This question tests the understanding of temperature scale conversions, specifically between the Kelvin and Celsius scales. The key is to know the relationship between the absolute zero point and the freezing point of water in both scales.
C) subtract 273 from the given temperature. This is the correct operation to convert a temperature from Kelvin to Celsius, as 0 °C corresponds to 273 K.
We are given two conditions relating a cube and a square: one about their dimensions (sum of edge lengths vs. perimeter) and another about their numerical values (volume vs. area). We need to set up equations based on these conditions and solve for the side lengths of the cube and square, eventually finding the perimeter of the square.
Let the side length of the cube be \(a\) units and the side length of the square be \(x\) units.
Condition 1: Sum of the lengths of the edges of a cube is equal to three-fifths of the perimeter of a square.
A cube has 12 edges, each of length \(a\). So, the sum of the lengths of the edges is \(12a\).
The perimeter of a square with side length \(x\) is \(4x\).
According to the condition:
\[ 12a = \frac{3}{5} (4x) \] \[ 12a = \frac{12x}{5} \]Dividing both sides by 12, we get:
\[ a = \frac{x}{5} \quad \text{(Equation 1)} \]Condition 2: The numerical value of the volume of the cube is equal to the numerical value of the area of the square.
The volume of the cube is \(a^3\).
The area of the square is \(x^2\).
According to the condition:
\[ a^3 = x^2 \quad \text{(Equation 2)} \]Substitute Equation 1 into Equation 2:
Substitute \(a = \frac{x}{5}\) into \(a^3 = x^2\):
\[ \left(\frac{x}{5}\right)^3 = x^2 \] \[ \frac{x^3}{125} = x^2 \]Solve for \(x\):
Since \(x\) is a side length, \(x \neq 0\). We can divide both sides by \(x^2\):
\[ \frac{x^3}{125x^2} = \frac{x^2}{x^2} \] \[ \frac{x}{125} = 1 \] \[ x = 125 \text{ units} \]Calculate the perimeter of the square:
The perimeter of the square is \(4x\).
\[ \text{Perimeter} = 4 \times 125 \] \[ \text{Perimeter} = 500 \text{ units} \]C) 500 units
This question asks us to evaluate the strength of two arguments in response to the statement: "Should Yoga be made compulsory in schools?" We need to determine which argument(s) provide a strong, logical, and relevant reason for or against the statement.
A) Only argument I is strong. Argument I provides a valid and relevant reason in favor of making Yoga compulsory, focusing on the holistic development of children. Argument II is weak because its premise ("parents will not object") contradicts its conclusion ("No" to compulsory Yoga) and does not provide a coherent reason against the statement.
Consider a planet whose mass and radius are one-third the mass and radius of the earth. If g is the value of acceleration due to gravity on the earth, then its value on the planet will be __.
To solve this problem, we need to recall the formula for acceleration due to gravity on the surface of a planet. We will then use the given ratios of mass and radius of the planet relative to Earth to find the acceleration due to gravity on the planet.
The acceleration due to gravity (\(g\)) on the surface of a planet is given by the formula:
\[ g = \frac{GM}{R^2} \]where:
Let \(g_E\), \(M_E\), and \(R_E\) be the acceleration due to gravity, mass, and radius of the Earth, respectively. So, for Earth:
\[ g_E = \frac{GM_E}{R_E^2} \]Let \(g_P\), \(M_P\), and \(R_P\) be the acceleration due to gravity, mass, and radius of the planet, respectively. So, for the planet:
\[ g_P = \frac{GM_P}{R_P^2} \]According to the problem statement, the mass of the planet is one-third the mass of the Earth, and the radius of the planet is one-third the radius of the Earth. Therefore:
\[ M_P = \frac{1}{3} M_E \] \[ R_P = \frac{1}{3} R_E \]Substitute these values into the equation for \(g_P\):
\[ g_P = \frac{G \left(\frac{1}{3} M_E\right)}{\left(\frac{1}{3} R_E\right)^2} \] \[ g_P = \frac{G \frac{1}{3} M_E}{\frac{1}{9} R_E^2} \]Simplify the expression:
\[ g_P = \frac{1}{3} \times \frac{9}{1} \times \frac{GM_E}{R_E^2} \] \[ g_P = 3 \times \frac{GM_E}{R_E^2} \]Since \(g_E = \frac{GM_E}{R_E^2}\), we can substitute \(g_E\) into the equation for \(g_P\):
\[ g_P = 3 g_E \]Given that \(g\) is the value of acceleration due to gravity on Earth, \(g_E = g\). Therefore, the value on the planet will be \(3g\).
Correct Option: B) 3 g
During free fall, an object accelerates due to gravity. We need to analyze how different forms of energy change as the object moves downwards.
Correct Option: D) Kinetic energy
As an object falls, its speed increases due to gravity. Since kinetic energy is directly proportional to the square of its speed (\(KE = \frac{1}{2}mv^2\)), an increase in speed leads to an increase in kinetic energy.
Uma arranges for a terrace party. She divides the total expected expenditure into 5 broad segments. Here is the distribution:
| Expense segments | % Expense |
| 1 | 30 |
| 2 | 20 |
| 3 | 10 |
| 4 | 25 |
| 5 | 15 |
If she spends Rs 850 on segment 3, what will be her total expense on segment 2 and segment 5?
