All practice questions

What is the exact numerical value of the trigonometric expression

\frac{12 \sin 30^\circ \cos 30^\circ \tan 60^\circ}{\sin^2 45^\circ + \cos^2 45^\circ + \tan^2 60^\circ}

?

Show answer & explanation

Answer: 2.25

Answer

The exact numerical value of the expression is 2.25.

Substituting the exact values

\sin 30^\circ = \frac{1}{2}

,

\cos 30^\circ = \frac{\sqrt{3}}{2}

, and

\tan 60^\circ = \sqrt{3}

into the numerator gives

12 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} \times \sqrt{3} = 9

. Using the Pythagorean identity

\sin^2 45^\circ + \cos^2 45^\circ = 1

and

\tan^2 60^\circ = 3

, the denominator evaluates to

1 + 3 = 4

. Dividing 9 by 4 yields the exact value of 2.25.

Step-by-Step Solution

1

Evaluate the trigonometric ratios for special angles and apply trigonometric identities.

\sin 30^\circ = \frac{1}{2}

,

\cos 30^\circ = \frac{\sqrt{3}}{2}

,

\tan 60^\circ = \sqrt{3}

, and

\sin^2 45^\circ + \cos^2 45^\circ = 1

.

Standard special angle values and the fundamental Pythagorean identity

\sin^2 x + \cos^2 x = 1

simplify the expression.

2

Substitute these values into the numerator of the expression.

Numerator

= 12 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} \times \sqrt{3} = 9

.

Multiplying the terms:

\sqrt{3} \times \sqrt{3} = 3

, and

12 \times \frac{1}{4} \times 3 = 9

.

3

Substitute these values into the denominator of the expression.

Denominator

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

.

The sum

\sin^2 45^\circ + \cos^2 45^\circ = 1

combined with

\tan^2 60^\circ = 3

equals 4.

4

Divide the calculated numerator by the denominator.

\frac{9}{4} = 2.25

.

Dividing 9 by 4 gives the final decimal value 2.25.

Key Concept

Evaluation of trigonometric expressions using special angles (30°, 45°, 60°) and fundamental identities

Question 7822Question →

A stone is thrown into the air with an initial velocity of $20\text{ m/s}$ at an angle of $60^\circ$ to the horizontal. What is the magnitude of its velocity at the highest point of its trajectory?

0\text{ m/s}

10\text{ m/s}

17.3\text{ m/s}

20\text{ m/s}

Show answer & explanation

Answer: $10\text{ m/s}$

Answer

The magnitude of the velocity at the highest point is

10\text{ m/s}

.

At the peak of a projectile's trajectory, the vertical velocity component vanishes (

v_y = 0\text{ m/s}

), but horizontal motion continues at a constant speed (

v_x = u \cos \theta

). Substituting

u = 20\text{ m/s}

and

\theta = 60^\circ

yields

v_x = 20 \times 0.5 = 10\text{ m/s}

.

Step-by-Step Solution

1

Identify the velocity components at the apex

At maximum height, the vertical velocity component is

v_y = 0\text{ m/s}

, while the horizontal component remains constant at

v_x = u_x = u \cos \theta

.

Horizontal acceleration is zero when air resistance is neglected.

2

Calculate the horizontal component of velocity

v_x = 20 \times \cos(60^\circ) = 20 \times 0.5 = 10\text{ m/s}

.

The trigonometric cosine function gives the horizontal projection of the initial velocity vector.

3

Determine total velocity magnitude at the apex

v = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + 0^2} = 10\text{ m/s}

.

Since the vertical component is zero, the total velocity at the peak equals the horizontal component.

Key Concept

Velocity components at maximum height in projectile motion

Question 7823Question →

An investor deposited $\text{₦}60,000$ into a financial scheme for $2$ years at an annual interest rate of $r\%$ . If the difference between the compound interest (compounded annually) and the simple interest earned over the $2$ -year period is $\text{₦}384$ , calculate the value of $r$ .

Show answer & explanation

Answer: 8

Answer

The interest rate r is 8%

For a two-year investment period, the difference between compound interest (compounded annually) and simple interest equals the interest earned in the second year on the first year's interest, which is

P \left(\frac{r}{100}\right)^2

. Setting

60,000 \times \frac{r^2}{10,000} = 384

yields

6r^2 = 384

, giving

r^2 = 64

and

r = 8\%

.

Step-by-Step Solution

1

Express the simple interest for 2 years in terms of r

I_{\text{SI}} = 1,200r

Simple interest is calculated directly on the principal amount for the full term.

2

Express the compound interest for 2 years in terms of r

I_{\text{CI}} = 1,200r + 6r^2

Compound interest includes interest earned on the first year's interest.

3

Set up the equation for the difference between compound and simple interest

(1,200r + 6r^2) - 1,200r = 384 \implies 6r^2 = 384

The difference between CI and SI over 2 years isolates the interest-on-interest component.

