Which of the following bills presented along with the Annual Financial Statement provides details on the imposition, abolition, remission, alteration or regulation of taxes proposed in the Budget in India?
Ordinance Replacing Bill
Finance Bill
Money Bill
Ordinary Bill
The question asks to identify the specific type of bill that accompanies the Annual Financial Statement (Budget) and details proposals related to taxation (imposition, abolition, remission, alteration, or regulation of taxes) in India. Understanding the different types of bills in the Indian parliamentary system, especially those related to finance, is crucial here.
B) Finance Bill — The Finance Bill is introduced in Parliament every year to give effect to the financial proposals of the Government of India for the next financial year. It contains provisions for the imposition, abolition, remission, alteration, or regulation of taxes, as proposed in the Budget (Annual Financial Statement).
If the roots of the quadratic equation x2−kx+169=0 are equal, find the value of k.
To find the value of \(k\) for which the roots of a quadratic equation are equal, we use the discriminant condition. For a quadratic equation of the form \(ax^2 + bx + c = 0\), the roots are equal if and only if the discriminant \(\Delta = b^2 - 4ac\) is equal to zero.
Identify coefficients: The given quadratic equation is \(x^2 - kx + 169 = 0\). Comparing this with the standard form \(ax^2 + bx + c = 0\), we identify the coefficients:
Apply the discriminant condition: For equal roots, the discriminant must be zero:
\[ \Delta = b^2 - 4ac = 0 \]Substitute the values: Substitute the identified coefficients into the discriminant equation:
\[ (-k)^2 - 4(1)(169) = 0 \]Solve for \(k\):
\[ k^2 - 676 = 0 \]\[ k^2 = 676 \]\[ k = \pm\sqrt{676} \]Calculate the square root: To find \(\sqrt{676}\), we can recognize that \(20^2 = 400\) and \(30^2 = 900\), so the number is between 20 and 30. Since the last digit is 6, the unit digit of the square root must be 4 or 6. Let's try 26:
\[ 26 \times 26 = 676 \]So, \(\sqrt{676} = 26\).Final value of \(k\):
\[ k = \pm 26 \]Correct Option: A) \(\pm 26\) is the correct value for \(k\) because when \(k = 26\) or \(k = -26\), the discriminant of the quadratic equation \(x^2 - kx + 169 = 0\) becomes zero, ensuring that the roots are equal.
The question asks to identify the component of xylem that does NOT participate in water transport. To answer this, we need to recall the structure and functions of the four main components of xylem tissue: xylem vessels, tracheids, xylem parenchyma, and xylem fibres.
C) Xylem fibres — Xylem fibres are sclerenchymatous cells that are primarily involved in providing mechanical support to the plant. They are dead cells with thick, lignified walls and do not play any role in the conduction of water or minerals.
The question asks to identify which product in the given chemical reaction contains barium. We need to analyze the reactants and products of the provided chemical equation to determine the composition of each substance.
Correct Option: B) \(\text{BaBr}_2\)
As identified in the step-by-step analysis, \(\text{BaBr}_2\) is one of the products formed, and its chemical formula clearly indicates the presence of barium (Ba).
The question asks to identify the national festival that concludes with the 'Beating the Retreat' ceremony. This ceremony is a specific military tradition observed in India.
D) Republic Day — The 'Beating the Retreat' ceremony officially marks the end of the Republic Day festivities in India. It is held on January 29th, three days after Republic Day.
The question asks about the relationship between the energy lost as heat in an electrical circuit and various circuit parameters. This phenomenon is described by Joule's Law of Heating.
A) Directly proportional to the resistance. According to Joule's Law, \(H = I^2 R t\), which clearly shows that the heat generated (\(H\)) is directly proportional to the resistance (\(R\)) when current (\(I\)) and time (\(t\)) are constant.
\(\frac{9^{4}\times10^{3}\times4^{-4}\times16^{\frac{1}{2}}}{5^{6}\times3^{6}}\) equals:
\(\frac{1}{400}\)
\(\frac{4}{9000}\)
\(\frac{1}{9000}\)
\(\frac{9}{1000}\)
To solve this problem, we need to simplify the given expression by converting all numbers to their prime factor bases (2, 3, 5) and then applying the rules of exponents. We will combine terms with the same base by adding or subtracting their powers.
