All practice questions

Question 61Question →

A solid brass cube with an edge length of $10\text{ cm}$ at $15^\circ\text{C}$ is heated to a temperature of $115^\circ\text{C}$ . If the linear expansivity of brass is $2.0 \times 10^{-5}\text{ K}^{-1}$ , what is the increase in the volume of the cube in $\text{cm}^3$ ?

Show answer & explanation

Answer: 6

Answer

The increase in the volume of the brass cube is

6.0\text{ cm}^3

.

The volume expansion of a solid is given by

\Delta V = V_1 \gamma \Delta T

. The initial volume of the cube is

V_1 = (10\text{ cm})^3 = 1000\text{ cm}^3

and the temperature change is

\Delta T = 115^\circ\text{C} - 15^\circ\text{C} = 100\text{ K}

. Because the expansion occurs in three dimensions, the volume expansivity is

\gamma = 3\alpha = 3 \times 2.0 \times 10^{-5} = 6.0 \times 10^{-5}\text{ K}^{-1}

. Substituting these values yields

\Delta V = 1000 \times 6.0 \times 10^{-5} \times 100 = 6.0\text{ cm}^3

.

Step-by-Step Solution

1

Calculate the initial volume of the cube

V_1 = (10\text{ cm})^3 = 1000\text{ cm}^3

The volume of a cube is calculated using

V = L^3

, where

L

is the edge length.

2

Determine the change in temperature

\Delta T = 115^\circ\text{C} - 15^\circ\text{C} = 100\text{ K}

The change in temperature is the difference between the final and initial temperatures.

3

Calculate the volume expansivity (cubic expansivity)

\gamma = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1}

Volume expansivity

\gamma

is three times the linear expansivity

\alpha

for isotropic solids.

4

Compute the increase in volume

\Delta V = V_1 \gamma \Delta T = 1000 \times (6.0 \times 10^{-5}) \times 100 = 6.0\text{ cm}^3

The formula for volume expansion is

\Delta V = V_1 \gamma \Delta T

.

Key Concept

Thermal expansion of solids: volume expansion (

\Delta V = V_1 \gamma \Delta T

) and the relationship between linear and volume expansivity (

\gamma = 3\alpha

).

Question 62Question →

A body is projected horizontally from the top of a cliff $45\text{ m}$ high. If it lands on flat ground at a horizontal distance of $120\text{ m}$ from the base of the cliff, what is the speed of the body just before it strikes the ground? (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 50

Answer

The speed of the body just before striking the ground is 50 m/s.

The time of fall is determined by the height of

45\text{ m}

, yielding

t = \sqrt{2h/g} = 3\text{ s}

. The horizontal speed is constant at

120 / 3 = 40\text{ m/s}

. The vertical velocity gained on impact is

v_y = gt = 30\text{ m/s}

. Combining these mutually perpendicular velocity components gives a final impact speed of

v = \sqrt{40^2 + 30^2} = 50\text{ m/s}

.

Step-by-Step Solution

1

Calculate the time of flight from the vertical height

t = 3 s

Vertical acceleration is constant under gravity while initial vertical velocity is zero.

2

Compute the constant horizontal component of velocity

v_x = 40 m/s

Horizontal speed is uniform because zero horizontal force acts on the projectile.

3

Compute the final vertical component of velocity at impact

v_y = 30 m/s

Vertical speed increases linearly with time due to gravitational acceleration.

4

Determine the magnitude of the resultant velocity vector

v = 50 m/s

The horizontal and vertical components are perpendicular, so their vector sum uses the Pythagorean theorem.

Key Concept

Horizontal Projection and Impact Velocity Vector

Estimated Time:1m 30s

Question 63Question →

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 64Question →

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 65Question →

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 66Question →

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 67Question →

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 68Question →

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 69Question →

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 70Question →

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 71Question →

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

Show answer & explanation

Answer: 408

Answer

The total surface area of the right triangular prism is

408\text{ cm}^2

.

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

6\text{ cm}

and

8\text{ cm}

, so the hypotenuse is

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

. The combined area of the two bases is

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

. The perimeter of the base is

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

, making the lateral surface area

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

. Adding the base areas and lateral area yields

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

.

Step-by-Step Solution

1

Determine the length of the hypotenuse of the triangular base.

Hypotenuse =

10\text{ cm}

.

