All practice questions

Evaluate the numerical expression $\frac{0.00054 \times 0.002}{0.00036}$ and state the final result as a decimal.

Show answer & explanation

Answer: 0.003

Answer

The correct value as a decimal is 0.003.

Converting all terms to scientific notation yields

\frac{(5.4 \times 10^{-4}) \times (2 \times 10^{-3})}{3.6 \times 10^{-4}} = \frac{10.8 \times 10^{-7}}{3.6 \times 10^{-4}} = 3 \times 10^{-3} = 0.003

.

Step-by-Step Solution

1

Convert each decimal in the expression to scientific notation (standard form)

0.00054 = 5.4 \times 10^{-4}

,

0.002 = 2 \times 10^{-3}

,

0.00036 = 3.6 \times 10^{-4}

Converting decimals with leading zeros to powers of 10 prevents errors in decimal point alignment during multiplication and division.

2

Simplify the numerator by multiplying coefficients and adding exponents

(5.4 \times 10^{-4}) \times (2 \times 10^{-3}) = 10.8 \times 10^{-7}

According to the laws of indices,

10^a \times 10^b = 10^{a+b}

, so

-4 + (-3) = -7

.

3

Divide the simplified numerator by the denominator

\frac{10.8 \times 10^{-7}}{3.6 \times 10^{-4}} = \left(\frac{10.8}{3.6}\right) \times 10^{-7 - (-4)} = 3.0 \times 10^{-3}

Dividing the coefficients gives

10.8 \div 3.6 = 3

, and subtracting the exponents gives

-7 - (-4) = -3

.

4

Express the result in standard decimal form

3.0 \times 10^{-3} = 0.003

Shifting the decimal point 3 positions to the left converts

10^{-3}

to standard decimal representation.

Key Concept

Simplifying Decimal Expressions using Standard Form and Laws of Indices

Question 7742Question →

A glass flask of volume $1000\text{ cm}^3$ is filled completely with mercury at a temperature of $10^\circ\text{C}$ . The linear expansivity of the glass is $9.0 \times 10^{-6}\text{ K}^{-1}$ and the real cubic expansivity of mercury is $1.8 \times 10^{-4}\text{ K}^{-1}$ . What volume of mercury (in $\text{cm}^3$ ) will overflow when the system is heated to $110^\circ\text{C}$ ?

Show answer & explanation

Answer: 15.3

Answer

The volume of mercury that overflows is

15.3\text{ cm}^3

.

The apparent expansion of the liquid equals its real expansion minus the expansion of the container. Since

\gamma_v = 3\alpha = 2.7 \times 10^{-5}\text{ K}^{-1} = 0.27 \times 10^{-4}\text{ K}^{-1}

, the apparent cubic expansivity is

\gamma_a = 1.8 \times 10^{-4} - 0.27 \times 10^{-4} = 1.53 \times 10^{-4}\text{ K}^{-1}

. Multiplying by initial volume (

1000\text{ cm}^3

) and temperature change (

100\text{ K}

) yields an overflow volume of

15.3\text{ cm}^3

.

Step-by-Step Solution

1

Calculate the volume expansivity of the glass vessel (

\gamma_v

)

\gamma_v = 3 \times 9.0 \times 10^{-6}\text{ K}^{-1} = 2.7 \times 10^{-5}\text{ K}^{-1} = 0.27 \times 10^{-4}\text{ K}^{-1}

The volumetric (cubic) expansivity of a solid container is three times its linear expansivity.

2

Determine the apparent cubic expansivity of mercury (

\gamma_a

)

\gamma_a = \gamma_r - \gamma_v = 1.8 \times 10^{-4} - 0.27 \times 10^{-4} = 1.53 \times 10^{-4}\text{ K}^{-1}

The apparent expansion of a liquid accounts for both the expansion of the liquid itself and the expansion of the containing vessel.

3

Calculate the overflow volume (apparent expansion

\Delta V_a

)

\Delta V_a = V_0 \times \gamma_a \times \Delta T = 1000 \times 1.53 \times 10^{-4} \times 100 = 15.3\text{ cm}^3

The volume of liquid that overflows corresponds directly to its apparent volume increase.

Key Concept

Real and Apparent Cubical Expansivity of Liquids

Estimated Time:1m 30s

Question 7743Question →

Given the matrices $A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -1 \\ 2 & k \end{pmatrix}$ , find the value of $k$ such that the determinant of the product matrix $AB$ is equal to 50.

Show answer & explanation

Answer: 3

Answer

The value of

k

is 3.

By applying the determinant product rule

\det(AB) = \det(A)\det(B)

, we find

\det(A) = (4)(3) - (2)(1) = 10

. Given

\det(AB) = 50

, it follows that

\det(B) = 50 / 10 = 5

. Since

\det(B) = (1)(k) - (-1)(2) = k + 2

, setting

k + 2 = 5

gives

k = 3

.

