All practice questions

Question 81Question →

Monochromatic light of frequency $9.0 \times 10^{14}\text{ Hz}$ is incident on a clean potassium emitter surface having a threshold frequency of $5.0 \times 10^{14}\text{ Hz}$ . Given that Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ and the elementary charge $e = 1.6 \times 10^{-19}\text{ C}$ , what is the stopping potential in volts required to completely arrest the photoelectric current?

Show answer & explanation

Answer: 1.65

Answer

The stopping potential required to arrest the emitted photoelectrons is

1.65\text{ V}

.

Applying Einstein's photoelectric equation

K_{\max} = h f - h f_0 = h(f - f_0)

gives a maximum kinetic energy of

2.64 \times 10^{-19}\text{ J}

. Dividing this by the electronic charge

e = 1.6 \times 10^{-19}\text{ C}

yields the stopping potential

V_s = 1.65\text{ V}

.

Step-by-Step Solution

1

Determine the net energy available for photoelectron kinetic energy using Einstein's photoelectric equation

K_{\max} = h(f - f_0) = 6.6 \times 10^{-34}\text{ J}\cdot\text{s} \times (9.0 \times 10^{14} - 5.0 \times 10^{14})\text{ Hz} = 2.64 \times 10^{-19}\text{ J}

Photoelectron emission occurs only when photon energy

hf

exceeds the work function

W_0 = h f_0

, with the excess energy appearing as maximum kinetic energy.

2

Express stopping potential in terms of electron charge and maximum kinetic energy

V_s = \frac{K_{\max}}{e} = \frac{2.64 \times 10^{-19}\text{ J}}{1.6 \times 10^{-19}\text{ C}} = 1.65\text{ V}

The stopping potential

V_s

does work

e V_s

equal to the maximum kinetic energy

K_{\max}

of the fastest photoelectrons to bring them to rest.

Key Concept

Einstein's Photoelectric Equation and Stopping Potential Relationship

Question 82Question →

A $13.90\text{ g}$ sample of hydrated iron(II) tetraoxosulfate(VI), $\text{FeSO}_4 \cdot x\text{H}_2\text{O}$ , was heated strongly until all water of crystallization was driven off. The mass of the remaining anhydrous salt was $7.60\text{ g}$ . What is the value of $x$ ? $[\text{Fe} = 56, \text{S} = 32, \text{O} = 16, \text{H} = 1]$

Show answer & explanation

Answer: 7

Answer

The value of

x

in the hydrated salt formula

\text{FeSO}_4 \cdot x\text{H}_2\text{O}

is 7.

Heating the sample drives off all water of crystallization, leaving only anhydrous

\text{FeSO}_4

. The mass of water is

13.90\text{ g} - 7.60\text{ g} = 6.30\text{ g}

. Converting both components to moles gives

0.05\text{ mol}

of

\text{FeSO}_4

and

0.35\text{ mol}

of

\text{H}_2\text{O}

. The ratio

\frac{0.35}{0.05} = 7

, yielding

x = 7

.

Step-by-Step Solution

1

Find the mass of water lost

Mass of

\text{H}_2\text{O} = 13.90\text{ g} - 7.60\text{ g} = 6.30\text{ g}

The loss in mass upon heating represents the driven-off water of crystallization.

2

Calculate molar masses of anhydrous salt and water

Molar mass of

\text{FeSO}_4 = 152\text{ g/mol}

, Molar mass of

\text{H}_2\text{O} = 18\text{ g/mol}

Molar masses are required to convert the measured masses into mole quantities.

3

Calculate the amount in moles of both components

Moles of

\text{FeSO}_4 = \frac{7.60}{152} = 0.05\text{ mol}

; Moles of

\text{H}_2\text{O} = \frac{6.30}{18} = 0.35\text{ mol}

Stoichiometric coefficient

x

represents the mole ratio between water and anhydrous salt.

4

Determine the mole ratio

x = \frac{0.35\text{ mol}}{0.05\text{ mol}} = 7

Dividing the moles of water of crystallization by the moles of anhydrous salt yields the integer coefficient

x

.

Key Concept

Determining the formula of a hydrated salt from gravimetric data

Estimated Time:1m 30s

Question 83Question →

A weak monobasic acid has a concentration of $0.20\text{ mol dm}^{-3}$ in aqueous solution. At equilibrium, the hydrogen ion concentration, $[H^+]$ , is measured to be $1.0 \times 10^{-3}\text{ mol dm}^{-3}$ . What is the percentage degree of ionization of the acid in this solution?

Show answer & explanation

Answer: 0.5

Answer

The percentage degree of ionization of the weak acid is 0.5%.

The percentage degree of ionization measures the fraction of acid molecules that ionize in water, expressed as a percentage. By substituting

[H^+] = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

and initial concentration

C = 0.20\text{ mol dm}^{-3}

into

\text{Percentage ionization} = \left(\frac{[H^+]}{C}\right) \times 100\%

, we obtain

\left(\frac{0.001}{0.20}\right) \times 100\% = 0.5\%

.