4

Solve for r

r = 8

Dividing 384 by 6 gives 64, and taking the principal square root yields 8.

Key Concept

Difference between Compound Interest and Simple Interest for 2 years

Question 7824Question →

A liquid has a real cubic expansivity of $5.0 \times 10^{-4} \text{ K}^{-1}$ . When this liquid is heated inside a metallic vessel, its apparent cubic expansivity is determined to be $4.1 \times 10^{-4} \text{ K}^{-1}$ . What is the linear expansivity of the metallic vessel in $10^{-5} \text{ K}^{-1}$ ?

Show answer & explanation

Answer: 3

Answer

The linear expansivity of the metallic vessel is

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

.

Real cubic expansivity of a liquid accounts for both the expansion of the liquid relative to the container and the expansion of the container itself:

\gamma_r = \gamma_a + \gamma_v

. Subtracting the apparent cubic expansivity (

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

) from the real cubic expansivity (

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

) gives the vessel's cubic expansivity of

0.9 \times 10^{-4} \text{ K}^{-1} = 9.0 \times 10^{-5} \text{ K}^{-1}

. Dividing this value by 3 gives the vessel's linear expansivity:

\alpha = 3.0 \times 10^{-5} \text{ K}^{-1}

.

Step-by-Step Solution

1

Find the cubic expansivity of the vessel (

\gamma_v

)

\gamma_v = \gamma_r - \gamma_a = 5.0 \times 10^{-4} - 4.1 \times 10^{-4} = 0.9 \times 10^{-4} \text{ K}^{-1} = 9.0 \times 10^{-5} \text{ K}^{-1}

The real cubic expansivity of a liquid equals the sum of its apparent cubic expansivity and the cubic expansivity of the containing vessel (

\gamma_r = \gamma_a + \gamma_v

).

2

Calculate the linear expansivity of the vessel (

\alpha

)

\alpha = \frac{\gamma_v}{3} = \frac{9.0 \times 10^{-5}}{3} = 3.0 \times 10^{-5} \text{ K}^{-1}

For isotropic solids, volume (cubic) expansivity is three times the linear expansivity (

\gamma_v = 3\alpha

).

Key Concept

Relationship between real cubic expansivity, apparent cubic expansivity, and vessel expansivity

Question 7825Question →

In a survey of $90$ high school students regarding their participation in sports clubs, $40$ play Badminton ( $B$ ), $35$ play Volleyball ( $V$ ), and $42$ engage in Swimming ( $S$ ). It was found that $14$ play both Badminton and Volleyball, $12$ play both Volleyball and Swimming, and $15$ play both Badminton and Swimming. If $8$ students participate in none of these three sports, how many students participate in all three sports?

6

14

10

18

Show answer & explanation

Answer: $6$

Answer

The number of students participating in all three sports is

6

.

First, determine the number of students in the union of the three sports sets by subtracting the

8

non-participating students from the universal set of

90

, giving

82

. Then apply the Principle of Inclusion-Exclusion:

82 = 40 + 35 + 42 - (14 + 12 + 15) + x

, where

x

is the number of students participating in all three sports. Simplifying gives

82 = 76 + x

, which yields

x = 6

.

Step-by-Step Solution

1

Calculate the cardinality of the union of the three sets.

n(B \cup V \cup S) = n(U) - n((B \cup V \cup S)') = 90 - 8 = 82

Subtracting the students who do not participate in any sport from the total universal set gives the number of students participating in at least one sport.

2

Set up the Principle of Inclusion-Exclusion formula for three sets.

n(B \cup V \cup S) = n(B) + n(V) + n(S) - [n(B \cap V) + n(V \cap S) + n(B \cap S)] + n(B \cap V \cap S)

This fundamental relation accounts for overlapping subsets in a three-set system.

3

Substitute the known numerical values into the formula and solve for

n(B \cap V \cap S)

.

82 = 40 + 35 + 42 - (14 + 12 + 15) + n(B \cap V \cap S) \implies 82 = 117 - 41 + n(B \cap V \cap S) \implies 82 = 76 + n(B \cap V \cap S) \implies n(B \cap V \cap S) = 82 - 76 = 6

Simplifying the arithmetic expression directly isolates the unknown intersection value.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Estimated Time:1m 15s

Question 7826Question →

A water storage tank is initially $\frac{1}{3}$ full. When $28\text{ litres}$ of water are added to the tank, it becomes $\frac{4}{5}$ full. What is the total capacity of the water tank?

35\text{ litres}

60\text{ litres}

84\text{ litres}

140\text{ litres}

Show answer & explanation

Answer: $60\text{ litres}$

Answer

The total capacity of the water tank is

60\text{ litres}

.

