The question asks about the population requirement for the formation of 'Ward Committees' in municipalities as mandated by the Indian Constitution. This falls under the provisions related to Urban Local Bodies (ULBs) or Municipalities.
A) three lakhs or more. This is directly in line with Article 243S(1) of the Indian Constitution, which mandates the formation of Ward Committees in municipalities with a population of three lakhs or more.
The question asks to identify a renewable source of energy from the given options. Renewable energy sources are those that replenish naturally over a relatively short period, unlike non-renewable sources which are finite and take millions of years to form.
Correct Option: B) Wind is a renewable source of energy because it is continuously replenished by natural processes.
To find the area of a rhombus, we need the lengths of both its diagonals. We are given the perimeter and one diagonal. We can use the perimeter to find the side length of the rhombus. Then, using the property that the diagonals of a rhombus bisect each other at right angles, we can form a right-angled triangle and use the Pythagorean theorem to find half of the other diagonal.
Find the side length of the rhombus:
The perimeter of a rhombus is given by \(4 \times \text{side}\).
Given perimeter \(P = 146 \text{ cm}\).
Use the properties of diagonals:
Let the diagonals be \(d_1\) and \(d_2\). We are given one diagonal, say \(d_1 = 55 \text{ cm}\).
The diagonals of a rhombus bisect each other at right angles. This means they divide the rhombus into four congruent right-angled triangles. Consider one such triangle. Its hypotenuse is the side of the rhombus, and its legs are half the lengths of the diagonals.
So, half of the given diagonal is:
\[ \frac{d_1}{2} = \frac{55}{2} = 27.5 \text{ cm} \]Find the length of the other diagonal:
Let half of the other diagonal be \(x\). Using the Pythagorean theorem in one of the right-angled triangles:
\[ (\text{side})^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 \]\[ (36.5)^2 = (27.5)^2 + x^2 \]\[ 1332.25 = 756.25 + x^2 \]\[ x^2 = 1332.25 - 756.25 \]\[ x^2 = 576 \]\[ x = \sqrt{576} = 24 \text{ cm} \]Therefore, the full length of the second diagonal \(d_2\) is \(2x\):\[ d_2 = 2 \times 24 = 48 \text{ cm} \]Calculate the area of the rhombus:
The area of a rhombus is given by the formula:
\[ \text{Area} = \frac{1}{2} \times d_1 \times d_2 \]\[ \text{Area} = \frac{1}{2} \times 55 \times 48 \]\[ \text{Area} = 55 \times 24 \]\[ \text{Area} = 1320 \text{ cm}^2 \]B) 1320 cm² is the calculated area of the rhombus.
When two triangles are similar, the ratio of their corresponding sides is equal, and the ratio of their perimeters is also equal to the ratio of their corresponding sides. We will use this property to find the perimeter of \(\Delta \text{ABC}\).
Given that \(\Delta \text{ABC}\) is similar to \(\Delta \text{DEF}\) (denoted as \(\Delta \text{ABC} \sim \Delta \text{DEF}\)).
The corresponding sides are in proportion:
\[ \frac{\text{AB}}{\text{DE}} = \frac{\text{BC}}{\text{EF}} = \frac{\text{AC}}{\text{DF}} \]We are given the following side lengths:
First, calculate the perimeter of \(\Delta \text{DEF}\):
\[ \text{Perimeter}(\Delta \text{DEF}) = \text{DE} + \text{EF} + \text{FD} \]\[ \text{Perimeter}(\Delta \text{DEF}) = 8 \text{ cm} + 10 \text{ cm} + 14 \text{ cm} \]\[ \text{Perimeter}(\Delta \text{DEF}) = 32 \text{ cm} \]For similar triangles, the ratio of their perimeters is equal to the ratio of their corresponding sides:
\[ \frac{\text{Perimeter}(\Delta \text{ABC})}{\text{Perimeter}(\Delta \text{DEF})} = \frac{\text{AB}}{\text{DE}} \]Substitute the known values into the equation:
\[ \frac{\text{Perimeter}(\Delta \text{ABC})}{32 \text{ cm}} = \frac{6 \text{ cm}}{8 \text{ cm}} \]Simplify the ratio on the right side:
\[ \frac{6}{8} = \frac{3}{4} \]Now, solve for \(\text{Perimeter}(\Delta \text{ABC})\):
\[ \text{Perimeter}(\Delta \text{ABC}) = 32 \text{ cm} \times \frac{3}{4} \]\[ \text{Perimeter}(\Delta \text{ABC}) = \frac{32 \times 3}{4} \text{ cm} \]\[ \text{Perimeter}(\Delta \text{ABC}) = 8 \times 3 \text{ cm} \]\[ \text{Perimeter}(\Delta \text{ABC}) = 24 \text{ cm} \]B) The perimeter of \(\Delta \text{ABC}\) is 24 cm, as calculated by using the property that the ratio of perimeters of similar triangles is equal to the ratio of their corresponding sides.
Simplify the following: \(\left(\frac{(xy^2z)^{-3}}{y^{-3}}\right)^2\)
\(x^6y^{18}z^6\)
\(\frac{1}{x^6y^6z^6}\)
\(\frac{1}{x^6y^{18}z^6}\)
\(x^6y^6z^6\)
To simplify the given expression, we will use the rules of exponents. Key rules include \((a^m)^n = a^{mn}\), \((ab)^m = a^m b^m\), \[\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}\], and \(a^{-m} = \frac{1}{a^m}\).
First, let's write down the given expression:
\[ \left(\frac{(xy^2z)^{-3}}{y^{-3}}\right)^2 \]Apply the power rule \((ab)^m = a^m b^m\) to the numerator \((xy^2z)^{-3}\):
\[ (xy^2z)^{-3} = x^{-3}(y^2)^{-3}z^{-3} \]Apply the power rule \((a^m)^n = a^{mn}\) to \((y^2)^{-3}\):
\[ (y^2)^{-3} = y^{2 \times (-3)} = y^{-6} \]Substitute this back into the numerator:
\[ x^{-3}y^{-6}z^{-3} \]Now, the expression inside the large parenthesis becomes:
\[ \frac{x^{-3}y^{-6}z^{-3}}{y^{-3}} \]Simplify the terms with the same base, \(y\), using the rule \(\frac{a^m}{a^n} = a^{m-n}\):
\[ y^{-6 - (-3)} = y^{-6 + 3} = y^{-3} \]So, the expression inside the large parenthesis simplifies to:
\[ x^{-3}y^{-3}z^{-3} \]Now, raise this entire expression to the power of 2:
\[ (x^{-3}y^{-3}z^{-3})^2 \]Apply the power rule \((abc)^m = a^m b^m c^m\) and \((a^m)^n = a^{mn}\) again:
\[ (x^{-3})^2 (y^{-3})^2 (z^{-3})^2 = x^{-3 \times 2} y^{-3 \times 2} z^{-3 \times 2} = x^{-6}y^{-6}z^{-6} \]Finally, use the rule \(a^{-m} = \frac{1}{a^m}\) to express the terms with positive exponents:
\[ x^{-6}y^{-6}z^{-6} = \frac{1}{x^6} \cdot \frac{1}{y^6} \cdot \frac{1}{z^6} = \frac{1}{x^6y^6z^6} \]B) The simplified expression is \(\frac{1}{x^6y^6z^6}\).
This problem involves calculating the work done by pipes filling a tank. We will determine the individual filling rates of pipes P and Q, calculate the portion of the tank filled when both pipes are open, and then find the remaining portion to be filled by pipe Q alone.
Calculate the filling rate of pipe P:
Pipe P can fill the tank in 28 minutes. So, in 1 minute, pipe P fills \(\frac{1}{28}\) of the tank.
Calculate the filling rate of pipe Q:
Pipe Q can fill the tank in 35 minutes. So, in 1 minute, pipe Q fills \(\frac{1}{35}\) of the tank.
Calculate the combined filling rate of pipes P and Q:
When both pipes are open, their combined rate is the sum of their individual rates:
Calculate the portion of the tank filled in the first 12 minutes:
Pipes P and Q are opened simultaneously for 12 minutes. So, the portion filled is:
Calculate the remaining portion of the tank to be filled:
The total tank is 1 (or \(\frac{35}{35}\)). The remaining portion is:
Calculate the time taken by pipe Q to fill the remaining portion:
After 12 minutes, pipe P is closed, and pipe Q fills the remaining \(\frac{8}{35}\) of the tank by itself. Pipe Q's rate is \(\frac{1}{35}\) of the tank per minute.
D) 8 minutes — As calculated, pipe Q will take 8 minutes to fill the remaining portion of the tank.
To find the median of a given set of data, we first need to arrange the data in ascending order. Once arranged, the median is the middle value. If there is an odd number of data points, the median is the single middle value. If there is an even number of data points, the median is the average of the two middle values.
Correct Option: C) 9
The induced electromotive force (emf) in a conductor moving in a magnetic field is governed by the principle of electromagnetic induction, specifically Faraday's Law and the Lorentz force. The key factor is the relative motion between the conductor and the magnetic field, and more precisely, the component of the conductor's velocity that is perpendicular to the magnetic field lines.
B) Zero — As derived, when the conductor moves parallel to the magnetic field, the angle between its velocity and the magnetic field is \(0^\circ\) or \(180^\circ\), for which \(\sin(\theta)\) is zero. Consequently, the induced emf \(E = BLv \sin(\theta)\) becomes zero.
This is a linear arrangement puzzle. We need to deduce the seating arrangement of all individuals based on the given clues and then count the total number of people.
Arrangement so far: P R
Since P is to the left of R, S must be to the right of R to maintain the two-person gap. Let '_' denote an empty seat.
Arrangement so far: P R _ _ S
Arrangement so far: P R _ _ S _ T
Arrangement so far: Q _ P R _ _ S _ T
Arrangement so far: Q _ P R _ _ S _ T _ U
The final arrangement is Q _ P R _ _ S _ T _ U. Let's count the positions:
1 (Q) + 1 (empty) + 1 (P) + 1 (R) + 1 (empty) + 1 (empty) + 1 (S) + 1 (empty) + 1 (T) + 1 (empty) + 1 (U)
Total positions = \(1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 11\)
The people seated are Q, P, R, S, T, U. The total number of positions occupied by people or empty seats between them is 11.
Correct Option: D) 11