All practice questions

Question 7661Question →

In an agricultural trial, two independent seed varieties, $V_1$ and $V_2$ , have germination probabilities $P(V_1) = p$ and $P(V_2) = p + 0.20$ . If the probability that at least one seed variety germinates is $0.76$ , what is the probability that only variety $V_2$ germinates?

0.36

0.48

0.52

0.24

Show answer & explanation

Answer: 0.36

Answer

The probability that only variety

V_2

germinates is

0.36

.

Using the addition law

P(V_1 \cup V_2) = P(V_1) + P(V_2) - P(V_1)P(V_2)

gives

0.76 = 2p + 0.20 - (p^2 + 0.20p)

. Solving

p^2 - 1.80p + 0.56 = 0

gives

p = 0.40

. Thus

P(V_1) = 0.40

and

P(V_2) = 0.60

. The probability that only variety

V_2

germinates is

P(V_2) \cdot P(V_1') = 0.60 \times 0.60 = 0.36

.

Step-by-Step Solution

1

Set up the probability addition law for independent events.

P(V_1 \cup V_2) = P(V_1) + P(V_2) - P(V_1 \cap V_2)

. Since

V_1

and

V_2

are independent,

P(V_1 \cap V_2) = P(V_1) \cdot P(V_2) = p(p + 0.20)

.

Independent events allow the intersection probability to be expressed as the product of their individual probabilities.

2

Substitute the given values into the addition law and solve for

p

.

0.76 = p + (p + 0.20) - p(p + 0.20) \implies p^2 - 1.80p + 0.56 = 0 \implies (p - 0.40)(p - 1.40) = 0

. Since

p \le 1

,

p = 0.40

.

Formulating a quadratic equation yields the value of

p

within valid probability bounds.

3

Calculate individual probabilities

P(V_1)

and

P(V_2)

.

P(V_1) = 0.40

and

P(V_2) = 0.40 + 0.20 = 0.60

.

Knowing

p

gives the exact germination probabilities for both varieties.

4

Find the probability that only variety

V_2

germinates.

P(\text{only } V_2) = P(V_2 \cap V_1') = P(V_2) \times [1 - P(V_1)] = 0.60 \times (1 - 0.40) = 0.60 \times 0.60 = 0.36

.

Only

V_2

germinating means

V_2

germinates and

V_1

fails to germinate.

Key Concept

Probability laws for independent compound events

Question 7662Question →

Match each temperature scale reference state on the left with its correct thermodynamic definition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Absolute zero matches with the state of minimum molecular kinetic energy. The ice point matches with the lower fixed point of pure melting ice at standard pressure. The steam point matches with the upper fixed point of pure boiling water steam at standard pressure. The triple point of water matches with the thermodynamic equilibrium state of ice, liquid water, and water vapour.

Each temperature scale reference point is correctly paired with its defining physical state: absolute zero represents minimum molecular kinetic energy, the ice point represents melting ice at standard pressure, the steam point represents steam from boiling water at standard pressure, and the triple point represents the three-phase equilibrium of water.

Step-by-Step Solution

1

Identify the definition of absolute zero.

Absolute zero (

0\text{ K}

) corresponds to the state of minimum internal molecular kinetic energy.

At

0\text{ K}

, thermal motion of particles theoretically ceases.

2

Identify the definition of the ice point.

The ice point (

0^\circ\text{C}

) corresponds to pure melting ice at standard atmospheric pressure.

It serves as the standard lower fixed point on the Celsius temperature scale.

3

Identify the definition of the steam point.

The steam point (

100^\circ\text{C}

) corresponds to steam from pure boiling water at standard atmospheric pressure.

It serves as the standard upper fixed point on the Celsius temperature scale.

4

Identify the definition of the triple point of water.

The triple point (

273.16\text{ K}

) is the unique thermodynamic state where ice, liquid water, and steam coexist in equilibrium.

It is used as a single fundamental reference point on the Kelvin thermodynamic scale.

Key Concept

Temperature Scale Fixed Points and Reference States

Estimated Time:1m 30s

Question 7663Question →

In Bohr's model of the hydrogen atom, the radius of a stationary orbit is proportional to $n^2$ , where $n$ is the principal quantum number. If the radius of the ground-state orbit ( $n = 1$ ) is $r_1$ , what is the radius of the orbit corresponding to the second excited state?

3r_1

4r_1

6r_1

9r_1

Show answer & explanation

Answer: $9r_1$

Answer

The radius of the electron's orbit in the second excited state is

9r_1

.

In Bohr's atomic model, the ground state corresponds to the quantum number

n = 1

. The excited states are numbered sequentially above the ground state: the first excited state is

n = 2

and the second excited state is

n = 3

. Since the orbital radius scales with the square of the principal quantum number (

r_n = n^2 r_1

), substituting

n = 3

yields

r_3 = 3^2 r_1 = 9r_1

. Thus, the option stating

9r_1

is correct.

