All practice questions

Question 201Question →

In a sports academy of $85$ athletes, $52$ participate in track events, $43$ participate in field events, and $12$ participate in neither track nor field events. How many athletes participate in both track and field events?

Show answer & explanation

Answer: 22

Answer

The number of athletes participating in both track and field events is 22.

To find the number of athletes in both events, first calculate the total number of athletes who take part in at least one event by subtracting the 12 non-participants from 85, giving 73. Adding the 52 track athletes and 43 field athletes yields 95, which double-counts those who participate in both. The difference between 95 and 73 is 22, representing the athletes in the intersection.

Step-by-Step Solution

1

Determine the cardinality of the union of track and field athletes

N(T \cup F) = 85 - 12 = 73

Athletes participating in neither event are outside the union of track and field sets.

2

Formulate the two-set inclusion-exclusion equation

N(T \cup F) = N(T) + N(F) - N(T \cap F)

Adding individual set cardinalities double-counts the intersection.

3

Substitute values and solve for the intersection

N(T \cap F) = 52 + 43 - 73 = 22

Rearranging the equation yields the number of athletes in both sets.

Key Concept

Cardinality of Sets and Principle of Inclusion-Exclusion

Question 202Question →

A constant net force of $20\text{ N}$ acts on an object of mass $4\text{ kg}$ that is initially at rest on a frictionless horizontal surface. What is the magnitude of the linear momentum of the object after $3\text{ s}$ ?

Show answer & explanation

Answer: 60

Answer

The magnitude of the linear momentum of the object after

3\text{ s}

is

60\text{ kg m s}^{-1}

.

According to Newton's Second Law in terms of momentum, force is the rate of change of linear momentum (

F = \frac{\Delta p}{\Delta t}

). Rearranging gives

\Delta p = F \Delta t

. Since the object is initially at rest (

p_i = 0\text{ kg m s}^{-1}

), the final momentum is

p_f = F \Delta t = 20\text{ N} \times 3\text{ s} = 60\text{ kg m s}^{-1}

.

Step-by-Step Solution

1

Apply the impulse-momentum theorem.

J = F \Delta t = \Delta p

The impulse of the net force acting on an object equals the change in its linear momentum.

2

Calculate the final linear momentum.

p_f = 20\text{ N} \times 3\text{ s} = 60\text{ kg m s}^{-1}

Since the object starts from rest, its initial momentum is zero (

p_i = 0

), so the final momentum is equal to the total impulse supplied.

Key Concept

Impulse-Momentum Theorem

Question 203Question →

A school committee of $4$ members is to be selected from $6$ male teachers and $4$ female teachers. If the committee must contain exactly $2$ male teachers and $2$ female teachers, how many different committees can be formed?

Show answer & explanation

Answer: 90

Answer

The total number of different committees that can be formed is

90

.

Selecting 2 female teachers from 4 yields

\binom{4}{2} = 6

ways. Selecting 2 male teachers from 6 yields

\binom{6}{2} = 15

ways. By the multiplication principle, the total number of ways to form the committee is

6 \times 15 = 90

.

Step-by-Step Solution

1

Find the number of ways to select 2 female teachers out of 4

$\binom{4}{2} = 6$

Selection order does not matter within the committee, so combination formula $nCr$ applies.

2

Find the number of ways to select 2 male teachers out of 6

$\binom{6}{2} = 15$

Selection order does not matter within the committee, so combination formula $nCr$ applies.

3

Multiply the possibilities for selecting male and female members

6 \times 15 = 90

According to the fundamental counting principle, independent group selections are multiplied.

Key Concept

Combinations with restricted subset selections (Product Rule of Counting)

Question 204Question →

A sine function is given by the equation $y = 5\sin\left(\frac{2}{3}x\right) - 2$ . What is the period of this trigonometric function in degrees?

Show answer & explanation

Answer: 540

Answer

The period of the trigonometric function is

540^\circ

.

For a general sine curve of the form

y = a\sin(bx + c) + d

, the period

T

in degrees is calculated as

T = \frac{360^\circ}{|b|}

. Given

y = 5\sin\left(\frac{2}{3}x\right) - 2

, the value of

b

is

\frac{2}{3}

. Dividing

360^\circ

by

\frac{2}{3}

gives

540^\circ

.

Step-by-Step Solution

1

Identify the coefficient

b

of

x

from the standard sine form

y = a\sin(bx + c) + d

.

b = \frac{2}{3}

The coefficient of

x

determines the angular frequency and affects the horizontal compression or stretch of the graph.

