All practice questions

Question 7901Question →

Match each kinetic theory parameter of an ideal gas on the left with its corresponding microscopic physical description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Temperature corresponds to the measure of average translational kinetic energy; Gas pressure corresponds to the rate of momentum transfer per unit area from wall collisions; Root-mean-square speed corresponds to the square root of the mean of squared speeds.

Temperature is directly linked to the average kinetic energy of molecules, gas pressure arises from wall collisions delivering impulse per unit area, and r.m.s. speed is the square root of mean square velocity.

Step-by-Step Solution

1

Identify the microscopic origin of temperature.

Temperature represents the average translational kinetic energy of gas molecules.

From the kinetic theory equation

\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_B T

, absolute temperature directly measures molecular kinetic energy.

2

Identify the microscopic origin of pressure.

Pressure is caused by molecular collisions with container walls.

Each collision transfers momentum to the wall; force is the time rate of momentum change, and force per unit area defines pressure.

3

Identify the definition of root-mean-square speed.

r.m.s. speed is the square root of the average of squared molecular speeds.

It accounts for the statistical distribution of molecular velocities in a gas sample.

Key Concept

Microscopic interpretation of macroscopic gas properties via Kinetic Theory of Matter

Question 7902Question →

A transverse progressive wave traveling along a stretched string is represented by the displacement equation $y = 0.05 \sin\left(20\pi t - \frac{\pi}{4} x\right)$ , where $x$ and $y$ are in meters and $t$ is in seconds. What is the phase difference between two points on the string separated by a distance of $2.0\text{ m}$ ?

\frac{\pi}{4}\text{ rad}

\frac{\pi}{2}\text{ rad}

\pi\text{ rad}

2\pi\text{ rad}

Show answer & explanation

Answer: $\frac{\pi}{2}\text{ rad}$

Answer

The phase difference between the two points is

\frac{\pi}{2}\text{ rad}

.

Comparing the given equation

y = 0.05 \sin\left(20\pi t - \frac{\pi}{4} x\right)

to the standard wave equation

y = A \sin(\omega t - kx)

, the wavenumber is

k = \frac{\pi}{4}\text{ rad m}^{-1}

. The phase difference between two points separated by

\Delta x = 2.0\text{ m}

is

\Delta \phi = k \Delta x = \frac{\pi}{4} \times 2.0 = \frac{\pi}{2}\text{ rad}

. Thus, the option stating

\frac{\pi}{2}\text{ rad}

is correct.

Step-by-Step Solution

1

Identify the wavenumber

k

from the standard wave equation

Comparing

y = A \sin(\omega t - kx)

with

y = 0.05 \sin\left(20\pi t - \frac{\pi}{4} x\right)

gives

k = \frac{\pi}{4}\text{ rad m}^{-1}

.

The coefficient of

x

in the wave equation represents the wavenumber

k = \frac{2\pi}{\lambda}

.

2

Calculate the phase difference using

\Delta \phi = k \Delta x

\Delta \phi = \left(\frac{\pi}{4}\text{ rad m}^{-1}\right) \times 2.0\text{ m} = \frac{\pi}{2}\text{ rad}.

Phase difference is directly proportional to the spatial separation between two points along the path of propagation.

Key Concept

Phase difference in a progressive wave

Estimated Time:1m 0s

Question 7903Question →

A thin flat metal plate has an initial surface area of $0.5\text{ m}^2$ at $10^\circ\text{C}$ . When the plate is heated to a final temperature of $110^\circ\text{C}$ , its surface area increases by $1.8 \times 10^{-3}\text{ m}^2$ . What is the coefficient of cubical expansivity of the metal?

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

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

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

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

Show answer & explanation

Answer: $5.4 \times 10^{-5}\text{ K}^{-1}$

Answer

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

The correct answer is derived by first finding the area expansivity

\beta = \frac{\Delta A}{A_0 \Delta T} = 3.6 \times 10^{-5}\text{ K}^{-1}

. Since

\beta = 2\alpha

, the linear expansivity

\alpha = 1.8 \times 10^{-5}\text{ K}^{-1}

. The cubical expansivity is

\gamma = 3\alpha = 5.4 \times 10^{-5}\text{ K}^{-1}

.

Step-by-Step Solution

1

Calculate the temperature change

\Delta T

\Delta T = 110^\circ\text{C} - 10^\circ\text{C} = 100\text{ K}

Expansion calculations depend on the temperature increase.

