To determine which physical quantity has the dimension \([\text{M L}^{-3}]\), we need to recall the definitions and dimensional formulas for each of the given options: Energy, Density, Pressure, and Work. We will derive or state the dimensional formula for each and compare it with the target dimension.
B) Density has the dimension \([\text{M L}^{-3}]\) because it is defined as mass per unit volume.
The question asks about the primary benefit of customising the top-level master slide in MS PowerPoint's Slide Master view compared to individual layout masters. Understanding the hierarchy and inheritance in Slide Master view is key to answering this.
Correct Option: B) Customising the top-level master ensures consistency of universal elements across all slide layouts
This is the primary benefit because any changes made to the top-level master (like background, fonts, logos, footers) automatically propagate to all slide layouts and thus to all slides in the presentation, ensuring a uniform look and feel.
To determine the relative direction of the Malwa Plateau with respect to the Chotanagpur Plateau, one needs to have a clear understanding of the geographical locations of these two plateaus on the map of India. Visualizing their positions helps in identifying the correct cardinal direction.
Correct Option: B) West
Samrath ranked 31st from the top and 13th from the bottom in his class. How many students are there in his class?
To find the total number of students in a class when a person's rank from both the top and the bottom is given, we use a simple formula. The key is to avoid double-counting the person whose rank is given.
Therefore, there are 43 students in the class.
Correct Option: D) 43
If secθ + tanθ = 2, then what is the value of 3secθ + 4?
\( \frac{31}{4} \)
\( \frac{15}{4} \)
\( \frac{17}{4} \)
\( \frac{33}{4} \)
The problem provides an equation involving secant and tangent functions. We need to find the value of an expression involving the secant function. The key to solving this problem is to use the fundamental trigonometric identity relating secant and tangent, which is \( \sec^2\theta - \tan^2\theta = 1 \). By manipulating this identity and the given equation, we can find the value of \( \sec\theta \) and then substitute it into the required expression.
We are given the equation:
\[ \sec\theta + \tan\theta = 2 \quad \ldots (1) \]Recall the fundamental trigonometric identity:
\[ \sec^2\theta - \tan^2\theta = 1 \]Factor the left side of the identity using the difference of squares formula \( a^2 - b^2 = (a-b)(a+b) \):
\[ (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1 \]Substitute the given value from equation (1) into this factored identity:
\[ (\sec\theta - \tan\theta)(2) = 1 \]Solve for \( \sec\theta - \tan\theta \):
\[ \sec\theta - \tan\theta = \frac{1}{2} \quad \ldots (2) \]Now we have a system of two linear equations with \( \sec\theta \) and \( \tan\theta \):
\[ \sec\theta + \tan\theta = 2 \quad \ldots (1) \]\[ \sec\theta - \tan\theta = \frac{1}{2} \quad \ldots (2) \]Add equation (1) and equation (2) to eliminate \( \tan\theta \):
\[ (\sec\theta + \tan\theta) + (\sec\theta - \tan\theta) = 2 + \frac{1}{2} \]\[ 2\sec\theta = \frac{4}{2} + \frac{1}{2} \]\[ 2\sec\theta = \frac{5}{2} \]Solve for \( \sec\theta \):
\[ \sec\theta = \frac{5}{2} \times \frac{1}{2} \]\[ \sec\theta = \frac{5}{4} \]Finally, substitute the value of \( \sec\theta \) into the expression \( 3\sec\theta + 4 \):
\[ 3\left(\frac{5}{4}\right) + 4 \]\[ = \frac{15}{4} + 4 \]\[ = \frac{15}{4} + \frac{16}{4} \]\[ = \frac{15 + 16}{4} \]\[ = \frac{31}{4} \]A) \( \frac{31}{4} \) is the correct value obtained after calculating \( \sec\theta \) and substituting it into the given expression.
This problem involves calculating the discount percentage needed to achieve a specific profit margin, given that the marked price is fixed at a certain percentage above the cost price. We will assume a cost price, calculate the marked price, then the selling price required for the desired gain, and finally the discount percentage.
Let the Cost Price (CP) of the item be \(₹100\).
