All practice questions

Question 6341Question →

A stone is thrown vertically downwards from the top of a $60\text{ m}$ high tower with an initial speed of $5\text{ m/s}$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the time taken, in seconds, for the stone to reach the ground.

Show answer & explanation

Answer: 3

Answer

The stone takes

3\text{ s}

to reach the ground.

Applying the equation

h = ut + \frac{1}{2}gt^2

with

h = 60\text{ m}

,

u = 5\text{ m/s}

, and

g = 10\text{ m/s}^2

yields

60 = 5t + 5t^2

. Dividing by

5

gives

t^2 + t - 12 = 0

, which factorizes into

(t + 4)(t - 3) = 0

. Rejecting

t = -4\text{ s}

leaves the correct time of

3\text{ s}

.

Step-by-Step Solution

1

Set up the vertical motion equation

Using

h = ut + \frac{1}{2}gt^2

, substitute

h = 60\text{ m}

,

u = 5\text{ m/s}

, and

g = 10\text{ m/s}^2

.

This equation directly relates displacement, initial speed, acceleration, and time.

2

Form and solve the quadratic equation for time

60 = 5t + 5t^2 \Rightarrow t^2 + t - 12 = 0 \Rightarrow (t - 3)(t + 4) = 0

, giving

t = 3\text{ s}

.

Time must be positive, so the physically meaningful solution is

t = 3\text{ s}

.

Key Concept

Vertical motion under gravity with non-zero initial downward velocity

Question 6342Question →

During the administrative audit, the executive director was fully exonerated _____ all allegations of financial misconduct because his actions were in strict compliance _____ established ministry guidelines. Which pair of prepositions correctly completes the sentence?

from / with

of / to

from / to

against / by

Show answer & explanation

Answer: from / with

Answer

The combination containing 'from / with' is correct because the verb 'exonerate' requires the preposition 'from' when referring to freeing an individual from blame, and the noun 'compliance' forms the fixed idiomatic phrase 'in compliance with'.

The correct response features 'from' paired with 'exonerated' and 'with' paired with 'compliance'. Both prepositions fulfill standard grammatical dependencies in formal English register.

Step-by-Step Solution

1

Determine the dependent preposition governed by the verb 'exonerated'.

The verb 'exonerate' takes 'from' (e.g., exonerated from blame/allegations).

In standard English grammar, one is acquitted of a crime but exonerated from a charge.

2

Identify the correct preposition completing the phrase 'in compliance _____'.

The correct preposition is 'with'.

'In compliance with' is a standard multi-word prepositional phrase meaning in agreement or accordance with a directive.

Key Concept

Dependent prepositions and multi-word prepositional phrases

Question 6343Question →

A copper rod has an initial length of $100\text{ cm}$ at $25^\circ\text{C}$ . When its temperature is increased to $75^\circ\text{C}$ , its length becomes $100.085\text{ cm}$ . What is the area expansivity of copper in $10^{-5}\text{ K}^{-1}$ ?

Show answer & explanation

Answer: 3.4

Answer

The area expansivity of copper is

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

, giving a numerical value of

3.4

in units of

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

.

Linear expansivity is obtained as

\alpha = \frac{\Delta L}{L_0 \Delta T} = \frac{0.085\text{ cm}}{100\text{ cm} \times 50\text{ K}} = 1.7 \times 10^{-5}\text{ K}^{-1}

. Since area expansivity

\beta

is related to linear expansivity by

\beta = 2\alpha

, multiplying by

2

yields

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

, or

3.4

in units of

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

.

Step-by-Step Solution

1

Calculate the temperature change

\Delta T = 75^\circ\text{C} - 25^\circ\text{C} = 50\text{ K}

Thermal expansion depends directly on the change in temperature.

2

Find the change in length

\Delta L = 100.085\text{ cm} - 100\text{ cm} = 0.085\text{ cm}

The expansion is the difference between the final length and original length.

3

Determine linear expansivity (

\alpha

)

\alpha = \frac{\Delta L}{L_0 \Delta T} = \frac{0.085}{100 \times 50} = 1.7 \times 10^{-5}\text{ K}^{-1}

Linear expansivity defines fractional change in length per degree temperature change.

4

Compute area expansivity (

\beta

)

\beta = 2\alpha = 2 \times (1.7 \times 10^{-5}) = 3.4 \times 10^{-5}\text{ K}^{-1}

Area (superficial) expansivity is equal to twice the linear expansivity.

Key Concept

Relationship between linear expansivity (

\alpha

) and area expansivity (

\beta = 2\alpha

).

Question 6344Question →

At a point on the Earth's surface, the total magnetic intensity is $4.0 \times 10^{-5}\text{ T}$ and the angle of dip is $60^\circ$ . What is the horizontal component of the Earth's magnetic field at this point?

2.0 \times 10^{-5}\text{ T}

3.5 \times 10^{-5}\text{ T}

4.0 \times 10^{-5}\text{ T}

8.0 \times 10^{-5}\text{ T}

Show answer & explanation

Answer: $2.0 \times 10^{-5}\text{ T}$

Answer

The horizontal component of the Earth's magnetic field is

2.0 \times 10^{-5}\text{ T}

.

The horizontal component

B_h

of the Earth's magnetic field is given by resolving the total magnetic intensity

B

along the horizontal direction using

B_h = B \cos \theta

. Substituting

B = 4.0 \times 10^{-5}\text{ T}

and

\theta = 60^\circ

yields

B_h = 4.0 \times 10^{-5} \times 0.5 = 2.0 \times 10^{-5}\text{ T}

.

Step-by-Step Solution

1

Identify the given values and formula

Total field

B = 4.0 \times 10^{-5}\text{ T}

, Angle of dip

\theta = 60^\circ

. Formula:

B_h = B \cos \theta

The horizontal component of Earth's magnetic field is obtained by resolving the total magnetic vector along the horizontal plane.

2

Substitute the given values into the equation

B_h = 4.0 \times 10^{-5}\text{ T} \times \cos(60^\circ) = 4.0 \times 10^{-5} \times 0.5 = 2.0 \times 10^{-5}\text{ T}

Since

\cos(60^\circ) = 0.5

, evaluating the product gives the exact horizontal component.

Key Concept

Resolution of Earth's Magnetic Field Components

Question 6345Question →

A galvanometer with an internal resistance of $5\text{ }\Omega$ produces a full-scale deflection when a current of $10\text{ mA}$ flows through it. Calculate the shunt resistance, in ohms, required to convert this galvanometer into an ammeter capable of measuring currents up to $50\text{ mA}$ .

Show answer & explanation

Answer: 1.25

Answer

The required shunt resistance is

1.25\text{ }\Omega

.

To extend the range of a galvanometer, a shunt resistor

S

is placed in parallel with it. The potential difference across the galvanometer equals the potential difference across the shunt:

I_g R_g = (I - I_g) S

. Substituting the given values

R_g = 5\text{ }\Omega

,

I_g = 10\text{ mA}

, and maximum current

I = 50\text{ mA}

yields

10\text{ mA} \times 5\text{ }\Omega = (50\text{ mA} - 10\text{ mA}) \times S

, solving to

S = \frac{50}{40} = 1.25\text{ }\Omega

.

Step-by-Step Solution

1

Find the current passing through the parallel shunt resistor

I_s = 40\text{ mA}

By Kirchhoff's current law, the total maximum current divides into the galvanometer current and the shunt current (

I = I_g + I_s

).

2

Calculate the required shunt resistance

S

S = 1.25\text{ }\Omega

Because the galvanometer and shunt resistor are connected in parallel, the potential difference across both branches is equal (

I_g R_g = I_s S

).

Key Concept

Conversion of a galvanometer to an ammeter using a low-resistance shunt in parallel

Question 6346Question →

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

\frac{\pi}{4}

\frac{3\pi}{4}

\frac{\pi}{2}

\pi

Show answer & explanation

Answer: $\frac{\pi}{4}$

Answer

The x-coordinate of the maximum stationary point is

\frac{\pi}{4}

.

To find stationary points, we differentiate

y = \sin x + \cos x

to get

\frac{dy}{dx} = \cos x - \sin x

. Setting

\frac{dy}{dx} = 0

gives

\sin x = \cos x

, or

\tan x = 1

. Within

0 \le x \le \pi

, the solution is

x = \frac{\pi}{4}

. Evaluating the second derivative

\frac{d^2y}{dx^2} = -\sin x - \cos x

at

x = \frac{\pi}{4}

yields

-\sqrt{2} < 0

, confirming a local maximum.

Step-by-Step Solution

1

Differentiate the function

y = \sin x + \cos x

with respect to

x

.

\frac{dy}{dx} = \cos x - \sin x

Stationary points occur where the first derivative

\frac{dy}{dx} = 0

.

2

Set the first derivative to zero and solve for

x

in the interval

0 \le x \le \pi

.

\cos x - \sin x = 0 \implies \sin x = \cos x \implies \tan x = 1 \implies x = \frac{\pi}{4}

Dividing both sides by

\cos x

gives

\tan x = 1

, which has the solution

x = \frac{\pi}{4}

in the given domain.

3

Verify the nature of the stationary point using the second derivative test.

\frac{d^2y}{dx^2} = -\sin x - \cos x \implies \left.\frac{d^2y}{dx^2}\right|_{x=\frac{\pi}{4}} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2} < 0

A negative second derivative confirms that

x = \frac{\pi}{4}

is a local maximum point.

Key Concept

Finding stationary points and determining their nature using derivatives of trigonometric functions.

Question 6347Question →

A coin lies at the bottom of a vessel filled with a liquid to a depth of $14.0\text{ cm}$ . If the refractive index of the liquid relative to air is $1.40$ , calculate the apparent upward displacement of the coin in centimeters when viewed vertically from directly above.

Show answer & explanation

Answer: 4

Answer

The apparent upward displacement of the coin is

4.0\text{ cm}

.

Refraction at the liquid-air boundary makes an object at real depth

h = 14.0\text{ cm}

appear at an apparent depth

h' = \frac{h}{n} = \frac{14.0}{1.40} = 10.0\text{ cm}

. The apparent upward displacement is the difference between real depth and apparent depth:

d = 14.0\text{ cm} - 10.0\text{ cm} = 4.0\text{ cm}

.

Step-by-Step Solution

1

Identify the given values and formula for refractive index in terms of depth.

Real depth

h = 14.0\text{ cm}

, refractive index

n = 1.40

. Formula:

n = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{h}{h'}

.

Light rays bending away from the normal upon leaving the denser liquid cause the coin to appear closer to the surface.

2

Calculate the apparent depth (

h'

).

h' = \frac{14.0\text{ cm}}{1.40} = 10.0\text{ cm}

.

Rearranging the refractive index formula gives

h' = \frac{h}{n}

.

3

Calculate the apparent upward displacement (

d

).

d = h - h' = 14.0\text{ cm} - 10.0\text{ cm} = 4.0\text{ cm}

.

The displacement is the distance between the actual position at the bottom and the virtual image position.

Key Concept

Real depth, apparent depth, and apparent displacement

Question 6348Question →

A glass flask is filled to a mark with a liquid at $0^\circ\text{C}$ . If the linear expansivity of the glass is $8.0 \times 10^{-6} \text{ K}^{-1}$ and the apparent cubic expansivity of the liquid is $1.80 \times 10^{-4} \text{ K}^{-1}$ , what is the real cubic expansivity of the liquid?

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

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

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

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

Show answer & explanation

Answer: $2.04 \times 10^{-4} \text{ K}^{-1}$

Answer

The real cubic expansivity of the liquid is

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

.

The real cubic expansivity of a liquid is given by

\gamma_r = \gamma_a + \gamma_v

. Since the container's linear expansivity

\alpha = 8.0 \times 10^{-6} \text{ K}^{-1}

, its cubic expansivity is

\gamma_v = 3\alpha = 2.40 \times 10^{-5} \text{ K}^{-1} = 0.24 \times 10^{-4} \text{ K}^{-1}

. Adding this to the apparent cubic expansivity

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

gives

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

.

Step-by-Step Solution

1

Calculate the cubic expansivity of the glass vessel (

\gamma_v

).

\gamma_v = 3 \times \alpha = 3 \times (8.0 \times 10^{-6} \text{ K}^{-1}) = 2.40 \times 10^{-5} \text{ K}^{-1} = 0.24 \times 10^{-4} \text{ K}^{-1}

Cubic expansivity of a solid material is three times its linear expansivity.

2

Apply the relationship between real expansivity (

\gamma_r

), apparent expansivity (

\gamma_a

), and vessel expansivity (

\gamma_v

).

\gamma_r = \gamma_a + \gamma_v

The real expansion of a liquid is the sum of its observed (apparent) expansion and the expansion of the containing vessel.

3

Substitute the values into the formula to find

\gamma_r

.

\gamma_r = 1.80 \times 10^{-4} \text{ K}^{-1} + 0.24 \times 10^{-4} \text{ K}^{-1} = 2.04 \times 10^{-4} \text{ K}^{-1}

Adding the apparent cubic expansivity and the vessel's cubic expansivity yields the true (real) cubic expansivity.

Key Concept

Real cubic expansivity of a liquid equals the sum of its apparent cubic expansivity and the cubic expansivity of the vessel (

\gamma_r = \gamma_a + 3\alpha_v

).

