All practice questions

Question 3861Question →

In English phonology, words ending in the suffix -ity place primary accentual emphasis on the syllable immediately preceding the suffix. How is primary stress correctly indicated for the word curiosity?

cu-ri-OS-i-ty

cu-RI-os-i-ty

CU-ri-os-i-ty

cu-ri-os-i-TY

Show answer & explanation

Answer: cu-ri-OS-i-ty

Answer

The correct stress rendering for 'curiosity' is cu-ri-OS-i-ty, where the primary stress is placed on the third syllable.

The option showing capitalization on the third syllable ('cu-ri-OS-i-ty') is correct because words ending in '-ity' automatically place primary stress on the syllable immediately preceding the suffix (the antepenultimate syllable).

Step-by-Step Solution

1

Identify the word structure and suffix

The word 'curiosity' is a five-syllable noun ending in the nominal suffix '-ity'.

Suffixes in English determine stress placement rules.

2

Apply the phonological rule for '-ity' suffixed words

The suffix '-ity' causes stress to fall on the syllable directly before it (the antepenultimate syllable).

Nouns with '-ity' consistently pull primary stress to the pre-suffix syllable (e.g., a-BIL-i-ty, e-QUAL-i-ty, cu-ri-OS-i-ty).

3

Locate the target stressed syllable

In 'cu-ri-os-i-ty', the antepenultimate syllable is 'OS'. Thus, the primary stress is on 'OS' (cu-ri-OS-i-ty).

Capitalizing 'OS' correctly marks primary syllable stress.

Key Concept

Suffix-conditioned word stress (-ity rule)

Estimated Time:1m 0s

Question 3862Question →

The newly appointed chief financial officer insisted that the company's financial strategies must be strictly aligned ______ established regulatory frameworks, rather than being subservient ______ individual executive preferences. Which pair of prepositions correctly completes the sentence?

with / to

to / with

for / towards

at / against

Show answer & explanation

Answer: with / to

Answer

The prepositions 'with' and 'to' correctly complete the sentence as 'aligned with' and 'subservient to'.

The expression 'aligned with' is the correct dependent prepositional phrase used to express agreement or harmony with a set of guidelines, while 'subservient to' correctly denotes being subordinate or secondary to something else.

Step-by-Step Solution

1

Determine the dependent preposition required by 'aligned'.

The word 'aligned' collocates with 'with' when expressing conformity to a rule, policy, or framework.

'Aligned with' is the standard idiomatic expression.

2

Determine the dependent preposition required by 'subservient'.

The adjective 'subservient' requires the preposition 'to' to denote yielding or subordination.

'Subservient to' is the established grammatical complement.

Key Concept

Dependent prepositions and adjective-preposition collocations

Question 3863Question →

A right pyramid has a square base with a perimeter of $24\text{ cm}$ and a vertical height of $10\text{ cm}$ . What is the volume of the pyramid?

120\text{ cm}^3

360\text{ cm}^3

80\text{ cm}^3

240\text{ cm}^3

Show answer & explanation

Answer: $120\text{ cm}^3$

Answer

The volume of the pyramid is

120\text{ cm}^3

.

The volume of a right pyramid is given by

V = \frac{1}{3} \times \text{Base Area} \times h

. First, find the side length of the square base by dividing the perimeter by 4:

s = 24 / 4 = 6\text{ cm}

. The base area is

s^2 = 36\text{ cm}^2

. Substituting into the volume formula gives

V = \frac{1}{3} \times 36 \times 10 = 120\text{ cm}^3

.

Step-by-Step Solution

1

Determine the side length of the square base from its perimeter.

Side length

s = \frac{24\text{ cm}}{4} = 6\text{ cm}

.

A square has four equal sides, so perimeter divided by 4 yields the side length.

2

Calculate the area of the square base.

Base Area

A = s^2 = 6^2 = 36\text{ cm}^2

.

The area of a square is given by the side length squared.

3

Apply the volume formula for a right pyramid.

Volume

V = \frac{1}{3} \times A \times h = \frac{1}{3} \times 36\text{ cm}^2 \times 10\text{ cm} = 120\text{ cm}^3

.

The volume of any pyramid is one-third of the base area multiplied by the vertical height.

Key Concept

Volume of a Right Pyramid

Estimated Time:1m 15s

Question 3864Question →

An alternating voltage source of $90\text{ V}$ (RMS) is connected in series across a circuit containing a resistor of resistance $12\ \Omega$ , an inductor of reactance $20\ \Omega$ , and a capacitor of reactance $11\ \Omega$ . What is the root-mean-square (RMS) current flowing through the circuit?

