All practice questions

Question 4241Question →

Complete the statement below by identifying the correct consonant sound represented by the underlined letter combination in each word.

Fill in the blanks below

In the English word 'chef', the digraph 'ch' produces the consonant sound represented phonetically as , whereas in the word 'choir', the digraph 'ch' produces the sound represented as .

Show answer & explanation

Answer

In 'chef', the digraph 'ch' represents the consonant sound /ʃ/, while in 'choir', 'ch' represents the consonant sound /k/.

The digraph 'ch' is pronounced as the fricative /ʃ/ (like 'sh') in 'chef' and as the plosive /k/ (hard 'k' sound) in 'choir'.

Step-by-Step Solution

1

Analyze the pronunciation of the digraph 'ch' in the word 'chef'.

The letters 'ch' in 'chef' produce the voiceless postalveolar fricative sound, transcribed in IPA as /ʃ/.

Words borrowed into English from French (such as 'chef', 'machine', 'chiffon') preserve the /ʃ/ pronunciation for 'ch'.

2

Analyze the pronunciation of the digraph 'ch' in the word 'choir'.

The letters 'ch' in 'choir' produce the voiceless velar plosive sound, transcribed in IPA as /k/.

Words of Greek derivation (such as 'choir', 'school', 'chemist', 'monarch') typically realize the digraph 'ch' as the hard consonant sound /k/.

Key Concept

Orthographic variations of consonant sounds for the digraph 'ch'

Estimated Time:45s

Question 4242Question →

Match each physical quantity on the left with its correct physical definition and scalar or vector classification on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Work Done matches with Scalar quantity defined as the product of force and displacement in the direction of the force; Acceleration matches with Vector quantity defined as the rate of change of velocity with time; Electric Potential matches with Scalar quantity defined as the work done per unit positive charge in bringing it from infinity; Force matches with Vector quantity defined as the rate of change of linear momentum with time.

Work Done matches the scalar definition involving force and displacement. Acceleration matches the vector definition representing rate of change of velocity. Electric Potential matches the scalar definition of work per unit charge. Force matches the vector definition representing rate of change of momentum.

Step-by-Step Solution

1

Classify each physical quantity as a scalar (has magnitude only) or a vector (has both magnitude and direction).

Work Done and Electric Potential are scalar quantities. Acceleration and Force are vector quantities.

Scalars require only numerical value and unit, whereas vectors require direction to be fully specified.

2

Match each physical quantity with its precise physical definition.

Work Done corresponds to force multiplied by displacement in the line of action; Acceleration corresponds to velocity change per unit time; Electric Potential corresponds to work done per unit charge; Force corresponds to rate of change of momentum.

Each definition uniquely identifies the fundamental physical relationship for that quantity.

Key Concept

Classification of physical quantities into scalars and vectors based on their directional properties and definitions

Question 4243Question →

The gravitational force $F$ between two point masses $m_1$ and $m_2$ separated by a distance $r$ is expressed by the equation $F = \frac{G m_1 m_2}{r^2}$ , where $G$ represents the universal gravitational constant. What is the dimensional formula for $G$ ?

M^{-1} L^3 T^{-2}

M L^3 T^{-2}

M^{-1} L^2 T^{-2}

M^{-1} L^3 T^{-1}

Show answer & explanation

Answer: $M^{-1} L^3 T^{-2}$

Answer

The dimensional formula for the universal gravitational constant

G

is

M^{-1} L^3 T^{-2}

.

Isolating

G

gives

G = \frac{F r^2}{m_1 m_2}

. Substituting the fundamental dimensions for force (

M L T^{-2}

), distance (

L

), and mass (

M

) yields

\frac{(M L T^{-2})(L^2)}{M^2} = M^{-1} L^3 T^{-2}

.

Step-by-Step Solution

1

Make

G

the subject of the formula in Newton's law of gravitation.

G = \frac{F r^2}{m_1 m_2}

To derive the dimensions of

G

, isolate it in terms of force, distance, and mass.

2

Substitute fundamental dimensions for force, distance, and mass.

