All practice questions

Question 881Question →

A research rocket is launched vertically upwards from rest with a constant acceleration of $5.0\text{ m/s}^2$ . At an altitude of $250\text{ m}$ , its engine suddenly fails and the rocket continues to move vertically upward under gravity alone. Calculate the total time, in seconds, taken by the rocket from launch until it reaches its maximum height. (Take acceleration due to gravity $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 15

Answer

The total time taken from launch to reach maximum height is

15\text{ s}

.

The motion occurs in two phases. In phase 1, accelerating uniformly from rest at

5.0\text{ m/s}^2

over

250\text{ m}

yields a velocity of

50\text{ m/s}

in

10\text{ s}

. In phase 2, moving upward under gravity alone (

10\text{ m/s}^2

) reduces the velocity from

50\text{ m/s}

to rest (

0\text{ m/s}

) in

5\text{ s}

. Adding the durations of both phases gives

10\text{ s} + 5\text{ s} = 15\text{ s}

.

Step-by-Step Solution

1

Calculate the rocket's velocity and elapsed time at the moment of engine failure.

Velocity

v_1 = 50\text{ m/s}

and time

t_1 = 10\text{ s}

.

The rocket accelerates uniformly from rest at

5.0\text{ m/s}^2

over a distance of

250\text{ m}

.

2

Calculate the duration of the unpowered upward motion until vertical velocity becomes zero.

Unpowered flight time

t_2 = 5\text{ s}

.

After engine failure, the rocket acts as a free projectile moving upward against gravity (

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

) with an initial velocity of

50\text{ m/s}

.

3

Sum the time intervals of both stages.

Total time

t_{\text{total}} = 10\text{ s} + 5\text{ s} = 15\text{ s}

.

The total motion consists of two distinct stages: powered acceleration followed by gravitational deceleration.

Key Concept

Multi-stage vertical motion under constant acceleration followed by free-fall under gravity

Question 882Question →

The energy density $u$ (defined as energy per unit volume) stored in an electrostatic field is related to the permittivity of free space $\epsilon_0$ and the electric field strength $E$ by the dimensional formula $u = k \epsilon_0^x E^y$ , where $k$ is a dimensionless constant. What is the value of the numerical exponent $y$ ?

Show answer & explanation

Answer: 2

Answer

The value of the exponent

y

is 2.

By writing the dimensions of energy density

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

, permittivity

[M^{-1} L^{-3} T^4 I^2]

, and electric field strength

[M L T^{-3} I^{-1}]

, equating powers of electric current

I

yields

2x - y = 0

(or

y = 2x

). Substituting this into the equation for powers of mass

M

,

-x + y = 1

, yields

-x + 2x = 1

, so

x = 1

and

y = 2

.

Step-by-Step Solution

1

Derive the dimensional formulas for energy density

u

, permittivity

\epsilon_0

, and electric field

E

.

[u] = M L^{-1} T^{-2}, [\epsilon_0] = M^{-1} L^{-3} T^4 I^2, [E] = M L T^{-3} I^{-1}.

Expressing quantities in terms of base dimensions (M, L, T, I) is required for dimensional homogeneity.

2

Form the dimensional equation

u = k \epsilon_0^x E^y

and combine powers.

M L^{-1} T^{-2} = M^{-x+y} L^{-3x+y} T^{4x-3y} I^{2x-y}.

Applies the principle of dimensional consistency across the formula.

3

Equate corresponding powers of base dimensions to set up equations for

x

and

y

.

For I: 2x - y = 0; for M: -x + y = 1.

Base unit exponents on both sides of a physically valid equation must match.

4

Solve the algebraic equations for the unknown exponent

y

.

x = 1, y = 2.

Substituting y = 2x into -x + y = 1 directly gives x = 1 and y = 2.

Key Concept

Dimensional Analysis and Dimensional Homogeneity

Estimated Time:1m 30s

Question 883Question →

What is the positive value of $x$ that satisfies the equation $2^{2x+1} - 9 \cdot 2^x + 4 = 0$ ?

Show answer & explanation

Answer: 2

Answer

The positive value of

x

that satisfies the equation is 2.

Applying the law of indices

a^{m+n} = a^m \cdot a^n

gives

2^{2x+1} = 2 \cdot (2^x)^2

. Setting

y = 2^x

yields the quadratic equation

2y^2 - 9y + 4 = 0

. Factoring this expression gives

(2y - 1)(y - 4) = 0

, which yields roots

y = \frac{1}{2}

and

y = 4

. Solving

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

gives

x = -1

, and solving

2^x = 4

gives

x = 2

. The positive value is

2

.