2

State the period formula in degrees for a sine function.

T = \frac{360^\circ}{|b|}

The standard sine function completes one full wavelength over

360^\circ

, so scaling the input by

b

changes the period to

\frac{360^\circ}{b}

.

3

Substitute

b = \frac{2}{3}

into the formula and evaluate.

T = \frac{360^\circ}{\frac{2}{3}} = 360^\circ \times \frac{3}{2} = 540^\circ

Dividing by a fraction is performed by multiplying by its reciprocal.

Key Concept

Period of Trigonometric Functions

Estimated Time:1m 30s

Question 205Question →

An entrepreneur bought a commercial printing machine for $\text{₦}300,000$ . If the machine depreciates in value at a compound rate of $15\%$ per annum, what is its value in Naira ( $\text{₦}$ ) at the end of $2$ years?

Show answer & explanation

Answer: 216750

Answer

The value of the machine at the end of 2 years is ₦216,750.

The value of the asset after 2 years is calculated using the reducing balance formula:

V_2 = V_0(1 - r)^2 = 300,000(0.85)^2 = \text{₦}216,750

.

Step-by-Step Solution

1

Identify given parameters

Initial principal value

V_0 = 300,000

, annual depreciation rate

r = 0.15

, duration

n = 2

years.

Establishing known variables simplifies formula substitution.

2

Apply compound depreciation formula

V_2 = V_0(1 - r)^2 = 300,000(1 - 0.15)^2

Asset values decrease multiplicatively per period under compound depreciation.

3

Compute the final depreciated value

V_2 = 300,000 \times (0.85)^2 = 300,000 \times 0.7225 = 216,750

Multiplying the initial amount by the combined depreciation factor yields the residual asset value.

Key Concept

Compound Depreciation

Question 206Question →

A spherical planet has a radius of $7.2 \times 10^{6} \text{ m}$ and an acceleration due to gravity of $10 \text{ m/s}^2$ at its surface. What is the escape velocity for an object launched from the surface of this planet, in $\text{km/s}$ ?

Show answer & explanation

Answer: 12

Answer

The escape velocity from the surface of the planet is 12 km/s.

The escape velocity

v_e

from the surface of a spherical planet of radius

R

with surface gravity

g

is given by

v_e = \sqrt{2gR}

. Substituting

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

and

R = 7.2 \times 10^6 \text{ m}

yields

v_e = \sqrt{2 \times 10 \times 7.2 \times 10^6} = \sqrt{1.44 \times 10^8} = 12000 \text{ m/s}

, which equals

12 \text{ km/s}

.

Step-by-Step Solution

1

Identify the relationship between surface gravity, planetary radius, and escape velocity

The escape velocity formula is

v_e = \sqrt{2gR}

.

Escape velocity is the minimum initial speed required for an object to overcome the gravitational pull of a celestial body.

2

Substitute the given numerical values into the formula

v_e = \sqrt{2 \times 10 \text{ m/s}^2 \times 7.2 \times 10^6 \text{ m}} = \sqrt{144 \times 10^6} \text{ m/s}

.

Plugging in the given values allows direct calculation of the velocity in standard SI units.

3

Simplify the square root and convert units to km/s

v_e = 12000 \text{ m/s} = 12 \text{ km/s}

.

Taking the square root of

144 \times 10^6

gives

12000 \text{ m/s}

, which corresponds to

12 \text{ km/s}

.

Key Concept

Escape Velocity from a Planet's Surface

Question 207Question →

A flexible balloon contains a sample of gas occupying a volume of $3.0\text{ m}^3$ at a temperature of $27^\circ\text{C}$ under constant pressure. If the gas is heated to $127^\circ\text{C}$ while keeping the pressure constant, what is the new volume of the gas in $\text{m}^3$ ?

Show answer & explanation

Answer: 4

Answer

The new volume of the gas is

4.0\text{ m}^3

.

According to Charles's Law (

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

), at constant pressure, volume is directly proportional to absolute temperature in Kelvin. Converting

27^\circ\text{C}

to

300\text{ K}

and

127^\circ\text{C}

to

400\text{ K}

yields a final volume

V_2 = 3.0 \times \frac{400}{300} = 4.0\text{ m}^3

.

Step-by-Step Solution

1

Convert the given initial and final temperatures from Celsius to Kelvin

T_1 = 27 + 273 = 300\text{ K}

and

T_2 = 127 + 273 = 400\text{ K}

Gas law calculations require absolute temperature in Kelvin.

2

Set up Charles's Law equation for constant pressure processes

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

At constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature.

3

Substitute the values into the equation and solve for the final volume

V_2

V_2 = 3.0 \times \frac{400\text{ K}}{300\text{ K}} = 4.0\text{ m}^3

Direct algebraic evaluation yields the final gas volume.

Key Concept

Charles's Law

Question 208Question →

A meteorological balloon filled with $0.50\text{ kg}$ of helium gas is launched at sea level, where the atmospheric pressure is $1.01 \times 10^5\text{ Pa}$ and the ambient temperature is $27^\circ\text{C}$ . The balloon ascends to a high altitude where the external ambient pressure decreases to $4.00 \times 10^4\text{ Pa}$ and the ambient temperature drops to $-23^\circ\text{C}$ . As the balloon expands, its elastic membrane exerts an additional pressure, causing the internal gas pressure to be $20\%$ higher than the surrounding ambient pressure. Assuming helium behaves as an ideal gas with a molar mass of $4.0\text{ g/mol}$ and the molar gas constant $R = 8.31\text{ J mol}^{-1}\text{K}^{-1}$ , calculate the final volume of helium gas inside the balloon at this altitude in $\text{m}^3$ .

Show answer & explanation

Answer: 5.41

Answer

The final volume of helium gas inside the balloon at altitude is

5.41\text{ m}^3

.

The ideal gas equation

PV = nRT

relates state variables. By determining

n = 125\text{ moles}

from mass and molar mass, absolute temperature

T_2 = 250\text{ K}

, and total internal pressure

P_2 = 4.80 \times 10^4\text{ Pa}

, the final volume

V_2

evaluates to

5.41\text{ m}^3

.

Step-by-Step Solution

1

Convert the mass of helium into moles

n = 125\text{ moles}

Mass

m = 0.50\text{ kg} = 500\text{ g}

divided by molar mass

M = 4.0\text{ g/mol}

gives

n = \frac{500}{4.0} = 125\text{ moles}

.

2

Convert final temperature to Kelvin

T_2 = 250\text{ K}

Gas laws strictly require thermodynamic temperature:

T_2 = -23 + 273 = 250\text{ K}

.

3

Calculate final internal pressure of the gas

P_2 = 4.80 \times 10^4\text{ Pa}

The gas pressure inside the balloon is

20\%

higher than external ambient pressure:

P_2 = 1.20 \times (4.00 \times 10^4) = 4.80 \times 10^4\text{ Pa}

.

4

Solve for the final volume using the ideal gas equation

V_2 = 5.41\text{ m}^3

Rearranging

P_2 V_2 = n R T_2

yields

V_2 = \frac{n R T_2}{P_2} = \frac{125 \times 8.31 \times 250}{4.80 \times 10^4} = 5.41015...\text{ m}^3 \approx 5.41\text{ m}^3

.

Key Concept

Ideal Gas Equation (

PV = nRT

)

Estimated Time:3m 0s

Question 209Question →

A water tank has a real depth of $16.0\text{ cm}$ . If the refractive index of water relative to air is $\frac{4}{3}$ , what is the apparent depth of the tank in centimetres when viewed normally from above?

Show answer & explanation

Answer: 12

Answer

The apparent depth of the water tank is

12.0\text{ cm}

.

When light passes from water to air, refraction causes the apparent depth to be smaller than the real depth by a factor equal to the refractive index

n

. Substituting a real depth of

16.0\text{ cm}

and

n = \frac{4}{3}

into

\text{Apparent Depth} = \frac{\text{Real Depth}}{n}

yields

12.0\text{ cm}

.

Step-by-Step Solution

1

Identify the relationship between refractive index, real depth, and apparent depth.

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

Refraction at a plane boundary causes an object immersed in a denser medium to appear closer to the surface when viewed from a rarer medium.

2

Rearrange the equation to isolate Apparent Depth.

\text{Apparent Depth} = \frac{\text{Real Depth}}{n}

The unknown quantity requested by the question is the apparent depth.

3

Substitute the known numerical values into the equation.

\text{Apparent Depth} = \frac{16.0}{\frac{4}{3}} = 16.0 \times \frac{3}{4} = 12.0\text{ cm}

Dividing

16.0

by

\frac{4}{3}

gives

12.0

.

Key Concept

Real and Apparent Depth in Light Refraction

Question 210Question →

A worker strikes one end of a long solid aluminum pipeline. A detector at the opposite end records two sound signals—one traveling through the aluminum pipeline and the other through the surrounding air—separated by a time interval of $2.8\text{ s}$ . If the speed of sound in air is $340\text{ m s}^{-1}$ and the speed of sound in aluminum is $5100\text{ m s}^{-1}$ , what is the length of the pipeline in meters?

Show answer & explanation

Answer: 1020

Answer

The length of the pipeline is

1020\text{ m}

.

Sound travels significantly faster through solids like aluminum (

5100\text{ m s}^{-1}

) than through gases like air (

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

). The time taken for sound to travel a distance

L

through air is

t_{\text{air}} = \frac{L}{340}

, while through aluminum it is

t_{\text{metal}} = \frac{L}{5100}

. Setting their difference equal to

2.8\text{ s}

gives

\frac{L}{340} - \frac{L}{5100} = 2.8

, which solves to

L = 1020\text{ m}

.

Step-by-Step Solution

1

Formulate transit time expressions for both media

t_{\text{air}} = \frac{L}{340}

and

t_{\text{metal}} = \frac{L}{5100}

Time taken by a wave to travel distance

L

at constant speed

v

is

t = \frac{L}{v}

.

2

Set up the time difference equation

t_{\text{air}} - t_{\text{metal}} = 2.8\text{ s}

The sound wave travels faster through aluminum than air, so the air pulse arrives later by

2.8\text{ s}

.

3

Solve the algebraic equation for distance

L

L = 1020\text{ m}

Combining terms yields

\frac{14L}{5100} = 2.8

, which simplifies to

L = \frac{2.8 \times 5100}{14} = 1020\text{ m}

.

Key Concept

Propagation speed of sound waves in different physical media

Question 211Question →

If $(x, y)$ satisfies the simultaneous equations $x - y = 1$ and $x^2 + y^2 = 25$ , what is the value of the product $xy$ ?

Show answer & explanation

Answer: 12

Answer

12

Expanding

(x - y)^2

yields

x^2 - 2xy + y^2

. Substituting the given values

x - y = 1

and

x^2 + y^2 = 25

into this identity gives

1 = 25 - 2xy

. Rearranging yields

2xy = 24

, which solves to

xy = 12

. Alternatively, solving by substitution gives solution pairs

(4, 3)

and

(-3, -4)

, both yielding a product of

12

.

Step-by-Step Solution

1

Apply the algebraic expansion identity

(x - y)^2 = x^2 + y^2 - 2xy

Connects the difference of terms, the sum of their squares, and their product.

2

Substitute the values given in the system of equations

1^2 = 25 - 2xy

Replaces

x - y

with 1 and

x^2 + y^2

with 25.

3

Isolate and calculate the product

xy

2xy = 24 \implies xy = 12

Simplifies

1 = 25 - 2xy

to find the exact numerical value of

xy

.

Key Concept

Simultaneous Linear and Quadratic Equations

Question 212Question →

A projectile is launched from ground level over flat terrain with a constant horizontal velocity component of $10\text{ m/s}$ . At a height of $40\text{ m}$ above the ground, the magnitude of its vertical velocity component is equal to its horizontal velocity component. Taking $g = 10\text{ m/s}^2$ , calculate the maximum height reached by the projectile in meters.

Show answer & explanation

Answer: 45

Answer

The maximum height reached by the projectile is

45\text{ m}

.

At the height of

40\text{ m}

, the vertical velocity is equal to the horizontal velocity of

10\text{ m/s}

. Applying the vertical kinematic relation

v_y^2 = u_y^2 - 2gh

yields

10^2 = u_y^2 - 2(10)(40)

, which gives

u_y^2 = 900\text{ m}^2/\text{s}^2

. The maximum height attained above ground level occurs when

v_y = 0

, calculated as

H_{\text{max}} = \frac{u_y^2}{2g} = \frac{900}{20} = 45\text{ m}

.

Step-by-Step Solution

1

Identify the vertical component of velocity at the given height

v_y = 10\text{ m/s}

at

h = 40\text{ m}

The problem states that at

h = 40\text{ m}

, the vertical velocity component equals the constant horizontal component

u_x = 10\text{ m/s}

.

2

Determine the initial vertical launch velocity component

u_y

u_y^2 = 900\text{ m}^2/\text{s}^2 \implies u_y = 30\text{ m/s}

Applying the vertical motion equation

v_y^2 = u_y^2 - 2gh

gives

10^2 = u_y^2 - 2(10)(40) \implies u_y^2 = 100 + 800 = 900

.

3

Calculate the maximum height

H_{\text{max}}

H_{\text{max}} = 45\text{ m}

At maximum height, the vertical velocity becomes zero. Using

H_{\text{max}} = \frac{u_y^2}{2g} = \frac{900}{20} = 45\text{ m}

.

Key Concept

Vertical Kinematics and Maximum Height of a Projectile

Question 213Question →

A motorist travels from Town A to Town B at a constant speed of $60\text{ km/h}$ for $2\text{ hours}$ , and then continues from Town B to Town C at a constant speed of $90\text{ km/h}$ for $3\text{ hours}$ . What is the average speed of the motorist for the entire journey in $\text{km/h}$ ?

Show answer & explanation

Answer: 78

Answer

The average speed for the entire journey is

78\text{ km/h}

.

The correct average speed is

78\text{ km/h}

, obtained by dividing the total distance of

390\text{ km}

(

120\text{ km} + 270\text{ km}

) by the total duration of

5\text{ hours}

(

2\text{ h} + 3\text{ h}

).

Step-by-Step Solution

1

Find the distance traveled during each segment of the journey

First segment distance =

120\text{ km}

, second segment distance =

270\text{ km}

Distance equals speed multiplied by time (

D = v \times t

).

2

Determine the total distance traveled and total time taken

Total distance =

390\text{ km}

, Total time =

5\text{ hours}

Average rate requires the sum of all distances and the sum of all durations.

3

Calculate the average speed

Average speed =

78\text{ km/h}

Average speed is defined as total distance divided by total time.

Key Concept

Average Rate and Speed

Question 214Question →

A compound microscope in normal adjustment consists of an objective lens with a focal length of $1.5\text{ cm}$ and an eyepiece with a focal length of $5.0\text{ cm}$ . An object is placed at a distance of $1.6\text{ cm}$ in front of the objective lens. Calculate the distance, in centimeters, between the objective lens and the eyepiece.

Show answer & explanation

Answer: 29

Answer

The distance between the objective lens and the eyepiece is 29.0 cm.

Applying the thin lens formula to the objective lens yields

\frac{1}{1.5} = \frac{1}{1.6} + \frac{1}{v_o}

, giving an image distance

v_o = 24.0\text{ cm}

. Under normal adjustment, the intermediate image falls on the focal point of the eyepiece, making

u_e = f_e = 5.0\text{ cm}

. The total separation between the two lenses is

L = v_o + u_e = 24.0 + 5.0 = 29.0\text{ cm}

.

Step-by-Step Solution

1

Apply the thin lens formula to the objective lens to find the intermediate image position

v_o

.

\frac{1}{1.5} = \frac{1}{1.6} + \frac{1}{v_o} \Rightarrow \frac{1}{v_o} = \frac{2}{3} - \frac{5}{8} = \frac{1}{24}\text{ cm}^{-1} \Rightarrow v_o = 24.0\text{ cm}

The objective lens forms a real, inverted, magnified image at distance

v_o

from the objective.

2

Identify the object distance for the eyepiece

u_e

under normal adjustment.

u_e = f_e = 5.0\text{ cm}

For normal adjustment of a optical instrument, the final image is formed at infinity, requiring the intermediate image to sit exactly at the principal focus of the eyepiece.

3

Sum the intermediate image distance and eyepiece object distance to obtain total lens separation

L

.

L = v_o + u_e = 24.0\text{ cm} + 5.0\text{ cm} = 29.0\text{ cm}

The separation of lenses in a compound microscope is the distance from the objective to the intermediate image plus the distance from the intermediate image to the eyepiece.

Key Concept

Compound microscope optics and lens separation in normal adjustment

Question 215Question →

In a mathematics examination, a student is required to answer $5$ questions out of $8$ available questions. If the first $2$ questions are compulsory, in how many different ways can the student select the questions to answer?

Show answer & explanation

Answer: 20

Answer

The student can select the questions in 20 different ways.

With 2 compulsory questions, the student only has to select 3 more questions from the remaining 6 questions. The number of ways to select 3 items from 6 without regard to order is given by ^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20.

Step-by-Step Solution

1

Deduct compulsory questions from both the required total and available total

The student must choose 3 additional questions from the remaining 6 questions.

Compulsory questions are fixed and provide only 1 selection choice.

2

Apply the combination formula ^6C_3 to calculate selection ways

^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20

The order in which the student chooses the examination questions does not alter the group of questions selected.

Key Concept

Combinations with restricted/compulsory choices

Question 216Question →

Evaluate the limit $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$ . What is the numerical value of this limit?

Show answer & explanation

Answer: 0.25

Answer

The numerical value of the limit is 0.25 (or

\frac{1}{4}

).

When direct substitution into

\frac{\sqrt{x + 4} - 2}{x}

yields the indeterminate form

\frac{0}{0}

, rationalizing the numerator by multiplying by its conjugate

\sqrt{x + 4} + 2

simplifies the expression to

\frac{1}{\sqrt{x + 4} + 2}

. Taking the limit as

x \to 0

gives

\frac{1}{4} = 0.25

.

Step-by-Step Solution

1

Check direct substitution

Indeterminate form

\frac{0}{0}

Directly evaluating at

x = 0

yields zero in both numerator and denominator.

2

Multiply by the conjugate of the numerator

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

Rationalizing the radical in the numerator allows cancellation of the term causing the zero denominator.

3

Cancel common factors and evaluate limit

\frac{1}{\sqrt{0 + 4} + 2} = 0.25

Canceling

x

removes the zero factor, permitting direct evaluation.

Key Concept

Limits of indeterminate algebraic expressions using surd rationalization

Question 217Question →

A body of mass $4\text{ kg}$ moves with a constant velocity of $5\text{ m s}^{-1}$ . Calculate the kinetic energy of the body in Joules.

Show answer & explanation

Answer: 50

Answer

The kinetic energy of the body is

50\text{ J}

.

The kinetic energy of an object in linear motion is calculated using

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

. Substituting

m = 4\text{ kg}

and

v = 5\text{ m s}^{-1}

gives

E_k = \frac{1}{2} \times 4 \times 5^2 = 50\text{ J}

.

Step-by-Step Solution

1

Identify the given physical quantities

Mass

m = 4\text{ kg}

and velocity

v = 5\text{ m s}^{-1}

Extract values provided in the problem statement.

2

Apply the formula for kinetic energy

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

Kinetic energy is defined as half the product of mass and the square of velocity.

3

Substitute values and solve

E_k = \frac{1}{2} \times 4 \times (5)^2 = 50\text{ J}

Perform arithmetic evaluation to get the final energy in Joules.

Key Concept

Kinetic Energy

Question 218Question →

A metal block of mass $2.5\text{ kg}$ absorbs $1200\text{ J}$ of thermal energy, causing its temperature to rise by $15\text{ K}$ . What is the heat capacity of the block in $\text{J K}^{-1}$ ?

Show answer & explanation

Answer: 80

Answer

The heat capacity of the block is

80\text{ J K}^{-1}

.

Heat capacity

C

represents the quantity of heat energy required to raise the temperature of the entire object by

1\text{ K}

. Using the relation

C = \frac{Q}{\Delta T}

, substituting

Q = 1200\text{ J}

and

\Delta T = 15\text{ K}

gives

C = \frac{1200}{15} = 80\text{ J K}^{-1}

.

Step-by-Step Solution

1

Identify the given quantities from the problem.

Heat energy

Q = 1200\text{ J}

, temperature change

\Delta T = 15\text{ K}

, and mass

m = 2.5\text{ kg}

.

Clear identification of parameters is necessary to choose the correct formula.

2

Apply the formula for heat capacity.

C = \frac{Q}{\Delta T}

Heat capacity

C

measures the heat required to change the temperature of the entire body by

1\text{ K}

, regardless of mass per unit quantity.

3

Substitute the values to calculate the heat capacity.

C = \frac{1200}{15} = 80\text{ J K}^{-1}

Dividing total thermal energy absorbed by the resulting temperature rise yields the heat capacity.

Key Concept

Heat Capacity (

C = \frac{Q}{\Delta T}

)

Estimated Time:45s

Question 219Question →

A metallic container with an initial volume of $400 \text{ cm}^3$ at $15^\circ\text{C}$ is filled completely with paraffin. Upon heating the container and its contents to $65^\circ\text{C}$ , a volume of $18 \text{ cm}^3$ of paraffin spills over. Given that the linear expansivity of the metal container is $2.0 \times 10^{-5} \text{ K}^{-1}$ , what is the real cubic expansivity of the paraffin, expressed in units of $10^{-4} \text{ K}^{-1}$ ?