The problem provides the percentage distribution of total expected expenditure across five segments and the actual expense for one segment. We need to find the total expense for two other segments. The approach involves first determining the total expenditure using the given expense for segment 3 and its corresponding percentage. Once the total expenditure is known, we can calculate the expenses for segment 2 and segment 5 and then sum them up.
Let the total expected expenditure be \(T\). We know that 10% of \(T\) is Rs 850.
\[ 0.10 \times T = 850 \]\[ T = \frac{850}{0.10} \]\[ T = 8500 \]Segment 2 accounts for 20% of the total expenditure.
\[ \text{Expense for segment 2} = 0.20 \times T \]\[ \text{Expense for segment 2} = 0.20 \times 8500 \]\[ \text{Expense for segment 2} = 1700 \]Segment 5 accounts for 15% of the total expenditure.
\[ \text{Expense for segment 5} = 0.15 \times T \]\[ \text{Expense for segment 5} = 0.15 \times 8500 \]\[ \text{Expense for segment 5} = 1275 \]Alternatively, we can first find the combined percentage for segment 2 and segment 5 and then calculate the expense:
B) Rs 2,975 is the total expense on segment 2 and segment 5.
The question asks to identify the sector that received the highest Foreign Direct Investment (FDI) inflows in India over the last ten years, specifically as of 2018. This requires knowledge of FDI trends in India across various sectors.
C) Telecommunications — The telecommunications sector has historically attracted substantial FDI in India due to its vast market potential and continuous technological advancements. As of 2018, it remained one of the top sectors for FDI inflows, reflecting significant investments in infrastructure and services.
This problem involves concepts of speed, distance, and time. We are given two scenarios for Mobina's journey: one where she drives at a normal speed and another where she drives at a higher speed to compensate for a delay. The key is that the distance covered is the same in both scenarios, and the difference in time taken (or the delay) is given. We can set up equations based on the formula \(\text{Distance} = \text{Speed} \times \text{Time}\) and solve for the unknown distance.
Let the distance covered be \(D\) km.
Let the usual time taken to cover the distance be \(T\) hours.
In the first scenario (normal speed):
Speed (\(S_1\)) = 45 km/hr
Time (\(T_1\)) = \(T\) hours
So, \(D = S_1 \times T_1 = 45T\) (Equation 1)
In the second scenario (late start):
Mobina started 12 minutes late. 12 minutes in hours is \[\frac{12}{60} = \frac{1}{5}\] hours.
To reach on time, she had to cover the distance \(D\) in \(\left(T - \frac{1}{5}\right)\) hours.
New Speed (\(S_2\)) = 54 km/hr
New Time (\(T_2\)) = \(\left(T - \frac{1}{5}\right)\) hours
So, \(D = S_2 \times T_2 = 54 \left(T - \frac{1}{5}\right)\) (Equation 2)
Now, we equate Equation 1 and Equation 2 since the distance \(D\) is the same:
\[ 45T = 54 \left(T - \frac{1}{5}\right) \]Expand the right side of the equation:
\[ 45T = 54T - 54 \times \frac{1}{5} \]\[ 45T = 54T - \frac{54}{5} \]Rearrange the terms to solve for \(T\):
\[ \frac{54}{5} = 54T - 45T \]\[ \frac{54}{5} = 9T \]Solve for \(T\):
\[ T = \frac{54}{5 \times 9} \]\[ T = \frac{6}{5} \text{ hours} \]Now substitute the value of \(T\) back into Equation 1 to find the distance \(D\):
\[ D = 45T \]\[ D = 45 \times \frac{6}{5} \]\[ D = 9 \times 6 \]\[ D = 54 \text{ km} \]D) 54 km is the distance covered during the journey.
0.\(\overline{047619}\), when written as a vulgar fraction, is equal to
\( \frac{1}{19} \)
\( \frac{1}{23} \)
\( \frac{1}{17} \)
\( \frac{1}{21} \)
To convert a repeating decimal (recurring decimal) into a vulgar fraction, we use a standard algebraic method. We set the given decimal equal to a variable, multiply it by powers of 10 to align the repeating part, and then subtract the equations to eliminate the repeating part, allowing us to solve for the variable as a fraction.
Let the given repeating decimal be \(x\):
\[ x = 0.\overline{047619} \]This means the block of digits "047619" repeats indefinitely.
\[ x = 0.047619047619... \quad \text{(Equation 1)} \]Count the number of repeating digits. In this case, there are 6 repeating digits (0, 4, 7, 6, 1, 9).
Multiply Equation 1 by \(10^6\) (since there are 6 repeating digits) to shift the decimal point past one full repeating block:
\[ 10^6 x = 1000000x = 47619.047619... \quad \text{(Equation 2)} \]Subtract Equation 1 from Equation 2:
\[ 1000000x - x = 47619.047619... - 0.047619... \] \[ 999999x = 47619 \]Solve for \(x\):
\[ x = \frac{47619}{999999} \]Simplify the fraction. We need to find common factors for the numerator and denominator. We can test divisibility by the denominators given in the options.
Let's try dividing 999999 by 47619:
\[ \frac{999999}{47619} = 21 \]This means that \(999999 = 47619 \times 21\).
Substitute this back into the fraction for \(x\):
\[ x = \frac{47619}{47619 \times 21} \] \[ x = \frac{1}{21} \]D) \( \frac{1}{21} \) is the simplified vulgar fraction for \(0.\overline{047619}\).
The question asks to identify the host country for the 2019 Combined World Cup for shooting. This is a factual question related to sports events and their venues, requiring knowledge of international sporting event hosting.
Correct Option: D) India