Step-by-Step Solution

1

Calculate the determinant of matrix A

\det(A) = (4 \times 3) - (2 \times 1) = 12 - 2 = 10

The determinant of a 2x2 matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix} is ad - bc.

2

Apply the product property of determinants

\det(B) = \frac{\det(AB)}{\det(A)} = \frac{50}{10} = 5

For any two square matrices of the same dimension, \det(AB) = \det(A) \cdot \det(B).

3

Express the determinant of matrix B in terms of k and solve

\det(B) = (1)(k) - (-1)(2) = k + 2 = 5 \implies k = 3

Equating the calculated determinant formula for B to its numerical value of 5.

Key Concept

Determinant Product Property and 2x2 Matrix Determinant

Estimated Time:1m 30s

Question 7744Question →

Match each type of wave listed on the left with its correct classification and propagation characteristic on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Sound wave in air matches Mechanical longitudinal wave requiring a material medium; Radio wave in vacuum matches Electromagnetic transverse wave capable of traveling without a medium; Water ripple on a lake surface matches Mechanical transverse wave propagating along a liquid surface.

Each wave is correctly paired based on whether it needs a physical medium to propagate (mechanical waves require a medium, electromagnetic waves do not) and whether the displacement is parallel (longitudinal) or perpendicular (transverse) to the direction of energy propagation.

Step-by-Step Solution

1

Identify the medium requirement and vibration direction for a sound wave in air.

Sound waves require a material medium (air) and vibrate parallel to the direction of wave movement, making them mechanical longitudinal waves.

Classification depends on whether a physical medium is needed and how particles oscillate relative to energy transport.

2

Identify the medium requirement and vibration direction for a radio wave in a vacuum.

Radio waves can travel through empty space without a material medium and consist of field oscillations perpendicular to propagation, making them electromagnetic transverse waves.

Electromagnetic waves propagate via mutually perpendicular electric and magnetic field oscillations and require no medium.

3

Identify the medium requirement and vibration direction for a water ripple.

Ripples require a material medium (water) and displace surface water up and down perpendicular to wave travel, making them mechanical transverse waves.

Surface water waves exhibit transverse displacement characteristics in a physical liquid medium.

Key Concept

Classification of waves based on medium requirement (mechanical vs. electromagnetic) and particle vibration direction relative to propagation (transverse vs. longitudinal).

Question 7745Question →

A wheel is divided into $10$ equal sectors numbered $1$ through $10$ . In an initial experiment, the wheel is spun $400$ times, and a prime number is recorded $180$ times. Additional spins are to be conducted, all of which result in non-prime numbers. How many additional spins must be performed so that the overall experimental relative frequency of landing on a prime number equals the theoretical probability of landing on a prime number?

50

100

200

450

Show answer & explanation

Answer: $50$

Answer

50 additional spins

The correct value is 50. The theoretical probability of landing on a prime number from 1 to 10 is 4/10 = 2/5, since the prime numbers are 2, 3, 5, and 7. With 180 prime outcomes in 400 spins, adding N non-prime spins yields an experimental relative frequency of 180 / (400 + N). Setting 180 / (400 + N) = 2/5 yields 800 + 2N = 900, giving N = 50.

Step-by-Step Solution

1

Determine the theoretical probability of landing on a prime number.

The prime numbers between

1

and

10

are

2, 3, 5,

and

7

. Thus, there are

4

favorable outcomes out of

10

, giving

P(\text{theoretical}) = \frac{4}{10} = \frac{2}{5}

.

Theoretical probability is the ratio of favorable outcomes to the total number of equally likely outcomes.

2

Formulate the expression for experimental relative frequency after

N

additional non-prime spins.

The number of prime occurrences remains

180

, while the total number of spins becomes

400 + N

. Thus,

P(\text{experimental}) = \frac{180}{400 + N}

.

Since all

N

additional spins produce non-prime numbers, the count of prime outcomes does not increase, but the total number of trials increases by

N

.

3

Equate the experimental relative frequency to the theoretical probability and solve for

N

.

\frac{180}{400 + N} = \frac{2}{5} \implies 2(400 + N) = 180 \times 5 \implies 800 + 2N = 900 \implies 2N = 100 \implies N = 50.

Setting the experimental relative frequency equal to the theoretical probability allows determining the exact number of additional non-prime spins needed.

Key Concept

Theoretical vs Experimental Probability

Estimated Time:2m 0s

Question 7746Question →

A point $P(x, y)$ moves in the Cartesian plane such that it is at all times equidistant from two parallel lines given by the equations $3x + 4y - 12 = 0$ and $3x + 4y + 4 = 0$ . The locus of $P$ intersects the straight line $x - 2y - 8 = 0$ at the point $(a, b)$ . What is the value of $a - b$ ?