Question 6349Question →

A copper container with an initial volume of $800 \text{ cm}^3$ at $25^\circ\text{C}$ is completely filled with oil. The real cubic expansivity of the oil is $6.5 \times 10^{-4} \text{ K}^{-1}$ and the linear expansivity of copper is $1.5 \times 10^{-5} \text{ K}^{-1}$ . What volume of oil (in $\text{cm}^3$ ) will overflow when the temperature of the system is raised to $75^\circ\text{C}$ ?

Show answer & explanation

Answer: 24.2

Answer

The volume of oil that overflows is

24.2 \text{ cm}^3

.

When a container completely filled with liquid is heated, both liquid and container expand. The overflow volume equals the apparent volume expansion of the liquid

\Delta V_a = V_0 \gamma_a \Delta T

. The apparent cubic expansivity

\gamma_a

is obtained by subtracting the container's volume expansivity (

\gamma_v = 3\alpha = 4.5 \times 10^{-5} \text{ K}^{-1}

) from the liquid's real cubic expansivity (

\gamma_r = 6.5 \times 10^{-4} \text{ K}^{-1}

), giving

\gamma_a = 6.05 \times 10^{-4} \text{ K}^{-1}

. Multiplying by

V_0 = 800 \text{ cm}^3

and

\Delta T = 50 \text{ K}

gives

24.2 \text{ cm}^3

.

Step-by-Step Solution

1

Calculate the cubic expansivity of the copper container

\gamma_v = 3 \alpha = 3 \times (1.5 \times 10^{-5} \text{ K}^{-1}) = 4.5 \times 10^{-5} \text{ K}^{-1} = 0.45 \times 10^{-4} \text{ K}^{-1}

The volumetric expansion coefficient of a solid container is three times its linear expansivity.

2

Determine the apparent cubic expansivity of the oil

\gamma_a = \gamma_r - \gamma_v = 6.5 \times 10^{-4} \text{ K}^{-1} - 0.45 \times 10^{-4} \text{ K}^{-1} = 6.05 \times 10^{-4} \text{ K}^{-1}

The apparent expansion of a liquid accounts for the concurrent thermal expansion of the containing vessel.

3

Compute the temperature increase

\Delta T = 75^\circ\text{C} - 25^\circ\text{C} = 50 \text{ K}

Temperature change is the final temperature minus the initial temperature.

4

Calculate the volume of oil that overflows

\Delta V_a = V_0 \gamma_a \Delta T = 800 \text{ cm}^3 \times (6.05 \times 10^{-4} \text{ K}^{-1}) \times 50 \text{ K} = 24.2 \text{ cm}^3

The overflow volume is equal to the apparent increase in volume of the liquid.

Key Concept

Apparent and Real Expansion of Liquids

Question 6350Question →

Given the universal set $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 30\}$ , let $A = \{x \in \mathcal{U} : x \text{ is a multiple of } 4\}$ and $B = \{x \in \mathcal{U} : x \text{ is a perfect square}\}$ . What is the number of elements in $(A \cup B)'$ ?

20

18

10

22

Show answer & explanation

Answer: 20

Answer

The cardinality of

(A \cup B)'

is 20.

The universal set has 30 elements. Set

A

has 7 elements (multiples of 4 up to 30) and set

B

has 5 elements (perfect squares up to 30). The numbers 4 and 16 belong to both sets, so the intersection has 2 elements. By the principle of inclusion-exclusion, the union

A \cup B

contains

7 + 5 - 2 = 10

elements. The complement

(A \cup B)'

contains all elements in the universal set that are not in the union, giving

30 - 10 = 20

elements.

Step-by-Step Solution

1

Determine the elements and cardinality of the universal set

\mathcal{U}

.

\mathcal{U} = \{1, 2, 3, \dots, 30\}

, so

n(\mathcal{U}) = 30

.

The universal set bounds all possible elements under consideration.

2

List the elements of set

A

and set

B

, and find their individual cardinalities.

A = \{4, 8, 12, 16, 20, 24, 28\} \implies n(A) = 7

.

B = \{1, 4, 9, 16, 25\} \implies n(B) = 5

.

Identify multiples of 4 and perfect squares within the range 1 to 30.

3

Find the intersection

A \cap B

and compute the cardinality of the union

A \cup B

.

A \cap B = \{4, 16\} \implies n(A \cap B) = 2

.

n(A \cup B) = n(A) + n(B) - n(A \cap B) = 7 + 5 - 2 = 10

.