Subtracting the initial filled fraction (one-third) from the final filled fraction (four-fifths) yields seven-fifteenths. Since seven-fifteenths of the total capacity corresponds to 28 litres, dividing 28 by seven-fifteenths gives a total capacity of 60 litres.

Step-by-Step Solution

1

Find the fraction of the tank filled by adding the

28\text{ litres}

of water.

Fraction added =

\frac{4}{5} - \frac{1}{3} = \frac{12 - 5}{15} = \frac{7}{15}

.

Subtracting the initial fraction filled from the final fraction filled gives the fractional change.

2

Set up an equation equating the fractional change to the actual volume added.

\frac{7}{15} \times C = 28

, where

C

represents the total capacity of the tank.

The fractional change multiplied by total capacity equals the physical volume of water introduced.

3

Solve for the total capacity

C

.

C = 28 \times \frac{15}{7} = 4 \times 15 = 60\text{ litres}

.

Multiplying both sides by the reciprocal of

\frac{7}{15}

yields the total capacity.

Key Concept

Fractional Word Problems and Quantity Determination

Estimated Time:1m 30s

Question 7827Question →

A liquid has a real cubic expansivity of $7.5 \times 10^{-4}\text{ K}^{-1}$ . It is heated inside a metal vessel whose material has a linear expansivity of $2.0 \times 10^{-5}\text{ K}^{-1}$ . What is the ratio of the real cubic expansivity of the liquid to its apparent cubic expansivity?

\frac{25}{23}

\frac{75}{73}

\frac{25}{27}

\frac{75}{77}

Show answer & explanation

Answer: $\frac{25}{23}$

Answer

The ratio of the real cubic expansivity of the liquid to its apparent cubic expansivity is $\frac{25}{23}$.

The real cubic expansivity of a liquid $\gamma_r$ is related to its apparent cubic expansivity $\gamma_a$ and the cubic expansivity of the container $\gamma_v$ by the equation $\gamma_r = \gamma_a + \gamma_v$. For a vessel made of material with linear expansivity $\alpha$, $\gamma_v = 3\alpha$. Substituting $\alpha = 2.0 \times 10^{-5}\text{ K}^{-1}$ yields $\gamma_v = 6.0 \times 10^{-5}\text{ K}^{-1} = 0.6 \times 10^{-4}\text{ K}^{-1}$. The apparent expansivity is $\gamma_a = 7.5 \times 10^{-4} - 0.6 \times 10^{-4} = 6.9 \times 10^{-4}\text{ K}^{-1}$. The ratio $\frac{\gamma_r}{\gamma_a}$ is therefore $\frac{7.5 \times 10^{-4}}{6.9 \times 10^{-4}} = \frac{25}{23}$.

Step-by-Step Solution

1

Calculate the cubic expansivity of the metal vessel ($\gamma_v$) from its linear expansivity ($\alpha$).

$\gamma_v = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1} = 0.6 \times 10^{-4}\text{ K}^{-1}$

The volume expansion coefficient of a solid vessel is three times its linear expansion coefficient.

2

Determine the apparent cubic expansivity ($\gamma_a$) using the relationship between real expansivity, apparent expansivity, and vessel expansivity.

$\gamma_a = \gamma_r - \gamma_v = 7.5 \times 10^{-4}\text{ K}^{-1} - 0.6 \times 10^{-4}\text{ K}^{-1} = 6.9 \times 10^{-4}\text{ K}^{-1}$

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

3

Compute the ratio of real cubic expansivity to apparent cubic expansivity ($\frac{\gamma_r}{\gamma_a}$).

$\frac{\gamma_r}{\gamma_a} = \frac{7.5 \times 10^{-4}}{6.9 \times 10^{-4}} = \frac{75}{69} = \frac{25}{23}$

Dividing the given real cubic expansivity by the calculated apparent cubic expansivity gives the simplified fraction.

Key Concept

Thermal Expansion of Liquids: Relationship between Real and Apparent Expansivity

Estimated Time:1m 30s

Question 7828Question →

A hydrogen atom initially in its ground state with energy $E_1 = -13.6\text{ eV}$ absorbs a photon with energy $12.75\text{ eV}$ , raising the electron to an excited quantum level $n$ . The atom then undergoes de-excitation to lower energy levels. Taking Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ , speed of light $c = 3.0 \times 10^8\text{ m/s}$ , and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ , what is the shortest wavelength of light emitted during any of the possible downward transitions?