[G] = \frac{[F][r]^2}{[m_1][m_2]} = \frac{(M L T^{-2})(L^2)}{M \cdot M}

Force has dimensions

M L T^{-2}

, distance has dimension

L

, and mass has dimension

M

.

3

Simplify the powers of fundamental dimensions

M

,

L

, and

T

.

[G] = M^{1-2} L^{1+2} T^{-2} = M^{-1} L^3 T^{-2}

Applying exponent rules simplifies the combined base dimensions.

Key Concept

Dimensions of Physical Constants

Estimated Time:1m 15s

Question 4244Question →

Two point masses are separated by a distance $r$ and exert a gravitational force $F$ on each other. If the distance between them is doubled while keeping their masses constant, what is the new gravitational force between them?

\frac{F}{4}

\frac{F}{2}

2F

4F

Show answer & explanation

Answer: $\frac{F}{4}$

Answer

The new gravitational force between the two masses is

\frac{F}{4}

.

Gravitational force obeys an inverse-square law with respect to distance (

F \propto \frac{1}{r^2}

). When separation distance is multiplied by 2, the force decreases by a factor of

2^2 = 4

, yielding

\frac{F}{4}

.

Step-by-Step Solution

1

Write down Newton's Law of Universal Gravitation

F = \frac{G m_1 m_2}{r^2}

Establish the mathematical relationship governing gravitational force and separation distance.

2

Substitute the new distance

r' = 2r

into the gravitational force equation

F' = \frac{G m_1 m_2}{(2r)^2} = \frac{G m_1 m_2}{4r^2}

Evaluate how doubling the separation distance affects the magnitude of the force.

3

Express the new force

F'

in terms of the initial force

F

F' = \frac{1}{4}\left(\frac{G m_1 m_2}{r^2}\right) = \frac{F}{4}

Relate the calculated force directly to the original force

F

.

Key Concept

Inverse-Square Law of Gravitation

Estimated Time:45s

Question 4245Question →

In an experiment to determine the density of a solid sphere, the mass is measured as $(50.0 \pm 0.5)\text{ g}$ and the radius is measured as $(2.00 \pm 0.05)\text{ cm}$ . What is the percentage error in the calculated density of the sphere?

Show answer & explanation

Answer: 8.5

Answer

The percentage error in the calculated density of the sphere is

8.5\%

.

The density formula

\rho = \frac{3m}{4\pi r^3}

dictates that maximum relative error is given by

\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3\left(\frac{\Delta r}{r}\right)

. Evaluating the percentage errors gives

1.0\%

for mass and

2.5\%

for radius. Summing

1.0\% + 3(2.5\%)

yields

8.5\%

.

Step-by-Step Solution

1

Determine the functional dependence of density on measured quantities.

Density

\rho = \frac{m}{V} = \frac{3m}{4\pi r^3}

.

The volume of a sphere of radius

r

is

V = \frac{4}{3}\pi r^3

.

2

Formulate the maximum fractional error relationship.

\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3\left(\frac{\Delta r}{r}\right).

When combining uncertainties, fractional errors add, and exponents act as multiplying factors.

3

Calculate the percentage error in the mass measurement.

\frac{0.5}{50.0} \times 100\% = 1.0\%.

Percentage error in mass is the absolute uncertainty divided by the measured value multiplied by

100\%

.

4

Calculate the percentage error in the radius measurement.

\frac{0.05}{2.00} \times 100\% = 2.5\%.

Percentage error in radius is the absolute uncertainty divided by the measured value multiplied by

100\%

.

5

Calculate total percentage error in density.

Percentage error = 1.0\% + 3(2.5\%) = 8.5\%.

The radius contributes three times its relative error because volume depends on

r^3

.

Key Concept

Error propagation in derived quantities involving powers

Question 4246Question →

Xerophytic plants typically possess a thick waxy cuticle on their leaf surfaces as a morphological adaptation to minimize cuticular transpiration in arid environments.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because a thick waxy cuticle provides an impermeable physical barrier on the leaf epidermis, reducing non-stomatal water loss in plants adapted to dry habitats.