2.1\text{ A}

4.3\text{ A}

6.0\text{ A}

7.5\text{ A}

Show answer & explanation

Answer: $6.0\text{ A}$

Answer

The RMS current flowing through the circuit is

6.0\text{ A}

.

The total impedance

Z

of a series RLC circuit is given by

Z = \sqrt{R^2 + (X_L - X_C)^2}

. Substituting the given values gives

Z = \sqrt{12^2 + (20 - 11)^2} = \sqrt{144 + 81} = 15\ \Omega

. Dividing the RMS voltage (

90\text{ V}

) by this impedance yields an RMS current of

6.0\text{ A}

.

Step-by-Step Solution

1

Calculate the net reactance of the series AC circuit

X = X_L - X_C = 20\ \Omega - 11\ \Omega = 9\ \Omega

Inductive and capacitive reactances oppose each other in phase by

180^\circ

.

2

Determine the total impedance (Z) of the series RLC circuit using phasor addition

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\ \Omega

Resistance and net reactance are

90^\circ

out of phase, so impedance is calculated using the Pythagorean theorem.

3

Calculate the RMS current using Ohm's law for AC circuits

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{90\text{ V}}{15\ \Omega} = 6.0\text{ A}

The RMS current is equal to the RMS voltage divided by total circuit impedance.

Key Concept

Impedance and RMS Current in Series RLC Circuits

Question 3865Question →

The speed of light in medium $A$ is $1.5 \times 10^8\text{ m/s}$ and in medium $B$ is $3.0 \times 10^8\text{ m/s}$ . What is the critical angle for a light ray traveling from medium $A$ towards medium $B$ ?

30^\circ

45^\circ

60^\circ

90^\circ

Show answer & explanation

Answer: $30^\circ$

Answer

The critical angle for light traveling from medium

A

to medium

B

is

30^\circ

.

The critical angle

C

is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is

90^\circ

. Using the wave speed form of Snell's law,

\sin C = \frac{v_A}{v_B} = \frac{1.5 \times 10^8}{3.0 \times 10^8} = 0.5

. Taking the inverse sine yields

C = 30^\circ

.

Step-by-Step Solution

1

Determine the relationship between critical angle and wave speeds in the two media.

\sin C = \frac{n_B}{n_A} = \frac{v_A}{v_B}

By Snell's Law, critical angle occurs when the angle of refraction is

90^\circ

, giving

\sin C = \frac{n_2}{n_1}

. Since refractive index

n = \frac{c}{v}

, the ratio

\frac{n_B}{n_A}

simplifies to

\frac{v_A}{v_B}

.

2

Substitute the given wave speeds into the equation.

\sin C = \frac{1.5 \times 10^8\text{ m/s}}{3.0 \times 10^8\text{ m/s}} = 0.5

Light is traveling from the optically denser medium

A

(lower speed) to the less dense medium

B

(higher speed).

3

Calculate the inverse sine to find the critical angle

C

.

C = \arcsin(0.5) = 30^\circ

The angle whose sine is

0.5

is

30^\circ

.

Key Concept

Critical Angle and Total Internal Reflection in terms of Wave Speed

Question 3866Question →

Complete the sentence below by identifying the standard International Phonetic Alphabet (IPA) symbol for the underlined consonant sound in each word.

Fill in the blanks below

In the word 'measure', the underlined letter 's' represents the consonant sound , whereas in 'mission', the underlined letters 'ss' represent the consonant sound .

Show answer & explanation

Answer

The letter 's' in 'measure' represents the voiced post-alveolar fricative sound /ʒ/, while the letters 'ss' in 'mission' represent the voiceless post-alveolar fricative sound /ʃ/.

In English oral phonology, the letter 's' in 'measure' is pronounced with voicing as /ʒ/. In contrast, the digraph 'ss' in 'mission' is voiceless, yielding the sound /ʃ/.

Step-by-Step Solution

1

Determine the phonetic sound produced by the letter 's' in 'measure'.

The vocal cords vibrate during production, giving the voiced post-alveolar fricative sound represented by /ʒ/.

Words ending in '-sure' following a vowel (such as 'pleasure', 'leisure', 'measure') typically articulate the consonant sound /ʒ/.

2

Determine the phonetic sound produced by the letter combination 'ss' in 'mission'.

The vocal cords do not vibrate, giving the voiceless post-alveolar fricative sound represented by /ʃ/.