Rewrite the expression with prime bases:
The given expression is: \[ \frac{9^{4}\times10^{3}\times4^{-4}\times16^{\frac{1}{2}}}{5^{6}\times3^{6}} \] First, express all numbers as powers of their prime factors (2, 3, 5):
Substitute these into the expression:
\[ \frac{(3^2)^{4}\times(2\times5)^{3}\times(2^2)^{-4}\times(2^4)^{\frac{1}{2}}}{5^{6}\times3^{6}} \]Apply exponent rules \((a^m)^n = a^{mn}\) and \((ab)^n = a^n b^n\):
\[ \frac{3^{2\times4}\times2^{3}\times5^{3}\times2^{2\times(-4)}\times2^{4\times\frac{1}{2}}}{5^{6}\times3^{6}} \] \[ \frac{3^{8}\times2^{3}\times5^{3}\times2^{-8}\times2^{2}}{5^{6}\times3^{6}} \]Combine terms with the same base in the numerator using \(a^m \times a^n = a^{m+n}\):
Combine the powers of 2 in the numerator:
\[ 2^{3} \times 2^{-8} \times 2^{2} = 2^{3 + (-8) + 2} = 2^{3 - 8 + 2} = 2^{-3} \]The numerator becomes:
\[ 3^{8} \times 2^{-3} \times 5^{3} \]So the expression is now:
\[ \frac{3^{8}\times2^{-3}\times5^{3}}{5^{6}\times3^{6}} \]Combine terms with the same base by moving them to the numerator and applying \(a^m / a^n = a^{m-n}\) or \(1/a^n = a^{-n}\):
For base 3:
\[ \frac{3^{8}}{3^{6}} = 3^{8-6} = 3^{2} \]For base 5:
\[ \frac{5^{3}}{5^{6}} = 5^{3-6} = 5^{-3} \]For base 2, it's already in the numerator:
\[ 2^{-3} \]So the simplified expression is:
\[ 3^{2} \times 2^{-3} \times 5^{-3} \]Rewrite with positive exponents and simplify:
Recall that \(a^{-n} = 1/a^n\):
\[ 3^{2} \times \frac{1}{2^{3}} \times \frac{1}{5^{3}} \] \[ \frac{3^{2}}{2^{3} \times 5^{3}} \]Calculate the values:
\[ 3^{2} = 9 \] \[ 2^{3} = 8 \] \[ 5^{3} = 125 \]Substitute these values:
\[ \frac{9}{8 \times 125} \] \[ \frac{9}{1000} \]Correct Option: D) \(\frac{9}{1000}\)
A, B, C, D, E and F live on six different floors of the same building. The lowermost floor in the building is numbered 1, the floor above it is numbered 2 and so on till the topmost floor, which is numbered 6. Only two people live between A and F. Only three people live between B and C. C lives below B. E does not live below D. Only two people live below F.
Who lives immediately above F?
This is a logical reasoning problem involving arrangement on different floors. We need to deduce the exact floor on which each person lives based on the given clues and then identify who lives immediately above F.
This means F lives on floor 3. (Floors 1 and 2 are below F).
Floor 6:
Floor 5:
Floor 4:
Floor 3: F
Floor 2:
Floor 1:
Since F is on floor 3, A must be on floor 6 (3 + 2 + 1 = 6, or 3 - 2 - 1 = 0, which is not possible). So, A is on floor 6.
Floor 6: A
Floor 5:
Floor 4:
Floor 3: F
Floor 2:
Floor 1:
Combining clues 3 and 4, and looking at the remaining floors (1, 2, 4, 5):
So, B lives on floor 5 and C lives on floor 1.
Floor 6: A
Floor 5: B
Floor 4:
Floor 3: F
Floor 2:
Floor 1: C
The only remaining floors are 2 and 4. The only remaining people are D and E.
Therefore, D lives on floor 2 and E lives on floor 4.
Floor 6: A
Floor 5: B
Floor 4: E
Floor 3: F
Floor 2: D
Floor 1: C
From our final arrangement, F is on floor 3. The person immediately above F (on floor 4) is E.
Correct Option: B) E
The question asks about the process by which a sulphide ion is formed from a sulphur atom. This involves understanding the definitions of oxidation and reduction in terms of electron gain or loss.
Correct Option: D) Reduction — Sulphur gains two electrons to form a sulphide ion, \(\text{S}^{2-}\). The gain of electrons is defined as reduction.
If 0.3\(\overline{72}\) = \( \frac{x}{y} \), where x and y are coprime, then what is the value of (x + y)?
To solve this problem, we need to convert the given recurring decimal into a fraction in its simplest form. Then, we will identify the numerator (x) and the denominator (y) and check if they are coprime. Finally, we will calculate the sum (x + y).
Represent the recurring decimal as an equation:
Let \(N = 0.3\overline{72}\). This means \(N = 0.3727272...\)
Multiply to shift the non-repeating part to the left of the decimal:
Multiply \(N\) by 10 to move the non-repeating digit '3' to the left of the decimal point:
\[ 10N = 3.727272... \quad \text{(Equation 1)} \]
Multiply to shift one full repeating block to the left of the decimal:
Since the repeating block is '72' (two digits), multiply \(10N\) by \(10^2 = 100\):
\[ 100 \times (10N) = 100 \times (3.727272...) \]
\[ 1000N = 372.727272... \quad \text{(Equation 2)} \]
Subtract Equation 1 from Equation 2:
Subtracting Equation 1 from Equation 2 eliminates the repeating part:
\[ 1000N - 10N = 372.727272... - 3.727272... \]
\[ 990N = 369 \]
Solve for N:
\[ N = \frac{369}{990} \]
Simplify the fraction:
Both the numerator and the denominator are divisible by 9 (sum of digits of 369 is \(3+6+9=18\), which is divisible by 9; sum of digits of 990 is \(9+9+0=18\), which is divisible by 9).
\[ N = \frac{369 \div 9}{990 \div 9} = \frac{41}{110} \]
Identify x and y and check if they are coprime:
We have \[\frac{x}{y} = \frac{41}{110}\]. So, \(x = 41\) and \(y = 110\).
41 is a prime number. 110 is not divisible by 41 (\(110 = 2 \times 5 \times 11\)). Therefore, 41 and 110 are coprime.
Calculate (x + y):
\[ x + y = 41 + 110 = 151 \]
Correct Option: C) 151