The question asks to identify the spokesperson of the Indian Navy as of February 2018. This is a factual recall question related to current affairs or general knowledge about defense personnel appointments during a specific period.
Correct Option: A) Captain D. K. Sharma was the official spokesperson of the Indian Navy as of February 2018. He assumed the charge as the Chief PRO (Navy) in January 2018.
This problem can be solved by calculating the amount of work done (tank filled) by pipe C in the initial 3 hours and then determining the remaining work. Finally, calculate the time taken by pipe D to fill the remaining portion of the tank.
Pipe C can fill the tank in 4 hours. So, its filling rate is:
\[ \text{Rate of C} = \frac{1}{4} \text{ tank/hour} \]Pipe C operates for 3 hours. The amount of tank filled is:
\[ \text{Work done by C} = \text{Rate of C} \times \text{Time} = \frac{1}{4} \times 3 = \frac{3}{4} \text{ of the tank} \]The total tank capacity is 1. So, the remaining portion is:
\[ \text{Remaining tank} = 1 - \frac{3}{4} = \frac{1}{4} \text{ of the tank} \]Pipe D can fill the tank in 5 hours. So, its filling rate is:
\[ \text{Rate of D} = \frac{1}{5} \text{ tank/hour} \]Now, pipe D alone fills the remaining \(\frac{1}{4}\) of the tank. The time taken will be:
\[ \text{Time taken by D} = \frac{\text{Remaining tank}}{\text{Rate of D}} = \frac{\frac{1}{4}}{\frac{1}{5}} = \frac{1}{4} \times 5 = \frac{5}{4} \text{ hours} \]Correct Option: B) 1.25
To solve this problem, we need to follow the order of operations, often remembered by the acronym BODMAS/PEMDAS (Brackets/Parentheses, Orders/Exponents, Division and Multiplication, Addition and Subtraction). We will evaluate the innermost brackets first and work our way outwards.
First, evaluate the innermost parentheses in the left part of the expression:
\[ (9 - 2) = 7 \]Substitute this back into the left part:
\[ 52 - 7 = 45 \]Now, evaluate the innermost parentheses in the right part of the expression:
\[ (-2) \times (-2) = 4 \]Substitute this back into the curly braces of the right part:
\[ \{1 + 4\} = 5 \]Next, evaluate the square brackets of the right part:
\[ [3 \times 5] = 15 \]Finally, perform the division:
\[ 45 \div 15 = 3 \]A) The step-by-step calculation leads to the result of 3.
This is a blood relation question. We need to deduce the relationship between Nikhil and the boy based on the statement provided by the boy.
Correct Option: D) paternal uncle
The question asks for the year when football was officially included as a competitive game in the Olympics. This requires knowledge of Olympic history, specifically regarding the inclusion of football.
Correct Option: A) 1908. Football was officially included as a competitive game, with medals awarded, at the 1908 London Olympics.
To find the area of a triangle given its vertices, we can use the determinant formula (also known as the shoelace formula). This formula is efficient and directly calculates the area using the coordinates of the vertices.
Identify the coordinates:
Let the vertices be \(A(x_1, y_1) = (-4, -2)\), \(B(x_2, y_2) = (-3, -5)\), and \(C(x_3, y_3) = (3, -2)\).
Apply the area formula:
The area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by the formula:
Substitute the values:
Substitute the given coordinates into the formula:
Simplify the expression:
\[ \text{Area} = \frac{1}{2} |(-4)(-5 + 2) + (-3)(-2 + 2) + (3)(-2 + 5)| \]\[ \text{Area} = \frac{1}{2} |(-4)(-3) + (-3)(0) + (3)(3)| \]\[ \text{Area} = \frac{1}{2} |12 + 0 + 9| \]\[ \text{Area} = \frac{1}{2} |21| \]\[ \text{Area} = \frac{21}{2} \]\[ \text{Area} = 10.5 \text{ sq units} \]D) The calculated area of the triangle is 10.5 sq units, which matches option D.
A and B can do a piece of work together in 15 days while A alone can do it in 30 days. They start working together but B leaves 4 days before the completion of the work. For how many days did A and B work together?
This problem involves work and time concepts. We need to calculate the individual work rates of A and B, then determine how much work A completes in the last few days, and finally, how many days A and B worked together.
Calculate A's daily work rate:
A alone can do the work in 30 days.
So, A's 1-day work \( = \frac{1}{30} \).
Calculate the combined daily work rate of A and B:
A and B together can do the work in 15 days.
So, (A + B)'s 1-day work \( = \frac{1}{15} \).
Calculate B's daily work rate:
B's 1-day work = (A + B)'s 1-day work - A's 1-day work
\[ \text{B's 1-day work} = \frac{1}{15} - \frac{1}{30} = \frac{2 - 1}{30} = \frac{1}{30} \]So, B alone can complete the work in 30 days.
Determine the work done by A in the last 4 days:
B leaves 4 days before the completion of the work. This means A works alone for the last 4 days.
Work done by A in 4 days \[= 4 \times \frac{1}{30} = \frac{4}{30} = \frac{2}{15}\].
Calculate the remaining work:
Total work is 1. Work done by A in the last 4 days is \(\frac{2}{15}\).
Remaining work \[= 1 - \frac{2}{15} = \frac{15 - 2}{15} = \frac{13}{15}\].
Calculate the number of days A and B worked together:
The remaining work (\(\frac{13}{15}\)) was done by A and B working together.
Combined 1-day work of A and B \( = \frac{1}{15} \).
Number of days A and B worked together \[= \frac{\text{Remaining work}}{\text{(A + B)'s 1-day work}}\]
\[ \text{Number of days} = \frac{\frac{13}{15}}{\frac{1}{15}} = \frac{13}{15} \times 15 = 13 \text{ days} \]Correct Option: B) 13
The question asks about the relationship described by the Second Equation of Motion. We need to recall the three equations of motion and identify which physical quantities are related by the second one.
Correct Option: B) Position. The Second Equation of Motion, \(s = ut + \frac{1}{2}at^2\), directly relates the displacement (\(s\), which is a change in position) of an object to the time (\(t\)) it has been in motion, along with its initial velocity (\(u\)) and acceleration (\(a\)).
The question asks what happens when two charged bodies with equal potential are connected by a conducting wire. The key concept here is that current flows due to a potential difference. If there is no potential difference, there will be no flow of charge.
C) no current flows — As explained, current flows only when there is a potential difference. If the potentials are equal, the potential difference is zero, and thus no current flows.
Who among the following won the ‘ITA Sterling Icon of Indian Entertainment’ award at the 17th edition of the Indian Television Academy Awards held in November 2017?
Jitendra
Ekta Kapoor
Siddharth Roy Kapur
Shobha Kapoor
The question asks to identify the winner of the 'ITA Sterling Icon of Indian Entertainment' award at the 17th Indian Television Academy Awards in November 2017. This requires knowledge of specific awards and their recipients in the Indian television industry.
Correct Option: B) Ekta Kapoor
Ekta Kapoor, a renowned Indian television and film producer, was awarded the 'ITA Sterling Icon of Indian Entertainment' at the 17th Indian Television Academy Awards in November 2017. This award recognized her immense contribution and influence in shaping the Indian entertainment landscape, particularly in television.