The shopkeeper fixes the marked price 52% above its cost price.
\[ \text{MP} = \text{CP} + 52\% \text{ of CP} \]\[ \text{MP} = 100 + \frac{52}{100} \times 100 \]\[ \text{MP} = 100 + 52 = ₹152 \]The shopkeeper wants to gain 14%.
\[ \text{SP} = \text{CP} + 14\% \text{ of CP} \]\[ \text{SP} = 100 + \frac{14}{100} \times 100 \]\[ \text{SP} = 100 + 14 = ₹114 \]Discount is the difference between the Marked Price and the Selling Price.
\[ \text{Discount Amount} = \text{MP} - \text{SP} \]\[ \text{Discount Amount} = 152 - 114 = ₹38 \]Discount percentage is calculated on the Marked Price.
\[ \text{Discount Percentage} = \left( \frac{\text{Discount Amount}}{\text{MP}} \right) \times 100\% \]\[ \text{Discount Percentage} = \left( \frac{38}{152} \right) \times 100\% \]\[ \text{Discount Percentage} = \left( \frac{1}{4} \right) \times 100\% \]\[ \text{Discount Percentage} = 25\% \]D) 25% is the correct answer. The step-by-step calculation shows that a 25% discount on the marked price of ₹152 will result in a selling price of ₹114, which gives a 14% gain on the cost price of ₹100.
The question asks to identify the correct statements regarding the Indian Navy's 'Indian Ocean Ship (IOS) Sagar' initiative, specifically focusing on the deployment of INS Sunayna in April 2025. We need to evaluate each statement based on factual information about this initiative.
B) Only 1 and 3 are correct. Statement 1 accurately describes the nature of the deployment and the involvement of FFCs. Statement 3 correctly outlines the duration and port calls of the mission. Statement 2 is incorrect because the primary training aspect of this specific initiative is not about bringing FFC personnel to Visakhapatnam for school-based training, but rather engaging with them during the deployment.
We will solve this problem by setting up equations based on the given conditions. We'll define the initial cost price (CP) and selling price (SP), then adjust these based on the hypothetical scenarios to find the original CP.
Let the original Cost Price (CP) of the article be \(P\).
Umesh sold the article at a loss of 22%. So, the original Selling Price (SP) is:
\[ \text{SP} = P - 0.22P = 0.78P \]In the hypothetical scenario, he bought the item at 5% less. So, the new Cost Price (CP') would be:
\[ \text{CP}' = P - 0.05P = 0.95P \]He would have gained 15% on this new CP'. So, the new Selling Price (SP') would be:
\[ \text{SP}' = \text{CP}' + 0.15 \times \text{CP}' = 1.15 \times \text{CP}' \]\[ \text{SP}' = 1.15 \times (0.95P) = 1.0925P \]According to the problem, the new selling price (SP') is ₹935 more than the original selling price (SP). So, we can write the equation:
\[ \text{SP}' = \text{SP} + 935 \]Substitute the expressions for SP' and SP into the equation:
\[ 1.0925P = 0.78P + 935 \]Now, solve for \(P\):
\[ 1.0925P - 0.78P = 935 \]\[ 0.3125P = 935 \]\[ P = \frac{935}{0.3125} \]\[ P = \frac{935}{\frac{3125}{10000}} \]\[ P = \frac{935 \times 10000}{3125} \]\[ P = \frac{935 \times 32}{1} \]\[ P = 29920 \]Therefore, the original cost price of the article is ₹2,992.
Correct Option: A) 2,992
The question asks to identify the individual appointed as the Managing Director of NAFED in October 2024. This is a factual recall question related to current affairs and appointments in significant agricultural organizations.
Based on current affairs knowledge, Deepak Agarwal was appointed as the Managing Director of the National Agricultural Cooperative Marketing Federation of India Ltd (NAFED) in October 2024.
Correct Option: B) Deepak Agarwal was indeed appointed as the Managing Director of NAFED in October 2024. This is a direct current affairs fact.
The question asks to identify the caves known for their high citadel. This requires knowledge of specific features associated with prominent ancient Indian cave complexes.
Correct Option: A) The Junagadh caves, specifically the Uparkot Caves, are carved into the Uparkot Fort, which is a high citadel in Junagadh, Gujarat. This makes them famous for their association with a high citadel.