The statement is true because xerophytic plants possess a thick, waxy cuticle over their leaf surfaces specifically to cut down on cuticular water loss and endure prolonged periods of dry conditions.

Step-by-Step Solution

1

Identify the ecological group and environment described.

The organism is a xerophyte, which lives in arid or drought-prone environments.

Environmental conditions determine the survival pressures acting on the organism.

2

Analyze the function of the thick waxy cuticle.

Wax is hydrophobic and prevents water movement across the epidermal cell layer.

Structural features that limit evaporation conserve limited internal water reserves.

3

Conclude whether this represents an authentic morphological adaptation.

Reducing cuticular transpiration via a thick waxy cuticle is a confirmed morphological adaptation in xerophytic plants.

The factual claim in the statement is fully correct.

Key Concept

Morphological Adaptations of Xerophytes to Water Conservation

Question 4247Question →

What is the value of the logarithmic expression $\frac{1}{\log_3 12} + \frac{1}{\log_4 12}$ ?

1

\log_{12} 7

\frac{7}{12}

12

Show answer & explanation

Answer: $1$

Answer

1

Using the change of base relationship

\frac{1}{\log_a b} = \log_b a

, the expression simplifies to

\log_{12} 3 + \log_{12} 4

. By the product rule of logarithms, this equals

\log_{12}(3 \times 4) = \log_{12} 12 = 1

.

Step-by-Step Solution

1

Apply the change of base rule

\frac{1}{\log_a b} = \log_b a

to each term.

\frac{1}{\log_3 12} = \log_{12} 3 \quad \text{and} \quad \frac{1}{\log_4 12} = \log_{12} 4

Expressing both terms with a common base of

12

enables the use of logarithmic laws.

2

Apply the product law of logarithms

\log_b M + \log_b N = \log_b (M \times N)

.

\log_{12} 3 + \log_{12} 4 = \log_{12} (3 \times 4) = \log_{12} 12

The sum of logarithms with identical bases equals the logarithm of the product of their arguments.

3

Simplify

\log_{12} 12

.

1

The logarithm of any base to itself is always

1

(

\log_a a = 1

).

Key Concept

Change of Base Property and Logarithm Addition Law

Question 4248Question →

A solid right circular cone has a base radius of $7\text{ cm}$ and a height of $12\text{ cm}$ . Taking $\pi = \frac{22}{7}$ , what is the volume of the cone?

616\text{ cm}^3

1848\text{ cm}^3

924\text{ cm}^3

308\text{ cm}^3

Show answer & explanation

Answer: $616\text{ cm}^3$

Answer

616\text{ cm}^3

The correct answer is obtained by applying the standard cone volume formula

V = \frac{1}{3}\pi r^2 h

. Substituting

r = 7

,

h = 12

, and

\pi = \frac{22}{7}

yields

V = 616\text{ cm}^3

.

Step-by-Step Solution

1

Identify the formula for the volume of a solid cone.

V = \frac{1}{3}\pi r^2 h

The volume of a cone is one-third the volume of a cylinder with the same base radius and height.

2

Substitute the given values into the formula.

V = \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 12

Given

r = 7\text{ cm}

,

h = 12\text{ cm}

, and

\pi = \frac{22}{7}

.

3

Simplify the expression to find the volume.

V = \frac{1}{3} \times \frac{22}{7} \times 49 \times 12 = 22 \times 7 \times 4 = 616\text{ cm}^3

Simplifying

\frac{49}{7} = 7

and

\frac{12}{3} = 4

leaves

22 \times 7 \times 4

.

Key Concept

Volume of a Right Circular Cone

Estimated Time:45s

Question 4249Question →

The test scores of five students in a mathematics quiz are $5, 8, 11, 12,$ and $14$ . What is the variance of these test scores?

10

\sqrt{10}

50

2.8

Show answer & explanation

Answer: $10$

Answer

The variance of the test scores is

10

.

The mean of the scores is

10

. Subtracting

10

from each score gives deviations of

-5, -2, 1, 2,

and

4

. Squaring these gives

25, 4, 1, 4,

and

16

, which sum to

50

. Dividing this sum by the number of data points (

5

) yields a variance of

10

.