Show answer & explanation

Answer: 9.6

Answer

The real cubic expansivity of the paraffin is

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

, which corresponds to a numerical value of

9.6

in units of

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

.

The real cubic expansivity of a liquid accounts for both the observed (apparent) expansion of the liquid and the expansion of the container holding it. By applying

\gamma_a = \frac{\Delta V_a}{V_0 \Delta T} = 9.0 \times 10^{-4} \text{ K}^{-1}

and

\gamma_v = 3\alpha = 0.6 \times 10^{-4} \text{ K}^{-1}

, we sum them to obtain the real cubic expansivity

\gamma_r = 9.6 \times 10^{-4} \text{ K}^{-1}

.

Step-by-Step Solution

1

Determine the temperature change (

\Delta T

) and the apparent change in volume (

\Delta V_a

).

\Delta T = 65^\circ\text{C} - 15^\circ\text{C} = 50 \text{ K}

and

\Delta V_a = 18 \text{ cm}^3

.

The overflow volume represents the apparent expansion of the liquid relative to the expanding container over the temperature rise.

2

Calculate the apparent cubic expansivity (

\gamma_a

) of the paraffin.

\gamma_a = \frac{\Delta V_a}{V_0 \Delta T} = \frac{18}{400 \times 50} = \frac{18}{20000} = 9.0 \times 10^{-4} \text{ K}^{-1}

.

Apparent expansivity relates the apparent volume expansion to the original volume and temperature increase.

3

Calculate the cubic expansivity of the metallic vessel (

\gamma_v

).

\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}

.

Cubic expansivity of an isotropic solid container is three times its linear expansivity.

4

Calculate the real cubic expansivity of the paraffin (

\gamma_r

).

\gamma_r = \gamma_a + \gamma_v = 9.0 \times 10^{-4} \text{ K}^{-1} + 0.6 \times 10^{-4} \text{ K}^{-1} = 9.6 \times 10^{-4} \text{ K}^{-1}

.

The real expansion of a liquid is the sum of its apparent expansion and the expansion of the containing vessel.

Key Concept

Real vs. Apparent Expansion of Liquids (

\gamma_r = \gamma_a + \gamma_v

where

\gamma_v = 3\alpha

)

Question 220Question →

A uniform wooden cube of edge length $0.20\text{ m}$ floats in water of density $1000\text{ kg/m}^3$ with $0.15\text{ m}$ of its vertical height submerged. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what minimum mass, in kilograms, must be placed on the top surface of the cube so that its upper face becomes just flush with the water surface?

Show answer & explanation

Answer: 2

Answer

The minimum mass required to submerge the cube completely flush with the water surface is

2.0\text{ kg}

.

By the Law of Flotation, a floating body displaces its own weight of fluid. Initially, the cube displaces a volume of

0.20\text{ m} \times 0.20\text{ m} \times 0.15\text{ m} = 0.006\text{ m}^3

of water, corresponding to an upthrust of

60\text{ N}

(or mass of

6.0\text{ kg}

). When completely submerged, the total volume displaced is

0.20^3 = 0.008\text{ m}^3

, providing a total upthrust of

80\text{ N}

(or mass equivalent of

8.0\text{ kg}

). The additional mass required on top is therefore the difference:

8.0\text{ kg} - 6.0\text{ kg} = 2.0\text{ kg}

.

Step-by-Step Solution

1

Calculate the cross-sectional area of the cube

A = (0.20\text{ m})^2 = 0.04\text{ m}^2

The base area is needed to find the volume of the block submerged and unsubmerged.

2

Calculate the volume of the cube above the water surface

V_{\text{above}} = 0.04\text{ m}^2 \times (0.20\text{ m} - 0.15\text{ m}) = 0.002\text{ m}^3

To push the cube level with the surface, the additional weight added on top must balance the extra upthrust created by submerging this remaining volume.

3

Calculate the additional mass required

m = \rho_{\text{water}} \times V_{\text{above}} = 1000\text{ kg/m}^3 \times 0.002\text{ m}^3 = 2.0\text{ kg}

By Archimedes' principle, the additional downward mass must equal the mass of the extra water displaced when fully submerged.

Key Concept

Archimedes' Principle and Law of Flotation

Answer

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