All practice questions

Which prepositions correctly complete the sentence below?

'The state governor declared that the proposed educational reforms were completely at variance _____ the aspirations of the local community and urged all stakeholders to desist _____ spreading misleading rumors.'

with / from

to / from

from / in

with / against

Show answer & explanation

Answer: with / from

Answer

The sentence is correctly completed by the pair 'with / from'.

The fixed prepositional phrase 'at variance with' expresses conflict or disagreement. Additionally, the verb 'desist' requires the preposition 'from' when specifying the action being stopped. Therefore, the option containing 'with / from' completes the sentence correctly.

Step-by-Step Solution

1

Determine the required preposition following the idiom 'at variance ____'.

The standard English idiom is 'at variance with', meaning in disagreement or opposition.

Fixed prepositional phrases require exact collocations.

2

Determine the dependent preposition governed by the verb 'desist ____'.

The verb 'desist' takes the complement preposition 'from'.

Certain verbs in formal English strictly govern specific prepositions.

Key Concept

Prepositional Idioms and Dependent Prepositions

Question 3842Question →

A cell of electromotive force (e.m.f.) $1.5\text{ V}$ and internal resistance $0.5\text{ }\Omega$ is connected in series with a $2.5\text{ }\Omega$ resistor. What is the terminal potential difference across the cell?

1.25\text{ V}

1.50\text{ V}

0.25\text{ V}

1.75\text{ V}

Show answer & explanation

Answer: $1.25\text{ V}$

Answer

The terminal potential difference across the cell is

1.25\text{ V}

.

When current flows through a closed circuit, voltage drops across both the external load resistor and the cell's internal resistance. Subtracting the lost voltage (

0.25\text{ V}

) from the electromotive force (

1.50\text{ V}

) yields a terminal potential difference of

1.25\text{ V}

.

Step-by-Step Solution

1

Calculate the total resistance of the circuit

R_{\text{total}} = R + r = 2.5\text{ }\Omega + 0.5\text{ }\Omega = 3.0\text{ }\Omega

The external load resistor and the internal resistance of the cell are connected in series.

2

Calculate the total electric current flowing through the circuit

I = \frac{E}{R + r} = \frac{1.5\text{ V}}{3.0\text{ }\Omega} = 0.5\text{ A}

Ohm's law for a complete circuit states that current equals total e.m.f. divided by total circuit resistance.

3

Determine the terminal potential difference across the cell

V = I \times R = 0.5\text{ A} \times 2.5\text{ }\Omega = 1.25\text{ V}

Terminal potential difference is the potential drop across the external resistance.

Key Concept

Terminal Potential Difference and Cell Internal Resistance

Question 3843Question →

An equilateral triangle has an area of $16\sqrt{3}\text{ cm}^2$ . If a square has the same perimeter as this equilateral triangle, what is the area of the square?

36\text{ cm}^2

48\text{ cm}^2

64\text{ cm}^2

144\text{ cm}^2

Show answer & explanation

Answer: $36\text{ cm}^2$

Answer

The area of the square is

36\text{ cm}^2

.

Using the area formula for an equilateral triangle

\text{Area} = \frac{\sqrt{3}}{4}s^2 = 16\sqrt{3}

, we determine the side length

s = 8\text{ cm}

. The perimeter of the triangle is

3 \times 8\text{ cm} = 24\text{ cm}

. Since the square has an equal perimeter of

24\text{ cm}

, each side of the square measures

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

. The area of the square is therefore

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

.

Step-by-Step Solution

1

Find the side length of the equilateral triangle.

Side length

s = 8\text{ cm}

.

The area of an equilateral triangle is given by

\text{Area} = \frac{\sqrt{3}}{4}s^2

. Setting

\frac{\sqrt{3}}{4}s^2 = 16\sqrt{3}

yields

s^2 = 64

, so

s = 8\text{ cm}

.

2

Calculate the perimeter of the equilateral triangle.

Perimeter

P = 24\text{ cm}

.

An equilateral triangle has 3 equal sides, so

P = 3 \times 8 = 24\text{ cm}

.