9.7 \times 10^{-8}\text{ m}

1.55 \times 10^{-26}\text{ m}

1.88 \times 10^{-6}\text{ m}

1.02 \times 10^{-7}\text{ m}

Show answer & explanation

Answer: $9.7 \times 10^{-8}\text{ m}$

Answer

9.7 \times 10^{-8}\text{ m}

The correct answer of

9.7 \times 10^{-8}\text{ m}

is obtained by finding that the absorbed photon of

12.75\text{ eV}

promotes the electron to the

n = 4

state (

E_4 = -0.85\text{ eV}

). The shortest wavelength photon is emitted in the transition with the greatest energy difference, which is

n = 4 \rightarrow n = 1

(

\Delta E = 12.75\text{ eV}

). Converting

12.75\text{ eV}

to

2.04 \times 10^{-18}\text{ J}

and substituting into

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

yields

9.7 \times 10^{-8}\text{ m}

.

Step-by-Step Solution

1

Determine the energy of the excited state

E_n

E_n = E_1 + \Delta E_{\text{absorbed}} = -13.6\text{ eV} + 12.75\text{ eV} = -0.85\text{ eV}

Absorbing energy elevates the atom from its ground state energy to a higher energy level.

2

Identify the principal quantum number

n

of the excited state

n = \sqrt{\frac{-13.6\text{ eV}}{-0.85\text{ eV}}} = \sqrt{16} = 4

For hydrogen,

E_n = \frac{E_1}{n^2}

.

3

Identify the transition giving the shortest wavelength

Transition from

n = 4 \rightarrow n = 1

with maximum energy change

\Delta E_{\text{max}} = 12.75\text{ eV}

Since

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

, the shortest wavelength occurs at maximum energy emission.

4

Convert energy to Joules and calculate wavelength

\Delta E = 12.75 \times 1.6 \times 10^{-19}\text{ J} = 2.04 \times 10^{-18}\text{ J}

;

\lambda = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{2.04 \times 10^{-18}} \approx 9.7 \times 10^{-8}\text{ m}

Applying the photon energy-wavelength relation

E = \frac{hc}{\lambda}

in SI units.

Key Concept

Energy Levels and Atomic Spectra

Estimated Time:3m 0s

Question 7829Question →

The cumulative frequency distribution table below shows the mass of cocoa beans (in kg) harvested by $40$ smallholder farmers in a agricultural cooperative:

Mass (kg)	Cumulative Frequency
$\leq 20$	$4$
$\leq 30$	$12$
$\leq 40$	$28$
$\leq 50$	$36$
$\leq 60$	$40$

Using linear interpolation from the cumulative frequency table, what is the median mass (in kg) of cocoa beans harvested by the farmers?

Show answer & explanation

Answer: 35

Answer

The median mass of cocoa beans harvested by the farmers is

35.0\text{ kg}

.

The total number of farmers is

N = 40

. The median corresponds to the

\frac{40}{2} = 20^{\text{th}}

position. From the cumulative frequency table, the

20^{\text{th}}

item lies within the

30 - 40\text{ kg}

class interval. Using the grouped median formula

\text{Median} = L + \left(\frac{\frac{N}{2} - c.f.}{f}\right) \times w

, where

L = 30

,

c.f. = 12

,

f = 28 - 12 = 16

, and

w = 10

, we obtain

\text{Median} = 30 + \left(\frac{20 - 12}{16}\right) \times 10 = 35.0\text{ kg}

.

Step-by-Step Solution

1

Find the median position in the cumulative frequency distribution

Median position

= \frac{N}{2} = \frac{40}{2} = 20^{\text{th}}

item

The median corresponds to the 50th percentile, which is half of the total cumulative frequency

N = 40

.

2

Locate the median class interval and extract its statistical parameters

Median class interval is

30 - 40\text{ kg}

, with lower limit

L = 30\text{ kg}

, preceding cumulative frequency

c.f. = 12

, class frequency

f = 28 - 12 = 16

, and width

w = 10\text{ kg}

The cumulative frequency just below

20

is

12

(at upper boundary

30

), and at upper boundary

40

it rises to

28

.

3

Substitute values into the grouped data median formula

\text{Median} = 30 + \left(\frac{20 - 12}{16}\right) \times 10 = 30 + 5 = 35.0\text{ kg}

Linear interpolation estimates the exact position of the median within the median class interval.

Key Concept

Calculation of Median from Cumulative Frequency Data

Question 7830Question →

If $P$ varies directly as the square root of $q$ and inversely as the square of $r$ , what is the percentage change in $P$ when $q$ is increased by $44\%$ and $r$ is decreased by $20\%$ ?

50\%

increase

125\%

increase

87.5\%

increase

23.2\%

decrease

Show answer & explanation

Answer: $87.5\%$ increase

Answer

An increase of

87.5\%

The variation relation is

P = k \frac{\sqrt{q}}{r^2}

. Increasing

q

by

44\%

scales

\sqrt{q}

by

\sqrt{1.44} = 1.2

. Decreasing

r

by

20\%

scales

r^2

by

(0.8)^2 = 0.64

. Dividing

1.2

by

0.64

gives

1.875

, meaning the new value

P'

is

187.5\%

of the original

P

, which corresponds to an increase of

87.5\%

.

