Tüm alıştırma soruları

Soru 4321Soru →

Six members of a board of directors are to be seated around a circular conference table. If two specific members refuse to sit next to each other, how many different seating arrangements are possible?

72

48

120

240

Cevabı ve açıklamayı göster

Cevap: 72

Cevap

72

To find the number of circular seating arrangements where two specific members do not sit next to each other, use the complement rule. First, the total unrestricted circular arrangements of 6 members is

(6 - 1)! = 5! = 120

. Next, calculate the arrangements where the two members sit together by treating them as 1 unit (giving 5 units in total). The circular arrangements of these 5 units is

(5 - 1)! = 4! = 24

, and the 2 members can swap seats in

2! = 2

ways, yielding

24 \times 2 = 48

arrangements together. Subtracting this from the total yields

120 - 48 = 72

.

Adım Adım Çözüm

1

Calculate total circular arrangements without restrictions

Total arrangements =

(6 - 1)! = 5! = 120

The number of ways to arrange

n

distinct items in a circle is

(n - 1)!

.

2

Calculate arrangements where the two specific members sit together

Restricted arrangements =

(5 - 1)! \times 2! = 4! \times 2 = 24 \times 2 = 48

Treat the two members as a single block, giving 5 units to arrange around a circle in

(5-1)!

ways, and multiply by

2!

for internal ordering of the pair.

3

Subtract the together arrangements from total arrangements

Ways apart =

120 - 48 = 72

Complementary counting gives the number of arrangements where the two members do not sit next to each other.

Anahtar Kavram

Circular Permutations with Restrictions

Tahmini Süre:1m 30s

Soru 4322Soru →

A solid cylinder of length $10\text{ cm}$ floats vertically at the interface of two immiscible liquids: oil of density $800\text{ kg/m}^3$ and water of density $1000\text{ kg/m}^3$ . If $4\text{ cm}$ of its length is submerged in water while the remaining $6\text{ cm}$ is submerged in oil, what is the density of the cylinder?

880\text{ kg/m}^3

920\text{ kg/m}^3

900\text{ kg/m}^3

400\text{ kg/m}^3

Cevabı ve açıklamayı göster

Cevap: $880\text{ kg/m}^3$

Cevap

The density of the cylinder is

880\text{ kg/m}^3

.

According to the Law of Flotation, a floating body displaces its own weight of the fluids in which it floats. For a body submerged across two immiscible liquids, the total buoyant force is the sum of the upthrusts from both fluids:

U = (\rho_w V_w + \rho_o V_o)g

. Equating this to the total weight

W = \rho_c V_{total} g

yields

\rho_c (0.10\text{ m}) = 1000(0.04\text{ m}) + 800(0.06\text{ m}) = 88

, which gives

\rho_c = 880\text{ kg/m}^3

.

Adım Adım Çözüm

1

Set up the condition for flotation.

Weight of the cylinder = Total upthrust exerted by both liquids.

For a body floating in equilibrium, its total weight is balanced by the sum of buoyant forces from all surrounding fluids.

2

Express the weight and upthrusts in terms of density, cross-sectional area

A

, length

L

, and acceleration due to gravity

g

.

\rho_c A L g = (\rho_w A h_w + \rho_o A h_o) g

The weight of the cylinder is

\rho_c V_{total} g

and upthrust from each liquid is

\rho_{liquid} V_{submerged} g

.

3

Cancel common factors

A

and

g

from both sides.

\rho_c L = \rho_w h_w + \rho_o h_o

Since the cylinder has a uniform cross-sectional area, volume ratio simplifies to length ratio.

4

Substitute the given values (

L = 0.10\text{ m}

,

h_w = 0.04\text{ m}

,

h_o = 0.06\text{ m}

,

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

,

\rho_o = 800\text{ kg/m}^3

).

\rho_c (0.10) = 1000(0.04) + 800(0.06) = 40 + 48 = 88\text{ kg/m}^2

Evaluating the weighted contribution of buoyancy from each fluid.

5

Solve for the density of the cylinder

\rho_c

.

\rho_c = \frac{88}{0.10} = 880\text{ kg/m}^3

Dividing both sides by the total length of

0.10\text{ m}

gives the density.

Anahtar Kavram

Archimedes' Principle for Floating Bodies in Immiscible Liquids

Soru 4323Soru →

Match each acoustic or vibrating system operating under boundary conditions on the left with its corresponding fundamental or harmonic frequency relationship on the right (where $v$ is sound speed in air, $T$ is string tension, $\mu$ is linear mass density, $L$ is length, and $r$ is internal pipe radius).