3

Determine the side length of the square.

Square side length

a = 6\text{ cm}

.

The square and triangle have equal perimeters (

24\text{ cm}

). Since a square has 4 equal sides,

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

.

4

Calculate the area of the square.

Area

= 36\text{ cm}^2

.

The area of a square is

a^2 = 6^2 = 36\text{ cm}^2

.

Key Concept

Perimeter and Area of Equilateral Triangles and Squares

Estimated Time:1m 30s

Question 3844Question →

If $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} - \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} = x\sqrt{15}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 2

Answer

The value of

x

is 2.

Rationalising both fractions yields

4 + \sqrt{15}

and

4 - \sqrt{15}

. Subtracting the second from the first gives

(4 + \sqrt{15}) - (4 - \sqrt{15}) = 2\sqrt{15}

. Comparing

2\sqrt{15}

with

x\sqrt{15}

gives

x = 2

.

Step-by-Step Solution

1

Rationalise the denominator of the first fraction

\frac{(\sqrt{5}+\sqrt{3})^2}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})} = \frac{5 + 2\sqrt{15} + 3}{5 - 3} = 4 + \sqrt{15}

Multiplying numerator and denominator by the conjugate of the denominator removes the surd from the denominator.

2

Rationalise the denominator of the second fraction

\frac{(\sqrt{5}-\sqrt{3})^2}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})} = \frac{5 - 2\sqrt{15} + 3}{5 - 3} = 4 - \sqrt{15}

Multiply by the conjugate

(\sqrt{5}-\sqrt{3})

to simplify the second surd term.

3

Subtract the simplified expressions

(4 + \sqrt{15}) - (4 - \sqrt{15}) = 4 - 4 + \sqrt{15} + \sqrt{15} = 2\sqrt{15}

Distribute the negative sign and combine like surd terms.

4

Solve for the unknown coefficient x

2\sqrt{15} = x\sqrt{15} \implies x = 2

Divide both sides of the equation by

\sqrt{15}

to isolate

x

.

Key Concept

Binomial Surd Rationalisation and Simplification

Estimated Time:1m 30s

Question 3845Question →

Complete the sentence below by supplying the correct prepositions to fill the blanks.

Fill in the blanks below

The investigative panel concluded that the managing director had acquiesced the fraudulent financial transactions and was entirely devoid remorse for his actions.

Show answer & explanation

Answer

The correct prepositions are 'in' for the first blank and 'of' for the second blank.

The verb 'acquiesce' requires the dependent preposition 'in' to express tacit agreement or compliance, while the adjective 'devoid' consistently pairs with 'of' to mean completely lacking.

Step-by-Step Solution

1

Determine the correct dependent preposition that follows the verb 'acquiesce'.

The verb 'acquiesce' strictly collocates with the preposition 'in' when expressing passive compliance or agreement.

Using 'to' or 'with' after 'acquiesce' is a common prepositional error resulting from native language interference or erroneous analogy with 'agree to'.

2

Identify the standard preposition paired with the adjective 'devoid'.

The adjective 'devoid' takes the fixed preposition 'of' to denote being completely destitute or lacking in a quality.

Phrases such as 'devoid with' or 'devoid from' violate standard grammatical collocations in formal English.

Key Concept

Dependent Prepositions and Fixed Prepositional Collocations

Question 3846Question →

Find the gradient of the normal to the curve $y = \frac{x + 1}{x - 1}$ at the point where $x = 3$ .

Show answer & explanation

Answer: 2

Answer

The gradient of the normal to the curve at

x = 3

is

2

.

Differentiating

y = \frac{x + 1}{x - 1}

via the quotient rule gives

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

. At

x = 3

, the tangent gradient is

m_t = \frac{-2}{4} = -\frac{1}{2}

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

m_n = -\frac{1}{m_t} = 2

.

Step-by-Step Solution

1

Differentiate the rational function with respect to

x

Applying the quotient rule

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}

where

u = x + 1

and

v = x - 1

yields

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

.

The first derivative determines the slope function of the tangent line to the curve.

2

Substitute

x = 3

into the derivative to find the tangent slope

m_t

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

.

Evaluating the derivative at the given

x

-coordinate provides the exact gradient of the tangent at that point.

3

Calculate the slope of the normal line

m_n

m_n = -\frac{1}{m_t} = -\frac{1}{-1/2} = 2

.

The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent slope.

Key Concept

The slope of the normal line to a curve at a given point is the negative reciprocal of the slope of the tangent line at that point (

m_n = -\frac{1}{m_t}

).

Question 3847Question →

A sector of a circle of radius $21\text{ cm}$ has a total perimeter of $64\text{ cm}$ . Calculate the area of the sector in $\text{cm}^2$ . (Take $\pi = \frac{22}{7}$ )

Show answer & explanation

Answer: 231

Answer

The area of the sector is

231\text{ cm}^2

.

The total perimeter of a sector is given by

P = 2r + l

. Given radius

r = 21\text{ cm}

and perimeter

P = 64\text{ cm}

, the arc length is

l = 64 - 2(21) = 22\text{ cm}

. The area of the sector is calculated using

A = \frac{1}{2} r l = \frac{1}{2} \times 21 \times 22 = 231\text{ cm}^2

.

Step-by-Step Solution

1

Find the arc length of the sector

The arc length

l = 22\text{ cm}

The total perimeter of a sector includes its arc length plus its two bounding radii (

P = 2r + l

). Subtracting

2r = 42\text{ cm}

from

64\text{ cm}

gives

l = 22\text{ cm}

.

2

Calculate the area of the sector

The area

A = 231\text{ cm}^2

Using the relation between arc length and sector area,

A = \frac{1}{2} r l = \frac{1}{2} \times 21 \times 22 = 231\text{ cm}^2

.

Key Concept

Perimeter and Area of a Sector of a Circle

Estimated Time:1m 30s

Question 3848Question →

The committee chairman noted that although several board members were initially skeptical of the restructuring policy, they eventually proved amenable _____ the proposed changes. Which preposition correctly completes the sentence above?

to

with

for

about

Show answer & explanation

Answer: to

Answer

The correct option is 'to' because the adjective 'amenable' requires the preposition 'to' when signifying open, responsive, or willing to yield.

In standard formal English, the adjective 'amenable' takes the dependent preposition 'to' when used to mean open, receptive, or agreeable to a plan or advice. Therefore, 'to' is the only grammatically correct choice.

Step-by-Step Solution

1

Identify the head adjective requiring a dependent preposition in the sentence context.

The word requiring a complement preposition is the predicate adjective 'amenable'.

Certain adjectives in English take obligatory fixed dependent prepositions to complete their meaning.

2

Recall standard prepositional collocation rules for 'amenable'.

The correct complement structure is 'amenable to something' (meaning responsive, willing, or agreeable to something).

Substituting other prepositions such as 'with', 'for', or 'about' violates standard grammatical usage in formal English.

Key Concept

Dependent Prepositions and Adjective-Preposition Collocations

Question 3849Question →

If $\sqrt{x + 4} - \sqrt{x - 1} = 1$ , what is the value of $x$ ?

5

17

3

10

Show answer & explanation

Answer: $5$

Answer

The value of

x

is

5

.

Isolating

\sqrt{x + 4}

gives

\sqrt{x + 4} = 1 + \sqrt{x - 1}

. Squaring both sides yields

x + 4 = 1 + 2\sqrt{x - 1} + x - 1

, which simplifies to

4 = 2\sqrt{x - 1}

. Dividing by 2 gives

2 = \sqrt{x - 1}

. Squaring both sides once more gives

4 = x - 1

, which leads to

x = 5

.

Step-by-Step Solution

1

Isolate one of the surd terms on the left side of the equation.

\sqrt{x + 4} = 1 + \sqrt{x - 1}

Rearranging terms prevents dealing with complex cross-products when squaring.

2

Square both sides of the equation.

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

Squaring eliminates the outer radical on the left side.

3

Simplify both sides and isolate the remaining radical term.