Step-by-Step Solution

1

Set up the joint variation equation

P = k \frac{\sqrt{q}}{r^2}

, where

k

is a constant of variation.

Direct variation puts

\sqrt{q}

in the numerator, and inverse variation puts

r^2

in the denominator.

2

Express new values

q'

and

r'

in terms of original variables

q' = 1.44q

and

r' = 0.80r

.

An increase of

44\%

gives

1 + 0.44 = 1.44

, and a decrease of

20\%

gives

1 - 0.20 = 0.80

.

3

Substitute new variables into the variation formula to find the new value

P'

P' = k \frac{\sqrt{1.44q}}{(0.80r)^2} = k \frac{1.2\sqrt{q}}{0.64r^2} = \frac{1.2}{0.64} \left(k \frac{\sqrt{q}}{r^2}\right) = 1.875P

.

Evaluating

\sqrt{1.44} = 1.2

and

(0.80)^2 = 0.64

gives the multiplier for

P

.

4

Calculate the percentage change in

P

\text{Percentage Change} = (1.875 - 1) \times 100\% = 87.5\%

increase.

Subtracting

1

converts the multiplier to a relative increase.

Key Concept

Joint Variation with Percentage Changes

Estimated Time:1m 30s

Question 7831Question →

A body is projected from ground level with an initial speed of $40\text{ m/s}$ at an angle of $30^\circ$ to the horizontal. Calculate the time taken, in seconds, for the body to reach its maximum height. (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 2

Answer

The time taken to reach maximum height is

2\text{ s}

.

The initial vertical velocity component is

u_y = u \sin(30^\circ) = 40 \times 0.5 = 20\text{ m/s}

. Under gravitational deceleration (

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

), the vertical speed drops to zero at maximum height after

t = \frac{u_y}{g} = \frac{20}{10} = 2\text{ s}

.

Step-by-Step Solution

1

Find the vertical component of initial velocity (

u_y

)

u_y = 40 \sin(30^\circ) = 20\text{ m/s}

Only the vertical component of initial velocity determines the time to reach maximum height.

2

Calculate the time to maximum height (

t

)

t = \frac{u_y}{g} = \frac{20}{10} = 2\text{ s}

At maximum height, vertical velocity

v_y = 0

, giving

t = \frac{u_y}{g}

.

Key Concept

Time to reach maximum height in projectile motion

Question 7832Question →

An electron in an excited state of an atom moves from an energy level of $-2.80\text{ eV}$ to a lower energy state of $-7.60\text{ eV}$ . What is the energy of the emitted photon in Joules? ( $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

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

4.80\text{ J}

1.66 \times 10^{-18}\text{ J}

1.22 \times 10^{-18}\text{ J}

Show answer & explanation

Answer: $7.68 \times 10^{-19}\text{ J}$

Answer

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

The emitted photon's energy is given by the difference between the initial higher state and the final lower state:

\Delta E = E_2 - E_1 = -2.80\text{ eV} - (-7.60\text{ eV}) = 4.80\text{ eV}

. Converting to Joules gives

4.80 \times 1.6 \times 10^{-19}\text{ J} = 7.68 \times 10^{-19}\text{ J}

.

Step-by-Step Solution

1

Calculate the energy difference between the initial and final states in electron-volts (eV)

\Delta E = E_{\text{initial}} - E_{\text{final}} = -2.80\text{ eV} - (-7.60\text{ eV}) = 4.80\text{ eV}

The energy of an emitted photon during a downward atomic transition equals the difference in energy between the two states.

2

Convert the photon energy from electron-volts (eV) to Joules (J)

E = 4.80 \times 1.6 \times 10^{-19}\text{ J} = 7.68 \times 10^{-19}\text{ J}

SI units require energy to be expressed in Joules, where

1\text{ eV} = 1.6 \times 10^{-19}\text{ J}

.

Key Concept

Energy level transitions and photon emission

Question 7833Question →

A straight line $L_1$ has the equation $3x - 4y + 5 = 0$ . A second line $L_2$ is parallel to $L_1$ and passes through the point $(6, 1)$ . What is the perpendicular distance between lines $L_1$ and $L_2$ ?

Show answer & explanation

Answer: 3.8

Answer

The perpendicular distance between lines

L_1

and

L_2

is

3.8

units.

The perpendicular distance between parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

. Line

L_2

is parallel to

3x - 4y + 5 = 0

, so its equation is

3x - 4y + C = 0

. Substituting

(6, 1)

gives

3(6) - 4(1) + C = 0

, leading to

C = -14

. Substituting

C_1 = 5

and

C_2 = -14

into the distance formula gives

d = \frac{|5 - (-14)|}{\sqrt{3^2 + (-4)^2}} = \frac{19}{5} = 3.8

.

