To find the magnification produced by a convex lens, we use the lens formula and the magnification formula. The lens formula relates the object distance (\(u\)), image distance (\(v\)), and focal length (\(f\)). The magnification formula relates the image distance, object distance, and the ratio of image height to object height.
Identify the given information:
The lens is a convex lens.
The object is placed at a distance \(2f\) from the lens. According to the sign convention for lenses, the object distance \(u\) is negative for real objects placed in front of the lens. So, \(u = -2f\).
Apply the lens formula:
The lens formula is given by:
Substitute the value of \(u\) into the lens formula:
\[ \frac{1}{v} - \frac{1}{(-2f)} = \frac{1}{f} \]\[ \frac{1}{v} + \frac{1}{2f} = \frac{1}{f} \]Solve for \(v\) (image distance):
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{2f} \]\[ \frac{1}{v} = \frac{2 - 1}{2f} \]\[ \frac{1}{v} = \frac{1}{2f} \]\[ v = 2f \]Apply the magnification formula:
The magnification \(m\) for a lens is given by:
Substitute the values of \(v\) and \(u\) into the magnification formula:
\[ m = \frac{2f}{-2f} \]\[ m = -1 \]Interpret the result:
The magnification \(m = -1\) indicates that the image is real (inverted, due to the negative sign) and of the same size as the object (due to the magnitude of 1). This is consistent with the properties of a convex lens when the object is placed at \(2f\), where the image is formed at \(2f\) on the other side, real, inverted, and of the same size.
B) m = -1. This is derived directly from the lens formula and magnification formula, indicating a real, inverted image of the same size as the object.
Which of the following represents the closest approximate value that should come in place of the question mark (?) in the following equation? (\(\sqrt{360}\) + 12 ÷ 6 × 3 - \(\sqrt{170}\)) ÷ 2 =?
To solve this problem, we need to approximate the square roots and then follow the order of operations (BODMAS/PEMDAS) to simplify the expression. BODMAS stands for Brackets, Orders (powers and square roots), Division, Multiplication, Addition, Subtraction.
A) 6 is the closest approximate value.
This problem involves calculating displacement based on a series of movements in different directions. We can solve this by breaking down each movement into its north-south and east-west components, or by visualizing the path and identifying the net displacement.
Therefore, point X is 60 m to the East of point Y.
C) 60 m, East is correct because point Y is 60 m West of point X, which means point X is 60 m East of point Y.
A student scored \( \frac{80}{80} \) marks in term 1 and \( \frac{75}{90} \) marks in term 2. What will be his percentage of final score, if the weightage given to the terms is 40% and 60%, respectively, correct to the nearest integer?
To find the final percentage score, we need to calculate the percentage score for each term first. Then, we apply the given weightage to each term's percentage score and sum them up to get the final weighted average percentage.
Calculate the percentage score for Term 1:
The student scored \( \frac{80}{80} \) in Term 1.
\[ \text{Percentage in Term 1} = \left( \frac{\text{Marks Scored}}{\text{Total Marks}} \right) \times 100 \] \[ \text{Percentage in Term 1} = \left( \frac{80}{80} \right) \times 100 = 1 \times 100 = 100\% \]Calculate the percentage score for Term 2:
The student scored \( \frac{75}{90} \) in Term 2.
\[ \text{Percentage in Term 2} = \left( \frac{\text{Marks Scored}}{\text{Total Marks}} \right) \times 100 \] \[ \text{Percentage in Term 2} = \left( \frac{75}{90} \right) \times 100 \] \[ \text{Percentage in Term 2} = \left( \frac{5 \times 15}{6 \times 15} \right) \times 100 = \frac{5}{6} \times 100 \] \[ \text{Percentage in Term 2} = \frac{500}{6} = 83.333...\% \]Apply the weightage to each term's percentage:
Weightage for Term 1 = 40% (or 0.40)
Weightage for Term 2 = 60% (or 0.60)
\[ \text{Weighted Score for Term 1} = 100\% \times 0.40 = 40\% \] \[ \text{Weighted Score for Term 2} = 83.333...\% \times 0.60 \] \[ \text{Weighted Score for Term 2} = \frac{5}{6} \times 100 \times 0.60 = \frac{5}{6} \times 60 = 5 \times 10 = 50\% \]Calculate the final percentage score:
\[ \text{Final Percentage Score} = \text{Weighted Score for Term 1} + \text{Weighted Score for Term 2} \] \[ \text{Final Percentage Score} = 40\% + 50\% = 90\% \]Round to the nearest integer:
The final percentage score is exactly 90%, which is already an integer.
Correct Option: B) 90%
The goal is to find which two mathematical signs, when interchanged in the given equation, make the equation correct. We will test each option by substituting the interchanged signs into the equation and then evaluating it using the BODMAS/PEMDAS rule.
The given equation is: \(28 - 28 \div 14 + 7 \times 2 = 16\)
Interchange \(\times\) and \(-\) in the original equation:
\[ 28 \times 28 \div 14 + 7 - 2 \]Now, evaluate using BODMAS:
Division first:
\[ 28 \times (28 \div 14) + 7 - 2 \] \[ 28 \times 2 + 7 - 2 \]Multiplication next:
\[ 56 + 7 - 2 \]Addition next:
\[ 63 - 2 \]Subtraction last:
\[ 61 \]Since \(61 \neq 16\), Option A is incorrect.