Show answer & explanation

Answer: 6

Answer

The value of

a - b

is 6.

The locus of a point moving equidistant between two parallel lines is the parallel line lying midway between them. Combining the parallel line equations

3x + 4y - 12 = 0

and

3x + 4y + 4 = 0

yields the locus line

3x + 4y - 4 = 0

. Solving the system formed by this locus line and

x - 2y - 8 = 0

gives

x = 4

and

y = -2

. Therefore,

a = 4

and

b = -2

, so

a - b = 4 - (-2) = 6

.

Step-by-Step Solution

1

Find the equation of the locus of point P

Locus equation:

3x + 4y - 4 = 0

The locus of points equidistant from two parallel lines

ax + by + c_1 = 0

and

ax + by + c_2 = 0

is a parallel line midway between them, given by

ax + by + \frac{c_1 + c_2}{2} = 0

.

2

Solve the simultaneous equations to find the intersection point

(a, b)

a = 4

and

b = -2

Substitute

x = 2y + 8

into

3x + 4y - 4 = 0

to get

3(2y + 8) + 4y - 4 = 0

, which yields

10y = -20

, so

y = -2

and

x = 4

.

3

Calculate the difference

a - b

6

Subtract

b = -2

from

a = 4

to obtain

4 - (-2) = 6

.

Key Concept

Locus of points equidistant from two parallel lines and intersection of straight lines

Question 7747Question →

A railway line is laid using steel rails, each of length $15\text{ m}$ , at a temperature of $20^\circ\text{C}$ . What minimum gap, in millimetres ( $\text{mm}$ ), must be left between consecutive rails so that they just touch without buckling when heated to $60^\circ\text{C}$ ? [Linear expansivity of steel = $1.2 \times 10^{-5}\text{ K}^{-1}$ ]

Show answer & explanation

Answer: 7.2

Answer

The minimum gap required between consecutive rails is

7.2\text{ mm}

.

The expansion in length

\Delta L

is given by

\Delta L = L_0 \alpha \Delta T

. Substituting the initial length

L_0 = 15\text{ m}

, linear expansivity

\alpha = 1.2 \times 10^{-5}\text{ K}^{-1}

, and temperature change

\Delta T = 40\text{ K}

gives

\Delta L = 7.2 \times 10^{-3}\text{ m}

, which corresponds to

7.2\text{ mm}

.

Step-by-Step Solution

1

Determine the change in temperature

\Delta T = 60^\circ\text{C} - 20^\circ\text{C} = 40\text{ K}

Thermal expansion is driven by the temperature difference between the final and initial states.

2

Set up the linear expansion formula

\Delta L = L_0 \alpha \Delta T

The linear expansion of a solid bar depends on its initial length, the material's linear expansivity, and the temperature change.

3

Calculate the expansion in metres and convert to millimetres

\Delta L = 15 \times (1.2 \times 10^{-5}) \times 40 = 7.2 \times 10^{-3}\text{ m} = 7.2\text{ mm}

Multiplying the value in metres by

1000

yields the required measurement in millimetres.

Key Concept

Linear Expansivity and Thermal Expansion of Solids

Estimated Time:1m 30s

Question 7748Question →

A harmonic wave traveling through an initial elastic medium is represented by the wave equation $y = 0.04 \cos\left(50\pi t - \frac{\pi}{4} x\right)$ , where $x$ and $y$ are measured in meters and $t$ is in seconds. When this wave passes into a second medium, its propagation speed doubles. What is the wavelength of the wave in the second medium?

4.0\text{ m}

8.0\text{ m}

16.0\text{ m}

32.0\text{ m}

Show answer & explanation

Answer: $16.0\text{ m}$

Answer

The wavelength of the wave in the second medium is

16.0\text{ m}

.

The wave's angular frequency

\omega = 50\pi\text{ rad/s}

corresponds to a source frequency of

f = 25\text{ Hz}

, and its wave number

k = \frac{\pi}{4}\text{ rad/m}

corresponds to an initial wavelength

\lambda_1 = 8.0\text{ m}

. The wave speed in the first medium is

v_1 = f \lambda_1 = 200\text{ m/s}

. In the second medium, the wave speed doubles to

v_2 = 400\text{ m/s}

. Because frequency is determined by the source and remains invariant during refraction across media boundaries (

f_2 = f_1 = 25\text{ Hz}

), the new wavelength becomes

\lambda_2 = \frac{v_2}{f} = \frac{400\text{ m/s}}{25\text{ Hz}} = 16.0\text{ m}

.

Step-by-Step Solution

1

Extract angular frequency

\omega

and wave number

k

from the general wave equation

y = A \cos(\omega t - k x)

.

\omega = 50\pi\text{ rad/s}

and

k = \frac{\pi}{4}\text{ rad/m}

.