Step-by-Step Solution

1

Calculate the mean (

\bar{x}

) of the data set

\bar{x} = \frac{5 + 8 + 11 + 12 + 14}{5} = \frac{50}{5} = 10

The mean is required to find the deviation of each score from the central value.

2

Compute the squared deviations from the mean

(x_i - \bar{x})^2

(5-10)^2 = 25

,

(8-10)^2 = 4

,

(11-10)^2 = 1

,

(12-10)^2 = 4

,

(14-10)^2 = 16

Squaring deviations ensures all negative differences become positive.

3

Find the sum of all squared deviations

\sum (x_i - \bar{x})^2 = 25 + 4 + 1 + 4 + 16 = 50

Summing the squared deviations measures total variation around the mean.

4

Divide the total sum of squared deviations by the number of data values (

N = 5

)

\text{Variance } (\sigma^2) = \frac{50}{5} = 10

Variance is defined as the average of the squared deviations.

Key Concept

Variance of Ungrouped Data

Estimated Time:1m 30s

Question 4250Question →

Given the universal set $\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ , set $A = \{1, 2, 4, 5, 8\}$ , and set $B = \{2, 3, 5, 7, 9\}$ , what is $(A \cup B)'$ ?

\{6, 10\}

\{1, 4, 8\}

\{3, 6, 7, 9, 10\}

\{2, 5\}

Show answer & explanation

Answer: $\{6, 10\}$

Answer

The set

\{6, 10\}

Combining set

A

and set

B

gives

A \cup B = \{1, 2, 3, 4, 5, 7, 8, 9\}

. The complement

(A \cup B)'

contains the elements of the universal set

\mathcal{E}

that are not in this union, which are

6

and

10

, giving

\{6, 10\}

.

Step-by-Step Solution

1

Find the union of set

A

and set

B

(

A \cup B

)

A \cup B = \{1, 2, 3, 4, 5, 7, 8, 9\}

The union combines all distinct elements present in set

A

, set

B

, or both.

2

Determine the complement of

(A \cup B)

with respect to the universal set

\mathcal{E}

(A \cup B)' = \mathcal{E} \setminus (A \cup B) = \{6, 10\}

The complement consists of all elements in the universal set

\mathcal{E}

that are not present in

A \cup B

.

Key Concept

Set Union and Set Complement

Question 4251Question →

A simple pendulum suspended in a terrestrial laboratory has a period of oscillation $T$ when fitted with a bob of mass $m$ . If the bob is replaced by another bob of mass $4m$ and the entire setup is moved to a high-altitude station where the acceleration due to gravity is $\frac{g}{4}$ , what is the new period of oscillation of the pendulum?

2T

4T

8T

\frac{T}{2}

Show answer & explanation

Answer: $2T$

Answer

The new period of oscillation is

2T

.

The period of a simple pendulum is given by

T = 2\pi \sqrt{\frac{L}{g}}

. It depends strictly on the length of the string

L

and the local gravitational acceleration

g

, making it independent of the bob's mass

m

. Replacing mass

m

with

4m

does not alter the period. When gravity decreases to

g' = \frac{g}{4}

, the new period becomes

T' = 2\pi \sqrt{\frac{L}{g/4}} = 2 \left(2\pi \sqrt{\frac{L}{g}}\right) = 2T

.

Step-by-Step Solution

1

Identify the formula for the period of a simple pendulum

The period

T

of a simple pendulum is given by

T = 2\pi \sqrt{\frac{L}{g}}

, where

L

is the pendulum length and

g

is the acceleration due to gravity.

This formula defines how physical parameters affect the oscillatory period of a pendulum under simple harmonic motion.

2

Analyze the effect of changing the bob's mass

Changing the mass of the bob from

m

to

4m

has no effect on the period.

The equation for the period of a simple pendulum contains no mass term, demonstrating that mass does not influence the period of small-angle oscillations.

