The question asks to identify the tissue responsible for fat storage in the human body. This requires knowledge of the different types of connective tissues and their primary functions.
Correct Option: B) Adipose tissue is a specialized type of loose connective tissue that functions primarily as a site for fat storage. It is composed of adipocytes, which are cells filled with large lipid droplets. Adipose tissue serves several crucial roles, including energy reserve, insulation against heat loss, and protection of organs.
The question asks to identify the acid present in an ant sting. This is a common general science fact often tested in competitive exams. Recalling the specific acid associated with ant stings is key to answering this question.
B) Methanoic acid is the acid present in ant stings (and also in nettle stings). It is also known as formic acid.
To find the ratio of the area of the rhombus to that of the square, we first need to express the diagonals of the rhombus in terms of a variable. Then, calculate the area of the rhombus. Next, determine the side length of the square using the longer diagonal of the rhombus and calculate the area of the square. Finally, find the ratio of these two areas.
Let the two diagonals of the rhombus be \(d_1\) and \(d_2\).
According to the problem, one diagonal is 65% of the other. Let \(d_1\) be the longer diagonal and \(d_2\) be the shorter diagonal.
So, \[d_2 = 65\% \text{ of } d_1 = \frac{65}{100} d_1 = \frac{13}{20} d_1\].
The area of a rhombus is given by the formula:
\[ \text{Area of Rhombus} = \frac{1}{2} \times d_1 \times d_2 \]Substitute the value of \(d_2\) into the area formula:
\[ \text{Area of Rhombus} = \frac{1}{2} \times d_1 \times \left(\frac{13}{20} d_1\right) = \frac{13}{40} d_1^2 \]A square is drawn using the longer diagonal as its side. The longer diagonal is \(d_1\).
So, the side of the square, \(s = d_1\).
The area of the square is given by the formula:
\[ \text{Area of Square} = s^2 = d_1^2 \]Now, find the ratio of the area of the rhombus to that of the square:
\[ \frac{\text{Area of Rhombus}}{\text{Area of Square}} = \frac{\frac{13}{40} d_1^2}{d_1^2} \]Cancel out \(d_1^2\) from the numerator and denominator:
\[ \frac{\text{Area of Rhombus}}{\text{Area of Square}} = \frac{13}{40} \]The ratio is \(13 : 40\).
B) 13 : 40 is the correct ratio of the area of the rhombus to that of the square.
To find the area of an isosceles triangle, we can use the formula for the area of a triangle, which is \( \frac{1}{2} \times \text{base} \times \text{height} \). Since the height is not given, we need to calculate it first. In an isosceles triangle, the altitude from the vertex between the equal sides to the base bisects the base and forms two right-angled triangles. We can use the Pythagorean theorem to find the height.
Identify the given values:
Draw an altitude:
Draw an altitude \( AD \) from vertex \( A \) to the base \( BC \). Since triangle \( ABC \) is an isosceles triangle with \( AB = AC \), the altitude \( AD \) bisects the base \( BC \). Therefore, \( BD = DC = \frac{BC}{2} \).
Calculate the length of \( BD \):
\[ BD = \frac{20 \, \text{cm}}{2} = 10 \, \text{cm} \]Apply the Pythagorean theorem:
Consider the right-angled triangle \( ADB \). We have the hypotenuse \( AB = 26 \, \text{cm} \) and one leg \( BD = 10 \, \text{cm} \). We need to find the other leg, which is the height \( AD \).
Calculate the area of triangle \( ABC \):
The area of a triangle is given by the formula \( \frac{1}{2} \times \text{base} \times \text{height} \).
Correct Option: B) 240 cm²
This problem involves calculating individual weights given average weights of groups. We will use the formula for average weight, which is the sum of weights divided by the number of items, to set up a system of equations and solve for the unknown weight of Q.
The average weight of P, Q, and R is 58 kg. Since there are 3 individuals, their total weight is:
\[ P + Q + R = 58 \times 3 = 174 \text{ kg} \quad \text{(Equation 1)} \]The average weight of P and Q is 54 kg. Since there are 2 individuals, their total weight is:
\[ P + Q = 54 \times 2 = 108 \text{ kg} \quad \text{(Equation 2)} \]The average weight of Q and R is 48 kg. Since there are 2 individuals, their total weight is:
\[ Q + R = 48 \times 2 = 96 \text{ kg} \quad \text{(Equation 3)} \]Substitute the value of \((P + Q)\) from Equation 2 into Equation 1:
\[ 108 + R = 174 \]\[ R = 174 - 108 \]\[ R = 66 \text{ kg} \]Substitute the value of \(R\) into Equation 3:
\[ Q + 66 = 96 \]\[ Q = 96 - 66 \]\[ Q = 30 \text{ kg} \]C) 30 kg is the correct weight of Q, as calculated by solving the system of equations.