Step-by-Step Solution

1

Determine the equation of line

L_2

The equation of

L_2

is

3x - 4y - 14 = 0

Lines parallel to

3x - 4y + 5 = 0

have the form

3x - 4y + C = 0

. Substituting the point

(6, 1)

gives

3(6) - 4(1) + C = 0 \implies C = -14

.

2

Apply the parallel line distance formula

d = \frac{|5 - (-14)|}{\sqrt{3^2 + (-4)^2}} = \frac{19}{5}

The perpendicular distance between parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is given by

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

.

3

Convert fraction to decimal form

3.8

Dividing

19

by

5

yields

3.8

.

Key Concept

Perpendicular Distance Between Parallel Lines

Question 7834Question →

Given the matrices $A = \begin{pmatrix} 3 & y \\ 2 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ x & 2 \end{pmatrix}$ , if $AB = \begin{pmatrix} 9 & 16 \\ 5 & 10 \end{pmatrix}$ , what is the value of $x + y$ ?

5

6

7

9

Show answer & explanation

Answer: $5$

Answer

The value of

x + y

is

5

.

Multiplying matrix

A

by matrix

B

using standard row-by-column multiplication yields

AB = \begin{pmatrix} 3 + xy & 12 + 2y \\ 2 + x & 10 \end{pmatrix}

. Setting this equal to

\begin{pmatrix} 9 & 16 \\ 5 & 10 \end{pmatrix}

gives

2 + x = 5 \implies x = 3

and

12 + 2y = 16 \implies y = 2

. Thus,

x + y = 3 + 2 = 5

.

Step-by-Step Solution

1

Compute the product matrix

AB

using matrix multiplication rules.

AB = \begin{pmatrix} 3(1) + y(x) & 3(4) + y(2) \\ 2(1) + 1(x) & 2(4) + 1(2) \end{pmatrix} = \begin{pmatrix} 3 + xy & 12 + 2y \\ 2 + x & 10 \end{pmatrix}

Matrix multiplication requires taking the dot product of rows from the first matrix and columns from the second matrix.

2

Equate the elements of

AB

with the given matrix

\begin{pmatrix} 9 & 16 \\ 5 & 10 \end{pmatrix}

to solve for

x

and

y

.

From row 2, column 1:

2 + x = 5 \implies x = 3

. From row 1, column 2:

12 + 2y = 16 \implies 2y = 4 \implies y = 2

.

Two matrices are equal if and only if their corresponding elements are equal.

3

Calculate the sum

x + y

.

x + y = 3 + 2 = 5

Substituting the values found for

x

and

y

into the requested expression.

Key Concept

Matrix Multiplication and Equality of Matrices

Alternative Method

Verification can be done by checking row 1, column 1:

3 + xy = 3 + (3)(2) = 9

, which matches the given matrix entry.

Estimated Time:1m 15s

Question 7835Question →

A convex polygon with $n$ sides has a total interior angle sum of $1440^\circ$ . If $(n - 4)$ of its interior angles each measure $150^\circ$ , and the remaining four interior angles are equal in measure, what is the measure of one of the remaining interior angles in degrees?

Show answer & explanation

Answer: 135

Answer

The measure of one of the remaining interior angles is

135^\circ

.

Using the interior angle sum formula

(n - 2) \times 180^\circ = 1440^\circ

, we find

n - 2 = 8

, which means the polygon has

n = 10

sides. The number of angles measuring

150^\circ

is

10 - 4 = 6

. Their total measure is

6 \times 150^\circ = 900^\circ

. The sum of the remaining four equal angles is

1440^\circ - 900^\circ = 540^\circ

. Dividing

540^\circ

by 4 gives

135^\circ

for each remaining interior angle.

Step-by-Step Solution

1

Calculate the total number of sides

n

of the convex polygon.

n = 10

The sum of interior angles of an

n

-sided polygon is given by

(n - 2) \times 180^\circ

. Setting

(n - 2) \times 180^\circ = 1440^\circ

gives

n - 2 = 8

, so

n = 10

.

2

Determine the number of interior angles that measure

150^\circ

each and find their combined sum.

6 angles totaling

900^\circ

There are

(n - 4) = 10 - 4 = 6

interior angles of

150^\circ

each. Their sum is

6 \times 150^\circ = 900^\circ

.

3

Calculate the sum of the remaining four equal interior angles.

540^\circ

Subtracting the sum of the known angles from the total interior angle sum yields

1440^\circ - 900^\circ = 540^\circ

.

4

Find the measure of one of the remaining four equal angles.

135^\circ

Dividing the remaining sum equally among the 4 angles gives

540^\circ / 4 = 135^\circ

.