2

Determine the coefficient of area expansivity

\beta

\beta = \frac{\Delta A}{A_0 \Delta T} = \frac{1.8 \times 10^{-3}}{0.5 \times 100} = 3.6 \times 10^{-5}\text{ K}^{-1}

The area expansion formula is

\Delta A = A_0 \beta \Delta T

.

3

Calculate the coefficient of linear expansivity

\alpha

\alpha = \frac{\beta}{2} = \frac{3.6 \times 10^{-5}}{2} = 1.8 \times 10^{-5}\text{ K}^{-1}

Area expansivity is twice the linear expansivity (

\beta = 2\alpha

).

4

Calculate the coefficient of cubical expansivity

\gamma

\gamma = 3\alpha = 3 \times (1.8 \times 10^{-5}) = 5.4 \times 10^{-5}\text{ K}^{-1}

Cubical expansivity is three times the linear expansivity (

\gamma = 3\alpha

).

Key Concept

Relationship between linear (

\alpha

), area (

\beta

), and volume (

\gamma

) expansivities:

\beta = 2\alpha

and

\gamma = 3\alpha

.

Estimated Time:1m 30s

Question 7904Question →

If $x$ is the smallest non-negative integer satisfying the modular congruence $7x + 4 \equiv 2 \pmod{13}$ , find the value of $(x^3 + 2x) \pmod{13}$ expressed as a canonical non-negative remainder.

Show answer & explanation

Answer: 6

Answer

The canonical non-negative remainder is 6.

Subtracting 4 from both sides of

7x + 4 \equiv 2 \pmod{13}

gives

7x \equiv -2 \equiv 11 \pmod{13}

. Multiplying by the modular inverse of 7 (which is 2, since

7 \times 2 = 14 \equiv 1 \pmod{13}

) yields

x \equiv 22 \equiv 9 \pmod{13}

. Evaluating

(9^3 + 2 \times 9) \pmod{13}

gives

(729 + 18) = 747

. Dividing 747 by 13 gives a quotient of 57 and a remainder of 6.

Step-by-Step Solution

1

Isolate the linear term in the congruence

7x \equiv -2 \equiv 11 \pmod{13}

Subtracting 4 from both sides simplifies the equation, and

-2 + 13 = 11

converts the negative remainder to positive form.

2

Solve for

x

by multiplying by the multiplicative inverse of 7 modulo 13

x \equiv 9 \pmod{13}

, so

x = 9

Since

7 \times 2 = 14 \equiv 1 \pmod{13}

, multiplying

7x \equiv 11 \pmod{13}

by 2 yields

x \equiv 22 \equiv 9 \pmod{13}

.

3

Evaluate

(x^3 + 2x) \pmod{13}

using modular reduction

6

9^3 = 729 = 56 \times 13 + 1 \equiv 1 \pmod{13}

and

2 \times 9 = 18 = 1 \times 13 + 5 \equiv 5 \pmod{13}

. Adding these gives

1 + 5 = 6

.

Key Concept

Solving linear modular congruences and modular polynomial evaluation

Estimated Time:1m 30s

Question 7905Question →

In $\Delta ABC$ , the side lengths are given as $a = 7\text{ cm}$ , $b = 5\text{ cm}$ , and $c = 3\text{ cm}$ . What is the measure of angle $A$ in degrees?

Show answer & explanation

Answer: 120

Answer

The measure of angle

A

is

120^\circ

.

Using the Cosine Rule formula

\cos A = \frac{b^2 + c^2 - a^2}{2bc}

, substituting

a = 7

,

b = 5

, and

c = 3

yields

\cos A = \frac{25 + 9 - 49}{30} = -\frac{1}{2}

. The inverse cosine of

-\frac{1}{2}

gives an obtuse angle of

120^\circ

.

Step-by-Step Solution

1

Apply the Cosine Rule for an unknown angle in terms of the three sides

\cos A = \frac{b^2 + c^2 - a^2}{2bc}

When all three side lengths of a non-right triangle are given (SSS), the Cosine Rule is required to solve for any internal angle.

2

Substitute

a = 7

,

b = 5

, and

c = 3

into the Cosine Rule formula and evaluate

\cos A = \frac{25 + 9 - 49}{2 \times 5 \times 3} = \frac{-15}{30} = -0.5

Evaluating the terms in the numerator and denominator simplifies the expression for

\cos A

.

3

Calculate the inverse cosine of

-0.5

to find angle

A

A = 120^\circ

Since the cosine value is negative, angle

A

is obtuse and lies in the second quadrant (

90^\circ < A < 180^\circ

).

