All practice questions

Question 181Question →

A water pump driven by an engine with an efficiency of $80\%$ raises water from an underground tank of depth $20\text{ m}$ and discharges it through a nozzle of cross-sectional area $10\text{ cm}^2$ at a steady speed of $10\text{ m/s}$ . Taking the density of water as $1000\text{ kg/m}^3$ and acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the minimum input power rating (in $\text{W}$ ) required for the engine?

Show answer & explanation

Answer: 3125

Answer

The minimum input power rating required for the engine is

3125\text{ W}

.

The engine must supply power to lift

10\text{ kg}

of water per second through a vertical height of

20\text{ m}

while accelerating it to

10\text{ m/s}

. The useful power output is

2000\text{ W}

(potential)

+ 500\text{ W}

(kinetic)

= 2500\text{ W}

. Accounting for an engine efficiency of

80\%

, the total input power is

\frac{2500}{0.80} = 3125\text{ W}

.

Step-by-Step Solution

1

Determine the mass of water discharged per unit time (mass flow rate).

\frac{dm}{dt} = \rho \times A \times v = 1000\text{ kg/m}^3 \times (10 \times 10^{-4}\text{ m}^2) \times 10\text{ m/s} = 10\text{ kg/s}

Water is moving through a cross-sectional area at a constant velocity.

2

Calculate the useful output power required to lift the water and impart kinetic energy.

P_{\text{out}} = \frac{dm}{dt} g h + \frac{1}{2} \frac{dm}{dt} v^2 = (10 \times 10 \times 20) + \left(\frac{1}{2} \times 10 \times 10^2\right) = 2000\text{ W} + 500\text{ W} = 2500\text{ W}

The engine must perform work against gravity to raise the water depth and provide kinetic energy for exit velocity.

3

Calculate the total input power using engine efficiency.

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{2500\text{ W}}{0.80} = 3125\text{ W}

Efficiency is the ratio of useful power output to total power input.

Key Concept

Work-Energy Theorem applied to fluid flow and Power-Efficiency relations

Question 182Question →

If $\frac{4 + 3\sqrt{2}}{3 - 2\sqrt{2}}$ is expressed in the simplified form $a + b\sqrt{2}$ , where $a$ and $b$ are integers, what is the value of $a + b$ ?

Show answer & explanation

Answer: 41

Answer

The value of

a + b

is 41.

To rationalise

\frac{4 + 3\sqrt{2}}{3 - 2\sqrt{2}}

, multiply both numerator and denominator by the conjugate

3 + 2\sqrt{2}

. The denominator becomes

3^2 - (2\sqrt{2})^2 = 9 - 8 = 1

. Expanding the numerator gives

(4)(3) + 4(2\sqrt{2}) + 3\sqrt{2}(3) + 3\sqrt{2}(2\sqrt{2}) = 12 + 8\sqrt{2} + 9\sqrt{2} + 12 = 24 + 17\sqrt{2}

. Comparing with

a + b\sqrt{2}

gives

a = 24

and

b = 17

, so

a + b = 24 + 17 = 41

.

Step-by-Step Solution

1

Multiply the numerator and denominator by the conjugate of the denominator.

The expression becomes

\frac{(4 + 3\sqrt{2})(3 + 2\sqrt{2})}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})}

.

Multiplying by the conjugate eliminates surds from the denominator using the difference of two squares identity

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

.

2

Simplify the denominator.

(3)^2 - (2\sqrt{2})^2 = 9 - (4 \times 2) = 9 - 8 = 1

.

Squaring

2\sqrt{2}

yields

2^2 \times (\sqrt{2})^2 = 4 \times 2 = 8

.

3

Expand the numerator.

(4 \times 3) + (4 \times 2\sqrt{2}) + (3\sqrt{2} \times 3) + (3\sqrt{2} \times 2\sqrt{2}) = 12 + 8\sqrt{2} + 9\sqrt{2} + 12 = 24 + 17\sqrt{2}

.

Applying the distributive law and grouping rational terms together and like surd terms together.

4

Identify the values of

a

and

b

and calculate

a + b

.

a = 24

,

b = 17

, so

a + b = 24 + 17 = 41

.

Matching coefficients of the simplified surd form

a + b\sqrt{2}

.

