The force exerted by the earth on the moon is:
(Consider G = 6 × 10-11 Nm2 kg-2; mass of the earth M = 6 × 1024 kg; mass of the moon = 8 × 1022 kg, and distance between earth and moon is 4 × 108 m).
1.8 × 1018 N
1.8 × 1020 N
1.8 × 1024 N
1.8 × 1022 N
To calculate the gravitational force between the Earth and the Moon, we will use Newton's Law of Universal Gravitation. This law states that the force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
The formula for gravitational force \(F\) is given by:
\[ F = G \frac{M m}{r^2} \]where:
To match the options, we can write \(18 \times 10^{19}\) as \(1.8 \times 10^1 \times 10^{19} = 1.8 \times 10^{(1+19)} = 1.8 \times 10^{20} \text{ N}\).
The calculated force is \(1.8 \times 10^{20} \text{ N}\).
B) \(1.8 \times 10^{20} \text{ N}\) is the correct force calculated using Newton's Law of Universal Gravitation with the given values.
The question asks to identify an event that is a consequence of atmospheric refraction. We need to understand what atmospheric refraction is and then evaluate each option based on this understanding.
Correct Option: D) Advance sunrise and delayed sunset is a direct consequence of atmospheric refraction, where light from the Sun bends as it passes through the Earth's atmosphere, making the Sun visible earlier than it actually rises and later than it actually sets.
The problem asks us to identify symbols that are immediately preceded by a number and immediately followed by another symbol in the given series. We need to scan the series from left to right and apply this condition to each symbol.
A) More than two
As identified in the step-by-step analysis, there are three such symbols ($ , * , &) that satisfy the given condition. Since three is more than two, option A is the correct answer.
If θ is an acute angle, find the denominator D, when
(cot θ − cosec θ)2 = \(\frac{1-\cos\theta}{D}\)
To find the denominator D, we need to simplify the left-hand side of the given equation, \((cot \theta - cosec \theta)^2\), using fundamental trigonometric identities and algebraic manipulations, and then compare it with the right-hand side, \(\frac{1-\cos\theta}{D}\).
Express cot \(\theta\) and cosec \(\theta\) in terms of sin \(\theta\) and cos \(\theta\):
\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \quad \text{and} \quad \csc \theta = \frac{1}{\sin \theta} \]Substitute these into the given expression:
\[ (\cot \theta - \csc \theta)^2 = \left( \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta} \right)^2 \]Combine the terms inside the parenthesis:
\[ = \left( \frac{\cos \theta - 1}{\sin \theta} \right)^2 \]Square the numerator and the denominator:
\[ = \frac{(\cos \theta - 1)^2}{(\sin \theta)^2} \] \[ = \frac{(1 - \cos \theta)^2}{\sin^2 \theta} \quad \text{(since} \ (a-b)^2 = (b-a)^2 \text{)} \]Use the identity \(\sin^2 \theta + \cos^2 \theta = 1\), which implies \(\sin^2 \theta = 1 - \cos^2 \theta\):
\[ = \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta} \]Factor the denominator using the difference of squares formula, \(a^2 - b^2 = (a-b)(a+b)\):
\[ = \frac{(1 - \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)} \]Cancel out the common term \((1 - \cos \theta)\) from the numerator and denominator (since \(\theta\) is an acute angle, \(\cos \theta \neq 1\)):
\[ = \frac{1 - \cos \theta}{1 + \cos \theta} \]Compare this result with the given right-hand side, \(\frac{1-\cos\theta}{D}\):
\[ \frac{1 - \cos \theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{D} \]By comparing the denominators, we find that \(D = 1 + \cos \theta\).
Correct Option: D) \(1 + \cos \theta\)
The question asks for the theme of 'Financial Literacy Week 2024' conducted by the Reserve Bank of India (RBI). To answer this, one needs to be aware of the specific theme announced by the RBI for that particular year's financial literacy campaign.
C) Make a Right Start : Become Financially Smart is the correct theme for Financial Literacy Week 2024, as announced by the Reserve Bank of India.
This problem involves work and time, where the work done by men and women is combined. We need to find the individual work rate of a man and a woman first by setting up equations based on the given information. Once we find the work rate of one woman, we can calculate the time taken for one woman to complete the entire work.
Let \(M\) be the amount of work one man can do in one day, and \(W\) be the amount of work one woman can do in one day.
From the first statement: 8 men and 12 women can build a wall in 10 days. This means their combined daily work rate is \(\frac{1}{10}\) of the wall.
\[ 8M + 12W = \frac{1}{10} \quad \text{(Equation 1)} \]From the second statement: 6 men and 8 women can build the same wall in 14 days. This means their combined daily work rate is \(\frac{1}{14}\) of the wall.
\[ 6M + 8W = \frac{1}{14} \quad \text{(Equation 2)} \]Now we have a system of two linear equations. We can solve for \(M\) and \(W\). Multiply Equation 1 by 3 and Equation 2 by 4 to eliminate \(M\):
Multiply Equation 1 by 3:
\[ 3 \times (8M + 12W) = 3 \times \frac{1}{10} \Rightarrow 24M + 36W = \frac{3}{10} \quad \text{(Equation 3)} \]Multiply Equation 2 by 4:
\[ 4 \times (6M + 8W) = 4 \times \frac{1}{14} \Rightarrow 24M + 32W = \frac{4}{14} = \frac{2}{7} \quad \text{(Equation 4)} \]Subtract Equation 4 from Equation 3:
\[ (24M + 36W) - (24M + 32W) = \frac{3}{10} - \frac{2}{7} \] \[ 4W = \frac{21 - 20}{70} \] \[ 4W = \frac{1}{70} \]Solve for \(W\):
\[ W = \frac{1}{70 \times 4} = \frac{1}{280} \]This means one woman can complete \(\frac{1}{280}\) of the wall in one day.
The time taken for 1 woman to build the wall alone is the reciprocal of her daily work rate:
\[ \text{Time} = \frac{1}{W} = \frac{1}{\frac{1}{280}} = 280 \text{ days} \]A) 280 days is the correct answer, as calculated above.
The question asks about the factors that determine the kinetic energy of a moving object. We need to recall the formula for kinetic energy and identify the variables involved.
Correct Option: B) Both (i) and (ii)
The problem asks us to find a relationship between two sets of letters based on their positions in the English alphabet. We need to identify the pattern of letter shifts from the given examples and then apply the same pattern to the target set of letters.
B) MPNO is the correct answer, as derived by applying the +5 shift to each letter of HKIJ.
The problem asks us to identify the letter-cluster pair that does not follow the same pattern as the others, based on the English alphabetical order. We need to find the relationship between the letters in each pair and then identify the odd one out.
B) UM – GZ is the pair that does not belong to the group. In this pair, the differences between the corresponding letter positions are not consistently 3. For U (21) and G (7), the difference is 14. For M (13) and Z (26), the difference is -13 (or 13 if cyclic). This clearly deviates from the pattern found in the other options.
This is a circular seating arrangement problem. We need to deduce the positions of all individuals around a circular table based on the given clues and then determine the relative position of D with respect to B.
B) Immediate left
Based on the final arrangement, D is at position 4 and B is at position 3. When facing the center, moving from B to D is one position to the left. Therefore, D is to the immediate left of B.