All practice questions

What is the value of $\frac{1}{\log_2 100} + \frac{1}{\log_5 100}$ ?

\frac{1}{2}

1

2

\frac{1}{7}

Show answer & explanation

Answer: $\frac{1}{2}$

Answer

\frac{1}{2}

Using the change of base identity

\frac{1}{\log_a b} = \log_b a

, the expression simplifies to

\log_{100} 2 + \log_{100} 5

. By the product rule of logarithms, this equals

\log_{100}(2 \times 5) = \log_{100} 10

. Since

100^{\frac{1}{2}} = 10

, the value is

\frac{1}{2}

.

Step-by-Step Solution

1

Apply the reciprocal change of base rule

\frac{1}{\log_a b} = \log_b a

.

\frac{1}{\log_2 100} = \log_{100} 2

and

\frac{1}{\log_5 100} = \log_{100} 5

.

Converting to a common base of

100

allows the use of logarithmic addition laws.

2

Combine the two logarithms using the product law

\log_c x + \log_c y = \log_c (xy)

.

\log_{100} 2 + \log_{100} 5 = \log_{100} (2 \times 5) = \log_{100} 10.

Adding logarithms with the same base is equivalent to taking the logarithm of the product of their arguments.

3

Evaluate

\log_{100} 10

.

Since

100^{\frac{1}{2}} = 10

,

\log_{100} 10 = \frac{1}{2}

.

The logarithm asks what power base

100

must be raised to in order to equal

10

.

Key Concept

Logarithms and Change of Base

Estimated Time:1m 30s

Question 7882Question →

A bag contains red, blue, and yellow marbles. The theoretical probability of selecting a red marble at random from the bag is $\frac{2}{5}$ . In a probability experiment, a marble is drawn from the bag, its color recorded, and then replaced. This procedure is repeated $250$ times, resulting in a red marble being drawn $115$ times. What is the absolute difference between the observed experimental frequency of red marbles and the theoretical expected frequency?

Show answer & explanation

Answer: 15

Answer

15

The theoretical expected frequency of selecting a red marble over

250

trials is found by multiplying the total trials by the theoretical probability:

250 \times \frac{2}{5} = 100

. The experimental frequency observed was

115

. Taking the absolute difference gives

|115 - 100| = 15

.

Step-by-Step Solution

1

Calculate the theoretical expected frequency of red marble outcomes.

Theoretical expected frequency =

250 \times \frac{2}{5} = 100

.

Expected frequency is determined by multiplying the number of trials (

N = 250

) by the theoretical probability (

P = \frac{2}{5}

).

2

Find the absolute difference between the observed experimental frequency and the theoretical expected frequency.

|115 - 100| = 15

.

The observed experimental frequency is

115

and the theoretical expected frequency is

100

, so the positive difference is

15

.

Key Concept

Experimental Frequency vs. Theoretical Expected Frequency

Estimated Time:1m 30s

Question 7883Question →

An unpolarized light beam with an initial intensity of $80\text{ W/m}^2$ passes through an ideal linear polarizer. What is the intensity of the transmitted light in $\text{W/m}^2$ ?

Show answer & explanation

Answer: 40

Answer

The intensity of the transmitted light is

40\text{ W/m}^2

.

When unpolarized light of initial intensity

I_0

encounters an ideal linear polarizing filter, the transmitted intensity

I

is always equal to half of the incident intensity (

I = \frac{1}{2}I_0

). Substituting

80\text{ W/m}^2

gives

I = 40\text{ W/m}^2

.

Step-by-Step Solution

1

Determine the fraction of unpolarized light intensity transmitted by a polarizer.

Transmitted intensity formula

I = \frac{1}{2} I_0

.

Unpolarized light consists of randomly oriented electric field vectors, resulting in an average transmission factor of one-half.

2

Substitute the incident intensity value into the formula.

I = \frac{80}{2} = 40\text{ W/m}^2

.

Direct mathematical calculation.

Key Concept

Polarization and intensity reduction of unpolarized light upon passing through a linear polarizer.

Question 7884Question →

A conductor wire with a cross-sectional area of $2.0 \times 10^{-6}\,\text{m}^2$ carries a steady current of $1.6\,\text{A}$ . If the conduction electron density of the material is $5.0 \times 10^{28}\,\text{m}^{-3}$ and the elementary charge is $1.6 \times 10^{-19}\,\text{C}$ , what is the average drift velocity of the electrons in the wire?