Step-by-Step Solution

1

Identify the principal quantum number

n

for the second excited state.

n = 3

The ground state corresponds to

n = 1

, the first excited state corresponds to

n = 2

, and the second excited state corresponds to

n = 3

.

2

Apply Bohr's radius formula for hydrogen-like atoms.

r_n = n^2 r_1

According to Bohr's postulates, the orbital radius is proportional to the square of the principal quantum number

n

.

3

Calculate

r_3

by substituting

n = 3

into the radius expression.

r_3 = 3^2 r_1 = 9r_1

Squaring

n = 3

gives

9

, making the radius 9 times the ground-state radius.

Key Concept

Bohr's quantization of orbital radius (

r_n \propto n^2

)

Estimated Time:1m 0s

Question 7664Question →

If $\frac{4\sqrt{3}}{\sqrt{7} + \sqrt{3}} + \frac{4\sqrt{7}}{\sqrt{7} - \sqrt{3}} = p + q\sqrt{21}$ , where $p$ and $q$ are integers, what is the value of $p + q$ ?

Show answer & explanation

Answer: 6

Answer

The value of

p + q

is 6.

Rationalising each fraction yields

(\sqrt{21} - 3)

and

(7 + \sqrt{21})

. Summing these expressions gives

4 + 2\sqrt{21}

. Comparing this to

p + q\sqrt{21}

yields

p = 4

and

q = 2

, so

p + q = 6

.

Step-by-Step Solution

1

Rationalise the first term

\frac{4\sqrt{3}}{\sqrt{7} + \sqrt{3}}

by multiplying the numerator and denominator by the conjugate

(\sqrt{7} - \sqrt{3})

.

\frac{4\sqrt{3}(\sqrt{7} - \sqrt{3})}{(\sqrt{7})^2 - (\sqrt{3})^2} = \frac{4\sqrt{21} - 12}{7 - 3} = \frac{4\sqrt{21} - 12}{4} = \sqrt{21} - 3

Multiplying by the conjugate eliminates the surd from the denominator using the difference of two squares.

2

Rationalise the second term

\frac{4\sqrt{7}}{\sqrt{7} - \sqrt{3}}

by multiplying the numerator and denominator by the conjugate

(\sqrt{7} + \sqrt{3})

.

\frac{4\sqrt{7}(\sqrt{7} + \sqrt{3})}{(\sqrt{7})^2 - (\sqrt{3})^2} = \frac{28 + 4\sqrt{21}}{7 - 3} = \frac{28 + 4\sqrt{21}}{4} = 7 + \sqrt{21}

Conjugate rationalisation simplifies the second fraction into linear surd terms.

3

Add the two simplified expressions together and equate to

p + q\sqrt{21}

.

(\sqrt{21} - 3) + (7 + \sqrt{21}) = 4 + 2\sqrt{21}

Combining like surd terms yields the simplified form

p + q\sqrt{21}

.

4

Identify the values of

p

and

q

and evaluate

p + q

.

p = 4, q = 2 \implies p + q = 4 + 2 = 6

Equating the rational parts gives

p = 4

and the irrational coefficients gives

q = 2

.

Key Concept

Rationalisation of surds with binomial denominators

Alternative Method

Combine the two fractions directly over the common denominator

(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3}) = 4

: \frac{4\sqrt{3}(\sqrt{7} - \sqrt{3}) + 4\sqrt{7}(\sqrt{7} + \sqrt{3})}{4} = \frac{4\sqrt{21} - 12 + 28 + 4\sqrt{21}}{4} = \frac{16 + 8\sqrt{21}}{4} = 4 + 2\sqrt{21}.

Estimated Time:1m 30s

Question 7665Question →

The frequency distribution table below shows the daily profits (in thousands of Naira, $\text{₦}$ ) recorded by a sample of $40$ small-scale market traders:

Daily Profit ( $\text{₦'000}$ )	Frequency ( $f$ )
$10 - 14$	$6$
$15 - 19$	$10$
$20 - 24$	$12$
$25 - 29$	$8$
$30 - 34$	$4$

What is the mean daily profit of the traders, in thousands of Naira?

Show answer & explanation

Answer: 21.25

Answer

The mean daily profit of the traders is

21.25

thousand Naira.

The mean daily profit is found by dividing the sum of the product of each class midpoint and its frequency (

\sum fx = 850

) by the total frequency (

\sum f = 40

), yielding

21.25

.

Step-by-Step Solution

1

Find the class midpoints (

x

) for each interval

Midpoints are

12

,

17

,

22

,

27

, and

32

.

For grouped data, each class interval is represented by its midpoint.

2

Calculate the product of frequency and midpoint (

f \cdot x

) for each interval

Products are

72

,

170

,

264

,

216

, and

128

.

Multiplying class midpoint by class frequency estimates the sum of values within that class.

3

Calculate total frequency (

\sum f

) and total sum of products (

\sum fx

)

\sum f = 40

and

\sum fx = 850

.

The sum of frequencies gives the total number of observations, and the sum of products gives the estimated grand total.