Step-by-Step Solution

1

Use index laws to express the equation in terms of

2^x

2 \cdot (2^x)^2 - 9 \cdot (2^x) + 4 = 0

By the product law of indices,

2^{2x+1} = 2^{2x} \cdot 2^1 = 2 \cdot (2^x)^2

.

2

Substitute

y = 2^x

to form a quadratic equation

2y^2 - 9y + 4 = 0

Replacing

2^x

with a single variable simplifies the exponential equation into quadratic form.

3

Solve the quadratic equation for

y

y = \frac{1}{2}

or

y = 4

Factoring

2y^2 - 9y + 4 = 0

gives

(2y - 1)(y - 4) = 0

.

4

Substitute back

y = 2^x

to solve for

x

x = -1

or

x = 2

Since

2^x = \frac{1}{2} = 2^{-1}

,

x = -1

. Since

2^x = 4 = 2^2

,

x = 2

.

5

Select the positive value requested by the question

x = 2

x = 2

is positive, whereas

x = -1

is negative.

Key Concept

Reducing exponential equations to quadratic form using index laws

Estimated Time:1m 30s

Question 884Question →

Determine the smallest integer value of $x$ that satisfies the compound inequality $3 < 2x - 5 \le 11$ .

Show answer & explanation

Answer: 5

Answer

The smallest integer value of

x

that satisfies the inequality is 5.

Adding 5 across the compound inequality

3 < 2x - 5 \le 11

gives

8 < 2x \le 16

. Dividing by 2 yields

4 < x \le 8

. The integer values satisfying this range are 5, 6, 7, and 8. Therefore, the smallest integer solution is 5.

Step-by-Step Solution

1

Add 5 to all parts of the compound inequality

8 < 2x <= 16

Isolates the variable term in the middle segment.

2

Divide all parts of the compound inequality by 2

4 < x <= 8

Solves for x without changing inequality signs since 2 is positive.

3

Identify integer solutions within the range (4, 8]

x in {5, 6, 7, 8}

Since the inequality at 4 is strict (<), 4 is excluded, but 8 is included (<=).

4

Find the minimum integer value in the solution set

5

5 is the smallest integer strictly greater than 4.

Key Concept

Solving Compound Linear Inequalities

Question 885Question →

A steel rod and a brass rod are arranged such that the difference between their lengths remains constant at $10\text{ cm}$ at all temperatures. If the linear expansivity of steel is $1.2 \times 10^{-5}\text{ K}^{-1}$ and that of brass is $1.8 \times 10^{-5}\text{ K}^{-1}$ , what is the initial length of the steel rod in centimetres?

Show answer & explanation

Answer: 30

Answer

The initial length of the steel rod is

30\text{ cm}

.

For the length difference between two rods to remain constant regardless of temperature change, both rods must undergo equal absolute expansion ($\Delta L_1 = \Delta L_2$). Since $\Delta L = L_0 \alpha \Delta T$, this requires $L_1 \alpha_1 = L_2 \alpha_2$. Substituting $L_{\text{brass}} = L_{\text{steel}} - 10\text{ cm}$ and the given expansivity values gives $1.2 \times 10^{-5} L_{\text{steel}} = 1.8 \times 10^{-5} (L_{\text{steel}} - 10)$, which simplifies to $0.6 L_{\text{steel}} = 18$, giving $L_{\text{steel}} = 30\text{ cm}$.

Step-by-Step Solution

1

Relate the condition for a constant difference in length to individual expansions

$\Delta L_{\text{steel}} = \Delta L_{\text{brass}}$

If the difference between the two lengths is constant across temperature changes, both rods must increase in length by the exact same amount for any given temperature change.

2

Apply the linear thermal expansion formula to both rods

L_{\text{steel}} \alpha_{\text{steel}} = L_{\text{brass}} \alpha_{\text{brass}}

Since $\Delta L = L_0 \alpha \Delta T$, setting $\Delta L_{\text{steel}} = \Delta L_{\text{brass}}$ gives $L_{\text{steel}} \alpha_{\text{steel}} \Delta T = L_{\text{brass}} \alpha_{\text{brass}} \Delta T$. Cancelling $\Delta T$ yields $L_{\text{steel}} \alpha_{\text{steel}} = L_{\text{brass}} \alpha_{\text{brass}}$.

3

Substitute the length relationship into the equation

$L_{\text{steel}} (1.2 \times 10^{-5}) = (L_{\text{steel}} - 10) (1.8 \times 10^{-5})$

Because brass has a larger linear expansivity than steel, the brass rod must be shorter than the steel rod so that their products of length and expansivity remain equal, hence $L_{\text{brass}} = L_{\text{steel}} - 10\text{ cm}$.