Step-by-Step Solution

1

Identify the relationship between degree of ionization (\alpha), hydrogen ion concentration (

[H^+]

), and initial acid concentration (

C

).

\alpha = \frac{[H^+]}{C}

For a weak monobasic acid ionizing according to

HA \rightleftharpoons H^+ + A^-

, the concentration of ionized hydrogen ions equals

\alpha C

.

2

Calculate the fractional degree of ionization (\alpha).

\alpha = \frac{1.0 \times 10^{-3}\text{ mol dm}^{-3}}{0.20\text{ mol dm}^{-3}} = 0.005

Dividing the equilibrium hydrogen ion concentration by the initial acid concentration gives the fraction of acid molecules that ionized.

3

Convert the fractional degree of ionization into a percentage.

\text{Percentage ionization} = 0.005 \times 100\% = 0.5\%

Multiplying the decimal fraction by 100 converts the degree of ionization into percentage form.

Key Concept

Degree of Ionization of Weak Acids

Estimated Time:1m 30s

Question 84Question →

At 31st December 2025, a sole trader's trial balance showed Trade Debtors of ₦50,000. Additional bad debts of ₦2,000 are to be written off, and a provision for doubtful debts is to be created at 5% on the remaining trade debtors. What is the net trade debtors figure to be presented in the Statement of Financial Position?

Show answer & explanation

Answer: 45600

Answer

The net trade debtors figure to be presented in the Statement of Financial Position is ₦45,600.

To calculate net trade debtors for the Statement of Financial Position, first subtract the ₦2,000 additional bad debts from the gross debtors of ₦50,000 to get ₦48,000 adjusted debtors. Next, calculate 5% of ₦48,000, which gives a provision for doubtful debts of ₦2,400. Subtracting ₦2,400 from ₦48,000 yields the net debtors figure of ₦45,600.

Step-by-Step Solution

1

Deduct the additional bad debts from gross trade debtors

Adjusted Debtors = ₦50,000 - ₦2,000 = ₦48,000

Bad debts identified at year-end must be written off against gross debtors first before calculating the required percentage provision.

2

Calculate the 5% provision for doubtful debts on the adjusted debtors balance

Provision = 5% × ₦48,000 = ₦2,400

The provision for doubtful debts is estimated based on the net collectible debtors.

3

Deduct the provision for doubtful debts from the adjusted debtors balance

Net Debtors = ₦48,000 - ₦2,400 = ₦45,600

Net debtors are reported in the Statement of Financial Position after subtracting the provision for doubtful debts.

Key Concept

Adjustment for bad debts written off and provision for doubtful debts in final accounts

Question 85Question →

The following financial information was extracted from the books of Unity Recreation Club for the year ended 31st December 2025:

- Subscriptions received in cash: ₦640,000
- Subscriptions accrued as at 1st January 2025: ₦50,000
- Subscriptions received in advance as at 1st January 2025: ₦35,000
- Subscriptions accrued as at 31st December 2025: ₦65,000
- Subscriptions received in advance as at 31st December 2025: ₦40,000
- Rent paid in cash: ₦180,000
- Rent prepaid as at 31st December 2025: ₦20,000
- General expenses paid: ₦210,000
- Depreciation on equipment for the year: ₦45,000

What is the net surplus (in Naira) to be reported in the Income and Expenditure Account for the year ended 31st December 2025?

Show answer & explanation

Answer: 235000

Answer

The net surplus to be reported in the Income and Expenditure Account for the year ended 31st December 2025 is ₦235,000.

The correct net surplus of ₦235,000 is derived by calculating total revenue earned minus total revenue expenses incurred during the year. Subscription Income earned equals ₦640,000 + ₦35,000 (opening advance) - ₦50,000 (opening accrual) + ₦65,000 (closing accrual) - ₦40,000 (closing advance) = ₦650,000. Total expenditure equals Rent of ₦160,000 (₦180,000 - ₦20,000 prepayment) + General Expenses of ₦210,000 + Depreciation of ₦45,000 = ₦415,000. Net Surplus = ₦650,000 - ₦415,000 = ₦235,000.

Step-by-Step Solution

1

Determine the Subscription Income for the year using accrual adjustments

Subscription Income = ₦650,000

Opening advance (₦35,000) and closing accrual (₦65,000) are added to cash received (₦640,000), while opening accrual (₦50,000) and closing advance (₦40,000) are deducted.

2

Calculate the adjusted Rent Expense for the period

Rent Expense = ₦160,000

Prepaid rent at year-end (₦20,000) represents an unexpired cost for the next period and must be deducted from cash paid (₦180,000).