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Pipe closed at one end matches

f = \frac{v}{4L}

; Pipe open at both ends matches

f = \frac{v}{2L}

; Stretched string in second harmonic matches

f = \frac{1}{L}\sqrt{\frac{T}{\mu}}

; Pipe closed at one end with end-correction matches

f = \frac{v}{4(L + 0.6r)}

.

Each system is correctly matched based on wave mechanics boundary conditions: closed pipes produce quarter-wave fundamental modes (

\lambda = 4L

), open pipes produce half-wave fundamental modes (

\lambda = 2L

), the second harmonic of a string doubles the fundamental frequency

f_1 = \frac{1}{2L}\sqrt{T/\mu}

to yield

f_2 = \frac{1}{L}\sqrt{T/\mu}

, and end-correction increases the effective length of a closed pipe to

L + 0.6r

.

Adım Adım Çözüm

1

Analyze boundary conditions for an ideal closed pipe

Displacement node at closed end, antinode at open end. Length

L = \frac{\lambda}{4} \Rightarrow \lambda = 4L

. Frequency

f = \frac{v}{\lambda} = \frac{v}{4L}

.

Determines the fundamental mode frequency formula for a closed pipe without end correction.

2

Analyze boundary conditions for an ideal open pipe

Displacement antinodes at both open ends. Length

L = \frac{\lambda}{2} \Rightarrow \lambda = 2L

. Frequency

f = \frac{v}{2L}

.

Determines the fundamental mode frequency formula for an open pipe.

3

Calculate the second harmonic frequency of a stretched string

For wave speed

c = \sqrt{\frac{T}{\mu}}

, fundamental

f_1 = \frac{c}{2L}

. Second harmonic is

f_2 = 2f_1 = 2\left(\frac{1}{2L}\sqrt{\frac{T}{\mu}}\right) = \frac{1}{L}\sqrt{\frac{T}{\mu}}

.

Determines the frequency of the first overtone / second harmonic for a vibrating string fixed at both ends.

4

Apply end correction to a closed pipe

End correction

e = 0.6r

adds to physical length

L

at the open top end, giving

L_{eff} = L + 0.6r

. Fundamental frequency is

f = \frac{v}{4L_{eff}} = \frac{v}{4(L + 0.6r)}

.

Accounts for the antinode extending slightly beyond the open end of a real tube.

Anahtar Kavram

Boundary conditions, standing waves, harmonics in strings and air columns, and end-correction in resonance pipes

Soru 4324Soru →

For the domain $0^\circ \le x \le 360^\circ$ , solve the trigonometric equation $2\sin x + \sqrt{3} = 0$ . Which of the following gives the complete set of values for $x$ ?

60^\circ

and

120^\circ

120^\circ

and

240^\circ

240^\circ

and

300^\circ

210^\circ

and

330^\circ

Cevabı ve açıklamayı göster

Cevap: $240^\circ$ and $300^\circ$

Cevap

x = 240^\circ

and

x = 300^\circ

Rearranging

2\sin x + \sqrt{3} = 0

gives

\sin x = -\frac{\sqrt{3}}{2}

. The reference angle for which sine equals

\frac{\sqrt{3}}{2}

is

60^\circ

. Since sine is negative in the third and fourth quadrants, the solutions are

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

and

360^\circ - 60^\circ = 300^\circ

.

Adım Adım Çözüm

1

Isolate the trigonometric function in the equation

\sin x = -\frac{\sqrt{3}}{2}

Subtract

\sqrt{3}

from both sides and divide by

2

.

2

Find the basic reference angle

Reference angle

\alpha = 60^\circ

\sin(60^\circ) = \frac{\sqrt{3}}{2}

.

3

Determine the relevant quadrants

Quadrants III and IV

The sine function is negative in the third and fourth quadrants.

4

Calculate the solutions within the interval

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

Quadrant III:

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

; Quadrant IV:

x = 360^\circ - 60^\circ = 300^\circ

Apply quadrant reduction formulas for Quadrant III (

180^\circ + \alpha

) and Quadrant IV (

360^\circ - \alpha

).

Anahtar Kavram

Solving simple trigonometric equations using reference angles and quadrant rules

Soru 4325Soru →

A stretched string of length $0.5\text{ m}$ fixed at both ends vibrates in its fundamental mode. If the speed of transverse waves along the string is $200\text{ m/s}$ , calculate the fundamental frequency of the string in hertz.