Matching coefficients in the standard wave equation provides the temporal and spatial frequencies of the wave.

2

Calculate the frequency

f

and wavelength

\lambda_1

in the first medium.

f = \frac{\omega}{2\pi} = \frac{50\pi}{2\pi} = 25\text{ Hz}

and

\lambda_1 = \frac{2\pi}{k} = \frac{2\pi}{\pi/4} = 8.0\text{ m}

.

Fundamental wave relationships connect angular frequency to frequency and wave number to wavelength.

3

Determine the wave speed

v_1

in the first medium and

v_2

in the second medium.

v_1 = f \lambda_1 = 25 \times 8.0 = 200\text{ m/s}

. Therefore,

v_2 = 2 \times v_1 = 400\text{ m/s}

.

The problem states that wave propagation speed doubles upon entering the second medium.

4

Calculate the wavelength

\lambda_2

in the second medium using constant frequency

f = 25\text{ Hz}

.

\lambda_2 = \frac{v_2}{f} = \frac{400}{25} = 16.0\text{ m}

.

When a wave crosses a boundary between two media, its frequency is determined solely by the source and remains constant.

Key Concept

Frequency invariance across media boundaries and the wave speed equation

v = f \lambda

Estimated Time:1m 30s

Question 7749Question →

The first, third, and seventh terms of a non-constant arithmetic progression (AP) form the first three consecutive terms of a geometric progression (GP). If the first term of the AP is $4$ , what is the sum of the first $4$ terms of the geometric progression?

60

28

124

92

Show answer & explanation

Answer: 60

Answer

The sum of the first

4

terms of the geometric progression is

60

.

The first three terms of the GP are

T_1 = 4

,

T_3 = 4 + 2d

, and

T_7 = 4 + 6d

. Equating

(4 + 2d)^2 = 4(4 + 6d)

yields

4d^2 - 8d = 0

, giving

d = 2

. The first four terms of the GP are

4, 8, 16, 32

, which sum to

4 + 8 + 16 + 32 = 60

.

Step-by-Step Solution

1

Express the terms of the arithmetic progression in terms of first term

a

and common difference

d

.

First term

T_1 = 4

, third term

T_3 = 4 + 2d

, seventh term

T_7 = 4 + 6d

.

The

n^{\text{th}}

term of an AP is defined as

T_n = a + (n-1)d

.

2

Set up the geometric progression condition

(T_3)^2 = T_1 \cdot T_7

to solve for

d

.

(4 + 2d)^2 = 4(4 + 6d) \implies 16 + 16d + 4d^2 = 16 + 24d \implies 4d^2 - 8d = 0 \implies d = 2

(since the AP is non-constant,

d \neq 0

).

Three terms

x, y, z

form a GP if and only if

y^2 = xz

.

3

Determine the terms and common ratio

r

of the GP.

First term

G_1 = 4

, second term

G_2 = 4 + 2(2) = 8

. Thus,

r = \frac{8}{4} = 2

.

The common ratio

r

is the quotient of consecutive terms of the GP.

4

Calculate the sum of the first

4

terms of the GP using

S_n = \frac{a(r^n - 1)}{r - 1}

.

S_4 = \frac{4(2^4 - 1)}{2 - 1} = \frac{4(16 - 1)}{1} = 60

.

Formula for the sum of the first

n

terms of a geometric progression.

Key Concept

Arithmetic and Geometric Progression Inter-relationships

Question 7750Question →

The centripetal acceleration $a$ of a particle moving in a circular path depends on its linear speed $v$ and the radius $r$ of the path according to the formula $a = k v^x r^y$ , where $k$ is a dimensionless constant. Using dimensional analysis, what is the numerical value of the product $x \cdot y$ ?

Show answer & explanation

Answer: -2

Answer

The numerical value of the product

x \cdot y

is

-2

.

By dimensional analysis, centripetal acceleration has dimensions

[a] = \text{L T}^{-2}

, velocity

[v] = \text{L T}^{-1}

, and radius

[r] = \text{L}

. Substituting into

a = k v^x r^y

gives

\text{L T}^{-2} = \text{L}^{x+y} \text{T}^{-x}

. Equating exponents of time yields

-x = -2 \implies x = 2

. Equating exponents of length gives

x + y = 1 \implies 2 + y = 1 \implies y = -1

. Consequently,

x \cdot y = 2 \cdot (-1) = -2

.

Step-by-Step Solution

1

Identify the base dimensions for each physical quantity.

[a] = \text{M}^0 \text{L}^1 \text{T}^{-2}

,

[v] = \text{M}^0 \text{L}^1 \text{T}^{-1}

, and

[r] = \text{M}^0 \text{L}^1 \text{T}^0

.

Dimensional analysis requires substituting fundamental dimensions of mass, length, and time.

2

Set up the dimensional homogeneity equation.