3

Sum all revenue expenditures for the period

Total Expenditure = ₦415,000

Total revenue expenses comprise Rent Expense (₦160,000), General Expenses (₦210,000), and non-cash Depreciation (₦45,000).

4

Deduct Total Expenditure from Total Income to calculate Net Surplus

Net Surplus = ₦235,000

Surplus is the excess of total revenue income (₦650,000) over total revenue expenditure (₦415,000).

Key Concept

Income and Expenditure Account Net Surplus Determination

Question 86Question →

A metal plate is initially in the form of a rectangle $ABCD$ measuring $14\text{ cm}$ by $10\text{ cm}$ , where $AB = 14\text{ cm}$ . A semicircular piece with diameter $AB$ is cut out from side $AB$ . On the opposite side $CD$ , an isosceles triangular plate with base $CD$ and height $24\text{ cm}$ is attached externally. Taking $\pi = \frac{22}{7}$ , what is the total area of the resulting plate in $\text{cm}^2$ ?

Show answer & explanation

Answer: 231

Answer

The total area of the resulting plate is

231\text{ cm}^2

.

The net area of the plate is found by starting with the area of the rectangle (

140\text{ cm}^2

), subtracting the area of the semicircular cutout (

77\text{ cm}^2

), and adding the area of the attached isosceles triangle (

168\text{ cm}^2

), resulting in

140 - 77 + 168 = 231\text{ cm}^2

.

Step-by-Step Solution

1

Calculate the area of the original rectangle ABCD

140 cm²

The area of a rectangle is calculated as

\text{length} \times \text{width} = 14\text{ cm} \times 10\text{ cm} = 140\text{ cm}^2

.

2

Calculate the area of the removed semicircular section

77 cm²

The radius of the semicircle is

r = \frac{14}{2} = 7\text{ cm}

. The area of a semicircle is

\frac{1}{2}\pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 7^2 = 77\text{ cm}^2

.

3

Calculate the area of the attached triangular section

168 cm²

The area of a triangle is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14\text{ cm} \times 24\text{ cm} = 168\text{ cm}^2

.

4

Combine the area components to determine the final net area

231 cm²

Subtract the removed semicircular area from the rectangle's area and add the attached triangle's area:

140 - 77 + 168 = 231\text{ cm}^2

.

Key Concept

Perimeter and Area of Composite Plane Figures

Question 87Question →

A bullet of mass $20\text{ g}$ is moving horizontally at a speed of $400\text{ m s}^{-1}$ . It strikes a stationary block of wood and emerges from the opposite side with a speed of $100\text{ m s}^{-1}$ . What is the magnitude of the work done by the bullet against the resistance of the block, in Joules?

Show answer & explanation

Answer: 1500

Answer

The magnitude of the work done by the bullet in penetrating the block is

1500\text{ J}

.

According to the Work-Energy Theorem, the net work done on an object equals the change in its kinetic energy. The reduction in kinetic energy as the bullet slows from

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

to

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

equals the work done against the resistive force of the wooden block.

Step-by-Step Solution

1

Convert the mass of the bullet into standard SI units (kilograms)

m = \frac{20}{1000} = 0.02\text{ kg}

The standard SI unit for mass in energy equations is the kilogram.

2

Determine the initial kinetic energy of the bullet before entering the block

E_{ki} = \frac{1}{2} \times 0.02\text{ kg} \times (400\text{ m s}^{-1})^2 = 1600\text{ J}

Kinetic energy is defined as

E_k = \frac{1}{2} m v^2

.

3

Determine the final kinetic energy of the bullet as it emerges from the block

E_{kf} = \frac{1}{2} \times 0.02\text{ kg} \times (100\text{ m s}^{-1})^2 = 100\text{ J}

The bullet loses speed upon passing through the block, reducing its kinetic energy.

4

Apply the Work-Energy Theorem to find the work done against resistive forces

W = \Delta E_k = 1600\text{ J} - 100\text{ J} = 1500\text{ J}

The net work done on the bullet equals its change in kinetic energy.

Key Concept

Work-Energy Theorem

Question 88Question →

A line $L_1$ is given by the equation $3x + 4y - 24 = 0$ , intersecting the x-axis at point $A$ and the y-axis at point $B$ . A second line $L_2$ has the equation $4x - 3y + k = 0$ , where $k > 0$ . If the perpendicular distance from the midpoint of the line segment $AB$ to $L_2$ is $5$ units, what is the value of $k$ ?

Show answer & explanation

Answer: 18

Answer

The value of

k

is

18

.

To find

k

, first determine the intercepts of

L_1

: setting

y=0

gives

A(8, 0)

and setting

x=0

gives

B(0, 6)

. The midpoint

M

of segment

AB

is

\left(\frac{8+0}{2}, \frac{0+6}{2}\right) = (4, 3)

. Using the perpendicular distance formula

d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

for point

M(4,3)

and line

4x - 3y + k = 0

, we get

d = \frac{|4(4) - 3(3) + k|}{\sqrt{4^2 + (-3)^2}} = \frac{|7 + k|}{5}

. Setting

d = 5

gives

|7 + k| = 25

. Since

k > 0

, solving

7 + k = 25

gives the correct value

k = 18

.