Cevabı ve açıklamayı göster

Cevap: 200

Cevap

The fundamental frequency of the vibrating string is

200\text{ Hz}

.

For a string fixed at both ends, the fundamental mode corresponds to a standing wave with half a wavelength spanning the length of the string (

L = \frac{\lambda}{2}

, or

\lambda = 2L

). Applying the wave relation

v = f\lambda

, the fundamental frequency is

f = \frac{v}{2L}

. Substituting

v = 200\text{ m/s}

and

L = 0.5\text{ m}

gives

f = \frac{200}{2(0.5)} = 200\text{ Hz}

.

Adım Adım Çözüm

1

Identify the relationship between frequency, wave speed, and string length for the fundamental mode.

For a string fixed at both ends, the wavelength of the fundamental harmonic is

\lambda = 2L

, giving the frequency formula

f = \frac{v}{2L}

.

The fundamental standing wave pattern contains nodes at both fixed ends and a single antinode at the center.

2

Substitute the given numerical values into the formula.

f = \frac{200\text{ m/s}}{2 \times 0.5\text{ m}} = \frac{200}{1} = 200\text{ Hz}

.

Dividing the wave speed by twice the length of the string yields the frequency in hertz.

Anahtar Kavram

Fundamental frequency of a vibrating string fixed at both ends

Soru 4326Soru →

A screw jack with a thread pitch of $0.5\text{ cm}$ is operated using a Tommy bar of length $35\text{ cm}$ . If the machine has an efficiency of $40\%$ , what is the minimum effort force required to raise a load of mass $880\text{ kg}$ ? (Take $g = 10\text{ m/s}^2$ and $\pi = \frac{22}{7}$ )

50\text{ N}

8\text{ N}

20\text{ N}

100\text{ N}

Cevabı ve açıklamayı göster

Cevap: $50\text{ N}$

Cevap

The minimum effort force required is

50\text{ N}

.

The effort distance per turn is the circumference

2\pi r = 2 \times \frac{22}{7} \times 35\text{ cm} = 220\text{ cm}

. Dividing by the pitch (

0.5\text{ cm}

) yields a Velocity Ratio of

440

. Applying the

40\%

efficiency gives a Mechanical Advantage of

\text{MA} = 0.40 \times 440 = 176

. Finally, dividing the load force (

8800\text{ N}

) by

\text{MA}

gives an effort force of

50\text{ N}

.

Adım Adım Çözüm

1

Calculate the total load force in Newtons

L = m \cdot g = 880\text{ kg} \times 10\text{ m/s}^2 = 8800\text{ N}

Effort overcomes weight force, which is mass multiplied by gravitational acceleration.

2

Calculate the Velocity Ratio (VR) of the screw jack

\text{VR} = \frac{2 \pi r}{p} = \frac{2 \times \frac{22}{7} \times 35\text{ cm}}{0.5\text{ cm}} = \frac{220\text{ cm}}{0.5\text{ cm}} = 440

The effort travels around the circumference of a circle of radius

r

, while the load moves vertically by one pitch length

p

per revolution.

3

Determine the Mechanical Advantage (MA) from efficiency

\text{MA} = \eta \times \text{VR} = 0.40 \times 440 = 176

Efficiency is defined as

\eta = \frac{\text{MA}}{\text{VR}}

, so

\text{MA} = \eta \cdot \text{VR}

.

4

Calculate the effort force E

E = \frac{L}{\text{MA}} = \frac{8800\text{ N}}{176} = 50\text{ N}

Mechanical advantage is the ratio of load to effort force

(\text{MA} = \frac{L}{E})

.

Anahtar Kavram

Relationship between Velocity Ratio, Mechanical Advantage, Efficiency, and Pitch in a Screw Jack

Soru 4327Soru →

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?

Cevabı ve açıklamayı göster

Cevap: 22

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Cardinality of Sets and Principle of Inclusion-Exclusion

Soru 4328Soru →

The polynomial $P(x) = 2x^3 + px^2 + qx - 6$ has $(x - 2)$ as a factor. When $P(x)$ is divided by $(x + 1)$ , the remainder is $-12$ . What is the value of $p + q$ ?

-2

-8

4

-4

Cevabı ve açıklamayı göster

Cevap: $-2$

Cevap

The value of

p + q

is

-2

.

Using the Factor Theorem,

P(2) = 0

yields

2p + q = -5

. Using the Remainder Theorem,

P(-1) = -12

yields

p - q = -4

. Solving these two equations simultaneously gives

p = -3

and

q = 1

. Therefore,

p + q = -3 + 1 = -2

.