\text{L}^1 \text{T}^{-2} = (\text{L T}^{-1})^x \cdot (\text{L})^y = \text{L}^{x+y} \text{T}^{-x}.

Since

k

is dimensionless, the net dimensions on both sides of the equation must be identical.

3

Solve for exponents

x

and

y

by equating powers of corresponding base units.

For \text{T}:

-x = -2 \implies x = 2

. For \text{L}:

x + y = 1 \implies 2 + y = 1 \implies y = -1

.

Equating coefficients of identical base dimensions gives a system of linear equations.

4

Multiply the derived values of

x

and

y

.

x \cdot y = 2 \cdot (-1) = -2

.

The question asks specifically for the product of exponents

x

and

y

.

Key Concept

Dimensional Homogeneity and Exponent Analysis

Question 7751Question →

Two capacitors with capacitances of $3.0\text{ }\mu\text{F}$ and $6.0\text{ }\mu\text{F}$ are connected in parallel across a direct-current source. What is the equivalent capacitance of this combination?

9.0\text{ }\mu\text{F}

2.0\text{ }\mu\text{F}

4.5\text{ }\mu\text{F}

18.0\text{ }\mu\text{F}

Show answer & explanation

Answer: $9.0\text{ }\mu\text{F}$

Answer

The equivalent capacitance of the parallel combination is

9.0\text{ }\mu\text{F}

.

When capacitors are connected in parallel, each capacitor experiences the full potential difference of the voltage source, and the total charge stored is the sum of individual charges (

Q_{total} = Q_1 + Q_2

). Thus, the equivalent capacitance is the direct sum of the individual capacitances:

C_{eq} = 3.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 9.0\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Identify the combination rule for parallel capacitors.

The total capacitance is given by

C_{eq} = C_1 + C_2

.

Capacitors in parallel share the same potential difference, so total charge stored is the sum of individual charges.

2

Substitute the given values into the parallel capacitance formula.

C_{eq} = 3.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 9.0\text{ }\mu\text{F}

.

Direct addition yields the total equivalent capacitance.

Key Concept

Parallel Combination of Capacitors

Estimated Time:45s

Question 7752Question →

In a mathematics examination paper consisting of $8$ questions, a candidate is required to answer $5$ questions in total. If the candidate must answer the first $2$ questions, in how many ways can the candidate select the remaining questions?

Show answer & explanation

Answer: 20

Answer

The candidate can select the questions in 20 ways.

Because the first 2 questions are mandatory, the choice is reduced to picking 3 additional questions from the remaining 6 questions. Since selection order is irrelevant, the number of distinct ways is $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

Step-by-Step Solution

1

Determine the remaining number of questions needed

3 questions

Since 2 out of the required 5 questions are compulsory, the candidate must choose 3 more.

2

Determine the available pool of remaining questions

6 questions

Subtracting the 2 compulsory questions from the 8 total questions leaves 6 questions available.

3

Calculate the combinations using the nCr formula

20 ways

The order in which questions are selected does not matter, so we use combinations: $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

Key Concept

Combinations with restricted or fixed choices

Question 7753Question →

The $2^{\text{nd}}$ , $5^{\text{th}}$ , and $14^{\text{th}}$ terms of a non-constant arithmetic progression (AP) form the first three consecutive terms of a geometric progression (GP). If the sum of the first $6$ terms of the AP is $72$ , calculate the common difference of the AP.

Show answer & explanation

Answer: 4

Answer

The common difference of the arithmetic progression is

4

.

Equating the square of the middle GP term

(a+4d)^2

to the product of the outer terms

(a+d)(a+13d)

yields

3d^2 = 6ad

, which simplifies to

d = 2a

. Substituting

2a = d

into the AP sum formula

S_6 = 3(2a+5d) = 72

gives

3(6d) = 72 \implies 18d = 72

, so

d = 4

.

Step-by-Step Solution

1

Express the

2^{\text{nd}}

,

5^{\text{th}}

, and

14^{\text{th}}

terms of the AP algebraically

T_2 = a + d

,

T_5 = a + 4d

,

T_{14} = a + 13d

The

n^{\text{th}}

term of an AP is given by

T_n = a + (n-1)d

.

2

Apply the consecutive terms property of a GP

(a + 4d)^2 = (a + d)(a + 13d)

For three consecutive terms of a GP, the square of the middle term equals the product of the first and third terms.

3

Simplify the quadratic equation to find the relationship between

a

and

d

a^2 + 8ad + 16d^2 = a^2 + 14ad + 13d^2 \implies 3d^2 = 6ad \implies d = 2a

Since the AP is non-constant,

d \neq 0

, allowing division by

3d

.

4

Formulate the sum of the first

6

terms of the AP

S_6 = 3(2a + 5d) = 72 \implies 2a + 5d = 24

The sum of the first

n

terms of an AP is

S_n = \frac{n}{2}[2a + (n-1)d]

.

5

Substitute

2a = d

into the linear equation and solve for

d

d + 5d = 24 \implies 6d = 24 \implies d = 4

Replacing

2a

with

d

reduces the equation to a single variable.

Key Concept

Relating non-consecutive terms of an Arithmetic Progression to form a Geometric Progression

Question 7754Question →

Given the matrix $A = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}$ , if $A^2 - 7A + kI = \mathbf{0}$ , where $I$ is the $2 \times 2$ identity matrix and $\mathbf{0}$ is the $2 \times 2$ zero matrix, what is the value of $k$ ?

Show answer & explanation

Answer: 10

Answer

The value of

k

is 10.

By matrix multiplication and algebraic evaluation,

A^2 - 7A = -10I

. Substituting into

A^2 - 7A + kI = \mathbf{0}

yields

-10I + kI = \mathbf{0}

, which gives

k = 10

. Alternatively, by the Cayley-Hamilton Theorem, any

2 \times 2

matrix

A

satisfies

A^2 - \text{tr}(A)A + \det(A)I = \mathbf{0}

. Here

\text{tr}(A) = 3 + 4 = 7

and

\det(A) = (3)(4) - (2)(1) = 10

, directly giving

k = \det(A) = 10

.

Step-by-Step Solution

1

Calculate the matrix product

A^2

A^2 = \begin{pmatrix} (3\cdot 3 + 2\cdot 1) & (3\cdot 2 + 2\cdot 4) \\ (1\cdot 3 + 4\cdot 1) & (1\cdot 2 + 4\cdot 4) \end{pmatrix} = \begin{pmatrix} 11 & 14 \\ 7 & 18 \end{pmatrix}

Squaring matrix

A

involves multiplying rows of

A

by columns of

A

.

2

Perform scalar multiplication for

7A

7A = \begin{pmatrix} 21 & 14 \\ 7 & 28 \end{pmatrix}

Each entry of matrix

A

is multiplied by the scalar 7.

3

Subtract

7A

from

A^2

A^2 - 7A = \begin{pmatrix} 11 - 21 & 14 - 14 \\ 7 - 7 & 18 - 28 \end{pmatrix} = \begin{pmatrix} -10 & 0 \\ 0 & -10 \end{pmatrix} = -10I

Subtracting corresponding entries yields a scalar multiple of the identity matrix.

4

Solve for the unknown scalar

k

-10I + kI = \mathbf{0} \implies k = 10

Setting

(k - 10)I = \mathbf{0}

implies

k - 10 = 0

, so

k = 10

.

Key Concept

Matrix Polynomial Equations and Cayley-Hamilton Theorem

Question 7755Question →

Match each physical quantity on the left with its corresponding SI unit expressed in terms of fundamental (base) units on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Frequency matches

\text{s}^{-1}

, Electric charge matches

\text{A}\cdot\text{s}

, Mass density matches

\text{kg}\cdot\text{m}^{-3}

, and Acceleration matches

\text{m}\cdot\text{s}^{-2}

.

Frequency (

f = 1/T

) is expressed in reciprocal seconds (

\text{s}^{-1}

). Electric charge (

Q = I \cdot t

) is current times time, giving

\text{A}\cdot\text{s}

. Mass density (

\rho = m/V

) is mass per volume, giving

\text{kg}\cdot\text{m}^{-3}

. Acceleration (

a = \Delta v / \Delta t

) is rate of velocity change, giving

\text{m}\cdot\text{s}^{-2}

.

Step-by-Step Solution

1

Identify the defining formula for each physical quantity

Frequency

f = \frac{1}{T}

, Charge

Q = I \cdot t

, Density

\rho = \frac{m}{V}

, Acceleration

a = \frac{\Delta v}{\Delta t}

.

Relating derived quantities to their defining equations allows reduction into fundamental quantities.

2

Substitute the SI base units for mass (

\text{kg}

), length (

\text{m}

), time (

\text{s}

), and current (

\text{A}

)

Frequency:

\text{s}^{-1}

; Charge:

\text{A}\cdot\text{s}

; Density:

\frac{\text{kg}}{\text{m}^3} = \text{kg}\cdot\text{m}^{-3}

; Acceleration:

\frac{\text{m/s}}{\text{s}} = \text{m}\cdot\text{s}^{-2}

.

This expresses each derived unit strictly in terms of fundamental SI units.

3

Match each physical quantity to its calculated base unit representation

Frequency

\rightarrow \text{s}^{-1}

, Electric charge

\rightarrow \text{A}\cdot\text{s}

, Mass density

\rightarrow \text{kg}\cdot\text{m}^{-3}

, Acceleration

\rightarrow \text{m}\cdot\text{s}^{-2}

.

Completes the pairing verification.