4

Apply the grouped mean formula

\bar{x} = \frac{\sum fx}{\sum f}

\bar{x} = \frac{850}{40} = 21.25.

Dividing total sum by total frequency gives the mean value.

Key Concept

Measures of Central Tendency for Grouped Data - Mean

Question 7666Question →

In commercial law, a contract under seal (specialty contract) requires valuable consideration to be legally binding on the parties involved.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. A contract under seal (specialty contract) is legally binding due to its formal execution by deed, even in the absence of valuable consideration.

The statement is false because a specialty contract (or contract under seal) derives its binding legal force from the solemnity of its form and execution, making valuable consideration unnecessary for its validity.

Step-by-Step Solution

1

Identify the specific type of contract mentioned in the statement.

The statement refers to a specialty contract (contract under seal).

Different types of contracts (simple contracts vs. contracts under seal) have distinct legal requirements for validity.

2

Examine the role of consideration in specialty contracts.

Contracts under seal do not require valuable consideration because their validity stems from their formal execution under deed.

General simple contracts require offer, acceptance, and consideration, whereas formal specialty contracts are an exception to the requirement of consideration.

3

Evaluate the accuracy of the statement.

The statement asserts that consideration is required for a specialty contract, which is legally inaccurate.

Since consideration is not necessary for specialty contracts, the statement is false.

Key Concept

Distinction between simple contracts and specialty contracts regarding the requirement of consideration

Question 7667Question →

An electron in a hydrogen atom transitions from an excited energy state of $-3.4\text{ eV}$ to the ground state of $-13.6\text{ eV}$ . What is the energy of the photon emitted during this transition?

10.2\text{ eV}

17.0\text{ eV}

-10.2\text{ eV}

3.4\text{ eV}

Show answer & explanation

Answer: $10.2\text{ eV}$

Answer

10.2\text{ eV}

When an electron drops to a lower energy state, the energy of the emitted photon equals the difference between the initial and final energy levels:

E_{photon} = E_{initial} - E_{final} = -3.4\text{ eV} - (-13.6\text{ eV}) = 10.2\text{ eV}

.

Step-by-Step Solution

1

Identify the initial and final energy states.

Initial state

E_i = -3.4\text{ eV}

and final ground state

E_f = -13.6\text{ eV}

.

The electron drops from the higher energy state to the lower state.

2

Calculate the photon energy using the transition formula

E_{photon} = E_i - E_f

.

E_{photon} = -3.4\text{ eV} - (-13.6\text{ eV}) = 10.2\text{ eV}

.

By energy conservation, the energy of the emitted photon must equal the energy loss of the electron.

Key Concept

Photon Emission in Atomic Energy Level Transitions

Estimated Time:45s

Question 7668Question →

A point $P(x, y)$ moves in the Cartesian plane such that its distance from the fixed point $A(2, 3)$ is equal to its distance from the fixed point $B(4, 1)$ . What is the equation of the locus of $P$ ?

Show answer & explanation

Answer: x - y - 1 = 0; x - y = 1; y = x - 1; x-y-1=0; x-y=1; y=x-1

Answer

The equation of the locus of

P

is

x - y - 1 = 0

(or

y = x - 1

).

The set of all points equidistant from two fixed points

A(2, 3)

and

B(4, 1)

forms the perpendicular bisector of segment

AB

. Equating the squared distances

(x-2)^2 + (y-3)^2 = (x-4)^2 + (y-1)^2

and simplifying yields the linear equation

x - y - 1 = 0

(or

y = x - 1

).

Step-by-Step Solution

1

Set up the distance equality condition using the distance formula.

\\sqrt{(x - 2)^2 + (y - 3)^2} = \\sqrt{(x - 4)^2 + (y - 1)^2}

The locus of a point

P(x,y)

equidistant from two points

A

and

B

satisfies

PA = PB

.

2

Square both sides to eliminate the square roots and expand the terms.

(x - 2)^2 + (y - 3)^2 = (x - 4)^2 + (y - 1)^2 \\implies x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 - 8x + 16 + y^2 - 2y + 1

Squaring removes the radical sign, allowing algebraic simplification.

3

Subtract

x^2 + y^2

from both sides and collect linear terms.

-4x - 6y + 13 = -8x - 2y + 17

The quadratic terms cancel out since the locus equidistant from two points is a linear equation (perpendicular bisector).

4

Rearrange all terms to one side and simplify.

(-4x + 8x) + (-6y + 2y) + (13 - 17) = 0 \\implies 4x - 4y - 4 = 0 \\implies x - y - 1 = 0

Dividing the entire linear equation by

4

gives the equation in simplest form.

Key Concept

Locus equidistant from two fixed points (Perpendicular Bisector of a line segment)

Alternative Method

Find the midpoint

M

of

AB

,

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

. Calculate the gradient of

AB

,

m_1 = \\frac{1 - 3}{4 - 2} = \\frac{-2}{2} = -1

. The perpendicular gradient is

m_2 = -\\frac{1}{-1} = 1

. Use point-slope form:

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

or

x - y - 1 = 0

.