Step-by-Step Solution

1

Determine the coordinates of points A and B

A = (8, 0)

and

B = (0, 6)

Setting

y = 0

in

3x + 4y - 24 = 0

yields

3x = 24 \implies x = 8

. Setting

x = 0

yields

4y = 24 \implies y = 6

.

2

Calculate the midpoint M of segment AB

M = (4, 3)

Using the midpoint formula

M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{8 + 0}{2}, \frac{0 + 6}{2}\right) = (4, 3)

.

3

Set up the perpendicular distance equation from M(4,3) to line L₂

d = \frac{|7 + k|}{5}

Applying the distance formula

d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

gives

d = \frac{|4(4) - 3(3) + k|}{\sqrt{4^2 + (-3)^2}} = \frac{|16 - 9 + k|}{\sqrt{25}} = \frac{|7 + k|}{5}

.

4

Solve for k given d = 5 and k > 0

k = 18

Equating distance to

5

gives

\frac{|7 + k|}{5} = 5 \implies |7 + k| = 25

. Since

k > 0

,

7 + k = 25

, which yields

k = 18

.

Key Concept

Perpendicular Distance from a Point to a Straight Line

Estimated Time:2m 30s

Question 89Question →

If $214_x = 312_5$ , where $x$ represents a positive integer base, find the value of $x$ .

Show answer & explanation

Answer: 6

Answer

The value of the base

x

is 6.

Expanding

312_5

to base 10 gives

3(25) + 1(5) + 2(1) = 82

. Expanding

214_x

gives

2x^2 + x + 4

. Equating both expressions yields

2x^2 + x - 78 = 0

. Factoring as

(2x + 13)(x - 6) = 0

gives the positive integer root

x = 6

.

Step-by-Step Solution

1

Convert the right side of the equation from base 5 to base 10

312_5 = 3 \times 5^2 + 1 \times 5^1 + 2 \times 5^0 = 75 + 5 + 2 = 82_{10}

Converting known non-decimal bases to base 10 provides a standard baseline for algebraic manipulation.

2

Expand the left side expression in terms of powers of

x

214_x = 2x^2 + x + 4

Positional values in base

x

correspond to powers of

x

(

x^2, x^1, x^0

).

3

Formulate and rearrange the resulting quadratic equation in standard form

2x^2 + x + 4 = 82 \implies 2x^2 + x - 78 = 0

Setting the base 10 expansions equal forms a quadratic equation.

4

Solve the quadratic equation for positive integer values of

x

x = 6

Factoring

(2x + 13)(x - 6) = 0

gives

x = 6

or

x = -6.5

. A base must be a positive integer larger than all individual digits appearing in the number (digits are 2, 1, 4, so

x > 4

).

Key Concept

Converting non-decimal numbers to base 10 using positional expansion to solve polynomial equations involving unknown bases.

Question 90Question →

A container holds a liquid mixture where water constitutes $\frac{3}{8}$ of the total volume. When $15\text{ liters}$ of pure water is added to the mixture, water then accounts for $50\%$ of the new total volume. What was the initial total volume of the mixture in liters?

Show answer & explanation

Answer: 60

Answer

The initial total volume of the mixture was 60 liters.

The correct answer is 60 liters. By modeling the initial volume of water as

\frac{3}{8}V

, adding 15 liters yields a new water volume of

\frac{3}{8}V + 15

out of a total volume of

V + 15

. Setting

\frac{3}{8}V + 15 = 0.5(V + 15)

and solving gives

V = 60

.

Step-by-Step Solution

1

Define the variable and write an algebraic expression for the initial amount of water.

If

V

is the initial total volume in liters, initial water volume =

\frac{3}{8}V

.

Water makes up

\frac{3}{8}

of the total initial volume.

2

Account for the addition of 15 liters of pure water to both water volume and total volume.

New water volume =

\frac{3}{8}V + 15

; New total volume =

V + 15

.

Adding pure water increases both the specific water volume and the total mixture volume by 15 liters.

3

Formulate an equation relating new water volume to 50% of the new total volume.

\frac{3}{8}V + 15 = 0.5(V + 15)

Water constitutes 50% (or

\frac{1}{2}

) of the updated mixture.

4

Solve the linear equation for

V

.

7.5 = \frac{1}{8}V \implies V = 60

Subtracting

\frac{3}{8}V

and

7.5

from both sides isolates

\frac{1}{8}V

on one side.