Adım Adım Çözüm

1

Apply the Factor Theorem for divisor

(x - 2)

2p + q = -5

Since

(x - 2)

is a factor of

P(x)

,

P(2) = 0

. Substituting

x = 2

gives

2(2)^3 + p(2)^2 + q(2) - 6 = 0 \implies 16 + 4p + 2q - 6 = 0 \implies 4p + 2q = -10 \implies 2p + q = -5

.

2

Apply the Remainder Theorem for divisor

(x + 1)

p - q = -4

Dividing

P(x)

by

(x + 1)

leaves a remainder of

-12

, so

P(-1) = -12

. Substituting

x = -1

gives

2(-1)^3 + p(-1)^2 + q(-1) - 6 = -12 \implies -2 + p - q - 6 = -12 \implies p - q = -4

.

3

Solve the simultaneous linear equations for

p

and

q

p = -3

and

q = 1

Adding the two equations

(2p + q) + (p - q) = -5 + (-4)

yields

3p = -9 \implies p = -3

. Substituting

p = -3

into

p - q = -4

gives

-3 - q = -4 \implies q = 1

.

4

Calculate the required expression

p + q

p + q = -2

Summing the calculated constants:

p + q = -3 + 1 = -2

.

Anahtar Kavram

Factor and Remainder Theorems for Polynomials

Tahmini Süre:2m 0s

Soru 4329Soru →

Find the equation of the locus of a point $P(x, y)$ that moves such that it is equidistant from the fixed points $A(-1, 3)$ and $B(5, -1)$ .

Cevabı ve açıklamayı göster

Cevap: 3x - 2y - 4 = 0; 3x-2y-4=0; 3x - 2y = 4; 3x-2y=4; 12x - 8y - 16 = 0; y = (3/2)x - 2; y = 1.5x - 2

Cevap

The equation of the locus is

3x - 2y - 4 = 0

(or

3x - 2y = 4

).

The locus of a point equidistant from two fixed points

A(-1, 3)

and

B(5, -1)

is the perpendicular bisector of the line segment joining them. Equating the squared distances

(x+1)^2 + (y-3)^2 = (x-5)^2 + (y+1)^2

and simplifying yields the linear equation

3x - 2y - 4 = 0

.

Adım Adım Çözüm

1

Set up the distance equality condition using the distance formula.

\\sqrt{(x - (-1))^2 + (y - 3)^2} = \\sqrt{(x - 5)^2 + (y - (-1))^2}

Since point

P(x, y)

is equidistant from

A

and

B

,

PA = PB

.

2

Square both sides to remove the radical signs.

(x + 1)^2 + (y - 3)^2 = (x - 5)^2 + (y + 1)^2

Squaring both sides eliminates square roots and simplifies polynomial expansion.

3

Expand all squared terms on both sides.

x^2 + 2x + 1 + y^2 - 6y + 9 = x^2 - 10x + 25 + y^2 + 2y + 1

Expanding allows gathering like terms.

4

Cancel

x^2

and

y^2

from both sides and collect all terms on one side.

(2x + 10x) + (-6y - 2y) + (10 - 26) = 0 \\Rightarrow 12x - 8y - 16 = 0

Combining like terms simplifies the locus equation into standard linear form.

5

Divide the entire equation by the common factor of 4.

3x - 2y - 4 = 0

Expressing the linear equation in its simplest form gives the perpendicular bisector of line segment

AB

.

Anahtar Kavram

Locus equidistant from two fixed points (Perpendicular Bisector)

Tahmini Süre:2m 0s

Soru 4330Soru →

A rigid steel container holds a fixed mass of gas at an initial pressure of $1.20 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . If the temperature of the gas is increased to $127^\circ\text{C}$ while keeping its volume constant, what is the final pressure of the gas?

1.60 \times 10^5\text{ Pa}

5.64 \times 10^5\text{ Pa}

0.90 \times 10^5\text{ Pa}

4.44 \times 10^5\text{ Pa}

Cevabı ve açıklamayı göster

Cevap: $1.60 \times 10^5\text{ Pa}$

Cevap

The final pressure of the gas is

1.60 \times 10^5\text{ Pa}

.

According to Gay-Lussac's law, at constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature (

P_1/T_1 = P_2/T_2

). Converting the temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 400\text{ K}

. Solving for

P_2

yields

P_2 = 1.20 \times 10^5\text{ Pa} \times \frac{400}{300} = 1.60 \times 10^5\text{ Pa}

.