3

Calculate the new period

T'

under the altered gravitational field

g' = \frac{g}{4}

T' = 2\pi \sqrt{\frac{L}{g/4}} = 2\pi \sqrt{\frac{4L}{g}} = 2 \times \left(2\pi \sqrt{\frac{L}{g}}\right) = 2T

.

Reducing the gravitational acceleration to one-fourth increases the square-root term

\sqrt{\frac{1}{1/4}} = 2

, thereby doubling the period.

Key Concept

Mass Independence and Gravity Dependence of Simple Pendulum Period

Question 4252Question →

A straight line $L$ passes through the point $(4, -2)$ and is perpendicular to the line defined by the equation $3x - 2y + 8 = 0$ . What is the $y$ -intercept of line $L$ ?

\frac{2}{3}

-8

4

-\frac{14}{3}

Show answer & explanation

Answer: $\frac{2}{3}$

Answer

The

y

-intercept of line

L

is

\frac{2}{3}

.

The given line

3x - 2y + 8 = 0

has a gradient of

\frac{3}{2}

. A perpendicular line must have a gradient equal to the negative reciprocal, which is

-\frac{2}{3}

. Substituting the point

(4, -2)

into

y - y_1 = m(x - x_1)

gives

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

, which simplifies to

y = -\frac{2}{3}x + \frac{2}{3}

. Setting

x = 0

yields the

y

-intercept of

\frac{2}{3}

.

Step-by-Step Solution

1

Find the gradient of the given line.

Expressing

3x - 2y + 8 = 0

in slope-intercept form

y = mx + c

gives

2y = 3x + 8 \implies y = \frac{3}{2}x + 4

. The gradient

m_1 = \frac{3}{2}

.

The gradient of the given line is required to determine the slope of line

L

.

2

Determine the gradient of line

L

.

Since line

L

is perpendicular to the given line,

m_L = -\frac{1}{m_1} = -\frac{1}{3/2} = -\frac{2}{3}

.

Perpendicular lines have gradients whose product is

-1

(

m_1 \cdot m_L = -1

).

3

Find the equation of line

L

using point-slope form.

Using

(x_1, y_1) = (4, -2)

and

m_L = -\frac{2}{3}

:

y - y_1 = m_L(x - x_1)

y - (-2) = -\frac{2}{3}(x - 4)

y + 2 = -\frac{2}{3}x + \frac{8}{3}

y = -\frac{2}{3}x + \frac{8}{3} - 2

y = -\frac{2}{3}x + \frac{2}{3}

To find the

y

-intercept, we need the complete equation of line

L

.

4

Identify the

y

-intercept.

Comparing

y = -\frac{2}{3}x + \frac{2}{3}

to

y = mx + c

, the

y

-intercept

c = \frac{2}{3}

.

The constant term

c

in

y = mx + c

represents the

y

-intercept.

Key Concept

Perpendicular lines in coordinate geometry have gradients that are negative reciprocals of each other (

m_1 \cdot m_2 = -1

).

Question 4253Question →

What is the indefinite integral $\int (3x^2 + 4\cos x) \, dx$ ?

x^3 + 4\sin x + C

x^3 - 4\sin x + C

x^3 + 4\sin x

6x + 4\sin x + C

Show answer & explanation

Answer: $x^3 + 4\sin x + C$

Answer

x^3 + 4\sin x + C

Integrating

3x^2

gives

x^3

via the power rule, and integrating

4\cos x

gives

4\sin x

. Combining these terms along with the required constant of integration

C

results in

x^3 + 4\sin x + C

.

Step-by-Step Solution

1

Integrate the polynomial term

3x^2

using the power rule

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

.

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

The power rule increases the exponent by 1 and divides by the new exponent.

2

Integrate the trigonometric term

4\cos x

using the standard integral

\int \cos x \, dx = \sin x

.

\int 4\cos x \, dx = 4\sin x

The antiderivative of cosine is positive sine.

3

Combine the results and append the constant of integration

C

.

x^3 + 4\sin x + C

All indefinite integrals must include an arbitrary constant

C

.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Question 4254Question →

A point $P$ lies inside a circle of radius $13\text{ cm}$ at a distance of $5\text{ cm}$ from the center $O$ . A chord $AB$ passes through point $P$ such that the ratio of segment $AP$ to segment $PB$ is $1:4$ . Calculate the total length of the chord $AB$ in centimeters.

Show answer & explanation

Answer: 30

Answer

The total length of chord

AB

is

30\text{ cm}

.

Using the Power of a Point property for an interior point

P

, the product of the chord segments is

AP \cdot PB = R^2 - OP^2 = 13^2 - 5^2 = 144

. Given

AP : PB = 1 : 4

, we write

AP = x

and

PB = 4x

, leading to

4x^2 = 144 \implies x = 6\text{ cm}

. Summing the two segments gives

AB = 6 + 24 = 30\text{ cm}

.

Step-by-Step Solution

1

Calculate the constant product of chord segments passing through interior point

P

.

AP \cdot PB = R^2 - OP^2 = 13^2 - 5^2 = 169 - 25 = 144

.

By the intersecting chords theorem, the product of segments created by an interior point

P

on any chord equals

(R - d)(R + d) = R^2 - d^2

.

2

Set up an algebraic equation using the segment ratio

AP : PB = 1 : 4

.

Let

AP = x

and

PB = 4x

, giving

(x)(4x) = 144 \implies 4x^2 = 144

.

Expressing both chord segments in terms of a single variable

x

allows direct calculation of the segment lengths.

3

Solve for

x

to find the individual segment lengths.

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

. Therefore,

AP = 6\text{ cm}

and

PB = 24\text{ cm}

.

Taking the positive square root gives the scale factor

x

since physical distances must be positive.

4

Sum the segment lengths to find the total chord length.

AB = AP + PB = 6 + 24 = 30\text{ cm}

.

The entire chord length is the sum of its two divided parts.

Key Concept

Intersecting Chords Theorem and Power of an Interior Point

Question 4255Question →

A particle executes simple harmonic motion along a straight line with an amplitude of $0.10\text{ m}$ . At what displacement from the equilibrium position is the kinetic energy of the particle equal to three times its potential energy?

0.05\text{ m}

0.087\text{ m}

0.071\text{ m}

0.025\text{ m}

Show answer & explanation

Answer: $0.05\text{ m}$

Answer

The displacement from the equilibrium position is

0.05\text{ m}

.

In simple harmonic motion, potential energy is

U = \frac{1}{2}kx^2

and kinetic energy is

K = \frac{1}{2}k(A^2 - x^2)

. Equating

K = 3U

yields

A^2 - x^2 = 3x^2

, which simplifies to

4x^2 = A^2

or

x = \frac{A}{2}

. For an amplitude of

0.10\text{ m}

, the displacement is

0.05\text{ m}

.

Step-by-Step Solution

1

Write the expressions for kinetic energy

K

and potential energy

U

in simple harmonic motion.

U = \frac{1}{2} k x^2

and

K = \frac{1}{2} k (A^2 - x^2)

, where

A

is amplitude and

x

is displacement.

These equations express the energy distribution at any displacement

x

.

2

Set up the condition given in the problem,

K = 3U

.

\frac{1}{2} k (A^2 - x^2) = 3 \times \left(\frac{1}{2} k x^2\right) \implies A^2 - x^2 = 3x^2

.

Canceling common factor

\frac{1}{2} k

simplifies the relationship between amplitude and displacement.

3

Solve for displacement

x

in terms of amplitude

A

.

A^2 = 4x^2 \implies x = \frac{A}{2}

.

Taking the square root of both sides gives the position where kinetic energy is three times potential energy.

4

Substitute the given amplitude

A = 0.10\text{ m}

to find

x

.

x = \frac{0.10\text{ m}}{2} = 0.05\text{ m}

.

Carrying out the calculation yields the final numerical displacement.

Key Concept

Conservation of Energy in Simple Harmonic Motion

Estimated Time:1m 15s

Question 4256Question →

Given that the determinant of the $3 \times 3$ matrix $M = \begin{pmatrix} 3 & 1 & 0 \\ 2 & x & -1 \\ 0 & 4 & 2 \end{pmatrix}$ is equal to $20$ , calculate the value of $x$ .

Show answer & explanation

Answer: 2

Answer

The value of

x

is

2

.

Expanding the matrix

M

along its top row yields

\det(M) = 3(2x + 4) - 1(4) = 6x + 8

. Setting this expression equal to

20

gives

6x + 8 = 20

, which simplifies to

6x = 12

, yielding

x = 2

.

Step-by-Step Solution

1

Expand the 3x3 matrix along the first row

\det(M) = 3(2x - (-4)) - 1(4 - 0) + 0

Cofactor expansion along a row containing a zero simplifies the computation of a 3x3 determinant.

2

Simplify the algebraic expression for the determinant

\det(M) = 6x + 8

Distribute the coefficients and combine like constant terms.

3

Solve the linear equation for x

x = 2

Subtract 8 from 20 to get 12, then divide by 6.

Key Concept

Determinant of a 3x3 Matrix via Cofactor Expansion

Question 4257Question →

A body accelerates uniformly from rest at a rate of $4\text{ m/s}^2$ for $6\text{ s}$ . What is the distance covered by the body during this time interval?

72\text{ m}

144\text{ m}

24\text{ m}

12\text{ m}

Show answer & explanation

Answer: $72\text{ m}$

Answer

The distance covered by the body is

72\text{ m}

.

Applying the equation of motion

s = ut + \frac{1}{2}at^2

with

u = 0\text{ m/s}

,

a = 4\text{ m/s}^2

, and

t = 6\text{ s}

yields

s = 0 + \frac{1}{2}(4)(6^2) = 72\text{ m}

.

Step-by-Step Solution

1

Identify the given kinematic parameters

Initial velocity

u = 0\text{ m/s}

, acceleration

a = 4\text{ m/s}^2

, and time interval

t = 6\text{ s}

.

Since the body starts from rest, its initial velocity is zero.

2

Select and set up the equation of motion for displacement

s = ut + \frac{1}{2}at^2 = (0)(6) + \frac{1}{2}(4)(6)^2

This formula directly relates displacement to initial velocity, acceleration, and time under constant acceleration.

3

Calculate the total distance

s = \frac{1}{2} \times 4 \times 36 = 72\text{ m}

Squaring

6\text{ s}

yields

36\text{ s}^2

, and multiplying by

2\text{ m/s}^2

gives

72\text{ m}

.

Key Concept

Linear motion under uniform acceleration

Estimated Time:45s

Question 4258Question →

A body is released from rest from the top of a cliff of height $h$ . If it covers a distance equal to $\frac{7}{16}h$ in the final second of its motion before hitting the ground, what is the total height $h$ of the cliff? (Take acceleration due to gravity $g = 10\text{ m s}^{-2}$ )

45\text{ m}

80\text{ m}

100\text{ m}

160\text{ m}

Show answer & explanation

Answer: $80\text{ m}$

Answer

80\text{ m}

Using the equation of motion under constant acceleration, the total distance fallen from rest in time

t

is

h = \frac{1}{2}gt^2

. The distance fallen up to time

(t-1)

is

h_{t-1} = \frac{1}{2}g(t-1)^2

. The distance traveled during the final second is

\Delta h = h - h_{t-1} = \frac{1}{2}g(2t-1)

. Equating this to

\frac{7}{16}h

yields

\frac{1}{2}g(2t-1) = \frac{7}{16}\left(\frac{1}{2}gt^2\right)

, which simplifies to

7t^2 - 32t + 16 = 0

. Factoring gives

t = 4\text{ s}

(rejecting

t = \frac{4}{7}\text{ s}

since time must exceed

1\text{ s}

). Substituting

t = 4\text{ s}

into

h = \frac{1}{2}(10)(4)^2

gives

80\text{ m}

.

Step-by-Step Solution

1

Express total height

h

in terms of total fall time

t

.

h = \frac{1}{2} g t^2 = 5t^2

Since the body starts from rest (

u = 0\text{ m s}^{-1}

), displacement under uniform acceleration

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

is given by

h = \frac{1}{2}gt^2

.