The spelling pattern '-ssion' produces the voiceless consonant sound /ʃ/ as seen in words like 'session' and 'admission'.

Key Concept

Phonetic contrast between voiced post-alveolar fricatives (/ʒ/) and voiceless post-alveolar fricatives (/ʃ/).

Question 3867Question →

A solid right triangular prism has a base that is a right-angled triangle with legs of length $6\text{ cm}$ and $8\text{ cm}$ . If the height of the prism is $15\text{ cm}$ , what is the total surface area of the prism, in $\text{cm}^2$ ?

Show answer & explanation

Answer: 408

Answer

The total surface area of the right triangular prism is

408\text{ cm}^2

.

To find the total surface area of a right triangular prism, compute the sum of the areas of its 2 triangular bases and its 3 rectangular side faces. The legs of the right-angled triangle are

6\text{ cm}

and

8\text{ cm}

, so the hypotenuse is

\sqrt{6^2 + 8^2} = 10\text{ cm}

. The combined area of the two bases is

2 \times (\frac{1}{2} \times 6 \times 8) = 48\text{ cm}^2

. The perimeter of the base is

6 + 8 + 10 = 24\text{ cm}

, making the lateral surface area

24 \times 15 = 360\text{ cm}^2

. Adding the base areas and lateral area yields

48 + 360 = 408\text{ cm}^2

.

Step-by-Step Solution

1

Determine the length of the hypotenuse of the triangular base.

Hypotenuse =

10\text{ cm}

.

The base is a right-angled triangle with legs

6\text{ cm}

and

8\text{ cm}

. By Pythagoras:

c = \sqrt{6^2 + 8^2} = 10\text{ cm}

.

2

Calculate the total area of the two parallel triangular bases.

Base area total =

48\text{ cm}^2

.

The area of one right triangle is

\frac{1}{2} \times 6 \times 8 = 24\text{ cm}^2

, so two bases have an area of

2 \times 24 = 48\text{ cm}^2

.

3

Calculate the lateral surface area of the three rectangular faces.

Lateral surface area =

360\text{ cm}^2

.

The lateral surface area is equal to the perimeter of the base times the height:

(6 + 8 + 10) \times 15 = 24 \times 15 = 360\text{ cm}^2

.

4

Sum the base area total and lateral surface area.

Total Surface Area =

408\text{ cm}^2

.

Total surface area =

48 + 360 = 408\text{ cm}^2

.

Key Concept

Total Surface Area of a Right Triangular Prism

Question 3868Question →

Two cylindrical metallic conductors, $X$ and $Y$ , are connected in series across a direct-current source. Wire $X$ has a diameter of $1.0\,\text{mm}$ and a free-electron density of $6.0 \times 10^{28}\,\text{m}^{-3}$ . Wire $Y$ has a diameter of $3.0\,\text{mm}$ and a free-electron density of $2.0 \times 10^{28}\,\text{m}^{-3}$ . What is the ratio of the drift velocity of free electrons in wire $X$ to that in wire $Y$ ?

3.0

0.33

1.0

9.0

Show answer & explanation

Answer: $3.0$

Answer

The ratio of the drift velocity in wire X to that in wire Y is 3.0.

Because the wires are connected in series, the same current passes through both (

I_X = I_Y

). Using the formula for electric current in terms of drift velocity

I = n A e v_d

, where cross-sectional area

A = \pi d^2 / 4

, we find

v_d \propto 1 / (n d^2)

. Taking the ratio yields

v_X / v_Y = (n_Y d_Y^2) / (n_X d_X^2) = (2.0 \times 10^{28} \times 3.0^2) / (6.0 \times 10^{28} \times 1.0^2) = 18 / 6 = 3.0

.

Step-by-Step Solution

1

Relate electric current to drift velocity and conductor geometry.

I = n A e v_d = n \left( \frac{\pi d^2}{4} \right) e v_d

Electric current

I

depends on free-electron density

n

, cross-sectional area

A

, elementary charge

e

, and electron drift velocity

v_d

.

2

Apply the series connection constraint.

I_X = I_Y \implies n_X d_X^2 v_X = n_Y d_Y^2 v_Y

In a series circuit, the steady current flowing through every conductor is identical.

3

Rearrange to solve for the drift velocity ratio

v_X / v_Y

.

\frac{v_X}{v_Y} = \frac{n_Y d_Y^2}{n_X d_X^2}

Isolating

v_X / v_Y

demonstrates inverse proportionality to electron density and the square of conductor diameter.