To find the smallest number by which 35280 must be divided so that the quotient is a perfect square, we need to perform prime factorization of 35280. A number is a perfect square if all the exponents in its prime factorization are even. We will identify the prime factors with odd exponents and divide the original number by the product of these factors to make the exponents even.
First, we find the prime factors of 35280.
\[ 35280 = 10 \times 3528 \]\[ 35280 = (2 \times 5) \times 3528 \]Now, factorize 3528:\[ 3528 = 2 \times 1764 \]\[ 1764 = 2 \times 882 \]\[ 882 = 2 \times 441 \]\[ 441 = 3 \times 147 \]\[ 147 = 3 \times 49 \]\[ 49 = 7 \times 7 \]So, the prime factorization of 3528 is:\[ 3528 = 2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 7 = 2^3 \times 3^2 \times 7^2 \]Combining with the factors of 10:\[ 35280 = (2 \times 5) \times (2^3 \times 3^2 \times 7^2) \]\[ 35280 = 2^4 \times 3^2 \times 5^1 \times 7^2 \]For a number to be a perfect square, all the exponents in its prime factorization must be even. In the prime factorization of 35280, we have:
The prime factor with an odd exponent is \(5^1\).
To make the exponent of 5 even (specifically, \(5^0\)), we need to divide 35280 by 5. When we divide 35280 by 5, the factor \(5^1\) will become \(5^0\), which is 1. The resulting quotient will have all even exponents for its prime factors, making it a perfect square.
\[ \frac{35280}{5} = \frac{2^4 \times 3^2 \times 5^1 \times 7^2}{5^1} = 2^4 \times 3^2 \times 7^2 \]This quotient \(2^4 \times 3^2 \times 7^2\) is a perfect square because all exponents (4, 2, 2) are even.The smallest number by which 35280 must be divided so that the quotient is a perfect square is 5.
A) 5 is the smallest number by which 35280 must be divided to obtain a perfect square quotient, as it is the prime factor with an odd exponent in the prime factorization of 35280.
The question asks about the location of the first British presidency established in India. This requires knowledge of early British colonial history and the establishment of their trading posts and administrative centers.
Correct Option: A) Surat was the site of the first permanent British factory and subsequently the first British presidency established in India in 1613.
We need to find the original number. We can represent the original number as a variable, say \(x\). Then, we apply the given percentage changes sequentially to \(x\) to find the final number. Finally, we set up an equation based on the information that the final number is 64 less than the original number and solve for \(x\).
Let the original number be \(x\).
First decrease by 20%: When a number is decreased by 20%, it becomes \(100\% - 20\% = 80\%\) of its original value. So, the number after the first decrease is:
\[ x_1 = x - 0.20x = 0.80x \]Then increase by 15%: Now, this new number (\(0.80x\)) is increased by 15%. So, it becomes \(100\% + 15\% = 115\%\) of \(0.80x\). The final number is:
\[ x_2 = 0.80x + (0.15 \times 0.80x) \] \[ x_2 = 0.80x (1 + 0.15) \] \[ x_2 = 0.80x \times 1.15 \] \[ x_2 = 0.92x \]Formulate the equation: We are given that the number so obtained (\(x_2\)) is 64 less than the original number (\(x\)).
\[ x_2 = x - 64 \]Solve for \(x\): Substitute the value of \(x_2\) from step 3 into the equation from step 4:
\[ 0.92x = x - 64 \]Rearrange the terms to solve for \(x\):
\[ 64 = x - 0.92x \] \[ 64 = (1 - 0.92)x \] \[ 64 = 0.08x \] \[ x = \frac{64}{0.08} \] \[ x = \frac{64}{\frac{8}{100}} \] \[ x = \frac{64 \times 100}{8} \] \[ x = 8 \times 100 \] \[ x = 800 \]Correct Option: A) 800
The question asks for the location of Sambhar Lake. This is a factual question related to Indian geography. To answer this, one needs to recall the geographical distribution of major lakes in India.
Correct Option: B) Rajasthan is the correct answer because Sambhar Lake, India's largest inland salt lake, is situated in the state of Rajasthan.
The question asks to identify the person known as the 'Father of the White Revolution' in India. This refers to a significant movement in India's dairy sector. We need to recall the key figures associated with major agricultural and economic revolutions in India.
Correct Option: B) Verghese Kurien is widely regarded as the 'Father of the White Revolution' in India due to his pioneering work and leadership in Operation Flood, which significantly boosted India's milk production and dairy industry.