Key Concept

Solving multi-step fraction and percentage mixture problems

Question 91Question →

In a geometric progression of positive terms, the sum of the first two terms is $12$ and the sum of the third and fourth terms is $48$ . What is the $6^{\text{th}}$ term of the progression?

Show answer & explanation

Answer: 128

Answer

The 6th term of the geometric progression is 128.

Dividing

ar^2(1+r) = 48

by

a(1+r) = 12

yields

r^2 = 4

, so

r = 2

for positive terms. Substituting

r = 2

into

a(1+r) = 12

gives

a = 4

. Using

T_n = a r^{n-1}

for

n=6

, we get

T_6 = 4 \times 2^5 = 128

.

Step-by-Step Solution

1

Set up algebraic equations for the given sums using first term

a

and common ratio

r

.

a(1+r) = 12

and

ar^2(1+r) = 48

The terms of a geometric progression are given by

T_n = a r^{n-1}

.

2

Divide the equation for the third and fourth terms by the equation for the first and second terms.

r^2 = 4 \implies r = 2

Dividing eliminates

a

and

(1+r)

, giving

r^2 = 4

. Since terms are positive,

r > 0

.

3

Substitute

r = 2

into

a(1+r) = 12

to solve for

a

.

a = 4

3a = 12

leads directly to

a = 4

.

4

Evaluate the 6th term using the formula

T_6 = a r^{5}

.

T_6 = 4 \times 2^5 = 128

Applying the general term formula

T_n = a r^{n-1}

with

n=6

.

Key Concept

Geometric Progression term relations and finding the common ratio from consecutive term pairs

Question 92Question →

Given the function $y = \frac{(x^2 + 1)^3}{3x - 5}$ , calculate the value of $\frac{dy}{dx}$ at $x = 2$ .

Show answer & explanation

Answer: -75

Answer

The value of

\frac{dy}{dx}

at

x = 2

is

-75

.

Applying the Quotient Rule

\frac{u'v - uv'}{v^2}

along with the Chain Rule to differentiate

u(x) = (x^2+1)^3

yields

u'(x) = 6x(x^2+1)^2

. Evaluating at

x=2

gives

u(2)=125

,

u'(2)=300

,

v(2)=1

, and

v'(2)=3

, leading to

\frac{300(1) - 125(3)}{1^2} = -75

.

Step-by-Step Solution

1

Set up the Quotient Rule framework

Let

u(x) = (x^2 + 1)^3

and

v(x) = 3x - 5

, so that

y = \frac{u(x)}{v(x)}

and

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

.

The function is expressed as a quotient of two differentiable terms.

2

Differentiate the numerator using the Chain Rule

u'(x) = 3(x^2 + 1)^2 \cdot \frac{d}{dx}(x^2 + 1) = 6x(x^2 + 1)^2

.

The numerator is a composite function requiring the inner derivative derivative of

x^2+1

to be multiplied.

3

Differentiate the denominator

v'(x) = 3

.

The derivative of a linear function

3x - 5

with respect to

x

is its coefficient 3.

4

Evaluate

u(x)

,

u'(x)

,

v(x)

, and

v'(x)

at

x = 2

u(2) = 125

,

u'(2) = 300

,

v(2) = 1

,

v'(2) = 3

.

Substituting

x = 2

into each evaluated component simplifies the numerical calculation.

5

Substitute numerical values into the Quotient Rule formula

\left.\frac{dy}{dx}\right|_{x=2} = \frac{(300)(1) - (125)(3)}{1^2} = \frac{300 - 375}{1} = -75

.

Completing the arithmetic calculation yields the final numerical derivative value.

Key Concept

Combining the Quotient Rule and Chain Rule for composite fractional functions

Estimated Time:1m 30s

Question 93Question →

A trader buys a quantity of goods. He sells $\frac{1}{4}$ of the total goods at a profit of $20\%$ , and $\frac{1}{2}$ of the remaining goods at a loss of $10\%$ . If the rest of the goods are sold at cost price, resulting in an overall net profit of $\text{N}1,500$ , what was the total cost price of the goods in Naira?

Show answer & explanation

Answer: 120000

Answer

The total cost price of the goods is 120,000 Naira.

Let

C

be the total cost price of the goods. The first portion sold is

\frac{1}{4}C

at a profit of

20\%

, giving a gain of

0.20 \times \frac{1}{4}C = 0.05C

. The remaining portion is

1 - \frac{1}{4} = \frac{3}{4}C

. Half of this remainder is

\frac{1}{2} \times \frac{3}{4}C = \frac{3}{8}C

, which is sold at a

10\%

loss, causing a loss of

0.10 \times \frac{3}{8}C = 0.0375C

. The final remaining

\frac{3}{8}C

is sold at cost price (0 profit). Thus, the net profit is

0.05C - 0.0375C = 0.0125C

. Setting

0.0125C = 1,500

and solving for

C

gives

C = \frac{1,500}{0.0125} = 120,000

Naira.