Key Concept

Rationalisation of binomial surd denominators using conjugates

Estimated Time:2m 0s

Question 183Question →

The total profit $P(x)$ , in thousands of Naira, obtained from producing and selling $x$ hundred units of a commodity is modeled by the function $P(x) = -x^3 + 6x^2 + 15x - 8$ , where $x \ge 0$ . What is the maximum profit achievable?

Show answer & explanation

Answer: 92

Answer

The maximum profit achievable is 92 thousand Naira.

To find the maximum profit, we find the stationary points of

P(x) = -x^3 + 6x^2 + 15x - 8

by taking the derivative

P'(x) = -3x^2 + 12x + 15

and setting it to 0. Solving

-3(x-5)(x+1) = 0

with

x \ge 0

yields

x = 5

. Testing the second derivative gives

P''(5) = -18 < 0

, confirming

x = 5

is a maximum. Substituting

x = 5

into

P(x)

yields

P(5) = 92

.

Step-by-Step Solution

1

Differentiate the profit function

P(x)

with respect to

x

P'(x) = -3x^2 + 12x + 15

Stationary points occur where the rate of change of profit (the derivative) is equal to zero.

2

Set

P'(x) = 0

and solve for

x

-3(x^2 - 4x - 5) = 0 \implies (x - 5)(x + 1) = 0

, giving critical values

x = 5

and

x = -1

Factoring the quadratic equation yields the critical values of production level.

3

Filter critical values based on physical domain constraints

x = 5

(reject

x = -1

since production

x \ge 0

)

Production quantities cannot be negative in physical real-life contexts.

4

Perform the second derivative test to confirm the nature of the stationary point

P''(x) = -6x + 12 \implies P''(5) = -6(5) + 12 = -18

Since

P''(5) < 0

, the function achieves a local maximum at

x = 5

.

5

Substitute

x = 5

back into original function

P(x)

P(5) = -(5)^3 + 6(5)^2 + 15(5) - 8 = -125 + 150 + 75 - 8 = 92

Evaluating

P(5)

gives the maximum total profit value.

Key Concept

Optimization and Maximum Values using First and Second Derivatives

Question 184Question →

Find the positive integer value of $n$ such that $^{n}P_4 = 42 \times {^{n}P_2}$ .

Show answer & explanation

Answer: 9

Answer

The positive integer value of

n

is 9.

Expanding

^{n}P_4

as

n(n-1)(n-2)(n-3)

and

^{n}P_2

as

n(n-1)

allows dividing out

n(n-1)

, leading to

(n-2)(n-3) = 42

. Expanding and factoring gives

n^2 - 5n - 36 = 0

, which yields

n = 9

as the only valid positive integer.

Step-by-Step Solution

1

Apply the permutation formula

^{n}P_r = \frac{n!}{(n-r)!}

^{n}P_4 = n(n-1)(n-2)(n-3)

and

^{n}P_2 = n(n-1)

By definition of permutations, selecting

r

items from

n

distinct items without replacement.

2

Substitute the expansions into the given relation

n(n-1)(n-2)(n-3) = 42 n(n-1)

Direct substitution into

^{n}P_4 = 42 \times {^{n}P_2}

.

3

Simplify by dividing out common non-zero terms

(n-2)(n-3) = 42

Since

n \ge 4

,

n(n-1) \neq 0

and can be safely divided from both sides.

4

Form and solve the quadratic equation

n^2 - 5n + 6 = 42 \implies n^2 - 5n - 36 = 0 \implies (n-9)(n+4) = 0

Expanding terms and factoring the resulting quadratic expression.

5

Determine the valid root

n = 9

Permutation total items

n

must satisfy

n \ge r \ge 0

, rejecting the negative root

n = -4

.

Key Concept

Algebraic equations involving permutations

Estimated Time:1m 15s

Question 185Question →

A stone is projected from ground level with an initial velocity of $40\text{ m/s}$ at an angle of $30^\circ$ to the horizontal. What is the total time of flight of the stone, in seconds? (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 4

Answer

The total time of flight of the stone is

4\text{ s}

.

The total time of flight

T

for a projectile launched over level ground is calculated using

T = \frac{2 u \sin \theta}{g}

. Substituting

u = 40\text{ m/s}

,

\theta = 30^\circ

, and

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

yields

T = \frac{2 \times 40 \times 0.5}{10} = 4\text{ s}

.