Apply the principle of inclusion-exclusion to avoid double-counting elements belonging to both sets.

4

Calculate the cardinality of the complement

(A \cup B)'

.

n((A \cup B)') = n(\mathcal{U}) - n(A \cup B) = 30 - 10 = 20

.

The complement set consists of all elements in the universal set that are not in

A \cup B

.

Key Concept

Complement of Set Union and Inclusion-Exclusion Principle

Question 6351Question →

In a agricultural survey of 120 farmers in a community, 65 grow maize, 50 grow yam, and 42 grow cassava. Furthermore, 24 grow both maize and yam, 18 grow both maize and cassava, and 15 grow both yam and cassava. If 8 farmers grow none of these three crops, find the number of farmers who grow all three crops.

Show answer & explanation

Answer: 12

Answer

12 farmers grow all three crops.

Using the 3-set inclusion-exclusion principle, the total number of farmers growing at least one crop is

120 - 8 = 112

. Expanding

|M \cup Y \cup C| = |M| + |Y| + |C| - (|M \cap Y| + |M \cap C| + |Y \cap C|) + |M \cap Y \cap C|

gives

112 = 65 + 50 + 42 - 24 - 18 - 15 + x

. Simplifying yields

112 = 100 + x

, which gives

x = 12

.

Step-by-Step Solution

1

Determine the cardinality of the union of all three sets.

|M \cup Y \cup C| = 112

Subtract the farmers who grow none of the crops from the universal set size (

120 - 8 = 112

).

2

Set up the Inclusion-Exclusion equation for three sets.

112 = 65 + 50 + 42 - (24 + 18 + 15) + x

Inclusion-exclusion states that

|A \cup B \cup C| = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)

.

3

Solve for the unknown value

x

representing the intersection of all three sets.

x = 12

Simplifying gives

112 = 100 + x

, which leads directly to

x = 12

.

Key Concept

Principle of Inclusion-Exclusion for 3 Sets

Question 6352Question →

A projectile is launched from horizontal ground with an initial speed of $60\text{ m/s}$ at an angle of $60^\circ$ to the horizontal. Taking $g = 10\text{ m/s}^2$ and neglecting air resistance, what is the speed of the projectile at its maximum height?

0\text{ m/s}

30\text{ m/s}

30\sqrt{3}\text{ m/s}

60\text{ m/s}

Show answer & explanation

Answer: $30\text{ m/s}$

Answer

The speed of the projectile at its maximum height is

30\text{ m/s}

.

In two-dimensional projectile motion, gravity acts purely in the vertical direction. At the apex (maximum height), the vertical velocity component

v_y

drops to zero. However, the horizontal velocity component

v_x = u \cos \theta

remains constant throughout the entire flight because air resistance is neglected. Consequently, the speed at maximum height equals

u \cos 60^\circ = 60 \times 0.5 = 30\text{ m/s}

.

Step-by-Step Solution

1

Resolve initial launch velocity into orthogonal horizontal and vertical components.

u_x = u \cos \theta = 60 \cos 60^\circ = 30\text{ m/s}

and

u_y = u \sin \theta = 60 \sin 60^\circ = 30\sqrt{3}\text{ m/s}

.

Projectile motion consists of independent horizontal and vertical motions.

2

Determine the velocity components at the apex (maximum height).

At the peak, vertical velocity

v_y = 0\text{ m/s}

while horizontal velocity remains

v_x = u_x = 30\text{ m/s}

.

Gravity acts vertically causing

v_y

to become zero at peak height, whereas no horizontal force acts, leaving

v_x

unchanged.

3

Calculate the magnitude of total velocity (speed) at maximum height.

v = \sqrt{v_x^2 + v_y^2} = \sqrt{30^2 + 0^2} = 30\text{ m/s}

.

Speed is the resultant magnitude of orthogonal velocity components.

Key Concept

Velocity at Maximum Height in Projectile Motion

Question 6353Question →

Match each atomic model with its key defining feature or experimental foundation.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Thomson Model matches with electrons embedded in a positive sphere; Rutherford Model matches with the dense positive nucleus discovered via alpha particle scattering; Bohr Model matches with quantized non-radiating energy levels; Quantum Wave Model matches with standing matter waves in electron orbits.

Each atomic model corresponds directly to its historic milestone: Thomson proposed electrons scattered inside a uniform sphere of positive charge; Rutherford discovered the compact positive nucleus through alpha scattering experiments; Bohr introduced quantized stationary states to account for emission spectral lines; and the Quantum Wave Model integrated de Broglie matter wave harmonics into orbital mechanics.

Step-by-Step Solution

1

Identify Thomson's Plum Pudding Model features

Describes electrons distributed inside a broad, uniform cloud/sphere of positive charge.

Formulated after J.J. Thomson discovered the electron before the nucleus was known.

2

Identify Rutherford's Nuclear Model features

Demonstrated that positive charge is concentrated in a tiny nucleus because alpha particles bounced back at large angles.

Geiger-Marsden alpha scattering experiment disproved the uniform charge distribution model.

3

Identify Bohr's Model features

Postulated quantized non-radiating stationary orbits where angular momentum is restricted to integral multiples of

h / 2\pi

.

Successfully explained the discrete wavelengths of the hydrogen emission spectrum.

4

Identify the Quantum Wave Model features

Explains quantization by treating orbiting electrons as de Broglie standing waves around the nucleus.

Orbits are stable only when an integral number of electron wavelengths fit along the orbital circumference.

Key Concept

Characteristics and experimental foundations of atomic models

Question 6354Question →

A ray of light traveling in air strikes the flat surface of a transparent glass slab at an angle of incidence of $60^\circ$ . If the refractive index of the glass slab relative to air is $\sqrt{3}$ , what is the angle of refraction inside the glass slab in degrees?

Show answer & explanation

Answer: 30

Answer

The angle of refraction inside the glass slab is

30^\circ

.

According to Snell's Law,

n = \frac{\sin i}{\sin r}

. Substituting

n = \sqrt{3}

and

i = 60^\circ

gives

\sqrt{3} = \frac{\sin 60^\circ}{\sin r}

. Since

\sin 60^\circ = \frac{\sqrt{3}}{2}

, rearranging gives

\sin r = \frac{\sqrt{3}/2}{\sqrt{3}} = 0.5

. Taking the inverse sine of

0.5

gives

r = 30^\circ

.

Step-by-Step Solution

1

Identify Snell's law formula relating the angle of incidence and angle of refraction.

n = \frac{\sin i}{\sin r}

Snell's law describes how light bends when crossing the boundary between two optical media.

2

Substitute the given values

i = 60^\circ

and

n = \sqrt{3}

into the equation.

\sqrt{3} = \frac{\sin 60^\circ}{\sin r}

Plugging in the known parameters allows us to isolate the unknown sine of the angle of refraction.

3

Substitute

\sin 60^\circ = \frac{\sqrt{3}}{2}

and rearrange for

\sin r

.

\sin r = \frac{\sqrt{3}/2}{\sqrt{3}} = 0.5

Canceling

\sqrt{3}

from both sides yields a simple numerical value for

\sin r

.

4

Take the inverse sine of

0.5

to find

r

.

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

The angle whose sine is

0.5

is

30^\circ

.

Key Concept

Snell's Law of Refraction

Question 6355Question →

Let the universal set $\mathcal{E} = \{x \in \mathbb{Z} : 1 \le x \le 25\}$ . Two subsets $A$ and $B$ of $\mathcal{E}$ are defined as $A = \{x \in \mathcal{E} : x \text{ is a perfect square}\}$ and $B = \{x \in \mathcal{E} : x \text{ is an odd number}\}$ . What is the cardinality of $(A \cup B)'$ ?

10

12

15

7

Show answer & explanation

Answer: 10

Answer

10

The correct answer is 10 because the universal set has 25 elements. The union

A \cup B

contains all 13 odd numbers in the range along with the 2 even perfect squares (

4

and

16

), giving

n(A \cup B) = 15

. Subtracting this from the universal set size gives

25 - 15 = 10

.

Step-by-Step Solution

1

Identify the elements of the universal set and subsets A and B

\mathcal{E} = \{1, 2, 3, \dots, 25\}

with

n(\mathcal{E}) = 25

.

A = \{1, 4, 9, 16, 25\}

B = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25\}

Explicitly listing elements clarifies overlapping regions.

2

Find the union set

A \cup B

and its cardinality

A \cup B = \{1, 3, 4, 5, 7, 9, 11, 13, 15, 16, 17, 19, 21, 23, 25\}

, so

n(A \cup B) = 15

.

The union includes all odd numbers from 1 to 25 plus the even perfect squares (4 and 16).

3

Calculate the cardinality of the complement

(A \cup B)'

n((A \cup B)') = n(\mathcal{E}) - n(A \cup B) = 25 - 15 = 10

.

The complement set contains all elements in

\mathcal{E}

that are neither odd nor perfect squares, which are the 10 non-square even numbers:

\{2, 6, 8, 10, 12, 14, 18, 20, 22, 24\}

.

Key Concept

Set Complements and De Morgan's Laws / Set Operations

Estimated Time:1m 30s

Question 6356Question →

Match each prose excerpt with the primary narrative technique or point of view it demonstrates.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The excerpt starting with 'I packed my woven bag...' matches First-person central point of view. The excerpt starting with 'Tunde paced around his office...' matches Third-person omniscient point of view. The excerpt beginning with 'Thoughts tumbled through her head...' matches Stream of consciousness technique. The excerpt starting with 'Dear Brother, I write this letter...' matches Epistolary narrative technique.

Each literary excerpt displays distinct linguistic markers matching its narrative device: 'I packed' uses first-person pronouns for central perspective; narrative omniscience reveals actions happening in two places at once; rapid internal thought associations characterize stream of consciousness; and direct letter correspondence defines the epistolary format.

Step-by-Step Solution

1

Analyze the first excerpt for grammatical voice and perspective.

Identified the use of 'I', 'my', and personal emotion, signifying First-person central point of view.

First-person central narrative directly presents the protagonist's personal experiences.

2

Evaluate the second excerpt for narrative scope and knowledge.

The narrator knows Tunde's lack of awareness as well as Chief Adebayo's simultaneous actions elsewhere, establishing Third-person omniscient perspective.

Omniscience allows an all-knowing narrator to report simultaneous events in different settings.

3

Examine the stylistic structure of the third excerpt.

The rapid, unstructured presentation of internal worries reflects the Stream of consciousness technique.

Stream of consciousness mimics continuous mental flux and impressionistic thought flow.

4

Identify the genre format of the fourth excerpt.

The salutation 'Dear Brother' and reference to a courier indicate an Epistolary narrative technique.

Epistolary style advances plot and character development through written correspondence.

Key Concept

Identification and Functional Analysis of Narrative Techniques and Point of View in Prose Literature

Question 6357Question →

A heavy-duty truck tire contains a fixed mass of air at an initial absolute pressure of $2.00 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . After traveling a long distance, friction causes the temperature of the air inside the tire to increase to $57^\circ\text{C}$ while its volume remains constant. What is the new absolute pressure of the air inside the tire in pascals ( $\text{Pa}$ )?

Show answer & explanation

Answer: 220000

Answer

The new absolute pressure of the air inside the tire is

220,000\text{ Pa}

(or

2.20 \times 10^5\text{ Pa}

).

According to Gay-Lussac's Law, at constant volume, pressure is directly proportional to absolute temperature (

P \propto T

). Converting temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 330\text{ K}

. Calculating

P_2 = P_1 \times \frac{T_2}{T_1}

gives

2.00 \times 10^5\text{ Pa} \times \frac{330}{300} = 220,000\text{ Pa}

.

Step-by-Step Solution

1

Convert the initial and final temperatures from Celsius to the absolute Kelvin scale.

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

and

T_2 = 57^\circ\text{C} + 273 = 330\text{ K}

.

All gas law equations require absolute temperatures in Kelvin.

2

Apply the Pressure Law (Gay-Lussac's Law) for a fixed volume of gas.

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

When volume is constant, gas pressure is directly proportional to absolute temperature.

3

Substitute the known values into the equation to calculate the final pressure

P_2

.

P_2 = 2.00 \times 10^5\text{ Pa} \times \frac{330\text{ K}}{300\text{ K}} = 2.20 \times 10^5\text{ Pa} = 220,000\text{ Pa}

.

Multiplying the initial pressure by the ratio of absolute temperatures gives the new pressure.

Key Concept

Pressure Law (Gay-Lussac's Law)

Question 6358Question →

A flexible research balloon is filled with $1.50\text{ m}^3$ of helium gas at a temperature of $27^\circ\text{C}$ . If the gas is heated at constant pressure until its temperature reaches $127^\circ\text{C}$ , what is the new volume of the balloon?

2.00\text{ m}^3

7.06\text{ m}^3

1.13\text{ m}^3

1.83\text{ m}^3

Show answer & explanation

Answer: $2.00\text{ m}^3$

Answer

The new volume of the balloon is

2.00\text{ m}^3

.

The answer of

2.00\text{ m}^3

correctly uses Charles's Law with absolute temperatures converted to Kelvin (

300\text{ K}

and

400\text{ K}

), showing that heating the gas causes a proportional volume expansion.

Step-by-Step Solution

1

Convert temperatures from Celsius to the Kelvin absolute scale.

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

,

T_2 = 127 + 273 = 400\text{ K}

.

Gas laws strictly require thermodynamic (absolute) temperature measured in Kelvin.

2

Apply Charles's Law for constant pressure processes.

\frac{V_1}{T_1} = \frac{V_2}{T_2} \implies V_2 = V_1 \times \frac{T_2}{T_1}

.

Volume is directly proportional to absolute temperature when pressure remains constant.

3

Substitute the known values and calculate

V_2

.

V_2 = 1.50\text{ m}^3 \times \frac{400\text{ K}}{300\text{ K}} = 2.00\text{ m}^3

.

Simplifying the fraction

\frac{400}{300} = \frac{4}{3}

gives

1.50 \times \frac{4}{3} = 2.00\text{ m}^3

.

Key Concept

Charles's Law (

V_1/T_1 = V_2/T_2

at constant pressure)

Estimated Time:1m 15s

Question 6359Question →

A point $P(x, y)$ moves in a plane such that the line segment joining the fixed points $A(1, 2)$ and $B(5, 6)$ subtends a right angle at $P$ . Which of the following equations represents the locus of $P$ ?

x^2 + y^2 - 6x - 8y + 17 = 0

x - y + 1 = 0

x^2 + y^2 - 6x - 8y - 7 = 0

x^2 + y^2 + 6x + 8y + 17 = 0

Show answer & explanation

Answer: $x^2 + y^2 - 6x - 8y + 17 = 0$

Answer

The equation of the locus of

P

is

x^2 + y^2 - 6x - 8y + 17 = 0

.

The locus of a point

P

that subtends a

90^\circ

angle at two fixed points

A(1, 2)

and

B(5, 6)

forms a circle with

AB

as diameter. Using the gradient condition for perpendicular lines,

\frac{y-2}{x-1} \times \frac{y-6}{x-5} = -1

, which simplifies to

x^2 + y^2 - 6x - 8y + 17 = 0

.

Step-by-Step Solution

1

Apply the perpendicularity condition for the line segments

AP

and

BP

.

Gradient of

AP = \frac{y - 2}{x - 1}

and gradient of

BP = \frac{y - 6}{x - 5}

. Since

\angle APB = 90^\circ

, their product must be

-1

:

\left(\frac{y - 2}{x - 1}\right) \cdot \left(\frac{y - 6}{x - 5}\right) = -1

.

Two perpendicular line segments have gradients whose product is

-1

.

2

Multiply out the denominators and numerators.

(y - 2)(y - 6) = -(x - 1)(x - 5) \implies y^2 - 8y + 12 = -(x^2 - 6x + 5).

Algebraic expansion of the equation obtained from the gradient product.

3

Rearrange all terms to one side to express in standard second-degree form.

x^2 + y^2 - 6x - 8y + 17 = 0.

Rearranging yields the Cartesian equation of the locus.

Key Concept

Locus of a point subtending a right angle at two fixed points

Estimated Time:1m 30s

Question 6360Question →

An electron in an atom transitions from an excited state with energy $-1.7 \times 10^{-19}\text{ J}$ to a lower state with energy $-5.0 \times 10^{-19}\text{ J}$ . What is the wavelength of the emitted photon in nanometers ( $\text{nm}$ )? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ and the speed of light $c = 3.0 \times 10^8\text{ m/s}$ )

Show answer & explanation

Answer: 600

Answer

The wavelength of the emitted photon is 600 nm.

The energy released during the atomic transition is

\Delta E = (-1.7 \times 10^{-19}\text{ J}) - (-5.0 \times 10^{-19}\text{ J}) = 3.3 \times 10^{-19}\text{ J}

. Substituting this into

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

yields

\lambda = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{3.3 \times 10^{-19}} = 6.0 \times 10^{-7}\text{ m}

. Converting to nanometers (

1\text{ m} = 10^9\text{ nm}

) gives

600\text{ nm}

.

Step-by-Step Solution

1

Calculate energy of the emitted photon

\Delta E = 3.3 \times 10^{-19}\text{ J}

The energy of the photon equals the difference between the initial higher energy level and the final lower energy level.

2

Calculate wavelength in meters using Planck's relation

\lambda = 6.0 \times 10^{-7}\text{ m}

Rearranging

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

gives

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

.

3

Convert wavelength to nanometers

600\text{ nm}

Multiply by

10^9\text{ nm/m}

to obtain the final value in nanometers.

Key Concept

Energy level transitions and photon emission wavelength

Answer

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