Key Concept

Expressing derived physical quantities in terms of SI fundamental (base) units

Estimated Time:45s

Question 7756Question →

Two identical air-filled parallel-plate capacitors, $C_1$ and $C_2$ , each of capacitance $C$ , are connected in series across a direct-current voltage source of potential difference $V$ . While the circuit remains connected to the voltage source, a dielectric slab of relative permittivity $\varepsilon_r = 3$ is fully inserted into $C_1$ , completely filling the space between its plates. What is the ratio of the electrostatic energy stored in $C_1$ after inserting the dielectric to the energy stored in $C_1$ before the insertion?

3 : 4

1 : 3

3 : 1

27 : 4

Show answer & explanation

Answer: $3 : 4$

Answer

The ratio of the energy stored in the first capacitor after dielectric insertion to before insertion is

3 : 4

(or

0.75

).

Initially, the two identical capacitors divide the total source voltage

V

equally, giving

V_1 = V/2

and initial energy

U_{1,i} = \frac{1}{8} C V^2

. When the dielectric of relative permittivity

3

is inserted, the capacitance of the first capacitor becomes

3C

. In a series circuit connected to a constant voltage source, the total charge becomes

Q = C_{eq}V = \frac{3}{4}CV

, which reduces the voltage across the modified capacitor to

V_1' = \frac{Q}{3C} = \frac{V}{4}

. The new stored energy is

U_{1,f} = \frac{1}{2}(3C)(\frac{V}{4})^2 = \frac{3}{32} C V^2

. Dividing

U_{1,f}

by

U_{1,i}

yields

\frac{3/32}{1/8} = \frac{3}{4}

.

Step-by-Step Solution

1

Calculate initial capacitance and voltage across

C_1

Initial capacitance

C_{1,i} = C

. Since

C_1

and

C_2

are identical and in series, initial potential difference across

C_1

is

V_{1,i} = \frac{V}{2}

.

Equal capacitors in series divide total voltage equally.

2

Calculate initial electrostatic energy stored in

C_1

U_{1,i} = \frac{1}{2} C_{1,i} V_{1,i}^2 = \frac{1}{2} C \left(\frac{V}{2}\right)^2 = \frac{1}{8} C V^2

.

Formula for energy stored in a capacitor is

U = \frac{1}{2} C V^2

.

3

Determine final capacitance of

C_1

and new voltage division

New capacitance

C_{1,f} = \varepsilon_r C = 3C

. Total equivalent capacitance

C_{eq} = \frac{3C \cdot C}{3C + C} = \frac{3}{4} C

. Total charge supplied

Q = C_{eq} V = \frac{3}{4} C V

. Final voltage across

C_1

is

V_{1,f} = \frac{Q}{C_{1,f}} = \frac{\frac{3}{4} C V}{3C} = \frac{V}{4}

.

Dielectric increases capacitance by factor

\varepsilon_r

, altering equivalent capacitance and potential distribution in series.

4

Calculate final electrostatic energy in

C_1

and compute the ratio

U_{1,f} = \frac{1}{2} (3C) \left(\frac{V}{4}\right)^2 = \frac{3}{32} C V^2

. Ratio

\frac{U_{1,f}}{U_{1,i}} = \frac{\frac{3}{32} C V^2}{\frac{1}{8} C V^2} = \frac{3}{4}

.

Divide final stored energy by initial stored energy.

Key Concept

Series combination of capacitors with dielectric insertion under constant battery voltage

Question 7757Question →

In $\Delta KLM$ , side $k = 5\text{ cm}$ , side $l = 5\sqrt{3}\text{ cm}$ , and $\angle K = 30^\circ$ . If $\angle L$ is an obtuse angle, what is the measure of $\angle L$ ?

60^\circ

120^\circ

135^\circ

150^\circ

Show answer & explanation

Answer: $120^\circ$

Answer

The measure of angle

L

is

120^\circ

.

Applying the Sine Rule

\frac{k}{\sin K} = \frac{l}{\sin L}

gives

\sin L = \frac{5\sqrt{3} \cdot \sin 30^\circ}{5} = \frac{\sqrt{3}}{2}

. The inverse sine gives an acute angle of

60^\circ

. Because the problem specifies that angle

L

is obtuse, we find its supplementary angle in the second quadrant:

180^\circ - 60^\circ = 120^\circ

.

Step-by-Step Solution

1

Apply the Sine Rule relating sides

k, l

and angles

K, L

.

\frac{k}{\sin K} = \frac{l}{\sin L}

The Sine Rule connects the ratio of side lengths to the sines of their opposite angles.

2

Substitute the given values into the Sine Rule equation.

\frac{5}{\sin 30^\circ} = \frac{5\sqrt{3}}{\sin L} \implies \frac{5}{0.5} = \frac{5\sqrt{3}}{\sin L} \implies 10 = \frac{5\sqrt{3}}{\sin L}

Known values are

k = 5\text{ cm}

,

l = 5\sqrt{3}\text{ cm}

, and

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

.

3

Solve for

\sin L

.

\sin L = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}

Isolating

\sin L

yields the principal ratio.

4

Determine the obtuse angle solution for

L

.

L = 180^\circ - 60^\circ = 120^\circ

Since

\sin L = \frac{\sqrt{3}}{2}

, the acute solution is

60^\circ

, so the supplementary obtuse solution is

180^\circ - 60^\circ = 120^\circ

.

Key Concept

Ambiguous Case of the Sine Rule

Estimated Time:1m 15s

Question 7758Question →

Find the value of $r$ such that ${^{8}P_r} = 6720$ .

Show answer & explanation

Answer: 5

Answer

The value of

r

is 5.

Expanding

{^{8}P_r}

into consecutive decreasing factors starting from 8 yields

8 \times 7 \times 6 \times 5 \times 4 = 6720

. Counting the number of factors multiplied (8, 7, 6, 5, 4) gives exactly 5 factors, so

r = 5

.

Step-by-Step Solution

1

Write the formula for permutation

{^{8}P_r}

as a product of descending integers.

{^{8}P_r} = 8 \times 7 \times 6 \times \dots \times (8 - r + 1)

By definition,

{^{n}P_r}

represents the product of

r

consecutive factors starting from

n

and decreasing by 1.

2

Perform sequential multiplication starting from 8 until reaching 6720.

Product of 5 factors:

8 \times 7 \times 6 \times 5 \times 4 = 6720

Multiplying factors gives:

8 \times 7 = 56

;

56 \times 6 = 336

;

336 \times 5 = 1680

;

1680 \times 4 = 6720

.

3

Count the number of terms in the product to find

r

.

r = 5

Since 5 consecutive integers were multiplied together to obtain 6720, the subset size

r

is 5.

Key Concept

Permutations of

n

distinct items taken

r

at a time

Estimated Time:1m 0s

Question 7759Question →

Which of the following is the inverse of the matrix $M = \begin{pmatrix} 5 & 2 \\ 7 & 3 \end{pmatrix}$ ?

\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

\begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix}

\begin{pmatrix} -5 & 2 \\ 7 & -3 \end{pmatrix}

\begin{pmatrix} 5 & 7 \\ 2 & 3 \end{pmatrix}

Show answer & explanation

Answer: $\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}$

Answer

\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

For any non-singular 2x2 matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

, the inverse is

\frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

. Here,

ad - bc = (5)(3) - (2)(7) = 1

, so the inverse matrix is

\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

.

Step-by-Step Solution

1

Calculate the determinant of matrix M

\det(M) = (5)(3) - (2)(7) = 15 - 14 = 1

For a 2x2 matrix

M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

, the determinant is given by

\det(M) = ad - bc

.

2

Find the adjugate matrix of M

\text{adj}(M) = \begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

The adjugate of a 2x2 matrix is obtained by swapping the main diagonal elements (

a

and

d

) and negating the off-diagonal elements (

b

and

c

).

3

Apply the matrix inverse formula

M^{-1} = \frac{1}{\det(M)} \text{adj}(M) = \frac{1}{1}\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

Multiplying the adjugate matrix by the reciprocal of the determinant gives the exact inverse matrix.

Key Concept

Inverse of a 2x2 Matrix

Question 7760Question →

In a secondary school class of 45 students, 28 study Chemistry, 25 study Physics, and 6 study neither of the two subjects. How many students study both Chemistry and Physics?

Show answer & explanation

Answer: 14; 14 students

Answer

14 students study both Chemistry and Physics.

First, find the number of students taking at least one subject by subtracting those taking neither from the total: 45 - 6 = 39. Then, apply the formula n(C ∪ P) = n(C) + n(P) - n(C ∩ P). Substituting the values gives 39 = 28 + 25 - n(C ∩ P), which simplifies to n(C ∩ P) = 53 - 39 = 14.

Step-by-Step Solution

1

Calculate the number of students taking at least one of the two subjects.

n(Chemistry ∪ Physics) = 45 - 6 = 39

Students who study neither subject must be subtracted from the total class population to find the cardinality of the union.

2

Set up the two-set inclusion-exclusion formula.

n(Chemistry ∪ Physics) = n(Chemistry) + n(Physics) - n(Chemistry ∩ Physics)

The sum of individual set cardinalities overcounts elements present in both sets.

3

Substitute the known values into the equation and solve for the intersection.

39 = 28 + 25 - n(Chemistry ∩ Physics) ⇒ n(Chemistry ∩ Physics) = 53 - 39 = 14

Subtracting 39 from 53 gives the exact number of students taking both subjects.

Key Concept

Two-set inclusion-exclusion principle and complement of a set

Estimated Time:1m 30s

Answer

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