The words Socialist, Secular and Integrity were added in the Preamble of the Indian Constitution by which amendment?
The question asks about the constitutional amendment that added the words 'Socialist', 'Secular', and 'Integrity' to the Preamble of the Indian Constitution. This is a direct factual question related to significant amendments in Indian constitutional history.
Correct Option: C) 42 nd
If x + \( \frac{1}{x} \) = 8, then the value of x³ + \( \frac{1}{x^3} \) is:
To find the value of \(x^3 + \frac{1}{x^3}\) given \(x + \frac{1}{x} = 8\), we can use the algebraic identity for the cube of a sum: \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\). By substituting \(a=x\) and \(b=\frac{1}{x}\), we can directly solve for the required expression.
We are given the equation:
\[ x + \frac{1}{x} = 8 \]We need to find the value of \(x^3 + \frac{1}{x^3}\).
Recall the algebraic identity for the cube of a sum:
\[ (a+b)^3 = a^3 + b^3 + 3ab(a+b) \]Let \(a = x\) and \(b = \frac{1}{x}\). Substitute these into the identity:
\[ \left(x + \frac{1}{x}\right)^3 = x^3 + \left(\frac{1}{x}\right)^3 + 3(x)\left(\frac{1}{x}\right)\left(x + \frac{1}{x}\right) \]Simplify the equation:
\[ \left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3(1)\left(x + \frac{1}{x}\right) \] \[ \left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right) \]Now, substitute the given value \(x + \frac{1}{x} = 8\) into the simplified equation:
\[ (8)^3 = x^3 + \frac{1}{x^3} + 3(8) \]Calculate the values:
\[ 512 = x^3 + \frac{1}{x^3} + 24 \]Isolate \(x^3 + \frac{1}{x^3}\) by subtracting 24 from both sides:
\[ x^3 + \frac{1}{x^3} = 512 - 24 \] \[ x^3 + \frac{1}{x^3} = 488 \]Correct Option: D) 488
This is a statement and assumption question. We need to identify which of the given assumptions are implicitly contained within the statement. An assumption is something that is taken for granted or supposed to be true without proof. It must directly follow from the statement.
Correct Option: C) Only assumption I is implicit
The question asks to identify a method that helps prevent Sexually Transmitted Diseases (STDs). We need to evaluate each given option based on its primary function and effectiveness in preventing the transmission of STDs.
D) Condoms are barrier methods that prevent the exchange of bodily fluids during sexual intercourse. This physical barrier is effective in preventing the transmission of many STDs, such as HIV, gonorrhea, chlamydia, and syphilis, in addition to preventing pregnancy. While not 100% effective, they significantly reduce the risk.
The following table shows the marks of five students in five different subjects in a school examination. Each subject carries a maximum of 100 marks. Study the table and answer the question.
| Subject | Students | ||||
| Rohan | Mohan | Radha | Ali | Johan | |
| Mathematics | 70 | 88 | 70 | 90 | 95 |
| Statistics | 74 | 78 | 75 | 70 | 85 |
| Hindi | 65 | 78 | 76 | 53 | 50 |
| English | 60 | 64 | 70 | 65 | 85 |
| Social Science | 58 | 54 | 62 | 46 | 66 |
Who among them has achieved the second highest aggregated marks?
To find the student with the second highest aggregated marks, we need to calculate the total marks obtained by each student across all five subjects. Since each subject carries a maximum of 100 marks, the total marks for each student will be the sum of their scores in Mathematics, Statistics, Hindi, English, and Social Science. After calculating the total for each student, we will rank them from highest to lowest and identify the student with the second highest total.
The highest aggregated marks are 381 (Johan). The second highest aggregated marks are 362 (Mohan).
All 30 students of a class are standing in a straight row facing the north. Komal is 8th from the right end while Priya is 14th from the left end. How many students are standing between Komal and Priya?
The problem involves finding the number of students between two individuals in a straight row. We are given the total number of students and the positions of two students from opposite ends of the row. The key is to determine if the positions overlap or not, and then calculate the number of students between them.
If Komal is 8th from the right end in a row of 30 students, her position from the left end can be calculated as:
\[ \text{Komal's position from left} = \text{Total students} - \text{Komal's position from right} + 1 \] \[ \text{Komal's position from left} = 30 - 8 + 1 = 22 + 1 = 23\text{rd} \]So, Komal is 23rd from the left end.
If Priya is 14th from the left end in a row of 30 students, her position from the right end can be calculated as:
\[ \text{Priya's position from right} = \text{Total students} - \text{Priya's position from left} + 1 \] \[ \text{Priya's position from right} = 30 - 14 + 1 = 16 + 1 = 17\text{th} \]So, Priya is 17th from the right end.
Priya is 14th from the left, and Komal is 23rd from the left. Since Priya's position (14) is less than Komal's position (23) from the left end, they are not overlapping in a way that would complicate the calculation (i.e., Komal is to the right of Priya).
We can use their positions from the same end (e.g., left end) to find the number of students between them.
Priya is 14th from the left.
Komal is 23rd from the left.
The number of students between them is the difference in their positions minus 1 (because we are excluding Komal and Priya themselves).
\[ \text{Students between them} = \text{Komal's position from left} - \text{Priya's position from left} - 1 \] \[ \text{Students between them} = 23 - 14 - 1 = 9 - 1 = 8 \]Alternatively, using their positions from the right end:
Komal is 8th from the right.
Priya is 17th from the right.
\[ \text{Students between them} = \text{Priya's position from right} - \text{Komal's position from right} - 1 \] \[ \text{Students between them} = 17 - 8 - 1 = 9 - 1 = 8 \]Thus, there are 8 students standing between Komal and Priya.
A) 8 is the correct answer. There are 8 students standing between Komal and Priya.
To determine the diameter of the image of the ball, we need to calculate the magnification produced by the lens. First, we'll find the focal length of the lens using its power. Then, we'll use the lens formula to find the image distance. Finally, we'll use the magnification formula to find the size of the image.
The power (\(P\)) of a lens is given by \(P = \frac{1}{f}\), where \(f\) is in meters. Given \(P = +5.0 \text{ D}\).
\[ f = \frac{1}{P} = \frac{1}{+5.0 \text{ D}} = +0.2 \text{ m} \]Converting to centimeters:
\[ f = +0.2 \times 100 \text{ cm} = +20 \text{ cm} \]Since the power is positive, it's a converging (convex) lens.
The lens formula is \[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\].
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \] \[ \frac{1}{v} = \frac{1}{+20 \text{ cm}} + \frac{1}{-40 \text{ cm}} \] \[ \frac{1}{v} = \frac{1}{20} - \frac{1}{40} \] \[ \frac{1}{v} = \frac{2 - 1}{40} = \frac{1}{40} \] \[ v = +40 \text{ cm} \]A positive \(v\) indicates that the image is formed on the other side of the lens (real image).
Magnification is given by \[M = \frac{h_i}{h_o} = \frac{v}{u}\].
\[ M = \frac{+40 \text{ cm}}{-40 \text{ cm}} = -1 \]We know \(M = \frac{h_i}{h_o}\).
\[ h_i = M \times h_o \] \[ h_i = -1 \times 10 \text{ cm} \] \[ h_i = -10 \text{ cm} \]The negative sign indicates that the image is inverted. The magnitude of the image diameter is \(10 \text{ cm}\).
Correct Option: A) 10 cm
The diameter of the image of the ball is 10 cm. The magnification is -1, meaning the image is real, inverted, and the same size as the object. This occurs when the object is placed at \(2f\) for a convex lens, as \(u = -40 \text{ cm}\) and \(f = 20 \text{ cm}\), so \(u = -2f\).
The problem provides a set of observations arranged in ascending order and states that the median is 63. We need to find the value of 'x'. The median of a data set depends on whether the number of observations is odd or even. For an even number of observations, the median is the average of the two middle terms.
C) 62
This problem involves calculating the work rate of pipes. We'll use the concept that if a pipe fills a tank in 't' hours, its filling rate is \(1/t\) of the tank per hour. Similarly, if a drain pipe empties a tank in 't' hours, its emptying rate is \(1/t\) of the tank per hour. When multiple pipes work together, their rates are added (for filling) or subtracted (for emptying) to find the combined rate.
Determine the filling rate of the first tap:
The first tap fills the tank in 10 hours. So, its filling rate is:
\[ \text{Rate}_1 = \frac{1}{10} \text{ tank/hour} \]Determine the filling rate of the second tap:
The second tap fills the tank in 15 hours. So, its filling rate is:
\[ \text{Rate}_2 = \frac{1}{15} \text{ tank/hour} \]Determine the combined rate of all three pipes:
When all three pipes (two filling, one emptying) are opened together, the tank is filled in 12 hours. Let the emptying rate of the third pipe (drainpipe) be \(1/x\) tank/hour, where \(x\) is the time it takes to empty the tank alone. Since it's a drainpipe, its contribution is negative.
The combined filling rate is:
\[ \text{Combined Rate} = \frac{1}{12} \text{ tank/hour} \]Set up the equation for combined rates:
The combined rate is the sum of the individual rates, with the drainpipe's rate being subtracted:
\[ \text{Rate}_1 + \text{Rate}_2 - \text{Rate}_3 = \text{Combined Rate} \]\[ \frac{1}{10} + \frac{1}{15} - \frac{1}{x} = \frac{1}{12} \]Solve for \(x\):
First, find a common denominator for the fractions on the left side. The least common multiple of 10 and 15 is 30.
\[ \frac{3}{30} + \frac{2}{30} - \frac{1}{x} = \frac{1}{12} \]\[ \frac{5}{30} - \frac{1}{x} = \frac{1}{12} \]\[ \frac{1}{6} - \frac{1}{x} = \frac{1}{12} \]Now, isolate \(1/x\):\[ \frac{1}{x} = \frac{1}{6} - \frac{1}{12} \]Find a common denominator for the right side, which is 12:\[ \frac{1}{x} = \frac{2}{12} - \frac{1}{12} \]\[ \frac{1}{x} = \frac{1}{12} \]Therefore,\[ x = 12 \text{ hours} \]The third pipe alone can empty the filled tank in 12 hours.
D) 12 hrs
The correct relation between v, u, and f for a spherical mirror is:
\( \frac{1}{f} \) = \( \frac{1}{v} \) − \( \frac{1}{u} \)
v = u + f
\( \frac{1}{f} \) + \( \frac{1}{u} \) = \( \frac{1}{v} \)
\( \frac{1}{f} \) = \( \frac{1}{u} \) + \( \frac{1}{v} \)
The question asks for the correct mirror formula, which relates the object distance (\(u\)), image distance (\(v\)), and focal length (\(f\)) for a spherical mirror. This is a fundamental formula in optics.
D) \[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\] is the correct mirror formula relating the focal length (\(f\)), object distance (\(u\)), and image distance (\(v\)) for a spherical mirror.