma_a = 3.4 \times 10^{-4}\text{ K}^{-1}

Apparent expansion is measured directly from the liquid overflow relative to initial volume and temperature rise.

3

Calculate the cubic expansivity of the metal container

\gamma_v = 6.0 \times 10^{-5}\text{ K}^{-1}

The difference between real cubic expansivity of the liquid and its apparent cubic expansivity equals the cubic expansivity of the container.

4

Calculate the linear expansivity of the container material

\alpha = 2.0 \times 10^{-5}\text{ K}^{-1}

For isotropic solids, volume expansivity is three times linear expansivity (\gamma_v = 3\alpha).

Key Concept

Thermal expansion of liquids in expanding vessels: Real expansivity equals apparent expansivity plus vessel cubic expansivity (\gamma_r = \gamma_a + 3\alpha).

Estimated Time:2m 0s

Question 7807Question →

If $y = (2x^2 + 1)^3$ , what is the numerical value of $\frac{dy}{dx}$ evaluated at $x = 1$ ?

27

54

108

36

Show answer & explanation

Answer: $108$

Answer

108

Differentiating

y = (2x^2 + 1)^3

using the chain rule yields

\frac{dy}{dx} = 3(2x^2 + 1)^2 \cdot 4x = 12x(2x^2 + 1)^2

. Substituting

x = 1

gives

12(1)(3)^2 = 108

.

Step-by-Step Solution

1

Identify inner and outer functions for the composite expression

Let

u = 2x^2 + 1

, so

y = u^3

.

The function requires the application of the Chain Rule.

2

Compute the derivatives of the outer and inner functions

\frac{dy}{du} = 3u^2 = 3(2x^2 + 1)^2

and

\frac{du}{dx} = 4x

.

Apply the power rule to both components.

3

Apply the Chain Rule formula

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

\frac{dy}{dx} = 3(2x^2 + 1)^2 \cdot (4x) = 12x(2x^2 + 1)^2

.

Multiply the outer derivative by the inner derivative.

4

Substitute

x = 1

into the derivative

\frac{dy}{dx}\Big|_{x=1} = 12(1)(2(1)^2 + 1)^2 = 12(3)^2 = 12 \cdot 9 = 108

.

Evaluate at the given x-value.

Key Concept

Chain Rule of Differentiation

Estimated Time:1m 0s

Question 7808Question →

A platinum resistance thermometer has a resistance of $10\,\Omega$ at the ice point ( $0^\circ\text{C}$ ) and $50\,\Omega$ at the steam point ( $100^\circ\text{C}$ ). When placed in a warm liquid bath, an uncalibrated digital ohmmeter reads $38\,\Omega$ . If the ohmmeter has a known positive zero error of $+4\,\Omega$ (indicating $4\,\Omega$ above the true resistance), what is the actual temperature of the liquid bath on the absolute scale in kelvins?

333\,\text{K}

343\,\text{K}

353\,\text{K}

60\,\text{K}

Show answer & explanation

Answer: $333\,\text{K}$

Answer

The actual temperature of the liquid bath on the absolute scale is

333\,\text{K}

.

Subtracting the zero error of

+4\,\Omega

from the raw meter reading of

38\,\Omega

yields the true thermometric resistance of

34\,\Omega

. Applying the temperature formula

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}

gives

\theta = \frac{34 - 10}{50 - 10} \times 100 = 60^\circ\text{C}

. Converting to absolute temperature gives

T = 60 + 273 = 333\,\text{K}

.

Step-by-Step Solution

1

Correct the measured resistance for zero error

True resistance

R_\theta = 38\,\Omega - 4\,\Omega = 34\,\Omega

A positive zero error means the meter reads higher than the true value, so the zero error must be subtracted.

2

Calculate the temperature on the Celsius scale using linear interpolation

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C} = \frac{34 - 10}{50 - 10} \times 100 = \frac{24}{40} \times 100 = 60^\circ\text{C}

The thermometric property varies linearly between the fixed points.

3

Convert the Celsius temperature to the thermodynamic (absolute) Kelvin scale

T = \theta + 273 = 60 + 273 = 333\,\text{K}

The conversion from Celsius to Kelvin requires adding

273

(or

273.15

).