1.0 \times 10^{-4}\,\text{m/s}

1.0 \times 10^{-6}\,\text{m/s}

1.6 \times 10^{-23}\,\text{m/s}

4.0 \times 10^{-16}\,\text{m/s}

Show answer & explanation

Answer: $1.0 \times 10^{-4}\,\text{m/s}$

Answer

The average drift velocity of the electrons is

1.0 \times 10^{-4}\,\text{m/s}

.

The correct answer is derived from the fundamental relationship

I = n A e v_d

. Solving for drift velocity gives

v_d = \frac{I}{n A e} = \frac{1.6}{5.0 \times 10^{28} \times 2.0 \times 10^{-6} \times 1.6 \times 10^{-19}} = 1.0 \times 10^{-4}\,\text{m/s}

.

Step-by-Step Solution

1

Identify the drift velocity formula relating current to charge carrier parameters

The electric current is given by

I = n A e v_d

, where

I

is current,

n

is electron density,

A

is cross-sectional area,

e

is elementary charge, and

v_d

is drift velocity.

This formula connects macroscopic electric current to microscopic charge dynamics.

2

Rearrange the equation to solve for drift velocity

v_d

v_d = \frac{I}{n A e}

Isolating the unknown variable

v_d

before substituting known values.

3

Substitute the given values into the expression

v_d = \frac{1.6}{(5.0 \times 10^{28}) \times (2.0 \times 10^{-6}) \times (1.6 \times 10^{-19})}

Inserting

I = 1.6\,\text{A}

,

n = 5.0 \times 10^{28}\,\text{m}^{-3}

,

A = 2.0 \times 10^{-6}\,\text{m}^2

, and

e = 1.6 \times 10^{-19}\,\text{C}

.

4

Evaluate the denominator and compute the final value of

v_d

Denominator

= (5.0 \times 2.0 \times 1.6) \times 10^{28 - 6 - 19} = 16.0 \times 10^3 = 1.6 \times 10^4

. Thus,

v_d = \frac{1.6}{1.6 \times 10^4} = 1.0 \times 10^{-4}\,\text{m/s}

.

Simplifying powers of ten gives the final numerical answer.

Key Concept

Relationship between Electric Current and Drift Velocity

Estimated Time:1m 30s

Question 7885Question →

A liquid has a density of $840\text{ kg m}^{-3}$ at $10^\circ\text{C}$ . It is heated inside a container whose material has a linear expansivity of $2.0 \times 10^{-5}\text{ K}^{-1}$ . If the measured apparent cubic expansivity of the liquid in this container is $4.4 \times 10^{-4}\text{ K}^{-1}$ , what is the density of the liquid at $110^\circ\text{C}$ in $\text{kg m}^{-3}$ ?

Show answer & explanation

Answer: 800

Answer

800 kg m^-3

To find the density of the liquid at the elevated temperature, we must use its real cubic expansivity

\gamma_r

. The vessel expands with a cubic expansivity of

\gamma_v = 3\alpha = 6.0 \times 10^{-5}\text{ K}^{-1}

. The real cubic expansivity of the liquid is

\gamma_r = \gamma_a + \gamma_v = 4.4 \times 10^{-4} + 0.6 \times 10^{-4} = 5.0 \times 10^{-4}\text{ K}^{-1}

. Using the density variation formula

\rho_2 = \frac{\rho_1}{1 + \gamma_r \Delta T}

, we evaluate

\frac{840}{1 + (5.0 \times 10^{-4} \times 100)} = \frac{840}{1.05} = 800\text{ kg m}^{-3}

.

Step-by-Step Solution

1

Calculate the cubic expansivity of the container material

\gamma_v = 6.0 \times 10^{-5}\text{ K}^{-1}

The volume (cubic) expansivity of a solid container is three times its linear expansivity (

\gamma_v = 3\alpha

).

2

Determine the real cubic expansivity of the liquid

\gamma_r = 5.0 \times 10^{-4}\text{ K}^{-1}

The real cubic expansivity is the sum of the apparent cubic expansivity and the cubic expansivity of the container (

\gamma_r = \gamma_a + \gamma_v

).

3

Determine the change in temperature

\Delta T = 100\text{ K}

Subtract the initial temperature from the final temperature:

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

.

4

Calculate the final density of the liquid at

110^\circ\text{C}

\rho_2 = 800\text{ kg m}^{-3}

Density varies inversely with volumetric expansion according to

\rho_2 = \frac{\rho_1}{1 + \gamma_r \Delta T}

.

Key Concept

Relationship between real expansivity, apparent expansivity, container expansion, and density change in fluids

Question 7886Question →

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 7887Question →

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 7888Question →

A rifle of mass $4.0\text{ kg}$ fires a bullet of mass $0.01\text{ kg}$ with a velocity of $400\text{ m s}^{-1}$ . What is the recoil velocity of the rifle?

-1.0\text{ m s}^{-1}

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

-100\text{ m s}^{-1}

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

Show answer & explanation

Answer: $-1.0\text{ m s}^{-1}$

Answer

The recoil velocity of the rifle is

-1.0\text{ m s}^{-1}

(or

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

in the direction opposite to the bullet).

The system starts at rest, so the initial total momentum is zero. By the law of conservation of linear momentum, the total final momentum must also be zero:

m_{rifle}v_{rifle} + m_{bullet}v_{bullet} = 0

. Substituting

m_{rifle} = 4.0\text{ kg}

,

m_{bullet} = 0.01\text{ kg}

, and

v_{bullet} = 400\text{ m s}^{-1}

gives

4.0 v_{rifle} + 4.0 = 0

, which yields

v_{rifle} = -1.0\text{ m s}^{-1}

. The negative sign indicates that the rifle moves in the direction opposite to the bullet.

Step-by-Step Solution

1

State the principle of conservation of linear momentum

Total Initial Momentum = Total Final Momentum = 0

Before firing, both the rifle and bullet are at rest.

2

Set up the linear momentum conservation equation

m_{rifle} v_{rifle} + m_{bullet} v_{bullet} = 0

The sum of the final momenta of the system components must equal zero.

3

Substitute the given numerical values into the equation and solve for recoil velocity

4.0 \cdot v_{rifle} + (0.01 \cdot 400) = 0 \implies 4.0 \cdot v_{rifle} + 4 = 0 \implies v_{rifle} = -1.0\text{ m s}^{-1}

Solving the linear algebraic equation yields the exact magnitude and direction of the recoil velocity.

Key Concept

Law of Conservation of Linear Momentum and Newton's Third Law of Motion

Estimated Time:45s

Question 7889Question →

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 7890Question →

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 7891Question →

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 7892Question →

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 7893Question →

Match each length measuring instrument listed on the left with its standard precision or suitable measurement application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Micrometer screw gauge matches with precision

0.01\text{ mm}

for fine wire diameter; Vernier caliper matches with precision

0.01\text{ cm}

for tube diameters; Metre rule matches with precision

0.1\text{ cm}

for laboratory lengths; Tape measure matches with flexible long-distance measuring.

Each instrument is correctly matched to its standard least count and characteristic physical measurement task: Micrometer screw gauge (

0.01\text{ mm}

), Vernier caliper (

0.01\text{ cm}

), Metre rule (

0.1\text{ cm}

), and Tape measure for long flexible lengths.

Step-by-Step Solution

1

Identify the least count and primary application for each measuring tool.

Micrometer screw gauge:

0.01\text{ mm}

(wires); Vernier caliper:

0.01\text{ cm}

(internal/external tube diameters); Metre rule:

0.1\text{ cm}

(standard lab objects); Tape measure: large flexible distances.

Matching instruments requires recalling their resolution (least count) and physical design features.

Key Concept

Instrument least counts and appropriate selection based on object dimensions.

Question 7894Question →

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 7895Question →

The table below shows the distribution of test scores obtained by a group of students:

Score	1	2	3	4	5
Frequency	3	$x$	4	2	1

If the mean score of the distribution is $2.5$ , what is the median score?

2

2.5

3

6

Show answer & explanation

Answer: $2$

Answer

The median score is

2

.

First, calculate the missing frequency

x

using the mean formula

\bar{x} = \frac{\sum fx}{\sum f} = \frac{28 + 2x}{10 + x} = 2.5

, which yields

x = 6

. Total number of scores is

N = 16

. The median is the mean of the

8\text{th}

and

9\text{th}

values. Looking at cumulative frequencies, score 1 covers positions 1–3 and score 2 covers positions 4–9. Thus, both the

8\text{th}

and

9\text{th}

scores are 2, giving a median of 2.

Step-by-Step Solution

1

Set up the equation for the mean using

\bar{x} = \frac{\sum fx}{\sum f}

.

Total frequency

\sum f = 3 + x + 4 + 2 + 1 = 10 + x

, and total score sum

\sum fx = 1(3) + 2(x) + 3(4) + 4(2) + 5(1) = 28 + 2x

.

The mean of an ungrouped frequency distribution is the sum of all score-frequency products divided by the total frequency.

2

Solve for the unknown frequency

x

.

2.5 = \frac{28 + 2x}{10 + x} \implies 2.5(10 + x) = 28 + 2x \implies 25 + 2.5x = 28 + 2x \implies 0.5x = 3 \implies x = 6

.

Equating the mean expression to

2.5

gives the value of

x

.

3

Find total number of observations

N

and determine the median position.

N = 10 + 6 = 16

. Since

N

is even, median position is between the

8\text{th}

and

9\text{th}

values.

For an even number of observations

N

, the median is the average of the

\frac{N}{2}\text{th}

and

(\frac{N}{2} + 1)\text{th}

items.

4

Locate the

8\text{th}

and

9\text{th}

items using cumulative frequencies.

Cumulative frequencies: Score 1 has 3; Score 2 reaches

3 + 6 = 9

. Both the

8\text{th}

and

9\text{th}

values are

2

. Therefore, median

= \frac{2 + 2}{2} = 2

.

Positions 4 through 9 correspond to the score

2

.

Key Concept

Measures of central tendency for ungrouped frequency distributions

Question 7896Question →

A body starting from rest accelerates uniformly along a straight path at a rate of $2.5\text{ m/s}^2$ . What is the distance covered by the body in $4.0\text{ s}$ ?

10\text{ m}

20\text{ m}

40\text{ m}

5.0\text{ m}

Show answer & explanation

Answer: $20\text{ m}$

Answer

The distance covered by the body in

4.0\text{ s}

is

20\text{ m}

.

Using the kinematic relation

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

with

u = 0\text{ m/s}

,

a = 2.5\text{ m/s}^2

, and

t = 4.0\text{ s}

gives

s = 0 + \frac{1}{2} \times 2.5 \times (4.0)^2 = 20\text{ m}

.

Step-by-Step Solution

1

Identify the given kinematic values.

Initial velocity

u = 0\text{ m/s}

, acceleration

a = 2.5\text{ m/s}^2

, and time

t = 4.0\text{ s}

.

The problem states the body starts from rest and undergoes uniform linear acceleration.

2

Select the appropriate formula for distance under uniform acceleration.

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

This equation connects initial velocity, acceleration, elapsed time, and total displacement.

3

Substitute the given values into the equation and compute displacement.

s = (0)(4.0) + \frac{1}{2}(2.5)(4.0)^2 = 0.5 \times 2.5 \times 16 = 20\text{ m}

.

Evaluating the expressions yields the distance traveled in meters.

Key Concept

Kinematic Equation for Linear Distance Under Uniform Acceleration

Estimated Time:45s

Question 7897Question →

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 7898Question →

A pneumatic cylinder in a hydraulic brake mechanism contains $0.050\text{ m}^3$ of an ideal gas at an initial pressure of $1.20 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . The gas is first compressed isothermally to a volume of $0.020\text{ m}^3$ , and then heated at constant volume until its pressure reaches $6.00 \times 10^5\text{ Pa}$ . What is the final temperature of the gas in degrees Celsius?

327^\circ\text{C}

54^\circ\text{C}

600^\circ\text{C}

1227^\circ\text{C}

Show answer & explanation

Answer: $327^\circ\text{C}$

Answer

The final temperature of the gas is

327^\circ\text{C}

.

First, converting the initial temperature gives

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

. For the isothermal compression, the pressure increases to

P_2 = P_1 \times \frac{V_1}{V_2} = 1.20 \times 10^5 \times \frac{0.050}{0.020} = 3.00 \times 10^5\text{ Pa}

. Next, during the constant volume heating step to

P_3 = 6.00 \times 10^5\text{ Pa}

, the pressure doubles, requiring the absolute temperature to double from

300\text{ K}

to

600\text{ K}

. Converting

600\text{ K}

back to Celsius yields

600 - 273 = 327^\circ\text{C}

.

Step-by-Step Solution

1

Convert the initial temperature from Celsius to Kelvin.

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

.

Gas laws strictly require thermodynamic temperatures expressed in Kelvin.

2

Apply Boyle's Law (

P_1 V_1 = P_2 V_2

) to determine the intermediate pressure

P_2

after isothermal compression at constant temperature

T_2 = 300\text{ K}

.

P_2 = \frac{P_1 V_1}{V_2} = \frac{(1.20 \times 10^5\text{ Pa})(0.050\text{ m}^3)}{0.020\text{ m}^3} = 3.00 \times 10^5\text{ Pa}

.

During an isothermal process, the product of pressure and volume remains constant.

3

Apply Pressure Law (Gay-Lussac's Law) for the second stage, where volume remains constant (

V_3 = V_2 = 0.020\text{ m}^3

) while pressure increases from

P_2 = 3.00 \times 10^5\text{ Pa}

to

P_3 = 6.00 \times 10^5\text{ Pa}

.

\frac{P_2}{T_2} = \frac{P_3}{T_3} \implies T_3 = T_2 \times \frac{P_3}{P_2} = 300\text{ K} \times \frac{6.00 \times 10^5\text{ Pa}}{3.00 \times 10^5\text{ Pa}} = 600\text{ K}

.

At constant volume, pressure is directly proportional to absolute temperature.

4

Convert the final temperature

T_3

back to degrees Celsius.

t_3 = 600 - 273 = 327^\circ\text{C}

.

The question explicitly requests the final temperature in degrees Celsius.

Key Concept

Multi-stage gas processes combining Boyle's Law and Pressure Law using the Ideal Gas Equation

Question 7899Question →

How many distinct solutions exist for the trigonometric equation $2\cos 2\theta = 1$ within the interval $0^\circ \le \theta \le 360^\circ$ ?

1

2

4

8

Show answer & explanation

Answer: 4

Answer

4 distinct solutions

The equation

\cos 2\theta = \frac{1}{2}

requires finding all angles whose cosine is

\frac{1}{2}

. Because the angle argument is

2\theta

, as

\theta

completes one full rotation (

0^\circ

to

360^\circ

),

2\theta

completes two full rotations (

0^\circ

to

720^\circ

). In two rotations, the cosine function takes the value

+\frac{1}{2}

exactly four times (twice per rotation), leading to four distinct solutions for

\theta

.

Step-by-Step Solution

1

Isolate the trigonometric function in the equation

\cos 2\theta = \frac{1}{2}

Dividing both sides by 2 isolates the cosine term.

2

Determine the working interval for the multiple angle

2\theta

0^\circ \le 2\theta \le 720^\circ

Since

0^\circ \le \theta \le 360^\circ

, multiplying the inequality by 2 gives the domain for

2\theta

.

3

Find all values of

2\theta

where cosine is positive

\frac{1}{2}

in the domain

[0^\circ, 720^\circ]

2\theta = 60^\circ, 300^\circ, 420^\circ, 660^\circ

Cosine is positive in Quadrants I and IV. In the first revolution (

0^\circ \text{ to } 360^\circ

),

2\theta = 60^\circ

and

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

. Adding

360^\circ

for the second revolution gives

420^\circ

and

660^\circ

.

4

Solve for

\theta

by dividing each angle by 2

\theta = 30^\circ, 150^\circ, 210^\circ, 330^\circ

Dividing all four values of

2\theta

by 2 gives four distinct values of

\theta

within

[0^\circ, 360^\circ]

.

Key Concept

Solving trigonometric equations with multiple angles over a specified domain

Question 7900Question →

In electrostatics, Coulomb's law states that the force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ in a vacuum is given by $F = \frac{q_1 q_2}{4\pi \varepsilon_0 r^2}$ , where $\varepsilon_0$ is the permittivity of free space. What are the fundamental dimensions of $\varepsilon_0$ expressed in terms of mass ( $M$ ), length ( $L$ ), time ( $T$ ), and electric current ( $I$ )?

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

M^{-1} L^{-3} T^2 I^2

M L^3 T^{-4} I^{-2}

M^{-1} L^{-2} T^4 I^2

Show answer & explanation

Answer: $M^{-1} L^{-3} T^4 I^2$

Answer

The dimensions of permittivity of free space

\varepsilon_0

are

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

.

The correct answer is derived by isolating

\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}

. Substituting

[q] = I T

,

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

, and

[r] = L

gives

[\varepsilon_0] = \frac{I^2 T^2}{M L^3 T^{-2}} = M^{-1} L^{-3} T^4 I^2

.

Step-by-Step Solution

1

Rearrange Coulomb's Law to isolate permittivity of free space

\varepsilon_0

\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}

Isolating

\varepsilon_0

allows us to substitute the dimensions of each constituent physical quantity.

2

Determine the dimensions of charge

q

, force

F

, distance

r

, and the constant

4\pi

[q] = I T

,

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

,

[r^2] = L^2

, and

[4\pi] = 1

(dimensionless)

Electric current is a base unit (

I

), so electric charge is current multiplied by time (

I T

). Force is mass times acceleration (

M L T^{-2}

).

3

Substitute the fundamental dimensions into the rearranged equation and simplify exponent powers

[\varepsilon_0] = \frac{(I T)(I T)}{(M L T^{-2})(L^2)} = \frac{I^2 T^2}{M L^3 T^{-2}} = M^{-1} L^{-3} T^4 I^2

Applying exponent laws:

T^2 / T^{-2} = T^{2 - (-2)} = T^4

,

1 / M = M^{-1}

, and

1 / L^3 = L^{-3}

.

Key Concept

Dimensional analysis of physical constants in electromagnetism

Answer

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