4

Solve for the length of the steel rod

$L_{\text{steel}} = 30\text{ cm}$

Dividing both sides by $10^{-5}$ gives $1.2 L_{\text{steel}} = 1.8 L_{\text{steel}} - 18$. Rearranging gives $0.6 L_{\text{steel}} = 18$, which yields $L_{\text{steel}} = \frac{18}{0.6} = 30\text{ cm}$.

Key Concept

Equal absolute linear expansion for constant length difference

Estimated Time:2m 0s

Question 886Question →

A point $P(x, y)$ moves in the Cartesian plane such that it maintains a constant distance of $10\text{ units}$ from a fixed point $C(2, -3)$ . If the locus of $P$ intersects the vertical line $x = 8$ at two points $A$ and $B$ , what is the distance between $A$ and $B$ ?

Show answer & explanation

Answer: 16

Answer

The distance between the intersection points A and B is 16 units.

The locus of point P moving at a constant distance of 10 units from C(2, -3) forms a circle

(x - 2)^2 + (y + 3)^2 = 100

. Substituting

x = 8

yields

(y + 3)^2 = 64

, giving

y = 5

and

y = -11

. The distance between the two points (8, 5) and (8, -11) along the vertical line

x = 8

is

5 - (-11) = 16

units.

Step-by-Step Solution

1

Determine the equation representing the locus of point P

(x - 2)^2 + (y + 3)^2 = 100

The locus of a point moving at a fixed distance from a fixed point is a circle with equation

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

.

2

Substitute

x = 8

into the locus equation to find the

y

-coordinates of the intersection points

(8 - 2)^2 + (y + 3)^2 = 100 \implies 36 + (y + 3)^2 = 100 \implies (y + 3)^2 = 64

The intersection points lie on both the locus circle and the vertical line

x = 8

.

3

Solve for the two possible values of

y

y + 3 = \pm 8 \implies y_1 = 5

and

y_2 = -11

Taking the square root gives both positive and negative solutions for the vertical coordinate.

4

Calculate the vertical distance between points

A(8, 5)

and

B(8, -11)

\text{Distance} = 5 - (-11) = 16\text{ units}

Since both points have the same

x

-coordinate (

x = 8

), the distance is simply the absolute difference between their

y

-coordinates.

Key Concept

Locus of a point at a constant distance from a fixed point (Circle)

Question 887Question →

A shell is launched from level ground into the air. It reaches a maximum height of $45\text{ m}$ above the ground and has a total horizontal range of $240\text{ m}$ . Taking $g = 10\text{ m/s}^2$ , calculate the magnitude of the initial launch velocity of the shell in $\text{m/s}$ .

Show answer & explanation

Answer: 50

Answer

The initial launch velocity of the shell is

50\text{ m/s}

.

Combining the expressions for maximum height

H = \frac{u^2\sin^2\theta}{2g}

and range

R = \frac{2u^2\sin\theta\cos\theta}{g}

gives

\tan\theta = \frac{4H}{R}

. With

H = 45\text{ m}

and

R = 240\text{ m}

, we get

\tan\theta = 0.75 = \frac{3}{4}

, which yields

\sin\theta = 0.6

. Substituting these into the height equation yields

45 = \frac{u^2(0.6)^2}{20}

, solving to

u = 50\text{ m/s}

.

Step-by-Step Solution

1

Express the launch angle in terms of maximum height and horizontal range

\tan\theta = \frac{4H}{R} = \frac{4 \times 45}{240} = 0.75

Dividing the maximum height formula

H = \frac{u^2\sin^2\theta}{2g}

by the horizontal range formula

R = \frac{2u^2\sin\theta\cos\theta}{g}

yields

\frac{H}{R} = \frac{1}{4}\tan\theta

.

2

Find the sine of the launch angle from the tangent value

\sin\theta = 0.6

For a right-angled triangle with

\tan\theta = \frac{3}{4}

, the hypotenuse is

\sqrt{3^2 + 4^2} = 5

, giving

\sin\theta = \frac{3}{5} = 0.6

.

3

Calculate the magnitude of the initial velocity

u

u = 50\text{ m/s}

Substituting values into

H = \frac{u^2\sin^2\theta}{2g}

gives

45 = \frac{u^2 (0.6)^2}{2(10)} \implies 900 = 0.36 u^2 \implies u = 50\text{ m/s}

.

Key Concept

Interdependence of Maximum Height, Range, and Launch Velocity in Projectile Motion

Estimated Time:1m 30s

Question 888Question →

A beam of cathode rays (electrons) traveling at a speed of $4.00 \times 10^{6}\text{ m s}^{-1}$ enters a region with a uniform magnetic field of $5.00 \times 10^{-4}\text{ T}$ directed perpendicular to the beam. Taking the specific charge of an electron ( $\frac{e}{m}$ ) to be $1.60 \times 10^{11}\text{ C kg}^{-1}$ , what is the radius of the circular path traced by the cathode rays, in centimeters?

Show answer & explanation

Answer: 5

Answer

The radius of the circular path followed by the cathode rays is 5.0 cm.

When cathode rays enter a uniform magnetic field at right angles, the magnetic force acts as a centripetal force causing the electron beam to trace a circular arc of radius

r = \frac{v}{(e/m)B}

. Substituting

v = 4.00 \times 10^6\text{ m s}^{-1}

,

e/m = 1.60 \times 10^{11}\text{ C kg}^{-1}

, and

B = 5.00 \times 10^{-4}\text{ T}

yields

r = 0.05\text{ m}

, which equals

5.0\text{ cm}

.

Step-by-Step Solution

1

Set the magnetic Lorentz force equal to the required centripetal force for circular motion.

e v B = \frac{m v^2}{r}

Cathode rays consist of moving electrons experiences a magnetic force perpendicular to both their velocity and the magnetic field.

2

Express the radius

r

in terms of speed

v

, magnetic field

B

, and specific charge

\frac{e}{m}

.

r = \frac{v}{\left(\frac{e}{m}\right) B}

Simplifying the force balance equation isolates the radius on one side.

3

Substitute the given numerical values into the expression for

r

.

r = \frac{4.00 \times 10^{6}\text{ m s}^{-1}}{\left(1.60 \times 10^{11}\text{ C kg}^{-1}\right) \times \left(5.00 \times 10^{-4}\text{ T}\right)} = 0.05\text{ m}

Calculating the denominator gives

8.00 \times 10^{7}\text{ C T kg}^{-1}

, leading to

0.05\text{ m}

.

4

Convert the radius from meters to centimeters as requested by the question.

r = 0.05\text{ m} \times 100\text{ cm m}^{-1} = 5.0\text{ cm}

1 meter equals 100 centimeters.

Key Concept

Deflection of Cathode Rays in a Magnetic Field

Estimated Time:1m 30s

Question 889Question →

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 890Question →

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 891Question →

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 892Question →

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 893Question →

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 894Question →

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 895Question →

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 896Question →

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

Question 897Question →

Points $P(k, 2)$ and $Q(3, 8)$ lie on a straight line $L_1$ . If $L_1$ is perpendicular to the line $L_2$ given by $4x + 3y - 12 = 0$ , what is the value of $k$ ?

Show answer & explanation

Answer: -5

Answer

The value of

k

is

-5

.

To determine

k

, the gradient of

L_2

(

4x + 3y - 12 = 0

) is found to be

-\frac{4}{3}

. Using the perpendicularity rule

m_1 \cdot m_2 = -1

, the gradient of

L_1

is

\frac{3}{4}

. Equating this to the slope formula

\frac{8 - 2}{3 - k}

gives

\frac{6}{3 - k} = \frac{3}{4}

, which simplifies to

k = -5

.

Step-by-Step Solution

1

Find the gradient

m_2

of line

L_2

m_2 = -\frac{4}{3}

Converting

4x + 3y - 12 = 0

to

y = mx + c

form gives

y = -\frac{4}{3}x + 4

.

2

Apply the perpendicular line condition to find

m_1

m_1 = \frac{3}{4}

Perpendicular lines have negative reciprocal gradients (

m_1 \cdot m_2 = -1

).

3

Express the gradient

m_1

using the coordinates of

P

and

Q

m_1 = \frac{6}{3 - k}

Using the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

for points

P(k, 2)

and

Q(3, 8)

.

4

Solve for

k

k = -5

Equating

\frac{6}{3 - k} = \frac{3}{4}

yields

3(3 - k) = 24 \implies 9 - 3k = 24 \implies k = -5

.

Key Concept

Perpendicular Lines and Gradient Formula

Question 898Question →

A ray of light travels from air into a liquid with a refractive index of $1.33$ . If the sine of the angle of incidence in air is $0.80$ , what is the sine of the angle of refraction in the liquid?

Show answer & explanation

Answer: 0.6

Answer

The sine of the angle of refraction in the liquid is 0.60.

According to Snell's law for light passing from air into a medium, the refractive index

n

is given by

n = \frac{\sin i}{\sin r}

. Rearranging this equation to solve for the sine of the angle of refraction yields

\sin r = \frac{\sin i}{n}

. Substituting

\sin i = 0.80

and

n = 1.33

(or

\frac{4}{3}

) gives

\sin r = \frac{0.80}{4/3} = 0.60

.

Step-by-Step Solution

1

Identify the given physical parameters and state Snell's law

Refractive index

n = 1.33

(or

\frac{4}{3}

),

\sin i = 0.80

. Snell's law:

n = \frac{\sin i}{\sin r}

Snell's law relates the ratio of the sines of the angles of incidence and refraction to the refractive index of the medium.

2

Rearrange the equation to express the sine of the angle of refraction

\sin r = \frac{\sin i}{n}

Algebraically isolating

\sin r

allows direct substitution of the known quantities.

3

Substitute the values and compute the result

\sin r = \frac{0.80}{4/3} = 0.60

Dividing

0.80

by

\frac{4}{3}

gives

0.60

.

Key Concept

Snell's Law of Refraction

Alternative Method

Convert decimal numbers into simple fractions:

n = \frac{4}{3}

and

\sin i = \frac{4}{5}

. Evaluating

\sin r = \frac{4/5}{4/3}

simplifies directly to

\frac{3}{5} = 0.60

.

Estimated Time:45s

Question 899Question →

How many integer values of $x$ satisfy the quadratic inequality $2x^2 - 7x - 4 \le 0$ ?

Show answer & explanation

Answer: 5

Answer

There are 5 integer values of x that satisfy the inequality.

Solving the quadratic inequality yields

-\frac{1}{2} \le x \le 4

. The integer solutions within this interval are

0, 1, 2, 3,

and

4

. Counting them gives a total of

5

valid integer values.

Step-by-Step Solution

1

Factor the quadratic expression.

(2x + 1)(x - 4) \le 0

Factoring allows determination of the critical boundary points.

2

Find the critical values (roots of the equation).

x = -\frac{1}{2}

and

x = 4

The roots divide the number line into test intervals.

3

Determine the solution set of the inequality.

-\frac{1}{2} \le x \le 4

Since the quadratic coefficient is positive, the quadratic expression is non-positive between its roots.

4

List and count the integers within the range.

The integers are

0, 1, 2, 3, 4

, making a total of

5

integers.

Counting only whole numbers in the closed interval

[ -0.5, 4 ]

.

Key Concept

Quadratic Inequalities and Integer Solution Counting

Question 900Question →

A metallic wire has a resistance of $12.0\,\Omega$ at $0\,^\circ\text{C}$ and a temperature coefficient of resistance of $4.0 \times 10^{-3}\,^\circ\text{C}^{-1}$ . The wire is uniformly stretched until its length increases by $25\%$ . Assuming the density and total volume of the wire remain constant during stretching, what is the resistance of the stretched wire at $50\,^\circ\text{C}$ ?

Show answer & explanation

Answer: 22.5

Answer

The resistance of the stretched wire at

50\,^\circ\text{C}

is

22.5\,\Omega

.

Stretching a wire by

25\%

increases its length by a factor of

1.25

and reduces its cross-sectional area by a factor of

1.25

(since volume is conserved). The resistance at

0\,^\circ\text{C}

scales as

(1.25)^2 = 1.5625

, giving

18.75\,\Omega

. Accounting for the temperature increase to

50\,^\circ\text{C}

via

R(T) = R'_0(1 + \alpha T)

yields

18.75 \times (1 + 4.0 \times 10^{-3} \times 50) = 18.75 \times 1.20 = 22.5\,\Omega

.

Step-by-Step Solution

1

Calculate the resistance of the wire at

0\,^\circ\text{C}

after uniform stretching.

R'_0 = 18.75\,\Omega

Uniform stretching by

25\%

increases length to

L' = 1.25 L_0

. Volume conservation (

V = A L

) requires area to decrease to

A' = A_0 / 1.25

. Since

R = \rho L / A

,

R'_0 = R_0 (L'/L_0)^2 = 12.0 \times (1.25)^2 = 18.75\,\Omega

.

2

Apply the temperature coefficient formula to calculate resistance at

50\,^\circ\text{C}

.

R(50) = 22.5\,\Omega

Using

R(T) = R'_0 (1 + \alpha T)

, substitute

R'_0 = 18.75\,\Omega

,

\alpha = 4.0 \times 10^{-3}\,^\circ\text{C}^{-1}

, and

T = 50\,^\circ\text{C}

to find

R(50) = 18.75 \times (1 + 0.20) = 22.5\,\Omega

.

Key Concept

Combined effects of dimensional deformation and temperature on electrical resistance

Answer

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