Step-by-Step Solution

1

Define the unknown total cost price

Let

C

represent the total cost price of the goods in Naira.

Establishing a variable allows for algebraic formulation of profits and losses.

2

Calculate profit from the first portion

\text{Profit}_1 = 20\% \text{ of } \frac{1}{4}C = 0.20 \times 0.25C = +0.05C

The trader sells a quarter of the total value at a 20% gain.

3

Determine the remaining quantity and calculate loss from the second portion

\text{Remaining} = C - \frac{1}{4}C = \frac{3}{4}C; \quad \text{Second Portion} = \frac{1}{2} \times \frac{3}{4}C = \frac{3}{8}C = 0.375C; \quad \text{Loss}_2 = 10\% \text{ of } 0.375C = -0.0375C

The second sale applies to half of what was left after the first sale, incurred as a loss.

4

Equate net profit to given numerical value and solve for total cost price

\text{Net Profit} = 0.05C - 0.0375C = 0.0125C = 1,500 \implies C = \frac{1,500}{0.0125} = 120,000

The rest of the goods were sold at cost price (zero profit/loss), so net profit equals gain from portion 1 minus loss from portion 2.

Key Concept

Fractions and Percentages of Quantities

Estimated Time:1m 30s

Question 94Question →

In a probability experiment, a card is drawn at random with replacement from a bag containing red, green, and blue cards. After conducting $250$ trials, a green card was drawn $85$ times. Given that the theoretical probability of drawing a green card is $0.30$ , calculate the positive difference between the experimental probability and the theoretical probability of drawing a green card.

Show answer & explanation

Answer: 0.04

Answer

The positive difference between the experimental probability and the theoretical probability is 0.04.

The experimental probability is calculated as the ratio of observed favorable outcomes to total trials: $\frac{85}{250} = 0.34$. Subtracting the given theoretical probability of $0.30$ yields a positive difference of $|0.34 - 0.30| = 0.04$.

Step-by-Step Solution

1

Determine experimental probability from trial data

Experimental probability = 85 / 250 = 0.34

Experimental probability is calculated as the ratio of observed favorable trials to the total number of trials executed.

2

Subtract theoretical probability from experimental probability

|0.34 - 0.30| = 0.04

Finding the positive difference requires subtracting the theoretical probability (0.30) from the experimental relative frequency (0.34).

Key Concept

Experimental probability is determined empirically by dividing the number of times an event occurs by the total number of trials, whereas theoretical probability is based on expected outcomes under ideal conditions.

Question 95Question →

A rectangular coil of $100$ turns with dimensions $0.10\text{ m}$ by $0.20\text{ m}$ is positioned perpendicular to a uniform magnetic field of $0.50\text{ T}$ . The coil is rotated through $90^\circ$ about an axis perpendicular to the field lines in a time interval of $0.040\text{ s}$ , bringing its plane parallel to the magnetic field. What is the magnitude of the average electromotive force (e.m.f.) induced in the coil in volts?

Show answer & explanation

Answer: 25

Answer

The magnitude of the average induced electromotive force in the coil is

25\text{ V}

.

According to Faraday's law, the induced electromotive force magnitude is

E = N \frac{\Delta \Phi}{\Delta t}

. The initial flux through each turn is

\Phi_1 = B A = 0.50\text{ T} \times 0.020\text{ m}^2 = 0.010\text{ Wb}

. When rotated parallel to the field, the final flux

\Phi_2

is

0\text{ Wb}

, so

\Delta \Phi = 0.010\text{ Wb}

. Substituting

N = 100

and

\Delta t = 0.040\text{ s}

yields an induced e.m.f. of

100 \times \frac{0.010}{0.040} = 25\text{ V}

.

Step-by-Step Solution

1

Calculate the cross-sectional area of the rectangular coil.

A = 0.10\text{ m} \times 0.20\text{ m} = 0.020\text{ m}^2

The surface area is required to find the magnetic flux passing through each turn.

2

Determine the change in magnetic flux through one turn of the coil.

Initial flux

\Phi_1 = B A = 0.50 \times 0.020 = 0.010\text{ Wb}

; Final flux

\Phi_2 = 0\text{ Wb}

; Change

\Delta \Phi = 0.010\text{ Wb}

When perpendicular to the magnetic field, maximum flux links the coil. Rotating it parallel reduces the flux linking the coil to zero.

3

Apply Faraday's law of electromagnetic induction to calculate induced e.m.f.

E = N \frac{\Delta \Phi}{\Delta t} = 100 \times \frac{0.010\text{ Wb}}{0.040\text{ s}} = 25\text{ V}

Faraday's law states that the induced e.m.f. magnitude equals the rate of change of total magnetic flux linkage.