Estimated Time:1m 30s

Question 7669Question →

Two identical isolated metal spheres, $X$ and $Y$ , carry initial charges of $+q$ and $-3q$ respectively and are separated by a fixed distance $r$ in a vacuum. The magnitude of the electrostatic force between them is $F$ . A third identical, uncharged metal sphere $Z$ is touched briefly to sphere $X$ , then touched briefly to sphere $Y$ , and finally placed at the midpoint between spheres $X$ and $Y$ . What is the magnitude of the net electrostatic force acting on sphere $Z$ in terms of $F$ ?

\dfrac{35}{12}F

\dfrac{5}{4}F

\dfrac{35}{48}F

\dfrac{35}{4}F

Show answer & explanation

Answer: $\dfrac{35}{12}F$

Answer

The magnitude of the net electrostatic force acting on sphere

Z

is

\dfrac{35}{12}F

.

When sphere Z touches sphere X, charge is shared equally so both carry

+\frac{q}{2}

. Next, when sphere Z touches sphere Y (charge

-3q

), total charge becomes

-\frac{5q}{2}

, dividing equally into

-\frac{5q}{4}

for each. At the midpoint (

r/2

from each sphere), sphere X (

+\frac{q}{2}

) attracts sphere Z (

-\frac{5q}{4}

) towards the left with force

\frac{5}{2}\frac{kq^2}{r^2}

. Sphere Y (

-\frac{5q}{4}

) repels sphere Z (

-\frac{5q}{4}

) towards the left with force

\frac{25}{4}\frac{kq^2}{r^2}

. Adding these co-directional forces yields

\frac{35}{4}\frac{kq^2}{r^2}

. Given the initial force

F = \frac{3kq^2}{r^2}

, we substitute

\frac{kq^2}{r^2} = \frac{F}{3}

to obtain

\frac{35}{12}F

.

Step-by-Step Solution

1

Determine the initial electrostatic force

F

between spheres

X

and

Y

.

F = k \frac{|(+q)(-3q)|}{r^2} = \frac{3kq^2}{r^2}

, which gives

\frac{kq^2}{r^2} = \frac{F}{3}

.

Coulomb's law defines force as proportional to the product of charges divided by the square of separation distance.

2

Calculate the charges on the spheres after sequential contacts.

When

Z

(

0

) touches

X

(

+q

), charge divides equally:

q_X' = +\frac{q}{2}

and

q_Z' = +\frac{q}{2}

. When

Z

(

+\frac{q}{2}

) touches

Y

(

-3q

), the combined charge is

+\frac{q}{2} - 3q = -\frac{5q}{2}

, which divides equally to give

q_Y' = -\frac{5q}{4}

and

q_Z'' = -\frac{5q}{4}

.

Identical conductors share total charge equally upon contact due to conservation of charge and symmetric potential.

3

Determine the forces exerted on sphere

Z

at the midpoint.

Separation distance from

Z

to both

X

and

Y

is

\frac{r}{2}

. Force from

X

on

Z

(attractive, pulling towards

X

):

F_{ZX} = k \frac{|(+\frac{q}{2})(-\frac{5q}{4})|}{(\frac{r}{2})^2} = k \frac{\frac{5q^2}{8}}{\frac{r^2}{4}} = \frac{5}{2}\frac{kq^2}{r^2}

. Force from

Y

on

Z

(repulsive, pushing away from

Y

toward

X

):

F_{ZY} = k \frac{|(-\frac{5q}{4})(-\frac{5q}{4})|}{(\frac{r}{2})^2} = k \frac{\frac{25q^2}{16}}{\frac{r^2}{4}} = \frac{25}{4}\frac{kq^2}{r^2}

.

Opposite charges attract and like charges repel. Midpoint separation distance is

r/2

.

4

Calculate the net force on

Z

and express it in terms of

F

.

Since both forces act in the same direction (towards sphere

X

),

F_{\text{net}} = F_{ZX} + F_{ZY} = (\frac{5}{2} + \frac{25}{4})\frac{kq^2}{r^2} = \frac{35}{4}\frac{kq^2}{r^2}

. Substituting

\frac{kq^2}{r^2} = \frac{F}{3}

yields

F_{\text{net}} = \frac{35}{4} \times \frac{F}{3} = \frac{35}{12}F

.

Forces in the same direction add vectorially.

Key Concept

Electrostatic Charge Sharing and Coulomb's Law Vector Superposition

Question 7670Question →

Historical developments in atomic physics led to several distinct models of atomic structure. Match each atomic model on the left with its defining structural feature or experimental basis on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Thomson's Model matches with embedded electrons in a positive sphere; Rutherford's Model matches with a dense positive nucleus from alpha scattering; Bohr's Model matches with quantized non-radiating orbits; Quantum Mechanical Model matches with electron probability orbitals.

Each historical atomic model is correctly matched to its fundamental feature: Thomson proposed electrons suspended in positive mass; Rutherford used alpha scattering to discover the compact nucleus; Bohr quantized electron orbits to explain spectral lines; and the Quantum Mechanical Model represents electrons via three-dimensional probability orbitals.

Step-by-Step Solution

1

Identify Thomson's contribution

Thomson proposed the 'plum pudding' model where negative electrons are embedded inside a uniform positive charge sphere.

This preceded the discovery of the atomic nucleus.

2

Identify Rutherford's contribution

Rutherford discovered the central positive nucleus through the alpha particle deflection experiment.

Large deflections meant most atomic mass and positive charge concentrated at a tiny core.

3

Identify Bohr's contribution

Bohr added quantum conditions to planetary orbits so electrons remain stable without continuously radiating energy.

Quantized angular momentum explains discrete emission line spectra.

4

Identify the Quantum Mechanical Model contribution

Modern quantum mechanics replaces fixed circular orbits with wave functions and 3D probability clouds (orbitals).

Heisenberg's uncertainty principle rules out precise circular orbits.

Key Concept

Evolution of Atomic Models

Question 7671Question →

A mechanical longitudinal wave of frequency $250\text{ Hz}$ propagates from Medium 1 into Medium 2. In Medium 1, the distance between two consecutive compressions is $1.40\text{ m}$ . Upon entering Medium 2, the wave speed increases by $40\%$ . What is the distance, in meters, between a compression and the immediately adjacent rarefaction in Medium 2?

Show answer & explanation

Answer: 0.98

Answer

The distance between a compression and the adjacent rarefaction in Medium 2 is

0.98\text{ m}

.

When a mechanical wave travels between media, its frequency remains unchanged. The wave speed equation

v = f\lambda

indicates that wavelength is directly proportional to wave speed. A

40\%

increase in speed increases the wavelength in Medium 2 from

1.40\text{ m}

to

1.40 \times 1.40\text{ m} = 1.96\text{ m}

. In a longitudinal wave, the distance between a compression and an adjacent rarefaction is half a wavelength, yielding

\frac{1.96\text{ m}}{2} = 0.98\text{ m}

.

Step-by-Step Solution

1

Identify the wavelength in Medium 1 from the compression spacing.

\lambda_1 = 1.40\text{ m}

The distance between two successive compressions in a longitudinal wave corresponds to one complete wavelength.

2

Calculate the wavelength in Medium 2 using the constant frequency principle across media boundaries.

\lambda_2 = 1.40 \times 1.40\text{ m} = 1.96\text{ m}

The frequency of a wave is determined by the source and does not change upon entering a new medium. Since

v = f\lambda

, a

40\%

increase in wave speed results in a proportional

40\%

increase in wavelength.

3

Find the distance between a compression and the adjacent rarefaction in Medium 2.

d = \frac{\lambda_2}{2} = \frac{1.96\text{ m}}{2} = 0.98\text{ m}

In any longitudinal wave, a compression and its adjacent rarefaction are out of phase by half a cycle, corresponding to half a wavelength.

Key Concept

Wave propagation across boundaries and spatial separation of compressions and rarefactions in longitudinal waves.

Question 7672Question →

A binary operation $\odot$ defined on the set of real numbers $\mathbb{R}$ is given by $a \odot b = a + b + kab$ , where $k$ is a non-zero constant. If the inverse of $2$ under $\odot$ is $-4$ , what is the value of $(3 \odot 1)^{-1}$ ?

-\frac{52}{3}

\frac{13}{4}

-\frac{28}{5}

-\frac{13}{3}

Show answer & explanation

Answer: $-\frac{52}{3}$

Answer

The value of

(3 \odot 1)^{-1}

is

-\frac{52}{3}

.

First, the identity element is determined by solving

a \odot e = a

, which gives

a + e + kae = a \implies e = 0

. Next, using the inverse property

x \odot x^{-1} = 0

, we get

x^{-1} = \frac{-x}{1 + kx}

. Given

2^{-1} = -4

, substituting gives

\frac{-2}{1 + 2k} = -4

, leading to

k = -\frac{1}{4}

. Evaluating

3 \odot 1

yields

3 + 1 - \frac{3}{4} = \frac{13}{4}

. Finally, applying the inverse formula to

\frac{13}{4}

gives

\frac{-\frac{13}{4}}{1 - \frac{13}{16}} = -\frac{52}{3}

.

Step-by-Step Solution

1

Find the identity element

e

under the operation

\odot

e = 0

By definition of identity element,

a \odot e = a \implies a + e + kae = a \implies e(1 + ka) = 0 \implies e = 0

for all real numbers

a

.

2

Derive the formula for the inverse

x^{-1}

of an element

x

x^{-1} = \frac{-x}{1 + kx}

By definition of inverse,

x \odot x^{-1} = e \implies x + x^{-1} + kxx^{-1} = 0 \implies x^{-1}(1 + kx) = -x \implies x^{-1} = \frac{-x}{1 + kx}

.

3

Use the given inverse condition

2^{-1} = -4

to find the constant

k

k = -\frac{1}{4}

Substituting

x = 2

into the inverse formula gives

\frac{-2}{1 + 2k} = -4 \implies -2 = -4(1 + 2k) \implies -2 = -4 - 8k \implies 8k = -2 \implies k = -\frac{1}{4}

.

4

Evaluate the operation

3 \odot 1

3 \odot 1 = \frac{13}{4}

Using the operation definition with

k = -\frac{1}{4}

:

3 \odot 1 = 3 + 1 + \left(-\frac{1}{4}\right)(3)(1) = 4 - \frac{3}{4} = \frac{13}{4}

.

5

Calculate the inverse of

\frac{13}{4}

under

\odot

-\frac{52}{3}

Using the inverse formula

y^{-1} = \frac{-y}{1 + ky}

for

y = \frac{13}{4}

:

y^{-1} = \frac{-\frac{13}{4}}{1 + \left(-\frac{1}{4}\right)\left(\frac{13}{4}\right)} = \frac{-\frac{13}{4}}{1 - \frac{13}{16}} = \frac{-\frac{13}{4}}{\frac{3}{16}} = -\frac{13}{4} \times \frac{16}{3} = -\frac{52}{3}

.

Key Concept

Binary Operations: Finding Identity Elements, Unknown Parameters, and Inverse Elements

Estimated Time:2m 0s

Question 7673Question →

What is the $y$ -intercept of the normal line to the curve $y = x^3 - 3x^2 + 4x - 1$ at the point where $x = 1$ ?

2

0

-1

1

Show answer & explanation

Answer: $2$

Answer

The

y

-intercept of the normal line is

2

.

Evaluating

y = x^3 - 3x^2 + 4x - 1

at

x = 1

yields

y = 1

. Differentiating gives

\frac{dy}{dx} = 3x^2 - 6x + 4

, which evaluates to

1

at

x = 1

. Since the normal is perpendicular to the tangent, its gradient is

-1

. Substituting into

y - 1 = -1(x - 1)

yields

y = -x + 2

, giving a

y

-intercept of

2

.

Step-by-Step Solution

1

Find the

y

-coordinate of the point of tangency.

At

x = 1

,

y = (1)^3 - 3(1)^2 + 4(1) - 1 = 1 - 3 + 4 - 1 = 1

. Point of contact is

(1, 1)

.

The point must lie on the curve.

2

Differentiate

y

with respect to

x

to find the gradient function.

\frac{dy}{dx} = 3x^2 - 6x + 4.

The first derivative represents the gradient of the tangent to the curve.

3

Calculate the gradient of the tangent and normal at

x = 1

.

Tangent gradient

m_t = 3(1)^2 - 6(1) + 4 = 1

. Normal gradient

m_n = -\frac{1}{m_t} = -1

.

The normal line is perpendicular to the tangent line, so

m_n \cdot m_t = -1

.

4

Determine the equation of the normal line and find its

y

-intercept.

Using

y - y_1 = m_n(x - x_1) \implies y - 1 = -1(x - 1) \implies y = -x + 2

. Setting

x = 0

gives

y = 2

.

The

y

-intercept occurs where the line crosses the

y

-axis (

x = 0

).

Key Concept

Tangents and Normals to Curves

Question 7674Question →

Two capacitors of capacitances $C_1 = 6\text{ }\mu\text{F}$ and $C_2 = 12\text{ }\mu\text{F}$ are connected in series across a $180\text{ V}$ direct current power supply. After the capacitors are fully charged, the supply is disconnected. A dielectric material of dielectric constant $K = 4$ is then inserted to completely fill the space between the plates of $C_1$ . What is the new potential difference across $C_1$ ?

30\text{ V}

60\text{ V}

135\text{ V}

15\text{ V}

Show answer & explanation

Answer: $30\text{ V}$

Answer

30\text{ V}

The initial equivalent capacitance of the series combination is

4\text{ }\mu\text{F}

, which charges each capacitor to

720\text{ }\mu\text{C}

. Because the circuit is disconnected from the battery, charge is conserved. Inserting a dielectric of constant

K = 4

increases

C_1

to

24\text{ }\mu\text{F}

, resulting in a potential difference of

V = \frac{Q}{C_1'} = \frac{720\text{ }\mu\text{C}}{24\text{ }\mu\text{F}} = 30\text{ V}

.

Step-by-Step Solution

1

Calculate the initial equivalent capacitance of the series network

C_{eq} = 4\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_{eq}} = \frac{1}{6} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}\text{ }\mu\text{F}^{-1}

.

2

Determine the charge stored on each capacitor prior to disconnection

Q = 720\text{ }\mu\text{C}

Total charge supplied by the

180\text{ V}

battery is

Q = C_{eq} V = 4\text{ }\mu\text{F} \times 180\text{ V} = 720\text{ }\mu\text{C}

. In series, each capacitor holds this same charge.