4 = 2\sqrt{x - 1} \implies 2 = \sqrt{x - 1}

Subtracting

x

from both sides simplifies the algebraic expression.

4

Square both sides again to solve for

x

.

4 = x - 1 \implies x = 5

Squaring eliminates the remaining radical term.

Key Concept

Solving Surd Equations by Rational Elimination

Question 3850Question →

Monochromatic light of wavelength $500\text{ nm}$ is incident normally on a plane diffraction grating. If the second-order principal maximum is observed at an angle of $30^\circ$ to the normal, calculate the number of lines per millimeter ruled on the grating.

Show answer & explanation

Answer: 500

Answer

500 lines/mm

Using the diffraction grating equation

d \sin\theta = n\lambda

with

n = 2

,

\lambda = 5.0 \times 10^{-7}\text{ m}

, and

\sin(30^\circ) = 0.5

gives a slit separation of

d = 2.0 \times 10^{-6}\text{ m}

. Converting to grating ruling density per millimeter yields

\frac{10^{-3}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 500\text{ lines/mm}

.

Step-by-Step Solution

1

Identify the given physical parameters and convert wavelength to standard SI meters.

Order

n = 2

, diffraction angle

\theta = 30^\circ

, wavelength

\lambda = 500 \times 10^{-9}\text{ m} = 5.0 \times 10^{-7}\text{ m}

.

Standard SI unit conversion is required to perform wave equation calculations accurately.

2

Apply the diffraction grating equation

d \sin\theta = n\lambda

to compute the grating spacing

d

.

d = \frac{2 \times 5.0 \times 10^{-7}\text{ m}}{\sin 30^\circ} = \frac{1.0 \times 10^{-6}}{0.5} = 2.0 \times 10^{-6}\text{ m}

.

The condition for principal constructive interference maxima is

d \sin\theta = n\lambda

.

3

Calculate the number of lines per millimeter by dividing

1\text{ mm}

(

10^{-3}\text{ m}

) by the grating spacing

d

.

N_{\text{mm}} = \frac{10^{-3}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 500\text{ lines/mm}

.

Grating density (lines per unit length) is the reciprocal of the slit separation

d

.

Key Concept

Diffraction Grating Principal Maxima Condition

Estimated Time:2m 0s

Question 3851Question →

Given the universal set $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 15\}$ , with subsets $P = \{x \in \mathcal{U} : x \text{ is a prime number}\}$ and $Q = \{x \in \mathcal{U} : x \text{ is an odd number}\}$ , what is the number of elements in the set $(P \cup Q)'$ ?

Show answer & explanation

Answer: 6; six; 6 elements

Answer

The number of elements in

(P \cup Q)'

is 6.

The set

P \cup Q

consists of all numbers from 1 to 15 that are either prime or odd:

\{1, 2, 3, 5, 7, 9, 11, 13, 15\}

. The complement

(P \cup Q)'

relative to

\mathcal{U}

contains all elements of

\mathcal{U}

that are neither prime nor odd, which are the even composite numbers:

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

. Counting these elements gives 6.

Step-by-Step Solution

1

List all elements of the universal set

\mathcal{U}

, subset

P

, and subset

Q

.

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

,

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

, and

Q = \{1, 3, 5, 7, 9, 11, 13, 15\}

.

Listing the explicit elements allows precise execution of set union and complement operations.

2

Find the union

P \cup Q

.

P \cup Q = \{1, 2, 3, 5, 7, 9, 11, 13, 15\}

.

The union combines all unique elements that belong to either set

P

, set

Q

, or both.

3

Determine the complement set

(P \cup Q)'

relative to

\mathcal{U}

and count its elements.

(P \cup Q)' = \{4, 6, 8, 10, 12, 14\}

, which contains 6 elements.

The complement set consists of all elements in

\mathcal{U}

that are not present in

P \cup Q

.

Key Concept

Complement of Set Union

Question 3852Question →

An electric heater of resistance $20\,\Omega$ carries a current of $5\,\text{A}$ when operated. What is the electrical power dissipated by the heater?

500\,\text{W}

100\,\text{W}

4\,\text{W}

2000\,\text{W}

Show answer & explanation

Answer: $500\,\text{W}$

Answer

500\,\text{W}

According to Joule's law of electric heating, electrical power

P

dissipated in a resistor of resistance

R

carrying current

I

is given by

P = I^2 R

. Substituting the given values

I = 5\,\text{A}

and

R = 20\,\Omega

gives

P = (5\,\text{A})^2 \times 20\,\Omega = 25 \times 20 = 500\,\text{W}

.

Step-by-Step Solution

1

Identify the given quantities from the problem statement

Current

I = 5\,\text{A}

and resistance

R = 20\,\Omega

These parameters are required to calculate the power dissipated.

2

Select the appropriate formula for power in terms of current and resistance

P = I^2 R

Joule's heating law specifies that power dissipation is given by

P = I^2 R

.

3

Substitute the values and evaluate the numerical result

P = (5)^2 \times 20 = 25 \times 20 = 500\,\text{W}

Squaring

5

yields

25

, and multiplying by

20

gives the final power output of

500\,\text{W}

.

Key Concept

Electrical Power Dissipation in a Resistor

Estimated Time:45s

Question 3853Question →

An electric current of $2.0\,\text{A}$ flows through a resistor of resistance $5.0\,\Omega$ for a duration of $10.0\,\text{seconds}$ . What is the total electrical energy, in Joules, dissipated by the resistor?

Show answer & explanation

Answer: 200

Answer

The total electrical energy dissipated by the resistor is

200\,\text{J}

.

According to Joule's law of heating, the electrical energy

E

converted into thermal energy when a current

I

flows through a resistor

R

for time

t

is given by

E = I^2 R t

. Substituting

I = 2.0\,\text{A}

,

R = 5.0\,\Omega

, and

t = 10.0\,\text{s}

gives

E = (2.0)^2 \times 5.0 \times 10.0 = 4.0 \times 5.0 \times 10.0 = 200\,\text{J}

.

Step-by-Step Solution

1

Identify the given physical values from the problem statement.

I = 2.0\,\text{A}

,

R = 5.0\,\Omega

,

t = 10.0\,\text{s}

.

These are the necessary parameters to compute electrical energy.

2

State the formula for electrical energy dissipated in a resistor (Joule's law).

E = I^2 R t

Electrical energy is equal to power multiplied by time, where power

P = I^2 R

.

3

Calculate the numerical value of the energy.

E = (2.0)^2 \times 5.0 \times 10.0 = 200\,\text{J}

Squaring the current gives

4.0\,\text{A}^2

, and multiplying by

5.0\,\Omega

and

10.0\,\text{s}

yields

200\,\text{J}

.

Key Concept

Joule's Law of Electrical Heating

Question 3854Question →

A $6\text{ }\mu\text{F}$ capacitor is connected in series with a parallel arrangement of a $2\text{ }\mu\text{F}$ capacitor and a $1\text{ }\mu\text{F}$ capacitor. If the entire circuit is connected across a $30\text{ V}$ d.c. power supply, what is the electric charge stored on the $2\text{ }\mu\text{F}$ capacitor?

20\text{ }\mu\text{C}

30\text{ }\mu\text{C}

40\text{ }\mu\text{C}

60\text{ }\mu\text{C}

Show answer & explanation

Answer: $40\text{ }\mu\text{C}$

Answer

The charge stored on the

2\text{ }\mu\text{F}

capacitor is

40\text{ }\mu\text{C}

.

To find the charge on the

2\text{ }\mu\text{F}

capacitor, we first combine the parallel capacitors (

2\text{ }\mu\text{F} + 1\text{ }\mu\text{F} = 3\text{ }\mu\text{F}

). Next, we combine this in series with the

6\text{ }\mu\text{F}

capacitor to get an overall capacitance

C_{eq} = 2\text{ }\mu\text{F}

. The total charge drawn from the

30\text{ V}

supply is

Q = 2\text{ }\mu\text{F} \times 30\text{ V} = 60\text{ }\mu\text{C}

. This total charge enters the parallel combination, giving a potential drop across the parallel branch of

V_p = 60\text{ }\mu\text{C} / 3\text{ }\mu\text{F} = 20\text{ V}

. Therefore, the charge on the

2\text{ }\mu\text{F}

capacitor is

2\text{ }\mu\text{F} \times 20\text{ V} = 40\text{ }\mu\text{C}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the parallel section

C_{p} = 2\text{ }\mu\text{F} + 1\text{ }\mu\text{F} = 3\text{ }\mu\text{F}

Capacitors in parallel add directly.

2

Calculate the total equivalent capacitance of the network

C_{eq} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2\text{ }\mu\text{F}

The

6\text{ }\mu\text{F}

capacitor and the

3\text{ }\mu\text{F}

parallel equivalent are in series.

3

Find the total charge supplied by the

30\text{ V}

source

Q_{total} = C_{eq} \times V = 2\text{ }\mu\text{F} \times 30\text{ V} = 60\text{ }\mu\text{C}

Total charge is the product of equivalent capacitance and total voltage.

4

Determine the potential difference across the parallel branch

V_{p} = \frac{Q_{total}}{C_{p}} = \frac{60\text{ }\mu\text{C}}{3\text{ }\mu\text{F}} = 20\text{ V}

The total charge flows through the series combination, creating a potential drop across the parallel combination equal to

Q_{total} / C_{p}

.

5

Calculate the charge on the

2\text{ }\mu\text{F}

capacitor

Q_{2\mu\text{F}} = C_{2\mu\text{F}} \times V_{p} = 2\text{ }\mu\text{F} \times 20\text{ V} = 40\text{ }\mu\text{C}

The charge on a specific capacitor in parallel is the product of its capacitance and the voltage across the parallel branch.

Key Concept

Charge and voltage distribution in mixed capacitor networks

Question 3855Question →

A sports field consists of a central rectangular section of length $100\text{ m}$ bounded on two opposite ends by semicircular regions, each having a radius of $35\text{ m}$ . What is the total area of the sports field? (Take $\pi = \frac{22}{7}$ )

7,350\text{ m}^2

8,925\text{ m}^2

10,850\text{ m}^2

14,700\text{ m}^2

Show answer & explanation

Answer: $10,850\text{ m}^2$

Answer

The total area of the sports field is

10,850\text{ m}^2

.

The sports field is composed of a central rectangle of dimensions

100\text{ m}

by

70\text{ m}

(since width is equal to the diameter

2 \times 35\text{ m} = 70\text{ m}

) and two semicircular ends of radius

35\text{ m}

. The rectangular area is

100 \times 70 = 7,000\text{ m}^2

. The two semicircles join to make one full circle with an area of

\frac{22}{7} \times 35^2 = 3,850\text{ m}^2

. Adding both parts gives

7,000 + 3,850 = 10,850\text{ m}^2

.

Step-by-Step Solution

1

Calculate the width of the rectangular region.

Width =

2 \times r = 2 \times 35\text{ m} = 70\text{ m}

.

The diameter of the semicircular ends forms the width of the central rectangle.

2

Calculate the area of the rectangular region.

Area of rectangle =

\text{length} \times \text{width} = 100\text{ m} \times 70\text{ m} = 7,000\text{ m}^2

.

Formula for the area of a rectangle is

\text{length} \times \text{width}

.

3

Calculate the combined area of the two semicircular ends.

Combined area =

\pi r^2 = \frac{22}{7} \times 35 \times 35 = 22 \times 5 \times 35 = 3,850\text{ m}^2

.

Two identical semicircles of radius

r

combine to form one full circle of radius

r

.

4

Add the rectangular area and the combined circular area.

Total Area =

7,000\text{ m}^2 + 3,850\text{ m}^2 = 10,850\text{ m}^2

.

The total area is the sum of the composite plane shapes.

Key Concept

Area of Composite Plane Figures

Question 3856Question →

A curve is defined by the equation $y = x^3 - 3x^2 + k$ , where $k$ is a constant. If the local minimum value of $y$ on the curve is $2$ , what is the value of $k$ ?