Key Concept

Faraday's Law of Electromagnetic Induction

Question 96Question →

Find the sum, in degrees, of all solutions to the trigonometric equation $\sqrt{3}\tan(2x) = 3$ in the interval $0^\circ \le x \le 180^\circ$ .

Show answer & explanation

Answer: 150

Answer

The sum of all solutions to the equation in the given interval is 150 degrees.

Isolating

\tan(2x)

gives

\sqrt{3}

. For

0^\circ \le 2x \le 360^\circ

,

\tan(2x) = \sqrt{3}

yields solutions at

2x = 60^\circ

and

2x = 240^\circ

. Dividing by 2 gives

x = 30^\circ

and

x = 120^\circ

. Adding these solutions yields

30^\circ + 120^\circ = 150^\circ

.

Step-by-Step Solution

1

Isolate the trigonometric function

\tan(2x) = \frac{3}{\sqrt{3}} = \sqrt{3}

Dividing both sides by \sqrt{3} simplifies the expression to a standard special angle ratio.

2

Determine the domain for the argument

2x

Since

0^\circ \le x \le 180^\circ

, multiplying the inequality by 2 gives

0^\circ \le 2x \le 360^\circ

.

This establishes the range of angles to search for

2x

within one complete turn.

3

Find all values of

2x

where tangent equals

\sqrt{3}

2x = 60^\circ

(1st quadrant) and

2x = 180^\circ + 60^\circ = 240^\circ

(3rd quadrant)

The tangent function is positive in Quadrants I and III with a reference angle of

60^\circ

.

4

Solve for

x

x = \frac{60^\circ}{2} = 30^\circ

and

x = \frac{240^\circ}{2} = 120^\circ

Dividing each angle by 2 yields the values of

x

lying within the domain

0^\circ \le x \le 180^\circ

.

5

Calculate the sum of the solutions

30^\circ + 120^\circ = 150^\circ

The question specifically requests the sum of all valid solutions.

Key Concept

Solving trigonometric equations using reference angles and domain transformation

Estimated Time:2m 0s

Question 97Question →

Find the real value of

x

that satisfies the exponential equation

\frac{8^{x + 2}}{4^{x - 1}} = 16^{x - 1}

Show answer & explanation

Answer: 4

Answer

The value of

x

is

4

.

Rewriting the terms in base 2 gives

2^{3(x+2)} / 2^{2(x-1)} = 2^{4(x-1)}

. Applying the quotient rule gives an exponent of

(3x + 6) - (2x - 2) = x + 8

on the left. Equating the exponents yields

x + 8 = 4x - 4

, which solves cleanly to

x = 4

.

Step-by-Step Solution

1

Express all terms using a common prime base of 2

The equation becomes

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

.

Converting to a common base enables the use of index laws to simplify the equation.

2

Apply the power-of-a-power and quotient laws of indices

The left side simplifies to

2^{3(x+2) - 2(x-1)} = 2^{x + 8}

and the right side is

2^{4x - 4}

.

When dividing powers of the same base, exponents are subtracted:

a^m \div a^n = a^{m-n}

.

3

Equate exponents and solve the linear equation

x + 8 = 4x - 4 \implies 3x = 12 \implies x = 4

.

Because the bases on both sides are identical, their respective exponents must be equal.

Key Concept

Solving exponential equations using common base conversion and laws of indices

Question 98Question →

A copper calorimeter of heat capacity $300\text{ J K}^{-1}$ contains $0.5\text{ kg}$ of water at an initial temperature of $25^\circ\text{C}$ . An electric heater rated at $800\text{ W}$ is immersed in the water to heat the system for $4\text{ minutes}$ . If heat is lost to the surrounding environment at a constant rate of $200\text{ W}$ throughout the heating duration, what is the final temperature of the water-calorimeter system in $^\circ\text{C}$ ? (Take the specific heat capacity of water as $4200\text{ J kg}^{-1}\text{K}^{-1}$ )

Show answer & explanation

Answer: 85

Answer

The final temperature of the system is

85^\circ\text{C}

.

The net thermal energy added to the system accounts for both the supplied electrical energy and the heat energy lost to the surroundings:

Q_{\text{net}} = (800 - 200)\text{ W} \times 240\text{ s} = 144,000\text{ J}

. The total heat capacity of the water-calorimeter system is

C_{\text{total}} = 300\text{ J K}^{-1} + (0.5\text{ kg} \times 4200\text{ J kg}^{-1}\text{K}^{-1}) = 2400\text{ J K}^{-1}

. The temperature increase is

\Delta T = \frac{144,000}{2400} = 60^\circ\text{C}

. Adding this to the initial temperature of

25^\circ\text{C}

gives the final temperature of

85^\circ\text{C}

.

Step-by-Step Solution

1

Convert heating time to standard SI units (seconds)

t = 4 \times 60\text{ s} = 240\text{ s}

Power is measured in Joules per second (Watts), so time must be in seconds.