3

Calculate the modified capacitance of

C_1

after dielectric insertion

C_1' = 24\text{ }\mu\text{F}

A dielectric of constant

K = 4

scales capacitance by

K

:

C_1' = K \times C_1 = 4 \times 6\text{ }\mu\text{F} = 24\text{ }\mu\text{F}

.

4

Calculate the final potential difference across

C_1

using charge conservation

V_1' = 30\text{ V}

Because the source is disconnected, charge

Q = 720\text{ }\mu\text{C}

on

C_1

remains constant. Therefore,

V_1' = \frac{Q}{C_1'} = \frac{720\text{ }\mu\text{C}}{24\text{ }\mu\text{F}} = 30\text{ V}

.

Key Concept

Effect of dielectrics and charge conservation in disconnected series capacitor circuits

Question 7675Question →

The derived SI unit of dynamic viscosity, the pascal-second ( $\text{Pa}\cdot\text{s}$ ), can be expressed in fundamental SI base units as $\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c$ . Determine the numerical value of the exponent of length, $b$ .

Show answer & explanation

Answer: -1

Answer

The numerical value of the exponent of length

b

is

-1

.

Dynamic viscosity is measured in pascal-seconds (

\text{Pa}\cdot\text{s}

). Substituting

1\text{ Pa} = 1\text{ N/m}^2 = 1\text{ kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

into the formula gives

1\text{ Pa}\cdot\text{s} = 1\text{ kg}^1 \cdot \text{m}^{-1} \cdot \text{s}^{-1}

. Comparing this with

\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c

, the exponent of length (meters) is

b = -1

.

Step-by-Step Solution

1

Express the newton in base SI units using

F = ma

.

\text{N} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}

.

Force is the product of mass and acceleration.

2

Derive the SI base unit expression for pressure (pascal,

\text{Pa}

).

\text{Pa} = \frac{\text{N}}{\text{m}^2} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

.

Pressure is defined as force per unit area.

3

Multiply the base unit expression of pressure by seconds.

\text{Pa}\cdot\text{s} = (\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}) \cdot \text{s}^1 = \text{kg}^1 \cdot \text{m}^{-1} \cdot \text{s}^{-1}

.

Dynamic viscosity is measured in pascal-seconds.

4

Extract the exponent of length (

b

) corresponding to the meter unit.

b = -1

.

The power of meters (

\text{m}

) in

\text{kg}^1 \cdot \text{m}^{-1} \cdot \text{s}^{-1}

is

-1

.

Key Concept

Deriving base SI units for derived physical quantities

Estimated Time:1m 30s

Question 7676Question →

A cyclist travels a total distance of $90\text{ km}$ . She completes the first $40\text{ km}$ of the journey at a constant speed of $20\text{ km/h}$ . If her average speed for the entire journey is $30\text{ km/h}$ , find her speed, in $\text{km/h}$ , over the remaining distance.

Show answer & explanation

Answer: 50

Answer

The speed over the remaining distance is

50\text{ km/h}

.

Average speed is total distance divided by total time. For a

90\text{ km}

trip with an average speed of

30\text{ km/h}

, the total trip time is

3\text{ hours}

. Covering the first

40\text{ km}

at

20\text{ km/h}

requires

2\text{ hours}

, leaving

1\text{ hour}

to cover the remaining

50\text{ km}

(

90 - 40

). Dividing

50\text{ km}

by

1\text{ hour}

results in a required speed of

50\text{ km/h}

.

Step-by-Step Solution

1

Calculate the total time for the complete trip

Total duration =

3\text{ hours}

Average speed is defined as total distance divided by total time:

T = \frac{D}{V_{\text{avg}}} = \frac{90}{30} = 3\text{ hours}

.

2

Calculate the time spent covering the first part of the journey

Time for first segment =

2\text{ hours}

Time taken for a segment is distance divided by speed:

t_1 = \frac{40}{20} = 2\text{ hours}

.

3

Find the remaining time and remaining distance

Remaining time =

1\text{ hour}

; Remaining distance =

50\text{ km}

Subtracting the first segment's time and distance from the totals gives

3 - 2 = 1\text{ hour}

and

90 - 40 = 50\text{ km}

.

4

Calculate the required speed for the second segment

Speed for remaining distance =

50\text{ km/h}

Speed is calculated by dividing remaining distance by remaining time:

\frac{50\text{ km}}{1\text{ hour}} = 50\text{ km/h}

.

Key Concept

Average Rate and Speed Calculations

Question 7677Question →

A coastal monitoring station at point $O$ tracks two vessels on horizontal water. Vessel $A$ is located $15\text{ km}$ from $O$ on a bearing of $070^\circ$ , while Vessel $B$ is located $20\text{ km}$ from $O$ on a bearing of $160^\circ$ . What is the direct distance between Vessel $A$ and Vessel $B$ in kilometers?

Show answer & explanation

Answer: 25

Answer

The direct distance between Vessel A and Vessel B is 25 km.

The difference between the two bearings (

160^\circ - 070^\circ = 90^\circ

) establishes that triangle

AOB

is a right-angled triangle at station

O

. Applying Pythagoras' theorem yields

AB = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ km}

.