2

Express the height fallen in the first

(t - 1)

seconds.

h' = \frac{1}{2} g (t - 1)^2 = 5(t - 1)^2

The distance covered up to one second before impact is the total distance fallen minus the distance covered in the final second.

3

Calculate the distance fallen in the final second and set up the equation.

\Delta h = h - h' = 5t^2 - 5(t - 1)^2 = 5(2t - 1)

. Given

\Delta h = \frac{7}{16}h

, we have

5(2t - 1) = \frac{7}{16}(5t^2)

.

The distance fallen during the last second is the difference between total height and height fallen up to

(t-1)

seconds.

4

Solve the quadratic equation for

t

.

7t^2 - 32t + 16 = 0 \implies (7t - 4)(t - 4) = 0 \implies t = 4\text{ s}

(since

t > 1\text{ s}

).

Simplifying

2t - 1 = \frac{7}{16}t^2

gives

7t^2 - 32t + 16 = 0

. The root

t = 4/7\text{ s}

is discarded as

t

must be greater than

1\text{ s}

.

5

Substitute

t = 4\text{ s}

back into the total height formula.

h = 5(4)^2 = 80\text{ m}

.

Calculating total height using

h = 5t^2

for

t = 4\text{ s}

gives

80\text{ m}

.

Key Concept

Free Fall under Gravity and Motion in the

n

-th Second

Question 4259Question →

What is the equation of the locus of a point $P(x, y)$ that moves in a plane such that its distance from the origin $(0,0)$ is always 5 units?

Show answer & explanation

Answer: x^2 + y^2 = 25; x^2+y^2=25; x²+y²=25

Answer

x^2 + y^2 = 25

By definition, the locus of a point moving at a fixed distance of 5 units from the origin

(0,0)

is a circle centered at

(0,0)

with radius 5. Substituting into the standard circle equation

x^2 + y^2 = r^2

yields

x^2 + y^2 = 5^2 = 25

.

Step-by-Step Solution

1

Apply the distance formula between a general point

P(x, y)

and the origin

(0,0)

.

d = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}

The distance between two points

(x_1, y_1)

and

(x_2, y_2)

is

\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

.

2

Equate the distance formula to the given fixed distance of 5 units and square both sides.

\sqrt{x^2 + y^2} = 5 \implies x^2 + y^2 = 25

Squaring both sides eliminates the square root to give the algebraic equation of the locus.

Key Concept

The locus of points at a constant distance

r

from a fixed point

(h, k)

forms a circle with equation

(x-h)^2 + (y-k)^2 = r^2

.

Estimated Time:45s

Question 4260Question →

A solid object of volume $0.002\text{ m}^3$ is completely immersed in water of density $1000\text{ kg/m}^3$ . What is the magnitude of the upthrust exerted on the object by the water? [Take $g = 10\text{ m/s}^2$ ]

Show answer & explanation

Answer: 20

Answer

The magnitude of the upthrust exerted on the object is

20\text{ N}

.

According to Archimedes' principle, any body completely or partially submerged in a fluid experiences an upward force (upthrust) equal to the weight of the fluid displaced. The weight of the displaced fluid is calculated using

U = V \cdot \rho \cdot g

. Substituting

V = 0.002\text{ m}^3

,

\rho = 1000\text{ kg/m}^3

, and

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

yields

U = 0.002 \times 1000 \times 10 = 20\text{ N}

.

Step-by-Step Solution

1

Identify the given quantities from the problem statement.

V = 0.002\text{ m}^3

,

\rho = 1000\text{ kg/m}^3

, and

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

.

These are the essential inputs required to calculate the weight of the displaced liquid.

2

Apply Archimedes' principle to find upthrust force.

U = V \rho g = 0.002 \times 1000 \times 10 = 20\text{ N}

.

Archimedes' principle states that the upthrust force equals the weight of the fluid displaced by the object.

Key Concept

Archimedes' Principle and Upthrust

Answer

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