Interchange \(+\) and \(\div\) in the original equation:
\[ 28 - 28 + 14 \div 7 \times 2 \]Now, evaluate using BODMAS:
Division first:
\[ 28 - 28 + (14 \div 7) \times 2 \] \[ 28 - 28 + 2 \times 2 \]Multiplication next:
\[ 28 - 28 + 4 \]Addition/Subtraction from left to right:
\[ 0 + 4 \] \[ 4 \]Since \(4 \neq 16\), Option B is incorrect.
Interchange \(\times\) and \(\div\) in the original equation:
\[ 28 - 28 \times 14 + 7 \div 2 \]Now, evaluate using BODMAS:
Multiplication first:
\[ 28 - (28 \times 14) + 7 \div 2 \] \[ 28 - 392 + 7 \div 2 \]Division next:
\[ 28 - 392 + 3.5 \]Addition/Subtraction from left to right:
\[ -364 + 3.5 \] \[ -360.5 \]Since \(-360.5 \neq 16\), Option C is incorrect.
Interchange \(+\) and \(-\) in the original equation:
\[ 28 + 28 \div 14 - 7 \times 2 \]Now, evaluate using BODMAS:
Division first:
\[ 28 + (28 \div 14) - 7 \times 2 \] \[ 28 + 2 - 7 \times 2 \]Multiplication next:
\[ 28 + 2 - (7 \times 2) \] \[ 28 + 2 - 14 \]Addition/Subtraction from left to right:
\[ 30 - 14 \] \[ 16 \]Since \(16 = 16\), Option D is correct.
Correct Option: D) \(+\) and \(-\)
To solve this problem, we first need to determine the cost price (CP) of the blanket using the given selling price (SP) and loss percentage. Once the cost price is known, we can calculate the new selling price required to achieve a 5% gain.
Calculate the Cost Price (CP):
Given:
The formula for selling price with a loss is:
\[ \text{SP} = \text{CP} \times \left(1 - \frac{\text{Loss percentage}}{100}\right) \]Substitute the given values:
\[ 1148 = \text{CP} \times \left(1 - \frac{30}{100}\right) \] \[ 1148 = \text{CP} \times \left(1 - 0.30\right) \] \[ 1148 = \text{CP} \times 0.70 \]Now, solve for CP:
\[ \text{CP} = \frac{1148}{0.70} \] \[ \text{CP} = 1640 \text{ Rs} \]Calculate the New Selling Price for a 5% Gain:
We want to achieve a gain of 5% on the Cost Price (CP = Rs 1,640).
The formula for selling price with a gain is:
\[ \text{SP}_2 = \text{CP} \times \left(1 + \frac{\text{Gain percentage}}{100}\right) \]Substitute the values:
\[ \text{SP}_2 = 1640 \times \left(1 + \frac{5}{100}\right) \] \[ \text{SP}_2 = 1640 \times \left(1 + 0.05\right) \] \[ \text{SP}_2 = 1640 \times 1.05 \] \[ \text{SP}_2 = 1722 \text{ Rs} \]Therefore, the blanket should be sold for Rs 1,722 to gain 5%.
C) Rs 1,722
The question asks to identify the institution founded by Prasanta Chandra Mahalanobis. This requires knowledge of prominent Indian institutions and their founders, particularly in the fields of science and education.
Correct Option: C) Indian Statistical Institute
The problem asks us to identify the logic connecting the given alphabet pairs and then apply the same logic to the options to find the correct pair. We need to find the positional difference between corresponding letters in the given pairs.
The pattern for the first pair is +2, +3, +4, +5, +6.
The pattern for the second pair is also +2, +3, +4, +5, +6.
Correct Option: A) RHINO : TKMSU
This pair follows the same logic (+2, +3, +4, +5, +6).
The question asks to identify the reproductive parts of a flower that contain germ cells. This requires knowledge of the structure of a flower and the function of each part in reproduction.
Correct Option: B) Stamens and pistils
The problem asks us to determine who is sitting immediately next to Q, apart from R, based on a set of seating arrangements for six students in a straight line. We will deduce the seating arrangement step-by-step using the given clues.
Identify the total number of students and their orientation: There are six students (P, Q, R, S, T, U) sitting in a straight line facing north. This means their left is our left, and their right is our right.
Place P and S at the ends: "P and S sit at the ends of the row." This gives us two possible initial arrangements:
Place T as an immediate neighbour of P: "T sits as an immediate neighbour of P."
Place U as an immediate neighbour of S: "U sits as an immediate neighbour of S."
Place Q based on the clue "Only one person sits between Q and S":
Final check and answer the question: We have two possible arrangements:
The question asks, "Apart from R, who is sitting immediately next to Q?"
We need to find a consistent answer. Let's re-evaluate the clues to see if one arrangement can be eliminated or if there's a unique solution.
Let's re-read "Only one person sits between Q and S."
In the correct arrangement S U Q R T P, Q is sitting between U and R. Therefore, apart from R, U is sitting immediately next to Q.
Correct Option: D) U is sitting immediately next to Q, apart from R, in the final arrangement S U Q R T P.