Key Concept

Polygon interior angle sum theorem

Estimated Time:1m 30s

Question 7836Question →

A metallic rod of initial length $2.0\text{ m}$ experiences a temperature increase of $50\text{ K}$ . If the linear expansivity of the metal is $1.5 \times 10^{-5}\text{ K}^{-1}$ , what is the expansion in length of the rod in millimetres?

Show answer & explanation

Answer: 1.5

Answer

The expansion in length of the rod is 1.5 mm.

The expansion in length is calculated using

\Delta L = L_0 \alpha \Delta T

. Substituting the values

L_0 = 2.0\text{ m}

,

\alpha = 1.5 \times 10^{-5}\text{ K}^{-1}

, and

\Delta T = 50\text{ K}

gives

\Delta L = 1.5 \times 10^{-3}\text{ m}

, which equals

1.5\text{ mm}

.

Step-by-Step Solution

1

Identify known variables from the problem statement.

L_0 = 2.0\text{ m}

,

\Delta T = 50\text{ K}

, and

\alpha = 1.5 \times 10^{-5}\text{ K}^{-1}

.

Listing given physical quantities clarifies which thermal expansion formula to apply.

2

Calculate the change in length in metres using

\Delta L = L_0 \alpha \Delta T

.

\Delta L = 2.0\text{ m} \times (1.5 \times 10^{-5}\text{ K}^{-1}) \times 50\text{ K} = 0.0015\text{ m}

.

Thermal expansion in one dimension is directly proportional to initial length, linear expansivity, and temperature change.

3

Convert the calculated expansion from metres to millimetres.

0.0015\text{ m} \times 1000\text{ mm/m} = 1.5\text{ mm}

.

The question explicitly requests the value in millimetres.

Key Concept

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

Question 7837Question →

If $y = \ln(2 + \sin(3x)) + e^{4x}$ , find the value of $\frac{dy}{dx}$ at $x = 0$ .

Show answer & explanation

Answer: 5.5

Answer

The value of

\frac{dy}{dx}

at

x = 0

is

5.5

.

Differentiating

\ln(2 + \sin(3x))

by the chain rule gives

\frac{3\cos(3x)}{2 + \sin(3x)}

, and differentiating

e^{4x}

gives

4e^{4x}

. Evaluating

\frac{3\cos(3x)}{2 + \sin(3x)} + 4e^{4x}

at

x = 0

yields

\frac{3(1)}{2 + 0} + 4(1) = 1.5 + 4 = 5.5

.

Step-by-Step Solution

1

Differentiate the logarithmic component

\ln(2 + \sin(3x))

using the chain rule

\frac{d}{dx}[\ln(2 + \sin(3x))] = \frac{3\cos(3x)}{2 + \sin(3x)}

By the chain rule,

\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx}

, where

u = 2 + \sin(3x)

and

\frac{du}{dx} = 3\cos(3x)

.

2

Differentiate the exponential component

e^{4x}

\frac{d}{dx}[e^{4x}] = 4e^{4x}

The standard rule for exponential differentiation states that

\frac{d}{dx}[e^{kx}] = k e^{kx}

.

3

Combine the results to state the derivative function

\frac{dy}{dx}

\frac{dy}{dx} = \frac{3\cos(3x)}{2 + \sin(3x)} + 4e^{4x}

The derivative of a sum of functions is the sum of their individual derivatives.

4

Evaluate

\frac{dy}{dx}

at

x = 0

\frac{dy}{dx}\Big|_{x=0} = \frac{3\cos(0)}{2 + \sin(0)} + 4e^0 = \frac{3(1)}{2 + 0} + 4(1) = 1.5 + 4 = 5.5

Substituting

x = 0

uses the values

\cos(0) = 1

,

\sin(0) = 0

, and

e^0 = 1

.

Key Concept

Differentiation of Logarithmic, Trigonometric, and Exponential Functions using the Chain Rule

Question 7838Question →

In a cathode-ray experiment, electrons are accelerated from rest through a potential difference of $100\text{ V}$ . Taking the specific charge ( $\frac{e}{m}$ ) of an electron to be $1.80 \times 10^{11}\text{ C/kg}$ , calculate the final speed of the electrons as they pass through the anode aperture, expressed in units of $10^6\text{ m/s}$ .

Show answer & explanation

Answer: 6

Answer

The final speed of the electrons is

6.0 \times 10^6\text{ m/s}

, which corresponds to a coefficient value of 6.0.

By applying conservation of energy, the kinetic energy acquired by electrons in a cathode-ray tube equals the electric work done on them (

eV = \frac{1}{2}mv^2

). Solving for velocity gives

v = \sqrt{2V(e/m)}

. Substituting

V = 100\text{ V}

and

e/m = 1.80 \times 10^{11}\text{ C/kg}

gives

v = 6.0 \times 10^6\text{ m/s}

.

Step-by-Step Solution

1

Set up the energy conservation relation for cathode ray electrons.