Key Concept

Linear interpolation of thermometric properties with instrument zero error correction

Estimated Time:2m 0s

Question 7809Question →

A progressive transverse wave traveling along a medium is described by the displacement equation $y = 0.05 \sin\left(160\pi t - 4\pi x\right)$ , where $x$ and $y$ are in meters and $t$ is in seconds. What is the ratio of the maximum particle velocity to the wave propagation velocity? (Take $\pi = 3.142$ .)

Show answer & explanation

Answer: 0.628

Answer

The ratio of the maximum particle velocity to the wave propagation velocity is 0.628.

For a progressive wave

y = A \sin(\omega t - k x)

, the wave advances at speed

v = \frac{\omega}{k} = \frac{160\pi}{4\pi} = 40\text{ m/s}

. Meanwhile, individual particles vibrate transversely with simple harmonic motion where maximum velocity is

v_{p,\text{max}} = A\omega = 0.05 \times 160\pi = 8\pi\text{ m/s}

. Taking the ratio gives

\frac{v_{p,\text{max}}}{v} = \frac{8\pi}{40} = 0.2\pi = 0.2 \times 3.142 = 0.6284

, which rounds to 0.628.

Step-by-Step Solution

1

Extract parameters from the standard wave equation

y = A \sin(\omega t - k x)

A = 0.05\text{ m}

,

\omega = 160\pi\text{ rad/s}

,

k = 4\pi\text{ rad/m}

Matching the given equation with standard progressive wave form yields the required wave constants.

2

Calculate the wave propagation velocity

v

v = \frac{\omega}{k} = \frac{160\pi}{4\pi} = 40\text{ m/s}

The speed at which the wave energy advances through the medium depends on angular frequency and wave number.

3

Calculate the maximum transverse particle velocity

v_{p,\text{max}}

v_{p,\text{max}} = A\omega = 0.05 \times 160\pi = 8\pi\text{ m/s} \approx 25.136\text{ m/s}

Particles perform simple harmonic motion, whose maximum speed is given by the product of amplitude and angular frequency.

4

Compute the ratio of maximum particle velocity to wave velocity

Ratio

= \frac{v_{p,\text{max}}}{v} = \frac{8\pi}{40} = 0.2\pi = 0.2 \times 3.142 = 0.6284

Dividing the maximum particle speed by the wave speed gives the desired dimensionless ratio.

Key Concept

Distinction between particle oscillation velocity and wave propagation velocity

Question 7810Question →

In the study of electrical discharge through gases and cathode ray behavior, specific physical setups and pressure conditions produce distinct observable phenomena. Match each experimental condition or observation on the left with its corresponding underlying physical mechanism or property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1. Extension of the Crookes dark space matches with cessation of gas ionization causing direct glass fluorescence. 2. Maltese cross shadow matches with rectilinear propagation of cathode rays. 3. Magnetic field deflection into circular arc matches with centripetal magnetic Lorentz force (

F = qvB

). 4. Luminous striations match with periodic excitation, ionization, and recombination of gas molecules.

Each matching pair directly connects an observable discharge tube phenomenon with its fundamental physical principle: extreme evacuation (

0.01\text{ mmHg}

) allows unimpeded electron stream travel to fluoresce glass; obstacle shadows confirm rectilinear propagation; transverse magnetic fields induce circular motion via

evB

; and periodic energy exchange of electrons with gas molecules yields striations.

Step-by-Step Solution

1

Analyze the pressure condition at

0.01\text{ mmHg}

At very low pressure (

0.01\text{ mmHg}

), gas collisions drop significantly, allowing cathode rays to reach the tube walls directly, extending the Crookes dark space throughout the tube and exciting glass fluorescence.

Mean free path increases beyond tube dimensions when gas density drops.

2

Analyze ray propagation using obstacle shadow formation

The sharp shadow cast by a Maltese cross demonstrates that cathode rays propagate in straight lines normal to the cathode.

Diffraction is negligible and ray trajectories do not bend around macroscopic obstacles.

3

Evaluate magnetic field interaction with cathode rays

The Lorentz force

F = q(\vec{v} \times \vec{B})

acts as a centripetal force (

evB = \frac{mv^2}{r}

), bending the negatively charged particle trajectory into a circle.

Moving electric charges experience magnetic forces perpendicular to velocity.

4

Identify the mechanism behind positive column striations

Striations represent repeating regions of inelastic electron collisions with gas atoms resulting in excitation and emission of light, followed by dark zones where electrons re-accelerate.

Quantized energy transfer during gas excitation creates spatial periodicity in luminescence.

Key Concept

Physical mechanisms of gaseous conduction across pressure stages and properties of cathode rays

Question 7811Question →

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z}^+ : 1 \le x \le 15\}$ . Consider two subsets of $\mathcal{U}$ defined as $A = \{x \in \mathcal{U} : x \text{ is prime}\}$ and $B = \{x \in \mathcal{U} : x \text{ is an odd integer greater than } 1\}$ . What is the number of elements in $(A \cap B)' \setminus (A \cup B)'$ ?

3

5

7

10

Show answer & explanation

Answer: $3$

Answer

3

Evaluating

(A \cap B)'

gives

\{1, 2, 4, 6, 8, 9, 10, 12, 14, 15\}

and evaluating

(A \cup B)'

gives

\{1, 4, 6, 8, 10, 12, 14\}

. Subtracting

(A \cup B)'

from

(A \cap B)'

leaves

\{2, 9, 15\}

, which has 3 elements.

Step-by-Step Solution

1

List the elements of the universal set

\mathcal{U}

and subsets

A

and

B

.

\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}

,

A = \{2, 3, 5, 7, 11, 13\}

, and

B = \{3, 5, 7, 9, 11, 13, 15\}

.

Explicitly writing the elements helps ensure set operations are performed accurately.

2

Find the intersection

A \cap B

and union

A \cup B

.

A \cap B = \{3, 5, 7, 11, 13\}

and

A \cup B = \{2, 3, 5, 7, 9, 11, 13, 15\}

.

These intermediate set operations are required to evaluate their complements.

3

Determine the complements

(A \cap B)'

and

(A \cup B)'

relative to

\mathcal{U}

.

(A \cap B)' = \{1, 2, 4, 6, 8, 9, 10, 12, 14, 15\}

and

(A \cup B)' = \{1, 4, 6, 8, 10, 12, 14\}

.

The complement of a set consists of all elements in

\mathcal{U}

not present in that set.

4

Calculate the set difference

(A \cap B)' \setminus (A \cup B)'

and count its elements.

(A \cap B)' \setminus (A \cup B)' = \{2, 9, 15\}

, which contains 3 elements.

Set difference removes all elements of

(A \cup B)'

from

(A \cap B)'

, leaving elements present in

A \cup B

but not in

A \cap B

.

Key Concept

Set Complements and Relative Difference

Estimated Time:1m 30s

Question 7812Question →

At which values of $x$ within the domain $0^\circ \le x \le 360^\circ$ do the graphs of $f(x) = 2\sin x$ and $g(x) = \tan x$ intersect?

0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ

60^\circ, 300^\circ

0^\circ, 30^\circ, 180^\circ, 330^\circ, 360^\circ

0^\circ, 60^\circ, 120^\circ, 180^\circ, 360^\circ

Show answer & explanation

Answer: $0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ$

Answer

The graphs intersect at

x = 0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ

.

Equating

2\sin x = \frac{\sin x}{\cos x}

yields

\sin x (2\cos x - 1) = 0

. Setting

\sin x = 0

gives

x = 0^\circ, 180^\circ, 360^\circ

, and setting

\cos x = \frac{1}{2}

gives

x = 60^\circ, 300^\circ

. Combining these gives the full solution set.

Step-by-Step Solution

1

Set the two trigonometric expressions equal to each other to find points of intersection.

2\sin x = \tan x

Intersection points occur where

f(x) = g(x)

.

2

Rewrite

\tan x

in terms of sine and cosine.

2\sin x = \frac{\sin x}{\cos x}

, for

x \neq 90^\circ, 270^\circ

Using the quotient identity

\tan x = \frac{\sin x}{\cos x}

allows simplification.

3

Rearrange the equation and factor out

\sin x

.

2\sin x \cos x - \sin x = 0 \implies \sin x(2\cos x - 1) = 0

Factoring prevents losing solutions that occur when

\sin x = 0

.

4

Solve each factor separately within

0^\circ \le x \le 360^\circ

.

Factor 1:

\sin x = 0 \implies x = 0^\circ, 180^\circ, 360^\circ

.
Factor 2:

2\cos x - 1 = 0 \implies \cos x = \frac{1}{2} \implies x = 60^\circ, 300^\circ

.

Cosine is positive in the first and fourth quadrants.

5

Combine all valid solutions.

x = 0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ

All five values satisfy the original equation and lie within the given domain.

Key Concept

Solving trigonometric equations by factoring and finding all roots in a given domain

Estimated Time:2m 0s

Question 7813Question →

Match each physical quantity to its correct SI unit.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electric current matches Ampere (A); Force matches Newton (N); Thermodynamic temperature matches Kelvin (K); Energy matches Joule (J).

Electric current and thermodynamic temperature are fundamental quantities measured in amperes and kelvins respectively. Force and energy are derived quantities whose units (newton and joule) are combinations of base SI units.

Step-by-Step Solution

1

Identify fundamental physical quantities and their base SI units.

Electric current is measured in amperes (A), and thermodynamic temperature is measured in kelvins (K). Both are fundamental SI units.

Fundamental units are basic units that are defined independently of other quantities.

2

Identify derived physical quantities and their derived SI units.

Force is measured in newtons (N), and energy is measured in joules (J). Both are derived SI units.

Derived units are obtained by combining fundamental SI base units according to physical equations.

Key Concept

Classification of fundamental and derived SI units

Estimated Time:45s

Question 7814Question →

A point $P(x, y)$ moves in the Cartesian plane such that it is always equidistant from two fixed points $A(-2, 3)$ and $B(4, 1)$ . If the locus of $P$ intersects the horizontal line $y = 5$ at the point $(k, 5)$ , what is the value of $k$ ?

Show answer & explanation

Answer: 2

Answer

The value of

k

is

2

.

The locus of points equidistant from

A(-2, 3)

and

B(4, 1)

is the perpendicular bisector of

AB

. The midpoint of

AB

is

(1, 2)

and the slope of

AB

is

-\frac{1}{3}

, giving a perpendicular slope of

3

. The equation of the locus is

3x - y = 1

. Substituting

y = 5

gives

3k - 5 = 1

, which yields

k = 2

.

Step-by-Step Solution

1

Identify the nature of the locus

The locus of points equidistant from two fixed points

A

and

B

is the perpendicular bisector of the line segment

AB

.

By definition, the set of points equidistant from two fixed points forms a straight line perpendicular to the segment joining the two points at its midpoint.

2

Find the midpoint of segment

AB

Midpoint

M = \left(\frac{-2+4}{2}, \frac{3+1}{2}\right) = (1, 2)

.

The perpendicular bisector passes through the midpoint of the line segment.

3

Calculate the gradient of

AB

and the perpendicular gradient

Gradient of

AB = \frac{1 - 3}{4 - (-2)} = -\frac{1}{3}

. Thus, the perpendicular gradient is

3

.

Perpendicular lines have gradients whose product is

-1

.

4

Derive the equation of the locus

y - 2 = 3(x - 1) \implies y = 3x - 1 \implies 3x - y = 1

.

Use the point-slope form of a line equation with point

(1, 2)

and slope

3

.

5

Determine the value of

k

at

y = 5

3k - 5 = 1 \implies 3k = 6 \implies k = 2

.

Substitute the point

(k, 5)

into the locus equation.

Key Concept

Perpendicular Bisector as a Locus

Estimated Time:1m 30s

Question 7815Question →

The sum of all the interior angles of a convex polygon, except for one, is $2190^\circ$ . If the measure of the remaining interior angle is strictly less than $180^\circ$ , how many sides does the polygon have?

13

14

15

16

Show answer & explanation

Answer: 15

Answer

The polygon has 15 sides.

The total sum of interior angles of an

n

-sided polygon is

(n - 2) \times 180^\circ

. Given that all angles except one sum to

2190^\circ

and the missing angle is between

0^\circ

and

180^\circ

, the total sum must be the smallest multiple of

180^\circ

greater than

2190^\circ

. The next multiple of

180^\circ

after

2190^\circ

is

2340^\circ

. Dividing

2340^\circ

by

180^\circ

gives

13

, so

n - 2 = 13

, which yields

n = 15

. The missing angle is

2340^\circ - 2190^\circ = 150^\circ

, which is valid.

Step-by-Step Solution

1

State the formula for the sum of interior angles of an

n

-sided convex polygon.

S = (n - 2) \times 180^\circ

, where

n

is an integer representing the number of sides (

n \ge 3

).

The sum of interior angles of any convex polygon is determined by dividing it into

(n - 2)

triangles.

2

Set up an inequality using the given partial sum and the constraint on the remaining interior angle.

Let the missing interior angle be

x

. Then

(n - 2) \times 180^\circ = 2190^\circ + x

, with

0^\circ < x < 180^\circ

.

Since the polygon is convex, every interior angle is non-reflex and greater than

0^\circ

.

3

Solve for the bounds of

(n - 2)

.

Dividing

2190^\circ < (n - 2) \times 180^\circ < 2190^\circ + 180^\circ

by

180^\circ

gives

12.166... < n - 2 < 13.166...

. Since

n - 2

must be an integer,

n - 2 = 13

.

The total interior angle sum must be an exact integer multiple of

180^\circ

.

4

Calculate

n

and verify the value of

x

.

n = 13 + 2 = 15

. The total sum is

13 \times 180^\circ = 2340^\circ

, making the remaining angle

x = 2340^\circ - 2190^\circ = 150^\circ

.

The remaining angle

150^\circ

satisfies

0^\circ < 150^\circ < 180^\circ

, confirming

n = 15

is correct.

Key Concept

Sum of Interior Angles of Convex Polygons

Question 7816Question →

Two identical isolated metal spheres carrying charges of $+8.0 \times 10^{-6}\text{ C}$ and $-2.0 \times 10^{-6}\text{ C}$ are brought into contact and then separated to a distance of $0.30\text{ m}$ in a vacuum. What is the magnitude of the electrostatic force of repulsion, in newtons ( $\text{N}$ ), between the spheres after contact? (Take Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{C}^{-2}$ )

Show answer & explanation

Answer: 0.9

Answer

The magnitude of the electrostatic force of repulsion between the spheres after contact is

0.9\text{ N}

.

When identical conducting spheres touch, their total electric charge is conserved and shared equally. The net charge is

+8.0\,\mu\text{C} + (-2.0\,\mu\text{C}) = +6.0\,\mu\text{C}

, giving each sphere a charge of

+3.0\,\mu\text{C}

. Applying Coulomb's law with a distance of

0.30\text{ m}

yields

0.9\text{ N}

.

Step-by-Step Solution

1

Calculate the net combined charge of the two identical spheres when brought into contact.

Q_{\text{total}} = q_1 + q_2 = (+8.0 \times 10^{-6}\text{ C}) + (-2.0 \times 10^{-6}\text{ C}) = +6.0 \times 10^{-6}\text{ C}

.

According to the principle of conservation of charge, charges add algebraically.

2

Determine the charge on each individual sphere after separation.

q' = \frac{Q_{\text{total}}}{2} = \frac{+6.0 \times 10^{-6}\text{ C}}{2} = +3.0 \times 10^{-6}\text{ C}

.

Identical conducting spheres share total charge equally when in contact.

3

Compute the force of repulsion using Coulomb's law.

F = \frac{k(q')^2}{r^2} = \frac{(9.0 \times 10^9)(3.0 \times 10^{-6})^2}{(0.30)^2} = \frac{9.0 \times 10^9 \times 9.0 \times 10^{-12}}{0.09} = 0.9\text{ N}

.

Coulomb's law defines the electrostatic force between two point charges.

Key Concept

Charge conservation, redistribution by conduction, and Coulomb's law

Question 7817Question →

An electron inside an excited atom drops from an energy state of $-1.20\text{ eV}$ to a lower energy state of $-4.50\text{ eV}$ . What is the frequency of the emitted electromagnetic radiation, in units of $10^{14}\text{ Hz}$ ? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J s}$ and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

Show answer & explanation

Answer: 8

Answer

The frequency of the emitted electromagnetic radiation is

8.0 \times 10^{14}\text{ Hz}

(or 8.0 in units of

10^{14}\text{ Hz}

).

The energy of the photon emitted during a downward transition between discrete atomic energy levels is equal to the energy difference between the initial and final states. Calculating

\Delta E = -1.20\text{ eV} - (-4.50\text{ eV}) = 3.30\text{ eV}

, converting to Joules gives

3.30 \times 1.6 \times 10^{-19}\text{ J} = 5.28 \times 10^{-19}\text{ J}

. Dividing this energy by Planck's constant

6.6 \times 10^{-34}\text{ J s}

yields a frequency of

8.0 \times 10^{14}\text{ Hz}

.

Step-by-Step Solution

1

Determine the energy of the emitted photon in electron-volts

\Delta E = 3.30\text{ eV}

The energy of the emitted photon equals the difference between the upper and lower atomic energy levels:

\Delta E = -1.20\text{ eV} - (-4.50\text{ eV}) = 3.30\text{ eV}

.

2

Convert photon energy from electron-volts to Joules

\Delta E = 5.28 \times 10^{-19}\text{ J}

Since Planck's constant is given in SI units (J s), energy must be converted to Joules using

1\text{ eV} = 1.6 \times 10^{-19}\text{ J}

.

3

Calculate photon frequency using Planck's relation

f = 8.0 \times 10^{14}\text{ Hz}

Applying

f = \frac{\Delta E}{h} = \frac{5.28 \times 10^{-19}\text{ J}}{6.6 \times 10^{-34}\text{ J s}}

yields

8.0 \times 10^{14}\text{ Hz}

.

Key Concept

Photon Emission and Energy Level Transitions

Question 7818Question →

A transverse wave of frequency $15\text{ Hz}$ propagates from a dense medium where its speed is $45\text{ m s}^{-1}$ into a lighter medium where its speed decreases to $30\text{ m s}^{-1}$ . What is the frequency of the wave in the second medium?

10\text{ Hz}

15\text{ Hz}

22.5\text{ Hz}

2.0\text{ Hz}

Show answer & explanation

Answer: $15\text{ Hz}$

Answer

The frequency of the wave in the second medium is

15\text{ Hz}

.

The correct answer is

15\text{ Hz}

because the frequency of a wave is determined solely by the source that creates it. When a wave passes from one medium into another, its speed and wavelength change proportionally, but the frequency remains constant.

Step-by-Step Solution

1

Identify the property of a wave that is source-dependent and invariant across boundaries.

The frequency (

f

) of a wave is determined entirely by the oscillating source generating the wave.

When a wave crosses the boundary between two different media, its speed (

v

) and wavelength (

λ

) change due to the properties of the new medium, but its frequency (

f

) remains constant.

2

Determine the frequency in the new medium.

f_{\text{second medium}} = f_{\text{initial}} = 15\text{ Hz}

.

Since frequency does not change across medium boundaries, the wave maintains a frequency of

15\text{ Hz}

.

Key Concept

Invariance of Wave Frequency Across Media Boundaries

Estimated Time:45s

Question 7819Question →

A progressive wave propagating through an initial medium is represented by the displacement equation $y = 0.05 \sin\left(100\pi t - \frac{2\pi}{5} x\right)$ , where $x$ and $y$ are in metres and $t$ is in seconds. Upon entering a second medium, the wave speed decreases by $20\%$ . What is the wavelength of the wave in the second medium?

4.0\text{ m}

5.0\text{ m}

6.25\text{ m}

2.0\text{ m}

Show answer & explanation

Answer: $4.0\text{ m}$

Answer

4.0\text{ m}

Comparing

y = 0.05 \sin\left(100\pi t - \frac{2\pi}{5} x\right)

with the standard wave equation

y = A \sin(\omega t - kx)

yields a wave number

k = \frac{2\pi}{5}\text{ rad/m}

, giving an initial wavelength

\lambda_1 = \frac{2\pi}{k} = 5.0\text{ m}

. When a wave refracts into another medium, its frequency stays constant, making wave speed and wavelength directly proportional. A

20\%

reduction in speed decreases the wavelength by

20\%

, giving

\lambda_2 = 5.0 \times 0.80 = 4.0\text{ m}

.

Step-by-Step Solution

1

Extract the angular frequency

\omega

and wave number

k

from the wave equation

\omega = 100\pi\text{ rad/s}

and

k = \frac{2\pi}{5}\text{ rad/m}

The standard progressive wave equation is

y = A \sin(\omega t - kx)

2

Calculate the wavelength

\lambda_1

in the first medium

$\lambda_1 = \frac{2\pi}{k} = \frac{2\pi}{\frac{2\pi}{5}} = 5.0\text{ m}$

The wave number is related to wavelength by

k = \frac{2\pi}{\lambda}

3

Determine the new wavelength

\lambda_2

in the second medium using the refraction property

$\lambda_2 = \lambda_1 \times (1 - 0.20) = 5.0 \times 0.80 = 4.0\text{ m}$

When a wave crosses a boundary between two media, its frequency

f

remains constant. Therefore, wave speed

v = f\lambda

is directly proportional to wavelength

\lambda

Key Concept

Constancy of wave frequency during refraction and wave equation parameters extraction

Estimated Time:2m 0s

Question 7820Question →

A glass vessel with a linear expansivity of $9.0 \times 10^{-6} \text{ K}^{-1}$ has a volume capacity of $300 \text{ cm}^3$ at $15^\circ\text{C}$ . It is completely filled with a liquid at this temperature. When the system is uniformly heated to $65^\circ\text{C}$ , a volume of $4.5 \text{ cm}^3$ of the liquid overflows. What is the real cubic expansivity of the liquid?

3.27 \times 10^{-4} \text{ K}^{-1}

3.09 \times 10^{-4} \text{ K}^{-1}

3.00 \times 10^{-4} \text{ K}^{-1}

2.73 \times 10^{-4} \text{ K}^{-1}

Show answer & explanation

Answer: $3.27 \times 10^{-4} \text{ K}^{-1}$

Answer

The real cubic expansivity of the liquid is

3.27 \times 10^{-4} \text{ K}^{-1}

.

The real cubic expansivity of a liquid equals the sum of its apparent cubic expansivity and the cubical expansivity of the containing vessel. The apparent cubic expansivity is

\frac{4.5}{300 \times 50} = 3.0 \times 10^{-4} \text{ K}^{-1}

. The cubical expansivity of the vessel is

3 \times 9.0 \times 10^{-6} = 2.7 \times 10^{-5} \text{ K}^{-1}

. Summing these yields

3.27 \times 10^{-4} \text{ K}^{-1}

.

Step-by-Step Solution

1

Calculate the temperature change (

\Delta T

) and the apparent cubic expansivity (

\gamma_a

) of the liquid.

\Delta T = 65^\circ\text{C} - 15^\circ\text{C} = 50 \text{ K}

.

\gamma_a = \frac{V_{\text{overflow}}}{V_0 \cdot \Delta T} = \frac{4.5}{300 \times 50} = 3.0 \times 10^{-4} \text{ K}^{-1}

.

Apparent expansion measures the overflow relative to initial volume and temperature rise.

2

Calculate the cubical expansivity of the glass vessel (

\gamma_v

).

\gamma_v = 3 \alpha_v = 3 \times (9.0 \times 10^{-6} \text{ K}^{-1}) = 2.7 \times 10^{-5} \text{ K}^{-1} = 0.27 \times 10^{-4} \text{ K}^{-1}

.

The volume (cubical) expansivity of an isotropic solid container is three times its linear expansivity.

3

Determine the real cubic expansivity of the liquid (

\gamma_r

) using the relation

\gamma_r = \gamma_a + \gamma_v

.

\gamma_r = 3.0 \times 10^{-4} \text{ K}^{-1} + 0.27 \times 10^{-4} \text{ K}^{-1} = 3.27 \times 10^{-4} \text{ K}^{-1}

.

Real volume expansion equals the observed apparent expansion plus the expansion of the container.

Key Concept

Relationship between real expansivity, apparent expansivity, and vessel expansivity:

\gamma_r = \gamma_a + \gamma_v

, where

\gamma_v = 3\alpha

.

Answer

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