4

Substitute the given values into the ratio expression.

\frac{v_X}{v_Y} = \frac{(2.0 \times 10^{28}\,\text{m}^{-3}) \times (3.0\,\text{mm})^2}{(6.0 \times 10^{28}\,\text{m}^{-3}) \times (1.0\,\text{mm})^2} = \frac{2.0 \times 9.0}{6.0 \times 1.0} = \frac{18.0}{6.0} = 3.0

Numerical calculation yields the simplified dimensionless ratio.

Key Concept

Drift Velocity and Current Density in Series Conductors

Question 3869Question →

A diver releases a bubble of air of volume $8.00\text{ cm}^3$ at a depth where the water pressure is $3.50 \times 10^5\text{ Pa}$ and the temperature is $7^\circ\text{C}$ . What is the volume of the air bubble, in $\text{cm}^3$ , just as it reaches the surface where the pressure is $1.00 \times 10^5\text{ Pa}$ and the temperature is $27^\circ\text{C}$ ?

Show answer & explanation

Answer: 30

Answer

The final volume of the air bubble at the surface is

30.0\text{ cm}^3

.

Using the combined gas law

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

with absolute temperatures

T_1 = 280\text{ K}

and

T_2 = 300\text{ K}

yields a final volume of

30.0\text{ cm}^3

.

Step-by-Step Solution

1

Convert temperatures from Celsius to Kelvin

T_1 = 7^\circ\text{C} + 273 = 280\text{ K}

and

T_2 = 27^\circ\text{C} + 273 = 300\text{ K}

Gas laws require absolute thermodynamic temperature in Kelvin.

2

Apply the combined gas law equation

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

The amount of gas in the bubble remains constant while pressure, volume, and temperature all change simultaneously.

3

Rearrange for the unknown volume

V_2

and substitute the values

V_2 = \frac{3.50 \times 10^5 \times 8.00 \times 300}{1.00 \times 10^5 \times 280} = 30.0\text{ cm}^3

Calculates the expanded volume of the air bubble at surface conditions.

Key Concept

Combined Gas Law

Question 3870Question →

A DC power source with an electromotive force (e.m.f.) of $E = 12.0\text{ V}$ and an internal resistance of $r = 1.0\text{ }\Omega$ is connected to an external load. The load consists of two parallel resistors, $R_1 = 3.0\text{ }\Omega$ and $R_2 = 6.0\text{ }\Omega$ , connected in series with an unknown resistor $R_x$ . A real voltmeter with an internal resistance of $R_v = 90.0\text{ }\Omega$ is placed directly across the terminals of the power source and reads $V = 10.8\text{ V}$ . Calculate the resistance of $R_x$ in ohms.

Show answer & explanation

Answer: 8

Answer

The resistance of

R_x

is

8.0\text{ }\Omega

.

Applying the relationship

V = E - Ir

yields a total current of

1.2\text{ A}

through the cell. The total external resistance connected across the terminals is

9.0\text{ }\Omega

. Subtracting the parallel admittance of the

90.0\text{ }\Omega

voltmeter yields a load resistance of

10.0\text{ }\Omega

. Subtracting the

2.0\text{ }\Omega

equivalent resistance of the parallel pair (

3.0\text{ }\Omega

and

6.0\text{ }\Omega

) gives

8.0\text{ }\Omega

for

R_x

.

Step-by-Step Solution

1

Calculate total current supplied by the cell

I = 1.2\text{ A}

Terminal voltage is related to battery e.m.f. and internal resistance by

V = E - Ir

.

2

Calculate equivalent resistance of the entire external circuit across terminals

R_{\text{ext}} = 9.0\text{ }\Omega

By Ohm's law,

R_{\text{ext}} = \frac{V}{I} = \frac{10.8\text{ V}}{1.2\text{ A}} = 9.0\text{ }\Omega

.

3

Determine the resistance of the main circuit load

R_L

R_L = 10.0\text{ }\Omega

The voltmeter is in parallel with

R_L

, giving

\frac{1}{R_{\text{ext}}} = \frac{1}{R_v} + \frac{1}{R_L}

.

4

Calculate the equivalent resistance

R_p

of the parallel combination of

R_1

and

R_2

R_p = 2.0\text{ }\Omega

R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{3.0 \times 6.0}{3.0 + 6.0} = 2.0\text{ }\Omega

.

5

Determine the value of

R_x

R_x = 8.0\text{ }\Omega

The main circuit load consists of

R_p

in series with

R_x

, so

R_L = R_p + R_x

.