Electrical work done

W = eV

equals kinetic energy

K = \frac{1}{2}mv^2

.

Electrons starting from rest gain kinetic energy equal to the electrical potential energy lost across the potential difference.

2

Rearrange the formula to isolate the electron velocity

v

.

v^2 = 2V\left(\frac{e}{m}\right) \implies v = \sqrt{2V\left(\frac{e}{m}\right)}

.

Isolating velocity allows direct calculation using the given specific charge

(\frac{e}{m})

and accelerating voltage

V

.

3

Substitute given values into the equation and compute

v

.

v = \sqrt{2 \times 100 \times 1.80 \times 10^{11}} = \sqrt{36 \times 10^{12}} = 6.0 \times 10^6\text{ m/s}

.

Evaluating the square root yields the speed in meters per second.

Key Concept

Electron acceleration in cathode ray tubes and specific charge relation

Question 7839Question →

A galvanometer with an internal resistance of $40\text{ }\Omega$ gives a full-scale deflection when a current of $10\text{ mA}$ passes through it. What resistance must be connected in series with the galvanometer to convert it into a voltmeter capable of measuring potential differences up to $10\text{ V}$ ?

960\text{ }\Omega

1000\text{ }\Omega

1040\text{ }\Omega

60\text{ }\Omega

Show answer & explanation

Answer: $960\text{ }\Omega$

Answer

The required multiplier resistance is

960\text{ }\Omega

.

To convert a galvanometer into a voltmeter, a multiplier resistor

R_m

is connected in series. The total resistance of the voltmeter combination is

R_{\text{total}} = R_g + R_m = \frac{V}{I_g} = \frac{10\text{ V}}{0.010\text{ A}} = 1000\text{ }\Omega

. Subtracting the galvanometer's internal resistance (

40\text{ }\Omega

) yields

R_m = 960\text{ }\Omega

.

Step-by-Step Solution

1

Convert the full-scale deflection current to amperes.

I_g = 10\text{ mA} = 10 \times 10^{-3}\text{ A} = 0.010\text{ A}

.

Standard SI units must be used for electrical calculations.

2

Apply the voltmeter multiplier conversion formula.

V = I_g(R_g + R_m) \implies R_m = \frac{V}{I_g} - R_g

.

The multiplier resistor

R_m

is connected in series with the galvanometer resistance

R_g

.

3

Substitute the known values into the equation.

R_m = \frac{10}{0.010} - 40 = 1000 - 40 = 960\text{ }\Omega

.

Subtracting internal resistance gives the external resistance needed for full-scale voltage rating.

Key Concept

Voltmeter Conversion using a Series Multiplier Resistor

Question 7840Question →

A metallic conductor wire of cross-sectional area $2.5 \times 10^{-6}\,\text{m}^2$ has a resistance of $10.0\,\Omega$ at $0\,^\circ\text{C}$ . The temperature coefficient of resistance of the material is $5.0 \times 10^{-3}\,^\circ\text{C}^{-1}$ . The conductor contains a free-electron density of $5.0 \times 10^{28}\,\text{m}^{-3}$ . When the wire is heated to $100\,^\circ\text{C}$ and connected across a potential difference of $60\,\text{V}$ , what is the drift velocity of the conduction electrons in the wire in millimeters per second ( $\text{mm/s}$ )? (Take elementary charge $e = 1.6 \times 10^{-19}\,\text{C}$ .)

Show answer & explanation

Answer: 0.2

Answer

The drift velocity of the conduction electrons is

0.2\,\text{mm/s}

.

The resistance increases from

10.0\,\Omega

to

15.0\,\Omega

when heated from

0\,^\circ\text{C}

to

100\,^\circ\text{C}

. Applying

60\,\text{V}

results in a current of

4.0\,\text{A}

. Combining this with the cross-sectional area and electron density gives a drift velocity of

2.0 \times 10^{-4}\,\text{m/s}

, which equals

0.2\,\text{mm/s}

.

Step-by-Step Solution

1

Calculate the resistance at the operating temperature (

100\,^\circ\text{C}

)

R_{100} = 15.0\,\Omega

Resistance varies with temperature according to

R_T = R_0(1 + \alpha \Delta T)

.

2

Find the current in the wire using Ohm's Law

I = 4.0\,\text{A}

Current is given by

I = V / R_{100}

.

3

Determine current density

J

J = 1.6 \times 10^6\,\text{A/m}^2

Current density is total current per unit cross-sectional area,

J = I / A

.

4

Calculate the electron drift velocity

v_d

v_d = 2.0 \times 10^{-4}\,\text{m/s} = 0.2\,\text{mm/s}

Drift velocity relates to current density by

v_d = J / (n e)

.

Key Concept

Temperature Dependence of Resistance and Microscopic Model of Electric Current

Estimated Time:2m 0s

Answer

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