We need to find the height of the flag. We can use trigonometry, specifically the tangent function, to relate the angles of elevation, the heights, and the distance from the observer to the pole. We will set up two right-angled triangles and solve a system of equations.
Let the height of the pole be \(h_p = 18 \text{ m}\).
Let the height of the flag be \(h_f\).
Let the total height from the ground to the top of the flag be \(H = h_p + h_f = 18 + h_f\).
Let the distance from the point on the ground to the base of the pole be \(x\).
The angle of elevation to the bottom of the flag (which is the top of the pole) is \(30^\circ\). From the right-angled triangle formed by the observer, the base of the pole, and the top of the pole:
\[ \tan(30^\circ) = \frac{\text{height of pole}}{\text{distance}} = \frac{h_p}{x} \]\[ \frac{1}{\sqrt{3}} = \frac{18}{x} \]\[ x = 18\sqrt{3} \text{ m} \quad (1) \]The angle of elevation to the top of the flag is \(60^\circ\). From the right-angled triangle formed by the observer, the base of the pole, and the top of the flag:
\[ \tan(60^\circ) = \frac{\text{total height}}{\text{distance}} = \frac{H}{x} \]\[ \sqrt{3} = \frac{18 + h_f}{x} \]\[ x = \frac{18 + h_f}{\sqrt{3}} \quad (2) \]Now, equate the expressions for \(x\) from equations (1) and (2):
\[ 18\sqrt{3} = \frac{18 + h_f}{\sqrt{3}} \]Multiply both sides by \(\sqrt{3}\):
\[ 18\sqrt{3} \times \sqrt{3} = 18 + h_f \]\[ 18 \times 3 = 18 + h_f \]\[ 54 = 18 + h_f \]Solve for \(h_f\):
\[ h_f = 54 - 18 \]\[ h_f = 36 \text{ m} \]Thus, the height of the flag is 36 m.
B) 36 m — This is the calculated height of the flag.
The problem presents a number series, and we need to find the next term. We will analyze the differences between consecutive terms to identify a pattern. If the first-level differences don't reveal a clear pattern, we will look at the differences of these differences (second-level differences).
Thus, the missing term in the series is 59.
Correct Option: A) 59
The question asks about the pattern of magnetic field lines around a straight current-carrying conductor. This is a fundamental concept in electromagnetism, governed by Ampere's Law and the Right-Hand Thumb Rule.
D) concentric circles. The magnetic field lines around a straight current-carrying conductor are concentric circles centered on the conductor, lying in planes perpendicular to the conductor. This is a direct consequence of the Right-Hand Thumb Rule and experimental observations.
The question asks about the valency of carbon. Valency refers to the combining capacity of an element, which is determined by the number of electrons in its outermost shell (valence electrons) and how it achieves a stable electron configuration (usually an octet).
B) Carbon is tetravalent. This is correct because carbon has 4 valence electrons and forms 4 covalent bonds to achieve a stable octet configuration.
To find the time taken, we need to use the fundamental formula relating speed, distance, and time. The key is to ensure all units are consistent before performing calculations. Here, the speed is given in km/h and the distance in meters, so we must convert one of them to match the other, or convert both to a common base unit (like meters per second).
Since the distance is in meters and the options are in seconds, it's best to convert the speed to meters per second (m/s). To convert km/h to m/s, multiply by \(\frac{5}{18}\).
\[ \text{Speed in m/s} = 8 \times \frac{5}{18} \text{ m/s} \]\[ \text{Speed in m/s} = \frac{40}{18} \text{ m/s} \]\[ \text{Speed in m/s} = \frac{20}{9} \text{ m/s} \]The relationship between speed, distance, and time is:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]C) 90 seconds is the correct answer. Reena will take 90 seconds to cover a distance of 200 meters at a speed of 8 km/h.
The question asks to identify the phenomenon that occurs when white light passes through a glass prism and splits into its constituent seven colours. We need to understand the definitions of the given optical phenomena.
Correct Option: B) dispersion is the phenomenon where white light splits into its constituent colours due to the refractive index of the medium being different for different wavelengths.
The problem asks us to find the values of X and Y such that the mathematical operation applied to the first number to get the second number is consistent across both pairs. We need to analyze the relationship between the given pair of numbers and then apply the same logic to find X and Y.
D) X = 6, Y = 488
To find the average weight of the entire group, we need to calculate the weighted average of the weights of men, women, and children. The percentages of each group will serve as their respective weights in the average calculation.
This problem involves work and time, and the distribution of wages based on the work done. We need to calculate the individual work rates, determine the work done by each person, and then distribute the total wages proportionally to the work done.
Aarav's 1-day work: \(W_A = \frac{1}{6}\) of the total work.
Babita's 1-day work: \(W_B = \frac{1}{8}\) of the total work.
Combined 1-day work of Aarav and Babita: \[W_{A+B} = \frac{1}{6} + \frac{1}{8}\]
\[ W_{A+B} = \frac{4}{24} + \frac{3}{24} = \frac{7}{24} \]Since they completed the work in 3 days with Chaitanya's help, Aarav and Babita also worked for 3 days.
Work done by Aarav in 3 days: \[3 \times \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\]
Work done by Babita in 3 days: \[3 \times \frac{1}{8} = \frac{3}{8}\]
Total work is 1 unit. Work done by Chaitanya is the remaining work after Aarav and Babita's contribution.
Work done by Chaitanya: \(W_C = 1 - (\text{Work by Aarav} + \text{Work by Babita})\)
\[ W_C = 1 - \left(\frac{1}{2} + \frac{3}{8}\right) \]\[ W_C = 1 - \left(\frac{4}{8} + \frac{3}{8}\right) \]\[ W_C = 1 - \frac{7}{8} = \frac{1}{8} \]Ratio of work done \[W_A : W_B : W_C = \frac{1}{2} : \frac{3}{8} : \frac{1}{8}\]
Multiply by the common denominator (8) to get integer ratios:
\[W_A : W_B : W_C = \left(\frac{1}{2} \times 8\right) : \left(\frac{3}{8} \times 8\right) : \left(\frac{1}{8} \times 8\right)\]
\(W_A : W_B : W_C = 4 : 3 : 1\)
The total payment is ₹2,400. Babita's share will be proportional to her work done.
Babita's share \[= \frac{\text{Babita's work ratio}}{\text{Sum of ratios}} \times \text{Total payment}\]
\[ \text{Babita's share} = \frac{3}{4+3+1} \times 2400 \]\[ \text{Babita's share} = \frac{3}{8} \times 2400 \]\[ \text{Babita's share} = 3 \times 300 = \text{₹}900 \]Correct Option: B) ₹900
The question asks about the application of specific high-performance computing systems, 'Arka' and 'Arunika'. This requires knowledge of India's supercomputing initiatives and their designated uses.
Correct Option: A) Weather forecasting. 'Arka' and 'Arunika' are high-performance computing systems developed by India, primarily used for weather forecasting, climate modeling, and related atmospheric research. They are instrumental in providing more accurate and timely weather predictions.