Show answer & explanation

Answer: 6

Answer

The value of the constant

k

is

6

.

To find the constant

k

, differentiate the curve equation to obtain

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

. Setting

\frac{dy}{dx} = 0

gives stationary points at

x = 0

and

x = 2

. Calculating the second derivative

\frac{d^2y}{dx^2} = 6x - 6

shows

\frac{d^2y}{dx^2} = 6 > 0

at

x = 2

, confirming that the local minimum occurs at

x = 2

. Substituting

x = 2

and the minimum value

y = 2

into

y = x^3 - 3x^2 + k

yields

2 = 8 - 12 + k

, which simplifies to

k = 6

.

Step-by-Step Solution

1

Differentiate

y = x^3 - 3x^2 + k

with respect to

x

.

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

Stationary points occur where the first derivative equals zero.

2

Solve

\frac{dy}{dx} = 0

for

x

.

x = 0

or

x = 2

These

x

-values locate the turning points on the curve.

3

Evaluate the second derivative

\frac{d^2y}{dx^2} = 6x - 6

at each stationary point.

At

x = 0

,

\frac{d^2y}{dx^2} = -6

(local maximum); at

x = 2

,

\frac{d^2y}{dx^2} = 6

(local minimum).

A positive second derivative indicates a local minimum point.

4

Substitute

x = 2

and

y = 2

into the curve equation

y = x^3 - 3x^2 + k

.

2 = (2)^3 - 3(2)^2 + k \implies 2 = -4 + k \implies k = 6

The local minimum value of

y

is attained at

x = 2

.

Key Concept

Stationary Points, Maxima, and Minima

Question 3857Question →

Which prepositions correctly complete the sentence below?

Fill in the blanks below

The university senate insisted that all postgraduate applications must conform the newly approved admission guidelines, irrespective the applicants' prior qualifications.

Show answer & explanation

Answer

The correct prepositions needed to complete the sentence are 'to' for the first blank and 'of' for the second blank.

The verb 'conform' requires the dependent preposition 'to' when indicating adherence to established standards or guidelines. Additionally, 'irrespective' forms a fixed prepositional phrase with 'of' ('irrespective of') to mean 'without taking into account'. Thus, 'to' and 'of' correctly complete the sentence.

Step-by-Step Solution

1

Determine the correct dependent preposition required after the verb 'conform' when referring to rules or standards.

The verb 'conform' collocates with the preposition 'to'.

In standard English usage, one conforms 'to' a set of rules, regulations, or guidelines.

2

Identify the standard multi-word prepositional phrase starting with 'irrespective'.

The fixed prepositional phrase is 'irrespective of'.

'Irrespective' requires the preposition 'of' to form a functioning compound preposition meaning 'regardless of'.

Key Concept

Prepositions and Prepositional Phrases

Estimated Time:1m 0s

Question 3858Question →

An athlete throws a javelin from ground level such that its initial vertical component of velocity is $40\text{ m/s}$ and its initial horizontal component of velocity is $30\text{ m/s}$ . What is the horizontal distance in metres covered by the javelin when it reaches a height of $35\text{ m}$ above the ground for the first time? (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 30

Answer

The horizontal distance covered by the javelin when it reaches a height of 35 m for the first time is 30 m.

Using

y = u_y t - \frac{1}{2}g t^2

with

u_y = 40\text{ m/s}

,

y = 35\text{ m}

, and

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

yields

35 = 40t - 5t^2

. Dividing by 5 gives

t^2 - 8t + 7 = 0

, which factors to

(t - 1)(t - 7) = 0

. The roots are

t = 1\text{ s}

(ascent) and

t = 7\text{ s}

(descent). For the first time,

t = 1\text{ s}

. The horizontal displacement is

x = u_x t = 30\text{ m/s} \times 1\text{ s} = 30\text{ m}

.

Step-by-Step Solution

1

Set up the vertical motion equation to find the time when height is 35 m

35 = 40t - 5t^2

Vertical displacement in projectile motion depends on the vertical initial velocity component and acceleration due to gravity.