Adım Adım Çözüm

1

Convert temperatures from Celsius to Kelvin

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

,

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

Gas laws require absolute temperature in Kelvin to maintain proportional relationships.

2

Apply Pressure Law (Gay-Lussac's Law) for constant volume

\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_2 = P_1 \times \frac{T_2}{T_1}

For a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature.

3

Substitute the known values to calculate the final pressure

P_2 = (1.20 \times 10^5\text{ Pa}) \times \frac{400\text{ K}}{300\text{ K}} = 1.60 \times 10^5\text{ Pa}

Multiplying the initial pressure by the temperature expansion factor yields the final pressure.

Anahtar Kavram

Gay-Lussac's Law (Pressure Law)

Soru 4331Soru →

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 $3\text{ s}$ ?

Cevabı ve açıklamayı göster

Cevap: 60

Cevap

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}

.

Adım Adım Çözüm

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.

Anahtar Kavram

Impulse-Momentum Theorem

Soru 4332Soru →

A straight wire of length $0.50\text{ m}$ carrying a current of $4.0\text{ A}$ is placed perpendicular to a uniform magnetic field of flux density $0.20\text{ T}$ . What is the magnitude of the magnetic force acting on the wire?

0.40\text{ N}

0.80\text{ N}

1.00\text{ N}

0.00\text{ N}

Cevabı ve açıklamayı göster

Cevap: $0.40\text{ N}$

Cevap

The magnetic force acting on the wire is

0.40\text{ N}

.

Using the formula

F = BIL \sin\theta

, when a current-carrying wire is perpendicular to the magnetic field,

\theta = 90^\circ

and

\sin 90^\circ = 1

. Substituting

B = 0.20\text{ T}

,

I = 4.0\text{ A}

, and

L = 0.50\text{ m}

yields

F = 0.20 \times 4.0 \times 0.50 = 0.40\text{ N}

.

Adım Adım Çözüm

1

Identify the given values and formula

L = 0.50\text{ m}

,

I = 4.0\text{ A}

,

B = 0.20\text{ T}

,

\theta = 90^\circ

, and

F = B I L \sin\theta

.

The magnetic force on a current-carrying conductor in a uniform magnetic field depends on field strength, current, length, and angle of orientation.

2

Substitute the values into the formula and calculate

F = 0.20 \times 4.0 \times 0.50 \times \sin(90^\circ) = 0.40\text{ N}

.

Since

\sin 90^\circ = 1

, the force is maximum for a given magnetic field and current.

Anahtar Kavram

Magnetic force on a current-carrying conductor (

F = BIL \sin\theta

)

Tahmini Süre:45s

Soru 4333Soru →

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?

Cevabı ve açıklamayı göster

Cevap: 90

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Combinations with restricted subset selections (Product Rule of Counting)

Soru 4334Soru →

Producer gas is manufactured industrially by passing air over red-hot coke, yielding a mixture consisting approximately of $1$ mole of $\text{CO}$ to $2$ moles of $\text{N}_2$ . Water gas is produced by passing steam over incandescent coke, yielding an equimolar mixture of $\text{CO}$ and $\text{H}_2$ . If equal volumes of producer gas and water gas are allowed to effuse through identical porous barriers under the same conditions of temperature and pressure, which of the following statements correctly compares their initial rates of effusion? $(\text{Atomic masses: } \text{H} = 1, \text{C} = 12, \text{N} = 14, \text{O} = 16)$

Water gas effuses faster than producer gas because its average molar mass (

15\text{ g mol}^{-1}

) is less than that of producer gas (

28\text{ g mol}^{-1}

).

Producer gas effuses faster than water gas because the rate of effusion is directly proportional to the average molar mass of the gaseous mixture.

Both gas mixtures effuse at the exact same rate because carbon(II) oxide and nitrogen gas are gaseous allotropes of carbon with identical molecular masses.

Water gas effuses at twice the rate of producer gas because one mole of any gas occupies

22.4\text{ dm}^3

at STP while producer gas occupies

44.8\text{ dm}^3

per mole of carbon.

Cevabı ve açıklamayı göster

Cevap: Water gas effuses faster than producer gas because its average molar mass ( $15\text{ g mol}^{-1}$ ) is less than that of producer gas ( $28\text{ g mol}^{-1}$ ).

Cevap

Water gas effuses faster than producer gas because its average molar mass (

15\text{ g mol}^{-1}

) is less than that of producer gas (

28\text{ g mol}^{-1}

).

Water gas consists of an equimolar mixture of carbon(II) oxide and hydrogen gas, giving an average molar mass of

\frac{28 + 2}{2} = 15\text{ g mol}^{-1}

. Producer gas consists of carbon(II) oxide and nitrogen gas in a

1:2

molar ratio, giving an average molar mass of

\frac{28 + 2(28)}{3} = 28\text{ g mol}^{-1}

. According to Graham's Law of effusion, the rate of effusion is inversely proportional to the square root of the molar mass (

R \propto \frac{1}{\sqrt{M}}

). Because water gas has a significantly lower average molar mass (

15\text{ g mol}^{-1}

) than producer gas (

28\text{ g mol}^{-1}

), water gas effuses faster.

Adım Adım Çözüm

1

Calculate the average molar mass of producer gas

Molar mass of

\text{CO} = 12 + 16 = 28\text{ g mol}^{-1}

, Molar mass of

\text{N}_2 = 2 \times 14 = 28\text{ g mol}^{-1}

. For a

1:2

mole ratio of

\text{CO} : \text{N}_2

,

\bar{M}_{\text{producer}} = \frac{1(28) + 2(28)}{1 + 2} = 28\text{ g mol}^{-1}

.

To compare effusion rates, the average molar mass of the gas mixture must first be determined.

2

Calculate the average molar mass of water gas

Molar mass of

\text{CO} = 28\text{ g mol}^{-1}

, Molar mass of

\text{H}_2 = 2 \times 1 = 2\text{ g mol}^{-1}

. For an equimolar (

1:1

) mixture of

\text{CO} : \text{H}_2

,

\bar{M}_{\text{water gas}} = \frac{1(28) + 1(2)}{1 + 1} = 15\text{ g mol}^{-1}

.

The average molar mass of water gas is required to apply Graham's Law.

3

Apply Graham's Law of Effusion to compare rates

\frac{R_{\text{water gas}}}{R_{\text{producer gas}}} = \sqrt{\frac{\bar{M}_{\text{producer}}}{\bar{M}_{\text{water gas}}}} = \sqrt{\frac{28}{15}} \approx 1.37. Since

\bar{M}_{\text{water gas}} < \bar{M}_{\text{producer}}

, water gas effuses faster.

Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass (

R \propto \frac{1}{\sqrt{M}}

).

Anahtar Kavram

Composition of industrial fuel gases (producer gas vs water gas) and application of Graham's Law of effusion to gas mixtures

Soru 4335Soru →

A train moving at a constant speed of $15\text{ m/s}$ towards a tall vertical cliff emits a whistle signal. The driver hears the echo of the whistle $2.0\text{ s}$ after it was sounded. If the speed of sound in air is $340\text{ m/s}$ , what was the distance between the train and the cliff at the instant the whistle was sounded?

355\text{ m}

340\text{ m}

710\text{ m}

325\text{ m}

Cevabı ve açıklamayı göster

Cevap: $355\text{ m}$

Cevap

The initial distance between the train and the cliff when the whistle was sounded was

355\text{ m}

.

In

2.0\text{ s}

, the sound wave travels

680\text{ m}

while the train advances

30\text{ m}

towards the cliff. The total path of the sound consists of the initial distance

D

to the cliff plus the return distance

(D - 30)\text{ m}

to the moving train. Equating

D + (D - 30) = 680\text{ m}

gives

2D = 710\text{ m}

, which yields an initial distance of

355\text{ m}

.

Adım Adım Çözüm

1

Calculate the total distance traveled by the sound wave in the given time interval.

s_{\text{sound}} = v_{\text{sound}} \times t = 340\text{ m/s} \times 2.0\text{ s} = 680\text{ m}

.

Sound travels at a constant speed in air over the total elapsed time of

2.0\text{ s}

.

2

Calculate the distance covered by the moving train during the same time interval.

s_{\text{train}} = v_{\text{train}} \times t = 15\text{ m/s} \times 2.0\text{ s} = 30\text{ m}

.

The train continues moving towards the cliff while the sound wave travels to the cliff and reflects back.

3

Formulate the geometric path equation for the sound wave.

Let

D

be the initial distance to the cliff. Sound travels

D

to the cliff and reflects back a distance of

(D - 30)\text{ m}

to reach the train. Therefore,

D + (D - 30) = 680\text{ m}

.

The echo is received by the driver at a position

30\text{ m}

closer to the cliff than where the sound was emitted.

4

Solve the linear equation for

D

.

2D - 30 = 680 \implies 2D = 710 \implies D = 355\text{ m}

.

Solving for

D

gives the exact distance at the moment the whistle was sounded.

Anahtar Kavram

Echo path geometry with a moving sound source

Soru 4336Soru →

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?

Cevabı ve açıklamayı göster

Cevap: 540

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Period of Trigonometric Functions

Tahmini Süre:1m 30s

Soru 4337Soru →

An organ pipe open at both ends has a length of $0.60\text{ m}$ . The fundamental frequency of resonance for this pipe is observed to be equal to the frequency of the first overtone of a stretched wire of length $0.40\text{ m}$ fixed at both ends. If the linear mass density of the wire is $4.0 \times 10^{-4}\text{ kg/m}$ and the speed of sound in air is $330\text{ m/s}$ , what is the tension in the wire?

1.21\text{ N}

4.84\text{ N}

9.68\text{ N}

19.36\text{ N}

Cevabı ve açıklamayı göster

Cevap: $4.84\text{ N}$

Cevap

The tension in the wire is

4.84\text{ N}

.

The fundamental frequency of the open pipe is

f = \frac{v}{2L} = \frac{330}{2(0.60)} = 275\text{ Hz}

. Since the wire vibrates in its first overtone (second harmonic,

n=2

), its frequency is

f = \frac{2 v_{\text{wire}}}{2 L_{\text{wire}}} = \frac{v_{\text{wire}}}{0.40} = 275\text{ Hz}

. This gives

v_{\text{wire}} = 110\text{ m/s}

. Using

v_{\text{wire}} = \sqrt{T/\mu}

, the tension is

T = \mu v_{\text{wire}}^2 = (4.0 \times 10^{-4})(110)^2 = 4.84\text{ N}

.

Adım Adım Çözüm

1

Calculate the fundamental frequency of the pipe open at both ends.

f_{\text{pipe}} = \frac{v_{\text{air}}}{2 L_{\text{pipe}}} = \frac{330}{2 \times 0.60} = 275\text{ Hz}

.

An open pipe supports a fundamental wavelength of

\lambda = 2L_{\text{pipe}}

.

2

Express the frequency of the first overtone of the stretched wire and equate it to the pipe frequency.

f_{\text{wire, 2}} = \frac{2 v_{\text{wire}}}{2 L_{\text{wire}}} = \frac{v_{\text{wire}}}{L_{\text{wire}}} = 275\text{ Hz}

.

For a wire fixed at both ends, the first overtone corresponds to the second harmonic (

n = 2

).

3

Solve for the speed of the transverse wave on the wire.

v_{\text{wire}} = 275 \times 0.40 = 110\text{ m/s}

.

Rearranging

f = v_{\text{wire}} / L_{\text{wire}}

gives

v_{\text{wire}} = f \cdot L_{\text{wire}}

.

4

Calculate the tension in the wire using the wave speed formula.

T = \mu v_{\text{wire}}^2 = (4.0 \times 10^{-4}) \times 110^2 = 4.84\text{ N}

.

The speed of a transverse wave on a stretched string is given by

v = \sqrt{T / \mu}

.

Anahtar Kavram

Resonance between air columns and vibrating strings under distinct boundary conditions

Tahmini Süre:2m 0s

Soru 4338Soru →

An electric heater rated at $1.0\text{ kW}$ is used to convert $0.50\text{ kg}$ of ice initially at $-10^\circ\text{C}$ completely into water at $50^\circ\text{C}$ . Assuming zero thermal energy loss to the surroundings, what is the total time required for this conversion?

(Take specific heat capacity of ice = $2100\text{ J kg}^{-1}\text{ K}^{-1}$ , specific latent heat of fusion of ice = $3.36 \times 10^5\text{ J kg}^{-1}$ , specific heat capacity of water = $4200\text{ J kg}^{-1}\text{ K}^{-1}$ )

283.5\text{ s}

273.0\text{ s}

1953.5\text{ s}

126.0\text{ s}

Cevabı ve açıklamayı göster

Cevap: $283.5\text{ s}$

Cevap

283.5\text{ s}

The correct response of

283.5\text{ s}

accurately accounts for all three distinct phases of thermal absorption: raising the temperature of solid ice from

-10^\circ\text{C}

to

0^\circ\text{C}

(

10,500\text{ J}

), melting the ice to water at constant temperature (

168,000\text{ J}

), and raising the liquid water temperature to

50^\circ\text{C}

(

105,000\text{ J}

). Dividing the total energy of

283,500\text{ J}

by the heater power of

1000\text{ W}

yields

283.5\text{ s}

.

Adım Adım Çözüm

1

Calculate the heat required to raise the temperature of ice from

-10^\circ\text{C}

to

0^\circ\text{C}

(

Q_1

)

Q_1 = m \cdot c_{\text{ice}} \cdot \Delta T_1 = 0.50 \times 2100 \times (0 - (-10)) = 10,500\text{ J}

Ice must reach its melting point at

0^\circ\text{C}

before any phase change can occur.

2

Calculate the latent heat required to melt ice at

0^\circ\text{C}

into water at

0^\circ\text{C}

(

Q_2

)

Q_2 = m \cdot L_f = 0.50 \times 3.36 \times 10^5 = 168,000\text{ J}

Phase change occurs at a constant temperature of

0^\circ\text{C}

using latent heat of fusion.

3

Calculate the heat required to raise the temperature of the resulting water from

0^\circ\text{C}

to

50^\circ\text{C}

(

Q_3

)

Q_3 = m \cdot c_{\text{water}} \cdot \Delta T_2 = 0.50 \times 4200 \times (50 - 0) = 105,000\text{ J}

Once completely melted, sensible heat is absorbed by liquid water up to the target temperature.

4

Calculate total heat energy required (

Q_{\text{total}}

) and convert power to watts

Q_{\text{total}} = 10,500 + 168,000 + 105,000 = 283,500\text{ J}

, and

P = 1.0\text{ kW} = 1000\text{ W}

Total energy is the sum of all individual stage energies.

5

Determine the time required (

t

)

t = \frac{Q_{\text{total}}}{P} = \frac{283,500}{1000} = 283.5\text{ s}

Power is defined as energy per unit time (

P = \frac{Q}{t}

).

Anahtar Kavram

Multi-stage thermal energy balance combining sensible heat (

Q = m c \Delta T

) and latent heat of fusion (

Q = m L_f

).

Alternatif Yöntem

Calculate energy per unit mass first:

q_{\text{total}} = c_{\text{ice}}\Delta T_1 + L_f + c_{\text{water}}\Delta T_2 = (2100 \times 10) + 336000 + (4200 \times 50) = 567,000\text{ J kg}^{-1}

. Then total energy

Q = 0.50 \times 567,000 = 283,500\text{ J}

, leading to

t = \frac{283,500}{1000} = 283.5\text{ s}

.

Tahmini Süre:3m 0s

Soru 4339Soru →

The viscous drag force $F$ acting on a spherical body moving through a fluid at speed $v$ is given by Stokes' law, $F = 6 \pi \eta r v$ , where $r$ is the radius of the sphere and $\eta$ is the coefficient of viscosity. What is the dimensional formula of the coefficient of viscosity $\eta$ ?

\text{M L}^{-1} \text{T}^{-1}

\text{M L T}^{-2}

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

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

Cevabı ve açıklamayı göster

Cevap: $\text{M L}^{-1} \text{T}^{-1}$

Cevap

The dimensional formula of the coefficient of viscosity is

\text{M L}^{-1} \text{T}^{-1}

.

Isolating the coefficient of viscosity from Stokes' law yields

\eta = \frac{F}{6\pi r v}

. Substituting the dimensions

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

,

[r] = \text{L}

, and

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

gives

[\eta] = \frac{\text{M L T}^{-2}}{\text{L}^2 \text{T}^{-1}} = \text{M L}^{-1} \text{T}^{-1}

.

Adım Adım Çözüm

1

Identify the dimensional formulas for force, radius, and velocity.

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

,

[r] = \text{L}

, and

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

.

Base physical quantities must be represented by their fundamental dimensions.

2

Rearrange Stokes' law to express

\eta

in terms of the other variables, treating

6\pi

as a dimensionless constant.

[\eta] = \frac{[F]}{[r][v]}

.

Numerical constants do not carry physical dimensions.

3

Substitute the base dimensions into the formula and simplify using index rules.

[\eta] = \frac{\text{M L T}^{-2}}{\text{L} \cdot \text{L T}^{-1}} = \frac{\text{M L T}^{-2}}{\text{L}^2 \text{T}^{-1}} = \text{M L}^{-1} \text{T}^{-1}

.

Subtracting denominator indices from numerator indices yields the final dimensional expression.

Anahtar Kavram

Dimensional analysis of physical constants and equations

Soru 4340Soru →

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?

Cevabı ve açıklamayı göster

Cevap: 216750

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Compound Depreciation

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