Step-by-Step Solution

1

Find the angle between the lines of sight to the two vessels

\angle AOB = 160^\circ - 070^\circ = 90^\circ

Subtracting the smaller bearing angle from the larger bearing angle from the same origin point gives the included angle.

2

Set up the equation for distance AB using Pythagoras' theorem

AB^2 = 15^2 + 20^2 = 225 + 400 = 625

Since the included angle is 90 degrees, the three points form a right-angled triangle where AB is the hypotenuse.

3

Calculate the principal square root of 625

AB = \sqrt{625} = 25\text{ km}

Taking the square root converts the squared distance into the direct linear distance between the vessels.

Key Concept

Calculating the distance between two points using bearings and right-angled triangle properties (Pythagoras' theorem).

Estimated Time:1m 30s

Question 7678Question →

A binary operation $\ast$ defined on the set of real numbers $\mathbb{R}$ is given by $a \ast b = a + b + 7$ . What is the identity element under this operation?

Show answer & explanation

Answer: -7

Answer

The identity element under the operation is

-7

.

The identity element

e

satisfies

a \ast e = a

for any real number

a

. Substituting into the definition gives

a + e + 7 = a

, which simplifies to

e = -7

.

Step-by-Step Solution

1

Set up the identity element equation using the definition

a \ast e = a

.

a + e + 7 = a

By definition, operating any element

a

with the identity element

e

yields

a

.

2

Subtract

a

from both sides of the equation.

e + 7 = 0

Isolating terms involving

e

.

3

Subtract

7

from both sides to solve for

e

.

e = -7

Determining the numerical value of the identity element.

Key Concept

Identity Element in Binary Operations

Question 7679Question →

Calculate the area of the finite region bounded by the parabola $y = 3x^2 - 12x + 9$ and the $x$ -axis.

Show answer & explanation

Answer: 4

Answer

The area of the bounded region is 4 square units.

Finding the x-intercepts of

y = 3x^2 - 12x + 9

gives

x = 1

and

x = 3

. Integrating

-y

from

1

to

3

gives

\left[-x^3 + 6x^2 - 9x\right]_{1}^{3} = 0 - (-4) = 4

square units.

Step-by-Step Solution

1

Determine the limits of integration by finding the x-intercepts of the curve.

x = 1

and

x = 3

The bounded region lies between the points where the curve intersects the x-axis (

y = 0

).

2

Set up the definite integral with the correct integrand sign.

A = \int_{1}^{3} (9 - 12x + 3x^2) \, dx = \int_{1}^{3} (-3x^2 + 12x - 9) \, dx

Since

y \le 0

on

[1, 3]

, negating the function ensures the calculated area is positive.

3

Integrate term-by-term and evaluate between upper limit 3 and lower limit 1.

\left[-x^3 + 6x^2 - 9x\right]_{1}^{3} = (0) - (-4) = 4

Applying the Fundamental Theorem of Calculus yields the exact value of 4.

Key Concept

Area bounded by a curve and the x-axis lying below the x-axis

Question 7680Question →

A line segment joins the points $A(1, 4)$ and $B(7, 10)$ . Point $P$ divides the line segment $AB$ internally in the ratio $1:2$ . A second line $L_2$ passes through $P$ and is perpendicular to $AB$ . If line $L_2$ intersects the y-axis at the point $(0, k)$ , find the value of $k$ .

Show answer & explanation

Answer: 9

Answer

The value of k is 9.

Using the section formula for internal division in a 1:2 ratio, the coordinates of point P are found to be (3, 6). The gradient of the segment AB is 1, which means the perpendicular line L_2 has a gradient of -1. Writing the equation of line L_2 passing through (3, 6) yields y = -x + 9. Evaluating at x = 0 gives the y-intercept k = 9.

Step-by-Step Solution

1

Calculate the coordinates of point P dividing segment AB internally in the ratio 1:2.

P = (3, 6)

Using the section formula x = (m x_2 + n x_1) / (m + n) and y = (m y_2 + n y_1) / (m + n) with ratio m:n = 1:2.

2

Calculate the gradient m_1 of the line segment AB.

m_1 = 1

Applying the gradient formula m = (y_2 - y_1) / (x_2 - x_1) gives (10 - 4) / (7 - 1) = 1.

3

Determine the gradient m_2 of the perpendicular line L_2.

m_2 = -1

Perpendicular lines satisfy m_1 * m_2 = -1, hence m_2 = -1 / 1 = -1.

4

Find the equation of line L_2 passing through P(3, 6) with gradient -1.

y = -x + 9

Using point-slope form y - y_1 = m(x - x_1) yields y - 6 = -1(x - 3).

5

Determine the y-intercept value k by setting x = 0.

k = 9

Substituting x = 0 into y = -x + 9 gives y = 9.

Key Concept

Section formula, perpendicular line gradients, and y-intercept determination

Estimated Time:3m 0s

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