2

Solve the quadratic equation for time

t

t^2 - 8t + 7 = 0 \implies t = 1\text{ s} \text{ or } t = 7\text{ s}

A projectile reaches a given non-peak height twice: once ascending and once descending.

3

Select the first time value and calculate horizontal distance

x = u_x \times t = 30 \times 1 = 30\text{ m}

Horizontal velocity remains constant throughout the flight, so distance is speed multiplied by time.

Key Concept

Independence of vertical and horizontal components in projectile motion

Question 3859Question →

A uniform metal wire of length $20\,\text{m}$ and total mass $0.034\,\text{kg}$ is manufactured from a material of density $8.5 \times 10^3\,\text{kg/m}^3$ and electrical resistivity $1.7 \times 10^{-8}\,\Omega\cdot\text{m}$ . What is the electrical resistance of the wire in ohms?

Show answer & explanation

Answer: 1.7

Answer

The electrical resistance of the wire is

1.7\,\Omega

.

Combining the density relation

V = \frac{m}{d}

with the geometric expression

V = A \cdot L

gives

A = \frac{m}{d L}

. Substituting this into Pouillet's law

R = \frac{\rho L}{A}

yields

R = \frac{\rho d L^2}{m}

. Evaluating with the given values:

R = \frac{(1.7 \times 10^{-8})(8.5 \times 10^3)(20)^2}{0.034} = 1.7\,\Omega

.

Step-by-Step Solution

1

Calculate the volume of the wire using mass and density

V = 4.0 \times 10^{-6}\,\text{m}^3

Volume is related to mass and density by

V = \frac{m}{d}

.

2

Calculate the cross-sectional area of the wire

A = 2.0 \times 10^{-7}\,\text{m}^2

For a cylindrical wire of uniform cross-section,

V = A \cdot L

, so

A = \frac{V}{L}

.

3

Apply resistivity formula to find electrical resistance

R = 1.7\,\Omega

Resistance is given by

R = \frac{\rho L}{A}

.

Key Concept

Relationship between Resistance, Mass, Density, and Resistivity

Estimated Time:2m 0s

Question 3860Question →

An aerosol canister containing an ideal gas at an initial pressure of $2.20 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ is accidentally thrown into a fire, causing the temperature of the gas to rise to $327^\circ\text{C}$ . Assuming the volume of the canister remains constant, what is the new pressure of the gas inside the canister?

1.10 \times 10^5\text{ Pa}

2.66 \times 10^6\text{ Pa}

4.40 \times 10^5\text{ Pa}

2.45 \times 10^5\text{ Pa}

Show answer & explanation

Answer: $4.40 \times 10^5\text{ Pa}$

Answer

The new pressure of the gas inside the canister is

4.40 \times 10^5\text{ Pa}

.

According to Pressure's Law, the pressure of a fixed mass of gas at constant volume is directly proportional to its absolute temperature in Kelvin (

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

). Converting the temperatures gives

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

and

T_2 = 327 + 273 = 600\text{ K}

. Multiplying the initial pressure by the ratio

\frac{600\text{ K}}{300\text{ K}} = 2

gives

4.40 \times 10^5\text{ Pa}

.

Step-by-Step Solution

1

Convert initial and final temperatures from degrees Celsius to Kelvin.

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

and

T_2 = 327^\circ\text{C} + 273 = 600\text{ K}

.

All calculations using gas laws require absolute temperature measured in Kelvin.

2

State and rearrange Pressure's Law (Gay-Lussac's Law) for constant volume.

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

.

For a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature.

3

Substitute the known values to calculate the final pressure

P_2

.

P_2 = 2.20 \times 10^5\text{ Pa} \times \frac{600\text{ K}}{300\text{ K}} = 4.40 \times 10^5\text{ Pa}

.

Since absolute temperature doubles from

300\text{ K}

to

600\text{ K}

, the final pressure must also double.

Key Concept

Pressure Law (Gay-Lussac's Law) and Absolute Temperature Conversion

Estimated Time:1m 30s

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