The base is a right-angled triangle with legs

6\text{ cm}

and

8\text{ cm}

. By Pythagoras:

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

.

2

Calculate the total area of the two parallel triangular bases.

Base area total =

48\text{ cm}^2

.

The area of one right triangle is

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

, so two bases have an area of

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

.

3

Calculate the lateral surface area of the three rectangular faces.

Lateral surface area =

360\text{ cm}^2

.

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

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

.

4

Sum the base area total and lateral surface area.

Total Surface Area =

408\text{ cm}^2

.

Total surface area =

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

.

Key Concept

Total Surface Area of a Right Triangular Prism

Question 72Question →

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

Show answer & explanation

Answer: 30

Answer

The final volume of the air bubble at the surface is

30.0\text{ cm}^3

.

Using the combined gas law

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

with absolute temperatures

T_1 = 280\text{ K}

and

T_2 = 300\text{ K}

yields a final volume of

30.0\text{ cm}^3

.

Step-by-Step Solution

1

Convert temperatures from Celsius to Kelvin

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

and

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

Gas laws require absolute thermodynamic temperature in Kelvin.

2

Apply the combined gas law equation

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

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

3

Rearrange for the unknown volume

V_2

and substitute the values

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

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

Key Concept

Combined Gas Law

Question 73Question →

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

Show answer & explanation

Answer: 8

Answer

The resistance of

R_x

is

8.0\text{ }\Omega

.

Applying the relationship

V = E - Ir

yields a total current of

1.2\text{ A}

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

9.0\text{ }\Omega

. Subtracting the parallel admittance of the

90.0\text{ }\Omega

voltmeter yields a load resistance of

10.0\text{ }\Omega

. Subtracting the

2.0\text{ }\Omega

equivalent resistance of the parallel pair (

3.0\text{ }\Omega

and

6.0\text{ }\Omega

) gives

8.0\text{ }\Omega

for

R_x

.

Step-by-Step Solution

1

Calculate total current supplied by the cell

I = 1.2\text{ A}

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

V = E - Ir

.

2

Calculate equivalent resistance of the entire external circuit across terminals

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

By Ohm's law,

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

.

3

Determine the resistance of the main circuit load

R_L

R_L = 10.0\text{ }\Omega

The voltmeter is in parallel with

R_L

, giving

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

.

4

Calculate the equivalent resistance

R_p

of the parallel combination of

R_1

and

R_2

R_p = 2.0\text{ }\Omega

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

.

5

Determine the value of

R_x

R_x = 8.0\text{ }\Omega

The main circuit load consists of

R_p

in series with

R_x

, so

R_L = R_p + R_x

.

Key Concept

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

Estimated Time:2m 30s

Question 74Question →

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

Show answer & explanation

Answer: 10

Answer

The value of

y

at the maximum stationary point is

10

.

To locate the maximum stationary point, set the first derivative

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

to zero, obtaining stationary values

x = 1

and

x = 2

. Testing in the second derivative

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

gives

-6

at

x = 1

, confirming a local maximum. Substituting

x = 1

into the original cubic equation gives

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

.

Step-by-Step Solution

1

Differentiate

y

with respect to

x

to find the gradient function.

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

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

2

Set the first derivative to zero and solve for

x

.

x = 1 \text{ or } x = 2

Factoring

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

yields the

x

-coordinates of the turning points.

3

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

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

A negative second derivative indicates a local maximum stationary point.

4

Substitute

x = 1

into the original function to determine

y

.

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

Evaluating the curve function at the maximum

x

-coordinate provides the corresponding maximum

y

-value.

Key Concept

Stationary Points, Maxima, and Minima

Question 75Question →

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

Show answer & explanation

Answer: 4

Answer

The value of

k

is

4

.

According to the Remainder Theorem, dividing

P(x)

by

x - 2

leaves a remainder of

P(2)

. Evaluating

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

gives

25 - 4k

. Setting

25 - 4k = 9

and solving yields

k = 4

.

Step-by-Step Solution

1

Apply the Remainder Theorem

The remainder when

P(x)

is divided by

x - 2

is

P(2)

.

By the Remainder Theorem, dividing a polynomial

P(x)

by

x - a

leaves a remainder equal to

P(a)

.

2

Substitute

x = 2

into

P(x)

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

Evaluating the polynomial at

x = 2

expresses the remainder in terms of

k

.

3

Equate

P(2)

to the given remainder and solve for

k

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

Setting the calculated expression equal to

9

forms a linear equation that yields

k = 4

.

Key Concept

Polynomial Remainder Theorem

Estimated Time:1m 30s

Question 76Question →

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

Show answer & explanation

Answer: 93.75

Answer

The maximum height reached by the object is

93.75\text{ m}

.

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

u_y = u \sin \theta

. Resolving the initial velocity gives

u = 50\text{ m/s}

and

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

. Substituting into

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

yields

93.75\text{ m}

.

Step-by-Step Solution

1

Find the magnitude of the initial velocity

u = 50\text{ m/s}

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

u_x = u \cos \theta

.

2

Calculate the initial vertical velocity component

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

Vertical component of velocity is calculated using

u_y = u \sin \theta

.

3

Calculate the maximum height

H = 93.75\text{ m}

At maximum height, vertical velocity is zero, giving

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

.

Key Concept

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

Question 77Question →

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

Show answer & explanation

Answer: 4

Answer

The value of

a + b

is

4

.

Integrating

6x^2

gives

2x^3

, and integrating

-4\sin(2x)

gives

2\cos(2x)

. Comparing

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

to

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

yields

a=2

and

b=2

, giving

a+b=4

.

Step-by-Step Solution

1

Integrate each term of the integrand

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

with respect to

x

.

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

, and

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

.

Applying the power rule for integration

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

and the standard trigonometric integral formula

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

.

2

Combine the calculated antiderivative terms and include the constant of integration

C

.

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

.

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

3

Compare the resulting expression with

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

to determine

a

and

b

.

a = 2

and

b = 2

.

Matching corresponding coefficients of

x^3

and

\cos(2x)

.

4

Compute

a + b

.

2 + 2 = 4

.

Evaluating the sum of the extracted coefficients.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Estimated Time:1m 30s

Question 78Question →

A radioactive parent nucleus with an initial mass number of $226$ undergoes a natural decay sequence in which it emits $3$ $\alpha$ -particles and $2$ $\beta^-$ -particles. What is the mass number of the resulting daughter nucleus?

Show answer & explanation

Answer: 214

Answer

The mass number of the resulting daughter nucleus is 214.

Emitting an

\alpha

-particle reduces the nuclear mass number

A

by

4

, while emitting a

\beta^-

-particle leaves the mass number unchanged. Emitting

3

\alpha

-particles reduces the mass number by

3 \times 4 = 12

. Subtracting

12

from the original mass number of

226

yields a final mass number of

214

.

Step-by-Step Solution

1

Identify the mass number change caused by alpha and beta emissions.

Each alpha particle decreases the mass number by 4, while each beta particle causes no change in mass number.

An alpha particle consists of 2 protons and 2 neutrons (mass number 4), whereas a beta particle is an electron with negligible mass in nucleon terms (mass number 0).

2

Calculate the total change in mass number.

Total mass number reduction = 3 * 4 = 12.

Three alpha particles are emitted in the decay sequence.

3

Calculate the mass number of the daughter nucleus.

Daughter mass number = 226 - 12 = 214.

Subtracting the total change in mass number from the original parent mass number.

Key Concept

Conservation of mass number in radioactive decay series

Question 79Question →

A $6.95\text{ g}$ sample of hydrated iron(II) tetraoxosulfate(VI), $\text{FeSO}_4 \cdot x\text{H}_2\text{O}$ , is dissolved in dilute tetraoxosulfate(VI) acid and made up to $250\text{ cm}^3$ in a volumetric flask. A $25.0\text{ cm}^3$ portion of this solution requires $25.0\text{ cm}^3$ of $0.020\text{ mol dm}^{-3}$ acidified potassium tetraoxomanganate(VII), $\text{KMnO}_4$ , for complete titration. Given the relative atomic masses $\text{Fe} = 56$ , $\text{S} = 32$ , $\text{O} = 16$ , and $\text{H} = 1$ , what is the integer value of $x$ , the number of molecules of water of crystallization per formula unit?

Show answer & explanation

Answer: 7

Answer

The integer value of x is 7.

By using the titration volume and concentration of acidified potassium tetraoxomanganate(VII), the amount of iron(II) ions in the sample is calculated. Knowing the total mole amount of anhydrous iron(II) tetraoxosulfate(VI) allows determination of the mass of the anhydrous salt component (3.80 g). Subtracting this from the initial hydrated sample mass (6.95 g) gives the mass of water of crystallization (3.15 g). Dividing the moles of water (0.175 mol) by the moles of anhydrous salt (0.025 mol) yields exactly 7 water molecules of crystallization per formula unit.

Step-by-Step Solution

1

Calculate the moles of

\text{KMnO}_4

consumed in the titration

n(\text{KMnO}_4) = 0.020 \text{ mol dm}^{-3} \times 0.0250 \text{ dm}^3 = 0.00050 \text{ mol}

Concentration and volume of titrant are provided.

2

Determine moles of

\text{Fe}^{2+}

present in the

25.0\text{ cm}^3

aliquot using the redox reaction stoichiometry

n(\text{Fe}^{2+})_{25\text{cm}^3} = 5 \times 0.00050 \text{ mol} = 0.0025 \text{ mol}

The mole ratio of

\text{MnO}_4^-

to

\text{Fe}^{2+}

in acidic redox titration is

1:5

according to

\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}

.

3

Calculate the total moles of

\text{FeSO}_4

in the original

250\text{ cm}^3

volumetric flask

n(\text{FeSO}_4)_{\text{total}} = 0.0025 \text{ mol} \times \left(\frac{250\text{ cm}^3}{25.0\text{ cm}^3}\right) = 0.025 \text{ mol}

The aliquot represents one-tenth of the total solution volume.

4

Calculate the mass of anhydrous

\text{FeSO}_4

in the sample

Molar mass of

\text{FeSO}_4 = 56 + 32 + (4 \times 16) = 152 \text{ g mol}^{-1}

. Mass

= 0.025 \text{ mol} \times 152 \text{ g mol}^{-1} = 3.80 \text{ g}

.

Converting moles of anhydrous salt to mass using molar mass.

5

Determine the mass and moles of water of crystallization

Mass of

\text{H}_2\text{O} = 6.95 \text{ g} - 3.80 \text{ g} = 3.15 \text{ g}

. Moles of

\text{H}_2\text{O} = \frac{3.15 \text{ g}}{18 \text{ g mol}^{-1}} = 0.175 \text{ mol}

.

Subtracting anhydrous mass from initial mass gives water of crystallization mass, converted to moles using molar mass of

\text{H}_2\text{O} = 18 \text{ g mol}^{-1}

.

6

Compute the hydration coefficient

x = \frac{n(\text{H}_2\text{O})}{n(\text{FeSO}_4)}

x = \frac{0.175 \text{ mol}}{0.025 \text{ mol}} = 7

The coefficient

x

represents the mole ratio of water of crystallization to anhydrous salt.

Key Concept

Quantitative determination of water of crystallization in hydrated salts via redox volumetric analysis

Question 80Question →

A mountain climber of mass $60\text{ kg}$ climbs a vertical height of $15\text{ m}$ in a time of $30\text{ s}$ . Taking the acceleration due to gravity $g = 10\text{ m s}^{-2}$ , calculate the average power expended by the climber in Watts.

Show answer & explanation

Answer: 300

Answer

The average power expended by the climber is

300\text{ W}

.

The work done in lifting a mass

m

through vertical height

h

is given by

W = mgh = 60 \times 10 \times 15 = 9000\text{ J}

. The average power is the rate of doing work,

P = \frac{W}{t} = \frac{9000\text{ J}}{30\text{ s}} = 300\text{ W}

.

Step-by-Step Solution

1

Identify the given values and formula for work done against gravity.

Mass

m = 60\text{ kg}

, vertical displacement

h = 15\text{ m}

, acceleration due to gravity

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

, and time

t = 30\text{ s}

. Work done formula:

W = mgh

.

The climber works against gravity to increase their potential energy by an amount equal to

mgh

.

2

Calculate the total work done.

W = 60\text{ kg} \times 10\text{ m s}^{-2} \times 15\text{ m} = 9000\text{ J}

.

Multiplying force (

mg

) by vertical distance (

h

) yields work done in Joules.

3

Calculate the average power.

P = \frac{W}{t} = \frac{9000\text{ J}}{30\text{ s}} = 300\text{ W}

.

Power is defined as work done divided by the time interval (

P = \frac{W}{t}

).

Key Concept

Power as the rate of doing work against gravity

Estimated Time:45s

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept