Question

The integral \[ 80 \int_0^{\frac{\pi}{4}} \frac{(\sin \theta + \cos \theta)}{(9 + 16 \sin 2\theta)} \, d\theta \] is equal to:

• A. \( 6 \log 4 \)
• B. \( 2 \log 3 \)
• C. \( 4 \log 3 \)
• D. \( 3 \log 4 \)

Hint:

To solve integrals involving trigonometric functions, consider using substitution and applying relevant trigonometric identities to simplify the expression.

Answer 1

The Correct Option is D

Problem:

Evaluate the integral:

\[ I = 80 \int_0^{\frac{\pi}{4}} \frac{\sin\theta + \cos\theta}{9 + 16\sin 2\theta} \, d\theta \]

Solution:

Use the identity:

\[ \sin 2\theta = 2\sin\theta \cos\theta = 1 - (\sin\theta - \cos\theta)^2 \]

Let:

\[ t = \sin\theta - \cos\theta \quad \Rightarrow \quad dt = (\cos\theta + \sin\theta) d\theta \]

Limits change from \( \theta = 0 \Rightarrow t = -1 \) and \( \theta = \frac{\pi}{4} \Rightarrow t = 0 \)

Integral becomes:

\[ I = 80 \int_{-1}^{0} \frac{1}{25 - 16t^2} \, dt = \frac{80}{16} \int_{-1}^{0} \frac{1}{\left( \frac{5}{4} \right)^2 - t^2} dt \]

Use the standard result:

\[ \int \frac{1}{a^2 - t^2} dt = \frac{1}{2a} \ln\left| \frac{a + t}{a - t} \right| \quad \text{where } a = \frac{5}{4} \]

Evaluate definite integral:

\[ I = 5 \cdot \left[ \frac{1}{2 \cdot \frac{5}{4}} \ln\left( \frac{a + t}{a - t} \right) \right]_{-1}^{0} = 5 \cdot \frac{4}{10} \left[ \ln(1) + \ln(3) \right] = 4 \ln 3 \]

Final Answer: \( \boxed{4 \log 3} \)

Answer 2

Problem:

A positive ion A and a negative ion B have the following properties:

Charge of A: \( q_1 = 6.67 \times 10^{-19} \, \text{C} \)

Charge of B: \( q_2 = 9.6 \times 10^{-10} \, \text{C} \)

Mass of A: \( m_1 = 19.2 \times 10^{-27} \, \text{kg} \)

Mass of B: \( m_2 = 9 \times 10^{-27} \, \text{kg} \)

Objective:

Find the value of \( P \) such that the ratio of the magnitudes of electrostatic to gravitational force is:

\( \dfrac{F_e}{F_g} = P \times 10^{13} \)

Step 1: Formula for ratio

\[ \frac{F_e}{F_g} = \frac{\dfrac{k q_1 q_2}{r^2}}{\dfrac{G m_1 m_2}{r^2}} = \frac{k q_1 q_2}{G m_1 m_2} \]

Step 2: Substitute constants and values

\[ k = 9 \times 10^9, \quad G = 6.67 \times 10^{-11} \] \[ \frac{F_e}{F_g} = \frac{9 \times 10^9 \times 6.67 \times 10^{-19} \times 9.6 \times 10^{-10}}{6.67 \times 10^{-11} \times 19.2 \times 10^{-27} \times 9 \times 10^{-27}} \]

Step 3: Simplify numerator and denominator

Numerator: \( 9 \times 6.67 \times 9.6 = 576.19 \), exponent: \( 10^{-20} \)
Denominator: \( 6.67 \times 19.2 \times 9 = 1152.58 \), exponent: \( 10^{-65} \)

\[ \frac{F_e}{F_g} = \frac{576.19 \times 10^{-20}}{1152.58 \times 10^{-65}} = \frac{576.19}{1152.58} \times 10^{45} \approx 0.5 \times 10^{45} = 5 \times 10^{44} \]

Step 4: Express in terms of \( 10^{13} \)

\[ 5 \times 10^{44} = P \times 10^{13} \Rightarrow P = \frac{5 \times 10^{44}}{10^{13}} = 5 \times 10^{31} \]

✅ Final Answer:

\[ \boxed{P = 5 \times 10^{31}} \]