2

Calculate the net rate of heat energy input to the system

P_{\text{net}} = 800\text{ W} - 200\text{ W} = 600\text{ W}

The net heating power is the input power minus the power dissipated as heat loss.

3

Calculate total net heat energy transferred to the system

Q_{\text{net}} = P_{\text{net}} \times t = 600\text{ W} \times 240\text{ s} = 144,000\text{ J}

Thermal energy transferred equals net power multiplied by time.

4

Compute the total heat capacity of the combined system (water + calorimeter)

C_{\text{total}} = C_{\text{calorimeter}} + (m_{\text{water}} \times c_{\text{water}}) = 300\text{ J K}^{-1} + (0.5\text{ kg} \times 4200\text{ J kg}^{-1}\text{K}^{-1}) = 2400\text{ J K}^{-1}

Heat capacity of water is mass multiplied by specific heat capacity, added to the calorimeter's given heat capacity.

5

Calculate the temperature rise of the system

\Delta T = \frac{Q_{\text{net}}}{C_{\text{total}}} = \frac{144,000\text{ J}}{2400\text{ J K}^{-1}} = 60^\circ\text{C}

Temperature change is total net heat supplied divided by total heat capacity.

6

Find the final temperature of the system

T_{\text{final}} = T_{\text{initial}} + \Delta T = 25^\circ\text{C} + 60^\circ\text{C} = 85^\circ\text{C}

Final temperature equals initial temperature plus temperature increase.

Key Concept

Conservation of thermal energy in calorimeter systems with continuous power loss

Estimated Time:1m 30s

Question 99Question →

A factory uses $8$ identical machines to produce $480$ items in $6$ hours. If $3$ of the machines break down, how many hours will it take the remaining machines to produce $600$ items at the same rate?

Show answer & explanation

Answer: 12

Answer

The remaining machines will take 12 hours to produce 600 items.

Each machine produces

\frac{480}{8 \times 6} = 10

items per hour. With

5

active machines, the combined rate is

50

items per hour. To reach

600

items, the time required is

\frac{600}{50} = 12

hours.

Step-by-Step Solution

1

Calculate the output rate per machine per hour

10 items per machine per hour

The combined rate of 8 machines is

480 \div 6 = 80

items per hour. Dividing by 8 machines gives

10

items/hour per machine.

2

Determine the new combined production rate

50 items per hour

With 3 machines out of service,

8 - 3 = 5

machines remain, working at a total rate of

5 \times 10 = 50

items per hour.

3

Calculate total hours required for the new target quantity

12 hours

Dividing the target quantity of

600

items by the rate of

50

items/hour yields

\frac{600}{50} = 12

hours.

Key Concept

Compound Proportion and Work Rate

Estimated Time:1m 30s

Question 100Question →

An electron of mass $9.1 \times 10^{-31}\text{ kg}$ and charge $1.6 \times 10^{-19}\text{ C}$ enters perpendicularly into a uniform magnetic field of flux density $2.0 \times 10^{-3}\text{ T}$ with a speed of $3.2 \times 10^6\text{ m/s}$ . What is the radius of the circular path followed by the electron, expressed in millimeters (mm)?

Show answer & explanation

Answer: 9.1

Answer

The radius of the circular path followed by the electron is 9.1 mm.

When a charge

q

enters a magnetic field

B

perpendicularly at speed

v

, the magnetic force

qvB

supplies the centripetal force

\frac{mv^2}{r}

. Rearranging for radius yields

r = \frac{mv}{qB}

. Substituting

m = 9.1 \times 10^{-31}\text{ kg}

,

v = 3.2 \times 10^6\text{ m/s}

,

q = 1.6 \times 10^{-19}\text{ C}

, and

B = 2.0 \times 10^{-3}\text{ T}

gives

r = 9.1 \times 10^{-3}\text{ m} = 9.1\text{ mm}

.

Step-by-Step Solution

1

Equate the magnetic force to the centripetal force for circular motion

qvB = \frac{mv^2}{r}

A charged particle moving perpendicularly to a magnetic field experiences a magnetic force that acts entirely as a centripetal force.

2

Rearrange the equation to make the orbital radius

r

the subject

r = \frac{mv}{qB}

Cancelling one factor of velocity

v

from both sides allows direct computation of

r

.

3

Substitute the physical values into the formula

r = \frac{(9.1 \times 10^{-31})(3.2 \times 10^6)}{(1.6 \times 10^{-19})(2.0 \times 10^{-3})} = 9.1 \times 10^{-3}\text{ m}

Calculates the radius in standard SI units (meters).

4

Convert the resulting radius from meters to millimeters

r = 9.1 \times 10^{-3}\text{ m} \times 1000\text{ mm/m} = 9.1\text{ mm}

The question explicitly requests the answer in millimeters.

Key Concept

Motion of a charged particle in a uniform magnetic field

Answer

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