Step-by-Step Solution

1

Find the vertical component of the launch velocity

u_y = 20\text{ m/s}

The vertical motion determines the time the projectile remains in the air.

2

Calculate the total time of flight

T = 4\text{ s}

Applying

T = \frac{2 u \sin \theta}{g} = \frac{2 \times 20}{10} = 4\text{ s}

gives the total duration before landing back at ground level.

Key Concept

Time of Flight in Projectile Motion

Estimated Time:45s

Question 186Question →

If $(x, y)$ satisfies the simultaneous equations $x + 2y = 5$ and $x^2 + y^2 = 10$ , what is the positive value of $x$ ?

Show answer & explanation

Answer: 3

Answer

The positive value of

x

is 3.

Isolating

x

in the linear equation gives

x = 5 - 2y

. Substituting this expression into

x^2 + y^2 = 10

yields

(5 - 2y)^2 + y^2 = 10

. Expanding gives

25 - 20y + 4y^2 + y^2 = 10 \implies 5y^2 - 20y + 15 = 0

. Dividing all terms by

5

produces

y^2 - 4y + 3 = 0

, which factors as

(y - 1)(y - 3) = 0

, so

y = 1

or

y = 3

. Substituting these into

x = 5 - 2y

gives

x = 3

when

y = 1

and

x = -1

when

y = 3

. The positive value of

x

is 3.

Step-by-Step Solution

1

Express

x

from the linear equation

x = 5 - 2y

Isolating

x

allows substitution into the quadratic equation.

2

Substitute into the quadratic equation

(5 - 2y)^2 + y^2 = 10

Eliminates variable

x

to create a single-variable equation in

y

.

3

Expand and simplify

5y^2 - 20y + 15 = 0 \implies y^2 - 4y + 3 = 0

Transforms the equation into standard quadratic form for easy factorization.

4

Solve for

y

y = 1 \text{ or } y = 3

Factoring

(y - 1)(y - 3) = 0

yields the two possible values for

y

.

5

Determine corresponding

x

values and select the positive one

x = 3

(from

y = 1

)

Evaluating

x = 5 - 2y

gives

x = 3

and

x = -1

; the positive result requested is 3.

Key Concept

Solving simultaneous linear and quadratic equations by substitution

Estimated Time:1m 30s

Question 187Question →

A man standing at a distance in front of a tall vertical cliff claps his hands and hears the echo after $1.2\text{ s}$ . If the speed of sound in air is $340\text{ m/s}$ , what is the distance between the man and the cliff in meters?

Show answer & explanation

Answer: 204

Answer

The distance between the man and the cliff is

204\text{ m}

.

An echo is formed when sound travels to an obstacle and reflects back. The time taken for the sound to travel to the cliff and back is

1.2\text{ s}

. The total distance covered by sound is

v \times t = 340\text{ m/s} \times 1.2\text{ s} = 408\text{ m}

. Since this distance covers two equal trips (to the cliff and back), the distance to the cliff is

408\text{ m} / 2 = 204\text{ m}

.

Step-by-Step Solution

1

State the relationship between sound speed, total echo time, and distance.

Total distance traveled by the sound is twice the distance to the cliff:

2d = v \times t

.

An echo involves sound traveling from the source to the reflecting barrier and back.

2

Substitute the given values into the equation.

2d = 340\text{ m/s} \times 1.2\text{ s} = 408\text{ m}

.

To find the total distance traversed by the sound wave.

3

Solve for the distance

d

.

d = \frac{408\text{ m}}{2} = 204\text{ m}

.

The one-way distance to the cliff is half of the total distance traveled by the echo.

Key Concept

Calculation of echo distance using

d = \frac{v t}{2}

Question 188Question →

If the expression $\frac{3 - \sqrt{5}}{3 + \sqrt{5}}$ is simplified and written in the form $a + b\sqrt{5}$ , where $a$ and $b$ are rational numbers, what is the numerical value of $a + b$ ?

Show answer & explanation

Answer: 2

Answer

The numerical value of

a + b

is

2

.

To express

\frac{3 - \sqrt{5}}{3 + \sqrt{5}}

in the standard form

a + b\sqrt{5}

, multiply both the numerator and denominator by the conjugate of the denominator, which is

(3 - \sqrt{5})

. The numerator expands to

(3 - \sqrt{5})^2 = 9 - 6\sqrt{5} + 5 = 14 - 6\sqrt{5}

. The denominator becomes

3^2 - (\sqrt{5})^2 = 9 - 5 = 4

. Dividing gives

\frac{14}{4} - \frac{6}{4}\sqrt{5} = \frac{7}{2} - \frac{3}{2}\sqrt{5}

. Hence,

a = \frac{7}{2}

and

b = -\frac{3}{2}

, making

a + b = \frac{7}{2} - \frac{3}{2} = \frac{4}{2} = 2

.

Step-by-Step Solution

1

Multiply numerator and denominator by the conjugate of the denominator

\frac{(3 - \sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}

To eliminate the surd from the denominator.

2

Expand both the numerator and the denominator

\frac{14 - 6\sqrt{5}}{4}

Using

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

for the numerator and difference of two squares

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

for the denominator.

3

Separate into rational component and radical coefficient

\frac{7}{2} - \frac{3}{2}\sqrt{5}

Simplifying fractions by dividing numerator and denominator by their greatest common divisor.

4

Calculate the sum

a + b

\frac{7}{2} - \frac{3}{2} = 2

Comparing

\frac{7}{2} - \frac{3}{2}\sqrt{5}

with

a + b\sqrt{5}

yields

a = \frac{7}{2}

and

b = -\frac{3}{2}

.

Key Concept

Rationalization of Binomial Denominators using Conjugates

Estimated Time:1m 30s

Question 189Question →

A metallic sphere weighs $5.0\text{ N}$ in air. When completely immersed in water, its apparent weight is $3.0\text{ N}$ . When completely immersed in an unknown liquid $X$ , its apparent weight is $3.4\text{ N}$ . What is the density of liquid $X$ in $\text{kg/m}^3$ ? (Take the density of water as $1000\text{ kg/m}^3$ and acceleration due to gravity $g = 10\text{ m/s}^2$ ).

Show answer & explanation

Answer: 800

Answer

The density of liquid

X

is

800\text{ kg/m}^3

.

The upthrust in water (

2.0\text{ N}

) gives the volume of the sphere as

2.0 \times 10^{-4}\text{ m}^3

. Using the upthrust in liquid X (

1.6\text{ N}

), the density of liquid X is calculated as

\rho_X = \frac{1.6}{(2.0 \times 10^{-4})(10)} = 800\text{ kg/m}^3

.

Step-by-Step Solution

1

Calculate upthrust in water

U_w = 5.0\text{ N} - 3.0\text{ N} = 2.0\text{ N}

Upthrust equals the loss in weight of the submerged body in water.

2

Determine the volume of the metallic sphere

V = \frac{U_w}{\rho_w g} = \frac{2.0}{1000 \times 10} = 2.0 \times 10^{-4}\text{ m}^3

According to Archimedes' principle, upthrust in water equals the weight of displaced water.

3

Calculate upthrust in liquid X

U_X = 5.0\text{ N} - 3.4\text{ N} = 1.6\text{ N}

Loss of weight in liquid X gives the upthrust exerted by liquid X.

4

Calculate the density of liquid X

\rho_X = \frac{U_X}{V g} = \frac{1.6}{(2.0 \times 10^{-4}) \times 10} = 800\text{ kg/m}^3

Rearranging

U_X = \rho_X V g

allows solving for the unknown fluid density.

Key Concept

Archimedes' Principle and Apparent Weight

Estimated Time:2m 0s

Question 190Question →

Find the value of $k$ if the point $P(k, 3)$ is equidistant from the points $A(1, 5)$ and $B(7, 1)$ .

Show answer & explanation

Answer: 4

Answer

The value of

k

is 4.

Using the distance formula, the squared distance

PA^2 = (k-1)^2 + (3-5)^2 = (k-1)^2 + 4

, and

PB^2 = (k-7)^2 + (3-1)^2 = (k-7)^2 + 4

. Equating

PA^2 = PB^2

gives

(k-1)^2 = (k-7)^2

. Expanding both sides yields

k^2 - 2k + 1 = k^2 - 14k + 49

. Subtracting

k^2

from both sides gives

12k = 48

, which leads to

k = 4

.

Step-by-Step Solution

1

Write the expressions for the squared distances

PA^2

and

PB^2

using the distance formula.

PA^2 = (k - 1)^2 + 4

and

PB^2 = (k - 7)^2 + 4

The distance formula between two points

(x_1, y_1)

and

(x_2, y_2)

is

d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2

.

2

Equate

PA^2

and

PB^2

since point

P

is equidistant from points

A

and

B

.

(k - 1)^2 + 4 = (k - 7)^2 + 4 \implies (k - 1)^2 = (k - 7)^2

Subtracting 4 from both sides simplifies the equality of squared distances.

3

Expand both sides and isolate

k

to find its numerical value.

k^2 - 2k + 1 = k^2 - 14k + 49 \implies 12k = 48 \implies k = 4

Canceling

k^2

terms yields a simple linear equation.

Key Concept

Equidistant points and the distance formula in coordinate geometry

Estimated Time:1m 30s

Question 191Question →

The maximum acceleration of a body oscillating in simple harmonic motion is $8\text{ m/s}^2$ . If the period of oscillation is $\pi\text{ s}$ , calculate the amplitude of the oscillation in meters.

Show answer & explanation

Answer: 2

Answer

The amplitude of the oscillation is

2.0\text{ m}

.

The correct answer of

2.0\text{ m}

is obtained by first deriving the angular frequency

\omega = \frac{2\pi}{T} = 2\text{ rad/s}

, and then using the relation

a_{\text{max}} = \omega^2 A

to solve for amplitude:

A = \frac{8}{2^2} = 2.0\text{ m}

.

Step-by-Step Solution

1

Calculate angular frequency (

\omega

) from the given period (

T

).

\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2\text{ rad/s}

Angular frequency specifies the rate of phase change in oscillations.

2

Apply the maximum acceleration formula for simple harmonic motion to determine amplitude (

A

).

a_{\text{max}} = \omega^2 A \implies 8 = 2^2 \times A \implies A = 2.0\text{ m}

In simple harmonic motion, maximum acceleration occurs at the extreme position and equals

\omega^2 A

.

Key Concept

Simple Harmonic Motion Acceleration and Period Relationship

Question 192Question →

A stone is projected vertically upwards from the top edge of a cliff $80\text{ m}$ high with an initial speed of $30\text{ m/s}$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the total time, in seconds, taken by the stone to reach the ground at the base of the cliff.

Show answer & explanation

Answer: 8

Answer

The total time taken by the stone to reach the ground at the base of the cliff is

8\text{ s}

.

Using the equation of motion

s = ut - \frac{1}{2}gt^2

with

s = -80\text{ m}

,

u = 30\text{ m/s}

, and

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

yields the quadratic equation

t^2 - 6t - 16 = 0

. Solving gives

t = 8\text{ s}

(ignoring the unphysical negative root

t = -2\text{ s}

). Alternatively, breaking the motion into two parts: time to reach maximum height (

30\text{ m/s} / 10\text{ m/s}^2 = 3\text{ s}

, covering

45\text{ m}

) plus time to fall from maximum height of

125\text{ m}

to the ground (

t = \sqrt{2(125)/10} = 5\text{ s}

), giving a total time of

3 + 5 = 8\text{ s}

.

Step-by-Step Solution

1

Set up the kinematic equation with appropriate vector signs

Displacement

s = -80\text{ m}

, initial velocity

u = +30\text{ m/s}

, acceleration

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

Since the ground is below the release point, displacement is negative when taking the upward direction as positive.

2

Substitute values into

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

-80 = 30t - 5t^2

Relates displacement, initial speed, time, and constant gravitational acceleration.

3

Form and solve the quadratic equation

5t^2 - 30t - 80 = 0 \implies t^2 - 6t - 16 = 0 \implies (t - 8)(t + 2) = 0

Simplifies the algebraic expression to find the time roots.

4

Select the physical root

t = 8\text{ s}

Time elapsed must be a positive quantity.

Key Concept

Kinematics of Vertical Motion under Gravity with Displacement from Elevation

Estimated Time:2m 0s

Question 193Question →

An ultrasonic rangefinder mounted on a drone sends a sound pulse vertically downward to measure its altitude above flat ground. If the echo is detected by the sensor $0.08\text{ s}$ after emission and the speed of sound in air is $340\text{ m s}^{-1}$ , what is the altitude of the drone in meters?

Show answer & explanation

Answer: 13.6

Answer

13.6 meters

The sound pulse emitted by the drone travels down to the ground and reflects back to the sensor. The relationship between speed

v

, total round-trip time

t

, and altitude

d

is given by

2d = v \times t

. Substituting

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

and

t = 0.08\text{ s}

yields

d = \frac{340 \times 0.08}{2} = 13.6\text{ m}

.

Step-by-Step Solution

1

Identify the total time taken by the sound pulse for the round trip.

Total round-trip time

t = 0.08\text{ s}

and speed of sound

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

.

Echo detection measures the time for sound to travel to a barrier and return.

2

Apply the echo distance relationship

2d = v \times t

to solve for altitude

d

.

d = \frac{340 \times 0.08}{2} = 13.6\text{ m}

.

Dividing the total path distance by 2 yields the one-way distance to the ground.

Key Concept

Calculation of distance using echoes and two-way sound wave propagation

Alternative Method

Determine the one-way travel time first:

t_{\text{one-way}} = \frac{0.08}{2} = 0.04\text{ s}

. Then calculate altitude directly using distance = speed × one-way time:

d = 340 \times 0.04 = 13.6\text{ m}

.

Estimated Time:1m 0s

Question 194Question →

A spherical particle of radius $3.0 \times 10^{-3}\text{ m}$ and density $8.0 \times 10^3\text{ kg/m}^3$ is released from rest and falls vertically through a tall column of a viscous liquid of density $2.0 \times 10^3\text{ kg/m}^3$ . If the coefficient of viscosity of the fluid is $0.40\text{ Pa}\cdot\text{s}$ and the acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the magnitude of its terminal velocity in $\text{m/s}$ .

Show answer & explanation

Answer: 0.3

Answer

The magnitude of the terminal velocity of the falling sphere is

0.3\text{ m/s}

.

When a body falls at terminal velocity through a viscous medium, its weight is balanced by the sum of buoyancy upthrust and Stokes' viscous drag force. Applying

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

with sphere radius

r = 0.003\text{ m}

, sphere density

\rho_s = 8000\text{ kg/m}^3

, fluid density

\rho_f = 2000\text{ kg/m}^3

, viscosity

\eta = 0.40\text{ Pa}\cdot\text{s}

, and

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

yields

v_t = 0.3\text{ m/s}

.

Step-by-Step Solution

1

Formulate the dynamic equilibrium condition at terminal velocity.

At terminal velocity, the net acceleration is zero, leading to the force balance equation

W = U + F_v

, where

W

is the gravitational weight of the sphere,

U

is the buoyant upthrust, and

F_v

is the retarding viscous force.

Terminal velocity occurs when the downward force of gravity is precisely balanced by the sum of upward resistive and buoyancy forces.

2

Substitute algebraic expressions for weight, upthrust, and Stokes' viscous drag.

W = \frac{4}{3}\pi r^3 \rho_s g

,

U = \frac{4}{3}\pi r^3 \rho_f g

, and

F_v = 6\pi \eta r v_t

.

Archimedes' principle defines the upthrust force equal to the weight of displaced liquid, while Stokes' law governs viscous resistance on spherical bodies.

3

Solve the equilibrium equation for terminal velocity

v_t

.

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

.

Equating

\frac{4}{3}\pi r^3 (\rho_s - \rho_f) g = 6\pi \eta r v_t

and simplifying cancels common factors of

\pi

and

r

.

4

Substitute the specified numerical parameters into the derived expression.

v_t = \frac{2 \times (3.0 \times 10^{-3})^2 \times (8000 - 2000) \times 10}{9 \times 0.40} = \frac{2 \times 9.0 \times 10^{-6} \times 6000 \times 10}{3.6} = \frac{1.08}{3.6} = 0.3\text{ m/s}

.

Direct calculation yields the exact value of terminal velocity.

Key Concept

Terminal Velocity, Stokes' Law, and Archimedes' Principle

Question 195Question →

A glass vessel has a linear expansivity of $1.0 \times 10^{-5} \text{ K}^{-1}$ and is filled with a liquid. If the apparent cubic expansivity of the liquid in this vessel is $1.5 \times 10^{-4} \text{ K}^{-1}$ , calculate the real cubic expansivity of the liquid in units of $10^{-4} \text{ K}^{-1}$ .

Show answer & explanation

Answer: 1.8

Answer

The real cubic expansivity of the liquid is

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

(giving

1.8

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 volume expansivity of the vessel (\gamma_r = \gamma_a + \gamma_v). First, convert the linear expansivity of the glass vessel to volume expansivity:

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

. Adding this to the apparent cubic expansivity (

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

) yields a real cubic expansivity of

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

.

Step-by-Step Solution

1

Determine the cubic expansivity of the vessel (\gamma_v)

\gamma_v = 3.0 \times 10^{-5} \text{ K}^{-1} = 0.3 \times 10^{-4} \text{ K}^{-1}

The volume (cubic) expansivity of a solid vessel is three times its linear expansivity (\gamma_v = 3\alpha).

2

Calculate the real cubic expansivity of the liquid (\gamma_r)

\gamma_r = 1.8 \times 10^{-4} \text{ K}^{-1}

Real cubic expansivity is the sum of apparent cubic expansivity and vessel cubic expansivity (\gamma_r = \gamma_a + \gamma_v).

Key Concept

Real and Apparent Cubic Expansivity of Liquids

Question 196Question →

A binary operation $\circ$ on the set of real numbers $\mathbb{R}$ is defined by $a \circ b = a + b + 2ab$ . If the identity element of the operation is $e$ , what is the value of $x$ such that the inverse of $x$ under $\circ$ is equal to $2$ ?

Show answer & explanation

Answer: -0.4

Answer

The value of

x

is

-0.4

.

First find the identity element

e

by solving

a \circ e = a

, which yields

a + e + 2ae = a \implies e(1 + 2a) = 0 \implies e = 0

. Next, by definition of an inverse,

x \circ x^{-1} = e

. Substituting

x^{-1} = 2

and

e = 0

gives

x \circ 2 = 0

. Expanding this using the binary operation rule yields

x + 2 + 2(x)(2) = 0 \implies 5x + 2 = 0 \implies x = -0.4

.

Step-by-Step Solution

1

Find the identity element

e

of the operation

\circ

e = 0

By definition of identity element,

a \circ e = a \implies a + e + 2ae = a

, which simplifies to

e(1 + 2a) = 0

, giving

e = 0

.

2

Set up the inverse equation using

x^{-1} = 2

x \circ 2 = 0

The inverse of

x

satisfies

x \circ x^{-1} = e

. Since

x^{-1} = 2

and

e = 0

,

x \circ 2 = 0

.

3

Solve for

x

x = -0.4

Expanding

x \circ 2 = 0

gives

x + 2 + 4x = 0 \implies 5x = -2 \implies x = -0.4

.

Key Concept

Identity and Inverse Elements in Binary Operations

Question 197Question →

The area of the region bounded by the parabola $y = kx - x^2$ (where $k > 0$ ) and the $x$ -axis between its $x$ -intercepts at $x = 0$ and $x = k$ is equal to $36$ square units. What is the value of the positive constant $k$ ?

Show answer & explanation

Answer: 6

Answer

The value of the positive constant

k

is 6.

The area bounded by

y = kx - x^2

and the x-axis from

x = 0

to

x = k

is obtained by integrating

kx - x^2

, which yields

\frac{k^3}{6}

. Setting

\frac{k^3}{6} = 36

gives

k^3 = 216

, whose cube root is

k = 6

.

Step-by-Step Solution

1

Set up the definite integral representing the area bounded by the curve and the x-axis between the intercepts

x = 0

and

x = k

.

\int_{0}^{k} (kx - x^2) \, dx = 36

The area under a curve

y = f(x)

above the x-axis from

x = a

to

x = b

is given by

\int_{a}^{b} f(x) \, dx

.

2

Find the antiderivative and evaluate it at the limits

x = k

and

x = 0

.

\left[ \frac{kx^2}{2} - \frac{x^3}{3} \right]_{0}^{k} = \left(\frac{k(k)^2}{2} - \frac{k^3}{3}\right) - 0 = \frac{k^3}{6}

Applying the integration power rule

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

and simplifying

\frac{k^3}{2} - \frac{k^3}{3} = \frac{k^3}{6}

.

3

Set the evaluated expression equal to 36 and solve for

k

.

\frac{k^3}{6} = 36 \implies k^3 = 216 \implies k = 6

Multiplying both sides by 6 yields

k^3 = 216

, and taking the cube root gives

k = 6

.

Key Concept

Definite Integral and Area Under Curve

Question 198Question →

The speed $v$ of a transverse wave traveling along a stretched string under tension $T$ with mass per unit length $\mu$ is given by $v = k T^x \mu^y$ , where $k$ is a dimensionless constant. What is the numerical value of the exponent $x$ ?

Show answer & explanation

Answer: 0.5

Answer

The numerical value of the exponent

x

is 0.5.

Applying the principle of dimensional homogeneity, the dimensions on both sides must match. Speed

[v] = L T^{-1}

, tension force

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

, and mass per unit length

[\mu] = M L^{-1}

. Substituting these into

v = k T^x \mu^y

yields

M^0 L^1 T^{-1} = M^{x+y} L^{x-y} T^{-2x}

. Comparing the exponents of

T

gives

-2x = -1

, leading to

x = 0.5

.

Step-by-Step Solution

1

Determine the dimensions of speed

v

, tension force

T

, and linear density

\mu

.

[v] = L T^{-1}

,

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

,

[\mu] = M L^{-1}

Tension is a force (

F=ma

) with dimensions

[M L T^{-2}]

, and

\mu

is mass per unit length (

m/l

) with dimensions

[M L^{-1}]

.

2

Substitute the dimensional formulas into the equation

v = k T^x \mu^y

and collect powers of base dimensions.

M^0 L^1 T^{-1} = M^{x+y} L^{x-y} T^{-2x}

Combining exponents for base dimensions

M

,

L

, and

T

allows applying the principle of dimensional homogeneity.

3

Equate the exponent of

T

on both sides to solve for

x

.

-2x = -1 \implies x = 0.5

The exponent of

T

on the left side is

-1

and on the right side is

-2x

.

Key Concept

Dimensional Analysis and Determination of Exponents

Estimated Time:1m 30s

Question 199Question →

A motorist travelling at a constant speed of $20\text{ m s}^{-1}$ directly towards a tall vertical cliff sounds a horn. If the motorist hears the echo of the horn $2.0\text{ s}$ later and the speed of sound in air is $340\text{ m s}^{-1}$ , what was the distance of the car from the cliff at the moment the horn was sounded?

Show answer & explanation

Answer: 360

Answer

The distance of the car from the cliff at the instant the horn was sounded was

360\text{ m}

.

When the motorist sounds the horn at an initial distance

D

from the cliff, the sound wave travels toward the cliff. In the

2.0\text{ s}

it takes for the echo to return, the car advances

40\text{ m}

toward the cliff (

20\text{ m s}^{-1} \times 2.0\text{ s}

). The returning echo meets the motorist at a distance of

(D - 40)\text{ m}

from the cliff. Consequently, the sound covers a total distance of

D + (D - 40) = 2D - 40\text{ m}

. Because the sound wave travels at

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

for

2.0\text{ s}

, the actual distance covered by sound is

340 \times 2.0 = 680\text{ m}

. Setting

2D - 40 = 680

gives

2D = 720\text{ m}

, which yields

D = 360\text{ m}

.

Step-by-Step Solution

1

Calculate the distance covered by the car while moving toward the cliff during the echo time interval.

d_{\text{car}} = 20\text{ m s}^{-1} \times 2.0\text{ s} = 40\text{ m}

.

The car continues to move closer to the cliff for the entire

2.0\text{ s}

period.

2

Set up an expression for the total distance covered by the sound wave.

d_{\text{sound}} = D + (D - 40) = 2D - 40\text{ m}

.

The sound travels forward a distance

D

to the cliff and reflects back to the car's updated location, which is

(D - 40)\text{ m}

from the cliff.

3

Calculate the distance travelled by sound using the given speed of sound.

d_{\text{sound}} = 340\text{ m s}^{-1} \times 2.0\text{ s} = 680\text{ m}

.

Sound propagates through air at

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

.

4

Equate the geometric path expression to the physical sound distance and solve for

D

.

2D - 40 = 680 \implies 2D = 720 \implies D = 360\text{ m}

.

Solving the equation yields the initial position of the car relative to the cliff.

Key Concept

Echo distance calculation with a moving observer

Question 200Question →

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.

Show answer & explanation

Answer: 200

Answer

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}

.

Step-by-Step Solution

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.

Key Concept

Fundamental frequency of a vibrating string fixed at both ends

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