Key Concept

Terminal potential difference, loading effect of measuring instruments, and resistor network analysis

Estimated Time:2m 30s

Question 3871Question →

A decorative wooden cone has a slant height of $13\text{ cm}$ and a vertical height of $12\text{ cm}$ . What is the volume of the cone in terms of $\pi$ ?

100\pi\text{ cm}^3

300\pi\text{ cm}^3

400\pi\text{ cm}^3

65\pi\text{ cm}^3

Show answer & explanation

Answer: $100\pi\text{ cm}^3$

Answer

The volume of the cone is

100\pi\text{ cm}^3

.

The radius is determined via the Pythagorean relation

r = \sqrt{13^2 - 12^2} = 5\text{ cm}

. Substituting

r = 5\text{ cm}

and

h = 12\text{ cm}

into

V = \frac{1}{3}\pi r^2 h

yields

V = \frac{1}{3}\pi(25)(12) = 100\pi\text{ cm}^3

.

Step-by-Step Solution

1

Find the base radius of the cone using the Pythagorean theorem.

r = \sqrt{l^2 - h^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5\text{ cm}

.

The radius, vertical height, and slant height of a right circular cone form a right-angled triangle where the slant height is the hypotenuse.

2

Calculate the volume of the cone using the formula

V = \frac{1}{3}\pi r^2 h

.

V = \frac{1}{3} \times \pi \times 5^2 \times 12 = \frac{1}{3} \times \pi \times 25 \times 12 = 100\pi\text{ cm}^3

.

The formula for the volume of any right circular cone requires multiplying one-third of the base area by the vertical height.

Key Concept

Volume of a Right Circular Cone

Question 3872Question →

A curve is defined by the equation $y = 2x^3 - 9x^2 + 12x + 5$ . What is the value of $y$ at its maximum stationary point?

Show answer & explanation

Answer: 10

Answer

The value of

y

at the maximum stationary point is

10

.

To locate the maximum stationary point, set the first derivative

\frac{dy}{dx} = 6x^2 - 18x + 12

to zero, obtaining stationary values

x = 1

and

x = 2

. Testing in the second derivative

\frac{d^2y}{dx^2} = 12x - 18

gives

-6

at

x = 1

, confirming a local maximum. Substituting

x = 1

into the original cubic equation gives

y = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 10

.

Step-by-Step Solution

1

Differentiate

y

with respect to

x

to find the gradient function.

\frac{dy}{dx} = 6x^2 - 18x + 12

Stationary points occur where the gradient of the curve is zero.

2

Set the first derivative to zero and solve for

x

.

x = 1 \text{ or } x = 2

Factoring

6(x - 1)(x - 2) = 0

yields the

x

-coordinates of the turning points.

3

Determine the nature of the stationary points using the second derivative test.

\frac{d^2y}{dx^2} = 12x - 18; \quad \text{at } x = 1, \frac{d^2y}{dx^2} = -6 < 0

A negative second derivative indicates a local maximum stationary point.

4

Substitute

x = 1

into the original function to determine

y

.

y = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 10

Evaluating the curve function at the maximum

x

-coordinate provides the corresponding maximum

y

-value.

Key Concept

Stationary Points, Maxima, and Minima

Question 3873Question →

A battery with an electromotive force (e.m.f.) of $12.0\text{ V}$ and an internal resistance of $1.0\text{ }\Omega$ is connected to a circuit containing two resistors, $6.0\text{ }\Omega$ and $3.0\text{ }\Omega$ , connected in parallel with each other. What is the potential difference across the $3.0\text{ }\Omega$ resistor?

8.0\text{ V}

12.0\text{ V}

4.0\text{ V}

3.6\text{ V}

Show answer & explanation

Answer: $8.0\text{ V}$

Answer

The potential difference across the

3.0\text{ }\Omega

resistor is

8.0\text{ V}

.

The parallel external resistors of

6.0\text{ }\Omega

and

3.0\text{ }\Omega

combine to give an equivalent resistance of

2.0\text{ }\Omega

. Including the internal resistance of

1.0\text{ }\Omega

, the total circuit resistance is

3.0\text{ }\Omega

. This leads to a total circuit current of

4.0\text{ A}

. Multiplying this current by the external equivalent resistance of

2.0\text{ }\Omega

gives the terminal voltage of

8.0\text{ V}

, which is identical across both parallel branches.

Step-by-Step Solution

1

Calculate the equivalent resistance (

R_p

) of the parallel combination.

R_p = \frac{6.0 \times 3.0}{6.0 + 3.0} = \frac{18.0}{9.0} = 2.0\text{ }\Omega

Resistors in parallel combine according to reciprocal addition.

2

Find the total circuit resistance (

R_{\text{total}}

) including the internal resistance (

r

).

R_{\text{total}} = R_p + r = 2.0\text{ }\Omega + 1.0\text{ }\Omega = 3.0\text{ }\Omega

The internal resistance of the power supply acts in series with the external equivalent load resistance.

3

Calculate the total current (

I

) drawn from the battery using Ohm's law for a complete circuit.

I = \frac{E}{R_{\text{total}}} = \frac{12.0\text{ V}}{3.0\text{ }\Omega} = 4.0\text{ A}

Current is determined by the total electromotive force divided by the total resistance of the circuit.

4

Determine the terminal potential difference (

V

) across the parallel network.

V = I \times R_p = 4.0\text{ A} \times 2.0\text{ }\Omega = 8.0\text{ V}

Components connected in parallel experience the same potential difference, which equals the terminal voltage of the battery.

Key Concept

Terminal potential difference in circuits with internal resistance

Question 3874Question →

When the polynomial $P(x) = 3x^3 - kx^2 + 4x - 7$ is divided by $x - 2$ , the remainder is $9$ . What is the value of $k$ ?

Show answer & explanation

Answer: 4

Answer

The value of

k

is

4

.

According to the Remainder Theorem, dividing

P(x)

by

x - 2

leaves a remainder of

P(2)

. Evaluating

P(2) = 3(2)^3 - k(2)^2 + 4(2) - 7

gives

25 - 4k

. Setting

25 - 4k = 9

and solving yields

k = 4

.

Step-by-Step Solution

1

Apply the Remainder Theorem

The remainder when

P(x)

is divided by

x - 2

is

P(2)

.

By the Remainder Theorem, dividing a polynomial

P(x)

by

x - a

leaves a remainder equal to

P(a)

.

2

Substitute

x = 2

into

P(x)

P(2) = 3(2)^3 - k(2)^2 + 4(2) - 7 = 25 - 4k

Evaluating the polynomial at

x = 2

expresses the remainder in terms of

k

.

3

Equate

P(2)

to the given remainder and solve for

k

25 - 4k = 9 \implies 4k = 16 \implies k = 4

Setting the calculated expression equal to

9

forms a linear equation that yields

k = 4

.

Key Concept

Polynomial Remainder Theorem

Estimated Time:1m 30s

Question 3875Question →

Match each defect of vision to its correct image formation characteristic and mode of optical correction.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Hypermetropia pairs with light focusing behind the retina (corrected with a converging lens). Myopia pairs with light focusing in front of the retina (corrected with a diverging lens). Presbyopia pairs with age-related loss of accommodation (corrected with bifocal/converging lenses). Astigmatism pairs with unequal corneal curvature in different planes (corrected with a cylindrical lens).

Each vision defect uniquely corresponds to a specific optical fault and corrective lens: Hypermetropia focuses images behind the retina (convex lens), Myopia focuses images in front of the retina (concave lens), Presbyopia involves age-related elasticity loss (bifocal/convex lens), and Astigmatism stems from asymmetric corneal curvature (cylindrical lens).

Step-by-Step Solution

1

Identify the structural defect and focus point for Hypermetropia.

Near point moves farther away, so light rays from standard near point focus behind the retina. A converging (convex) lens is required to bend light rays inward prior to entering the eye.

Hypermetropic eyes lack sufficient refractive power for near vision.

2

Identify the structural defect and focus point for Myopia.

Far point is reduced, so light rays from distant objects focus in front of the retina. A diverging (concave) lens is required to diverge incoming parallel rays.

Myopic eyes possess excessive refractive power or an elongated eyeball.

3

Identify the cause and correction for Presbyopia.

Caused by aging of ciliary muscles and hardening of the eye lens. Corrected with converging or bifocal lenses.

Presbyopia specifically relates to loss of elastic accommodation in older individuals.

4

Identify the cause and correction for Astigmatism.

Caused by non-uniform curvature of the cornea along horizontal vs vertical axes. Corrected using cylindrical lenses.

Cylindrical lenses compensate for asymmetrical focal lengths along specific axes.

Key Concept

Defects of Vision and Corrective Lenses

Question 3876Question →

An electron in a hydrogen atom drops from an excited state with energy $-0.85\text{ eV}$ to a lower energy state of $-3.40\text{ eV}$ . What is the wavelength of the photon emitted during this transition? ( $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ , $c = 3.0 \times 10^8\text{ m/s}$ , $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

4.85 \times 10^{-7}\text{ m}

2.91 \times 10^{-7}\text{ m}

7.76 \times 10^{-26}\text{ m}

6.18 \times 10^{14}\text{ m}

Show answer & explanation

Answer: $4.85 \times 10^{-7}\text{ m}$

Answer

The wavelength of the emitted photon is

4.85 \times 10^{-7}\text{ m}

.

The energy of the emitted photon is given by

\Delta E = -0.85\text{ eV} - (-3.40\text{ eV}) = 2.55\text{ eV}

. Converting to Joules gives

2.55 \times 1.6 \times 10^{-19}\text{ J} = 4.08 \times 10^{-19}\text{ J}

. Substituting into

\lambda = \frac{hc}{\Delta E}

yields

\lambda = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{4.08 \times 10^{-19}} = 4.85 \times 10^{-7}\text{ m}

.

Step-by-Step Solution

1

Calculate the energy difference between the initial and final states

\Delta E = E_i - E_f = -0.85\text{ eV} - (-3.40\text{ eV}) = 2.55\text{ eV}

The energy of the emitted photon equals the energy difference between the two levels.

2

Convert the energy difference from electron-volts to Joules

\Delta E = 2.55 \times 1.6 \times 10^{-19}\text{ J} = 4.08 \times 10^{-19}\text{ J}

Standard SI units are required for calculations involving Planck's constant.

3

Calculate photon wavelength using

\lambda = \frac{hc}{\Delta E}

\lambda = \frac{(6.6 \times 10^{-34}\text{ J}\cdot\text{s})(3.0 \times 10^8\text{ m/s})}{4.08 \times 10^{-19}\text{ J}} = 4.85 \times 10^{-7}\text{ m}

Relating photon energy to wavelength using Planck's constant and speed of light.

Key Concept

Photon Emission and Atomic Transitions

Question 3877Question →

An object is placed $12\text{ cm}$ in front of a diverging lens having a focal length of $18\text{ cm}$ . What is the magnitude of the image distance from the lens?

7.2\text{ cm}

36.0\text{ cm}

30.0\text{ cm}

6.0\text{ cm}

Show answer & explanation

Answer: $7.2\text{ cm}$

Answer

The magnitude of the image distance is

7.2\text{ cm}

.

By the thin lens equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

, substituting

f = -18\text{ cm}

(due to the lens being diverging) and

u = +12\text{ cm}

gives

\frac{1}{v} = -\frac{1}{18} - \frac{1}{12} = -\frac{5}{36}\text{ cm}^{-1}

, which yields

v = -7.2\text{ cm}

. Thus, the image is located

7.2\text{ cm}

from the lens.

Step-by-Step Solution

1

Identify the given values and assign the correct signs according to the real-is-positive sign convention.

Object distance

u = +12\text{ cm}

; Focal length of diverging (concave) lens

f = -18\text{ cm}

.

Diverging lenses have virtual focus, so their focal length must be taken as negative.

2

Set up the thin lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

to solve for image distance

v

.

\frac{1}{-18} = \frac{1}{12} + \frac{1}{v} \implies \frac{1}{v} = -\frac{1}{18} - \frac{1}{12}

.

Rearranging the equation isolates the reciprocal of the image distance.

3

Find a common denominator and solve for

v

.

\frac{1}{v} = \frac{-2 - 3}{36} = -\frac{5}{36}\text{ cm}^{-1} \implies v = -\frac{36}{5} = -7.2\text{ cm}

.

The negative sign indicates that the image formed is virtual and located on the same side of the lens as the object.

4

Take the magnitude of the image distance.

|v| = 7.2\text{ cm}

.

The question asks for the distance, which is a scalar magnitude.

Key Concept

Thin Lens Formula & Sign Convention for Concave Lenses

Estimated Time:1m 30s

Question 3878Question →

An object is projected from ground level at an angle of $60^\circ$ to the horizontal. If the horizontal component of its initial velocity is $25\text{ m/s}$ , calculate the maximum height reached by the object in meters. (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 93.75

Answer

The maximum height reached by the object is

93.75\text{ m}

.

The maximum vertical height attained by a projectile depends on its vertical velocity component

u_y = u \sin \theta

. Resolving the initial velocity gives

u = 50\text{ m/s}

and

u_y = 25\sqrt{3}\text{ m/s}

. Substituting into

H = \frac{u_y^2}{2g}

yields

93.75\text{ m}

.

Step-by-Step Solution

1

Find the magnitude of the initial velocity

u = 50\text{ m/s}

The horizontal velocity component remains constant throughout flight and is given by

u_x = u \cos \theta

.

2

Calculate the initial vertical velocity component

u_y = 25\sqrt{3}\text{ m/s}

Vertical component of velocity is calculated using

u_y = u \sin \theta

.

3

Calculate the maximum height

H = 93.75\text{ m}

At maximum height, vertical velocity is zero, giving

H = \frac{u_y^2}{2g}

.

Key Concept

Resolution of velocity components in projectile motion and calculation of maximum height

Question 3879Question →

A gas sample enclosed in a container has an initial root-mean-square (r.m.s.) speed of $400\text{ m s}^{-1}$ at a temperature of $27^\circ\text{C}$ . If the gas is heated at constant volume until its temperature reaches $927^\circ\text{C}$ , what is the new r.m.s. speed of the gas molecules?

200\text{ m s}^{-1}

800\text{ m s}^{-1}

1600\text{ m s}^{-1}

2400\text{ m s}^{-1}

Show answer & explanation

Answer: $800\text{ m s}^{-1}$

Answer

The new root-mean-square speed of the gas molecules is

800\text{ m s}^{-1}

.

According to the kinetic theory of gases, the root-mean-square speed is directly proportional to the square root of the absolute temperature (

v_{\text{rms}} = \sqrt{3RT/M}

). Converting the temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 1200\text{ K}

. The ratio of absolute temperatures is

1200 / 300 = 4

. Taking the square root gives a factor of

2

, so the new r.m.s. speed is

400\text{ m s}^{-1} \times 2 = 800\text{ m s}^{-1}

.

Step-by-Step Solution

1

Convert initial and final temperatures from Celsius to Kelvin

T_1 = 27 + 273 = 300\text{ K}

and

T_2 = 927 + 273 = 1200\text{ K}

Kinetic theory equations require absolute temperature in Kelvin.

2

Apply the relationship between r.m.s. speed and absolute temperature

v_{\text{rms}} \propto \sqrt{T} \implies \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}

The mean kinetic energy of gas molecules is directly proportional to absolute temperature.

3

Substitute the values and calculate the final speed

v_2

v_2 = 400 \times \sqrt{\frac{1200}{300}} = 400 \times \sqrt{4} = 400 \times 2 = 800\text{ m s}^{-1}

Evaluating the square root factor yields the updated r.m.s. speed.

Key Concept

Root-mean-square speed of gas molecules is directly proportional to the square root of absolute temperature (

v_{\text{rms}} \propto \sqrt{T}

).

Estimated Time:1m 0s

Question 3880Question →

If $\int (6x^2 - 4\sin(2x)) \, dx = ax^3 + b\cos(2x) + C$ , where $a$ , $b$ , and $C$ are constants, what is the value of $a + b$ ?

Show answer & explanation

Answer: 4

Answer

The value of

a + b

is

4

.

Integrating

6x^2

gives

2x^3

, and integrating

-4\sin(2x)

gives

2\cos(2x)

. Comparing

2x^3 + 2\cos(2x) + C

to

ax^3 + b\cos(2x) + C

yields

a=2

and

b=2

, giving

a+b=4

.

Step-by-Step Solution

1

Integrate each term of the integrand

(6x^2 - 4\sin(2x))

with respect to

x

.

\int 6x^2 \, dx = \frac{6x^3}{3} = 2x^3

, and

\int -4\sin(2x) \, dx = -4 \cdot \left(-\frac{1}{2}\cos(2x)\right) = 2\cos(2x)

.

Applying the power rule for integration

\int x^n \, dx = \frac{x^{n+1}}{n+1}

and the standard trigonometric integral formula

\int \sin(kx) \, dx = -\frac{1}{k}\cos(kx)

.

2

Combine the calculated antiderivative terms and include the constant of integration

C

.

\int (6x^2 - 4\sin(2x)) \, dx = 2x^3 + 2\cos(2x) + C

.

Summing the individual term-by-term antiderivatives produces the complete indefinite integral.

3

Compare the resulting expression with

ax^3 + b\cos(2x) + C

to determine

a

and

b

.

a = 2

and

b = 2

.

Matching corresponding coefficients of

x^3

and

\cos(2x)

.

4

Compute

a + b

.

2 + 2 = 4

.

Evaluating the sum of the extracted coefficients.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Estimated Time:1m 30s

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