Key Concept

Using the Cosine Rule with three side lengths (SSS) to find an obtuse interior angle

Estimated Time:1m 30s

Question 7906Question →

Which of the following represents the solution set of real values of $x$ satisfying the inequality $4(1 - x) \le 3(x + 6)$ ?

x \ge -2

x \le -2

x \ge 2

x \le 2

Show answer & explanation

Answer: $x \ge -2$

Answer

The set of real values satisfying the inequality is

x \ge -2

.

Expanding the given inequality yields

4 - 4x \le 3x + 18

. Subtracting

3x

and

4

from both sides gives

-7x \le 14

. When dividing both sides by

-7

, the inequality sign must reverse direction, yielding

x \ge -2

.

Step-by-Step Solution

1

Expand both sides of the inequality

4 - 4x \le 3x + 18

Remove brackets to group like terms.

2

Rearrange terms by moving variable terms to the left side and constant terms to the right side

-4x - 3x \le 18 - 4 \implies -7x \le 14

Isolate the term containing the variable

x

.

3

Divide both sides by

-7

and reverse the inequality sign

x \ge -2

Dividing or multiplying an inequality by a negative number flips the direction of the inequality symbol.

Key Concept

Solving linear inequalities involving bracket expansion and division by negative numbers

Question 7907Question →

An electric lamp rated at $60\,\text{W}$ is kept switched on for $50\,\text{seconds}$ . What is the total electrical energy consumed by the lamp during this time?

3000\,\text{J}

300\,\text{J}

1.2\,\text{J}

110\,\text{J}

Show answer & explanation

Answer: $3000\,\text{J}$

Answer

3000\,\text{J}

Electrical energy is calculated using the formula

E = P \times t

. Substituting the given values

P = 60\,\text{W}

and

t = 50\,\text{s}

gives

E = 60 \times 50 = 3000\,\text{J}

.

Step-by-Step Solution

1

Identify the given physical quantities.

Power

P = 60\,\text{W}

and time

t = 50\,\text{s}

.

Power is given in watts (

\text{J/s}

) and time is given in seconds.

2

Apply the electrical energy formula

E = P \times t

.

E = 60\,\text{W} \times 50\,\text{s} = 3000\,\text{J}

.

Total electrical energy is the product of power dissipation and time duration.

Key Concept

Electrical Energy and Power

Estimated Time:45s

Question 7908Question →

A beam of light traveling in air is incident on a transparent liquid at the polarizing angle (Brewster's angle) of $53.1^\circ$ , where $\tan 53.1^\circ = 1.33$ . What is the critical angle for total internal reflection when light travels from this liquid into air?

\sin^{-1}(0.75)

\sin^{-1}(0.60)

\sin^{-1}(1.33)

53.1^\circ

Show answer & explanation

Answer: $\sin^{-1}(0.75)$

Answer

The critical angle for total internal reflection at the liquid-air boundary is

\sin^{-1}(0.75)

.

According to Brewster's law, the refractive index of the liquid is given by

n = \tan(53.1^\circ) = 1.33 = \frac{4}{3}

. When light travels from the denser liquid medium to the rarer air medium, the critical angle

\theta_c

for total internal reflection satisfies

\sin\theta_c = \frac{1}{n}

. Substituting

n = \frac{4}{3}

gives

\sin\theta_c = \frac{3}{4} = 0.75

, so

\theta_c = \sin^{-1}(0.75)

.

Step-by-Step Solution

1

Determine the refractive index of the liquid using Brewster's law.

n = \tan(53.1^\circ) = 1.33 = \frac{4}{3}

.

Brewster's law states that when light in air (

n_1 = 1

) is incident at the polarizing angle

\theta_B

on a medium of index

n

,

\tan\theta_B = n

.

2

Apply the total internal reflection condition for light passing from liquid to air.

\sin\theta_c = \frac{1}{n} = \frac{1}{4/3} = \frac{3}{4} = 0.75

.

Total internal reflection occurs at an interface when light travels from a denser medium (

n

) to a less dense medium (

1

) at an angle greater than

\theta_c

, where

\sin\theta_c = \frac{1}{n}

.

3

Solve for the critical angle

\theta_c

.

\theta_c = \sin^{-1}(0.75)

.

Taking the inverse sine of

0.75

yields the critical angle.

Key Concept

Synthesizing Brewster's law of polarization with total internal reflection critical angle

Question 7909Question →

A survey ship emits an ultrasonic sound pulse vertically downwards into the ocean. The sound wave travels through seawater at a speed of $1500\text{ m/s}$ , and its reflected echo from the seabed is detected by the ship's hydrophone $1.2\text{ s}$ after emission. What is the depth of the ocean floor at that location?

900\text{ m}

1800\text{ m}

450\text{ m}

2700\text{ m}

Show answer & explanation

Answer: $900\text{ m}$

Answer

900\text{ m}

Sound emitted by the ship travels down to the seabed and reflects back to the hydrophone, taking

1.2\text{ s}

for the complete round trip. Using the relation

\text{Depth} = \frac{v \times t}{2}

, the depth is

\frac{1500 \times 1.2}{2} = 900\text{ m}

.

Step-by-Step Solution

1

Calculate total distance traveled by the sound wave

Total distance

s = v \times t = 1500\text{ m/s} \times 1.2\text{ s} = 1800\text{ m}

Distance is the product of speed and total time elapsed.

2

Determine the depth of the seabed

Depth

d = \frac{s}{2} = \frac{1800\text{ m}}{2} = 900\text{ m}

An echo involves the sound traveling down to the ocean floor and back up, so the depth is half the total distance.

Key Concept

Echo Reflection and Distance Calculation

Estimated Time:45s

Question 7910Question →

An electric cell of electromotive force $E$ and internal resistance $r$ is connected across a parallel combination of two resistors with resistances $6.0\text{ }\Omega$ and $12.0\text{ }\Omega$ . The potential difference across the parallel combination is $4.0\text{ V}$ . When the $12.0\text{ }\Omega$ resistor is removed from the circuit, the current supplied by the cell becomes $0.75\text{ A}$ . What is the internal resistance $r$ of the cell in ohms?

Show answer & explanation

Answer: 2

Answer

The internal resistance of the cell is

2.0\text{ }\Omega

.

Analyzing the circuit under both states yields two simultaneous equations for the e.m.f.

E

in terms of internal resistance

r

:

E = 4.0 + 1.0r

and

E = 4.5 + 0.75r

. Solving these equations gives

r = 2.0\text{ }\Omega

.

Step-by-Step Solution

1

Calculate the equivalent resistance of the parallel resistor network.

R_p = 4.0\text{ }\Omega

Using the parallel resistor formula:

R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{6.0 \times 12.0}{6.0 + 12.0} = 4.0\text{ }\Omega

.

2

Determine the initial total current delivered by the cell.

I_1 = 1.0\text{ A}

From the terminal voltage across the parallel load:

I_1 = \frac{V_1}{R_p} = \frac{4.0\text{ V}}{4.0\text{ }\Omega} = 1.0\text{ A}

.

3

Formulate the e.m.f. equation for the initial circuit state.

E = 4.0 + 1.0r

Applying the equation

E = V + Ir

gives

E = 4.0 + (1.0)r

.

4

Formulate the e.m.f. equation after removing the

12.0\text{ }\Omega

resistor.

E = 4.5 + 0.75r

With external load

R_2 = 6.0\text{ }\Omega

and current

I_2 = 0.75\text{ A}

,

E = I_2(R_2 + r) = 0.75(6.0 + r) = 4.5 + 0.75r

.

5

Solve for the internal resistance

r

by equating the two e.m.f. expressions.

r = 2.0\text{ }\Omega

Equating the two expressions for

E

:

4.0 + 1.0r = 4.5 + 0.75r \implies 0.25r = 0.50 \implies r = 2.0\text{ }\Omega

.

Key Concept

Internal Resistance and Multi-State Circuit Analysis

Estimated Time:2m 0s

Question 7911Question →

Two cylindrical wires, X and Y, are made of the same uniform conducting material. Wire X has length $L$ , diameter $d$ , and an electrical resistance of $12\,\Omega$ . Wire Y has length $2L$ and diameter $2d$ . What is the resistance of wire Y?

3\,\Omega

6\,\Omega

12\,\Omega

24\,\Omega

Show answer & explanation

Answer: $6\,\Omega$

Answer

The resistance of wire Y is

6\,\Omega

.

The resistance of a uniform conductor is given by

R = \frac{4\rho L}{\pi d^2}

. For wire X,

R_X = 12\,\Omega

. For wire Y with length

2L

and diameter

2d

, the new resistance becomes

R_Y = \frac{4\rho(2L)}{\pi (2d)^2} = \frac{8\rho L}{4\pi d^2} = \frac{1}{2}\left(\frac{4\rho L}{\pi d^2}\right) = \frac{1}{2} R_X = 6\,\Omega

.

Step-by-Step Solution

1

Express the resistance of a cylindrical conductor in terms of length

L

and diameter

d

.

R = \rho \frac{L}{A} = \rho \frac{L}{\pi (d/2)^2} = \frac{4\rho L}{\pi d^2}

The cross-sectional area

A

of a circular wire with diameter

d

is given by

A = \frac{\pi d^2}{4}

.

2

Write the resistance formula for wire X using its given value.

R_X = \frac{4\rho L}{\pi d^2} = 12\,\Omega

Wire X has length

L

and diameter

d

.

3

Substitute the parameters of wire Y (

L_Y = 2L

and

d_Y = 2d

) into the resistance formula.

R_Y = \frac{4\rho (2L)}{\pi (2d)^2} = \frac{8\rho L}{4\pi d^2} = \frac{2\rho L}{\pi d^2}

Doubling diameter increases the cross-sectional area by a factor of

2^2 = 4

.

4

Relate the resistance of wire Y to the resistance of wire X and calculate the final numerical value.

R_Y = \frac{1}{2} \left(\frac{4\rho L}{\pi d^2}\right) = \frac{1}{2} R_X = \frac{12\,\Omega}{2} = 6\,\Omega

Since

R_Y = \frac{1}{2} R_X

, halving

12\,\Omega

yields

6\,\Omega

.

Key Concept

Dependence of Electrical Resistance on Conductor Dimensions

Estimated Time:1m 30s

Question 7912Question →

A glass flask with an internal volume of $400\text{ cm}^3$ is filled to the brim with a liquid at $25^\circ\text{C}$ . The system is heated to $75^\circ\text{C}$ , causing $9.0\text{ cm}^3$ of liquid to overflow. Given that the linear expansivity of glass is $1.0 \times 10^{-5}\text{ K}^{-1}$ , what is the real cubic expansivity of the liquid in units of $10^{-4}\text{ K}^{-1}$ ?

Show answer & explanation

Answer: 4.8

Answer

The real cubic expansivity of the liquid is

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

, which equals

4.8

in units of

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

.

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

\gamma_r = \gamma_a + \gamma_v

). Calculating the apparent cubic expansivity yields

\gamma_a = \frac{9.0}{400 \times 50} = 4.5 \times 10^{-4}\text{ K}^{-1}

. The cubic expansivity of the glass container is

\gamma_v = 3 \times 1.0 \times 10^{-5} = 0.3 \times 10^{-4}\text{ K}^{-1}

. Adding these values gives a real cubic expansivity of

\gamma_r = 4.8 \times 10^{-4}\text{ K}^{-1}

, which equals

4.8

in units of

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

.

Step-by-Step Solution

1

Calculate the temperature change

\Delta T = 75^\circ\text{C} - 25^\circ\text{C} = 50\text{ K}

Thermal expansion depends directly on the change in temperature.

2

Determine the apparent cubic expansivity of the liquid

\gamma_a = \frac{\Delta V_a}{V_0 \Delta T} = \frac{9.0\text{ cm}^3}{400\text{ cm}^3 \times 50\text{ K}} = 4.5 \times 10^{-4}\text{ K}^{-1}

Apparent cubic expansivity is determined from the volume of liquid that overflows relative to the initial volume and temperature increase.

3

Calculate the cubic expansivity of the glass container

\gamma_v = 3\alpha = 3 \times 1.0 \times 10^{-5}\text{ K}^{-1} = 0.3 \times 10^{-4}\text{ K}^{-1}

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

4

Compute the real cubic expansivity of the liquid

\gamma_r = \gamma_a + \gamma_v = 4.5 \times 10^{-4}\text{ K}^{-1} + 0.3 \times 10^{-4}\text{ K}^{-1} = 4.8 \times 10^{-4}\text{ K}^{-1}

Real expansivity accounts for both the observed apparent expansion of the liquid and the expansion of the vessel containing it.

Key Concept

Thermal Expansion of Liquids and Anomalous Expansion of Water

Question 7913Question →

The frequency distribution table below displays the examination marks of a group of candidates:

Mark Class	Frequency ( $f$ )
$10 - 19$	$5$
$20 - 29$	$9$
$30 - 39$	$16$
$40 - 49$	$12$
$50 - 59$	$8$

If an ogive (cumulative frequency curve) is constructed to represent this dataset, which of the following represents the correct coordinate pair for the point corresponding to the modal class?

(39.5, 30)

(29.5, 30)

(34.5, 30)

(39.5, 16)

Show answer & explanation

Answer: $(39.5, 30)$

Answer

The correct coordinate pair is

(39.5, 30)

.

The modal class is

30 - 39

because it has the highest frequency of

16

. The upper class boundary for this interval is

39.5

. Summing frequencies up to this class gives

5 + 9 + 16 = 30

. Therefore, the plotted point on the ogive must be

(39.5, 30)

.

Step-by-Step Solution

1

Identify the modal class interval from the frequency table.

The class with the highest frequency (

16

) is

30 - 39

.

The modal class is defined as the interval with the maximum frequency.

2

Determine the upper class boundary of the modal class.

Upper class boundary

= 39 + 0.5 = 39.5

.

Cumulative frequency (ogive) curves are plotted using the upper class boundaries on the horizontal axis.

3

Calculate the cumulative frequency up to and including the modal class.

Cumulative frequency

= 5 + 9 + 16 = 30

.

An ogive plots accumulated frequencies up to each boundary.

4

Form the coordinate pair

(x, y) = (\text{Upper Class Boundary}, \text{Cumulative Frequency})

.

(39.5, 30)

.

Points on an ogive take the form

(\text{Upper Boundary}, \text{Cumulative Frequency})

.

Key Concept

Plotting Cumulative Frequency Curves (Ogives)

Question 7914Question →

An object is projected from level ground with an initial speed of $30\text{ m/s}$ at an angle of $30^\circ$ to the horizontal. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the maximum height reached by the object in meters.

Show answer & explanation

Answer: 11.25

Answer

The maximum height reached by the object is

11.25\text{ m}

.

The vertical component of initial velocity is

u_y = 30 \sin(30^\circ) = 15\text{ m/s}

. At maximum height, vertical velocity becomes zero, so

H = \frac{u_y^2}{2g} = \frac{15^2}{20} = 11.25\text{ m}

.

Step-by-Step Solution

1

Calculate the vertical component of the initial velocity.

u_y = u \sin\theta = 30 \times \sin(30^\circ) = 15\text{ m/s}

Only the vertical component of initial velocity determines the maximum height.

2

Apply the vertical motion equation at maximum height where vertical velocity is zero.

H = \frac{u_y^2}{2g} = \frac{15^2}{2 \times 10} = 11.25\text{ m}

Using

v_y^2 = u_y^2 - 2gH

with

v_y = 0

gives

H = \frac{u_y^2}{2g}

.

Key Concept

Maximum height of a projectile

Estimated Time:45s

Question 7915Question →

The torque $\tau$ required to rotate a thin flat disk of radius $r$ at a constant angular velocity $\omega$ in a fluid of dynamic viscosity $\eta$ is expressed by the dimensional formula $\tau = k \eta^x \omega^y r^z$ , where $k$ is a dimensionless constant. What is the value of the sum of the exponents $x + y + z$ ?

Show answer & explanation

Answer: 5

Answer

The sum of the exponents

x + y + z

is 5.

By substituting the base dimensions into

\tau = k \eta^x \omega^y r^z

, we get

M L^2 T^{-2} = (M L^{-1} T^{-1})^x (T^{-1})^y L^z = M^x L^{-x+z} T^{-x-y}

. Equating exponents of

M

gives

x = 1

. Equating exponents of

T

gives

-1 - y = -2 \implies y = 1

. Equating exponents of

L

gives

-1 + z = 2 \implies z = 3

. Thus,

x + y + z = 1 + 1 + 3 = 5

.

Step-by-Step Solution

1

Determine the dimensions of torque, dynamic viscosity, angular velocity, and radius in base mechanical dimensions (M, L, T).

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

,

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

,

[\omega] = T^{-1}

, and

[r] = L

.

Dimensional analysis requires converting all parameters into base dimensions.

2

Apply the principle of dimensional homogeneity to set up exponential equations for each base dimension.

M^1 L^2 T^{-2} = M^x L^{-x+z} T^{-x-y}

.

Both sides of a physically valid equation must share identical net dimensions.

3

Solve for each exponent individually by comparing indices.

x = 1

,

y = 1

,

z = 3

.

Matching powers of M yields

x=1

, matching powers of T yields

y=1

, and matching powers of L yields

z=3

.

4

Sum the three calculated exponent values.

1 + 1 + 3 = 5

.

The question asks specifically for the value of

x + y + z

.

Key Concept

Dimensional analysis and dimensional homogeneity

Estimated Time:1m 30s

Question 7916Question →

A research vessel emits a high-frequency acoustic pulse vertically downward toward the seabed. The signal reflects off the ocean floor and is detected by the vessel's receiver $1.6\text{ s}$ after transmission. If the speed of sound in seawater is $1500\text{ m/s}$ , what is the depth of the ocean floor at this point?

2400\text{ m}

1200\text{ m}

600\text{ m}

937.5\text{ m}

Show answer & explanation

Answer: $1200\text{ m}$

Answer

The depth of the ocean floor is

1200\text{ m}

.

The correct answer is

1200\text{ m}

. Since an echo involves a two-way journey (from the vessel down to the ocean floor and back up), the sound wave takes half of the total time (

0.8\text{ s}

) to reach the bottom. Multiplying the speed of sound in seawater (

1500\text{ m/s}

) by

0.8\text{ s}

yields the correct depth of

1200\text{ m}

.

Step-by-Step Solution

1

Determine the time taken for the sound wave to travel one way to the ocean floor.

t_{\text{one-way}} = \frac{t_{\text{total}}}{2} = \frac{1.6\text{ s}}{2} = 0.8\text{ s}

An echo involves sound traveling from the transmitter to the reflecting surface and back to the receiver.

2

Calculate the depth using the speed of sound in seawater.

\text{Depth } d = v \times t_{\text{one-way}} = 1500\text{ m/s} \times 0.8\text{ s} = 1200\text{ m}

The distance traveled in one direction equals the speed of sound multiplied by the one-way travel time.

Key Concept

Echo location and depth sounding using two-way wave propagation

Estimated Time:1m 0s

Question 7917Question →

The $4^{\text{th}}$ term of an arithmetic progression (AP) is $15$ and the $9^{\text{th}}$ term is $35$ . What is the sum of the first $12$ terms of the progression?

300

276

600

228

Show answer & explanation

Answer: $300$

Answer

The sum of the first

12

terms of the progression is

300

.

Subtracting the

4^{\text{th}}

term from the

9^{\text{th}}

term gives

5d = 35 - 15 = 20

, so the common difference

d = 4

. Substituting

d = 4

into the

4^{\text{th}}

term expression

a + 3d = 15

yields

a = 3

. Using the sum formula

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

for

n = 12

,

a = 3

, and

d = 4

gives

S_{12} = 6 \times (6 + 44) = 300

.

Step-by-Step Solution

1

Set up the term equations using

T_n = a + (n-1)d

.

a + 3d = 15

and

a + 8d = 35

.

The

n^{\text{th}}

term formula for an AP is

T_n = a + (n-1)d

.

2

Solve the system of equations for

a

and

d

.

Subtracting the first equation from the second yields

5d = 20 \Rightarrow d = 4

. Substituting

d = 4

into

a + 3(4) = 15

gives

a = 3

.

Subtracting eliminates

a

to find the common difference

d

, which is then used to calculate the first term

a

.

3

Calculate the sum of the first

12

terms

S_{12}

.

S_{12} = \frac{12}{2}[2(3) + (12-1)4] = 6[6 + 44] = 6 \times 50 = 300

.

Applying the AP sum formula

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

with

n = 12

,

a = 3

, and

d = 4

.

Key Concept

Arithmetic Progression

n^{\text{th}}

term and sum calculations

Question 7918Question →

In a chemistry experiment, a student measured the mass of a substance as $12.8\text{ g}$ . If the actual mass of the substance is $12.5\text{ g}$ , calculate the percentage error in the student's measurement.

Show answer & explanation

Answer: 2.4

Answer

The percentage error in the student's measurement is

2.4\%

.

The percentage error is calculated by taking the absolute error (

0.3\text{ g}

), dividing it by the actual value (

12.5\text{ g}

), and multiplying the result by

100\%

, yielding

2.4\%

.

Step-by-Step Solution

1

Identify the actual value and the measured value

Actual value =

12.5\text{ g}

, Measured value =

12.8\text{ g}

Percentage error is computed relative to the true, actual value.

2

Calculate the absolute error

\text{Error} = |12.8 - 12.5| = 0.3\text{ g}

The error represents the difference between the measured value and the actual value.

3

Calculate the percentage error

\text{Percentage Error} = \frac{0.3}{12.5} \times 100\% = 2.4\%

Dividing the error by the actual value and multiplying by 100 converts the relative error into a percentage.

Key Concept

Percentage Error

Estimated Time:1m 30s

Question 7919Question →

In an archery competition, two archers, Kemi and Chidi, attempt to hit a target independently. The probability that Kemi hits the target is $\frac{1}{3}$ and the probability that Chidi hits the target is $\frac{2}{5}$ . What is the probability that at least one of them hits the target?

\frac{3}{5}

\frac{11}{15}

\frac{2}{15}

\frac{13}{15}

Show answer & explanation

Answer: $\frac{3}{5}$

Answer

The probability that at least one archer hits the target is

\frac{3}{5}

.

For independent events, the probability of both events occurring is

P(K \cap C) = P(K) \times P(C) = \frac{1}{3} \times \frac{2}{5} = \frac{2}{15}

. Applying the addition law of probability

P(K \cup C) = P(K) + P(C) - P(K \cap C)

gives

\frac{1}{3} + \frac{2}{5} - \frac{2}{15} = \frac{9}{15} = \frac{3}{5}

.

Step-by-Step Solution

1

Identify event probabilities and independence

P(\text{Kemi}) = \frac{1}{3}

and

P(\text{Chidi}) = \frac{2}{5}

The problem specifies that the two attempts are independent events.

2

Calculate the joint probability (intersection) using the multiplication law for independent events

P(\text{Kemi} \cap \text{Chidi}) = P(\text{Kemi}) \times P(\text{Chidi}) = \frac{1}{3} \times \frac{2}{5} = \frac{2}{15}

For independent events, the probability of both occurring together is the product of their individual probabilities.

3

Apply the addition law of probability to find the union

P(\text{Kemi} \cup \text{Chidi}) = \frac{1}{3} + \frac{2}{5} - \frac{2}{15} = \frac{5 + 6 - 2}{15} = \frac{9}{15} = \frac{3}{5}

The addition law states

P(A \cup B) = P(A) + P(B) - P(A \cap B)

to prevent double counting the intersection.

Key Concept

Compound Events, Multiplication Law for Independent Events, and Addition Law of Probability

Estimated Time:1m 30s

Question 7920Question →

A research submarine moving underwater at a constant speed of $12.0\text{ m/s}$ directly toward a vertical underwater cliff face emits an ultrasonic acoustic pulse. The echo reflected from the cliff face is detected by the submarine's receiver $2.50\text{ s}$ after emission. If the speed of sound in seawater is $1500\text{ m/s}$ , what is the distance between the submarine and the cliff face at the exact moment the echo is detected?

Show answer & explanation

Answer: 1860

Answer

The distance between the submarine and the cliff face at the exact moment the echo is detected is

1860\text{ m}

.

During the

2.50\text{ s}

transit time of the acoustic signal, the sound covers a total path of

3750\text{ m}

(

1500\text{ m/s} \times 2.50\text{ s}

) while the submarine moves

30\text{ m}

closer to the cliff face (

12.0\text{ m/s} \times 2.50\text{ s}

). The total path of the sound consists of the outward journey to the cliff (

d + 30\text{ m}

) and the return journey to the submarine (

d

). Setting

(d + 30) + d = 3750

gives

2d + 30 = 3750

, leading to

d = 1860\text{ m}

.

Step-by-Step Solution

1

Calculate total sound travel distance and submarine displacement during the 2.50 s window.

Sound distance

d_{\text{sound}} = 1500\text{ m/s} \times 2.50\text{ s} = 3750\text{ m}

; Submarine displacement

d_{\text{sub}} = 12.0\text{ m/s} \times 2.50\text{ s} = 30.0\text{ m}

.

Both the acoustic wave and the submarine move continuously throughout the total elapsed transit time.

2

Establish the geometric equation for the sound path relative to the final distance d.

d_{\text{sound}} = 2d + d_{\text{sub}}

, where

d

is the remaining distance to the cliff face at detection time.

The sound pulse travels forward across the initial separation

(d + d_{\text{sub}})

and reflects back across the remaining separation

d

.

3

Solve the linear equation for the final separation distance d.

3750 = 2d + 30 \implies 2d = 3720 \implies d = 1860\text{ m}

.

Subtracting the submarine's forward displacement from the total sound path gives twice the distance to the obstacle at the instant of signal reception.

Key Concept

Echo distance calculations with moving receiver and source

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept