The curved surface area of a cylindrical pillar is 66 m² and its volume is 231 m³. Find the measure of its radius.
(Take π = \( \frac{22}{7} \))
To find the radius of the cylindrical pillar, we will use the given information about its curved surface area and volume. We will set up equations for both and then divide the volume equation by the curved surface area equation to eliminate the height and solve for the radius.
Let the radius of the cylindrical pillar be \(r\) and its height be \(h\).
The formula for the curved surface area (CSA) of a cylinder is \(2\pi rh\).
Given CSA = 66 m\(^2\).
\[ 2\pi rh = 66 \quad \text{(Equation 1)} \]The formula for the volume (V) of a cylinder is \(\pi r^2 h\).
Given V = 231 m\(^3\).
\[ \pi r^2 h = 231 \quad \text{(Equation 2)} \]To find \(r\), we can divide Equation 2 by Equation 1:
\[ \frac{\pi r^2 h}{2\pi rh} = \frac{231}{66} \]Simplify the left side of the equation:
\[ \frac{r}{2} = \frac{231}{66} \]Simplify the right side of the equation. Both 231 and 66 are divisible by 3 (2+3+1=6, 6+6=12). Also, both are divisible by 11 (231 = 11 * 21, 66 = 11 * 6).
\[ \frac{231}{66} = \frac{21 \times 11}{6 \times 11} = \frac{21}{6} \]Further simplify \(\frac{21}{6}\) by dividing both numerator and denominator by 3:
\[ \frac{21}{6} = \frac{7}{2} \]Now we have:
\[ \frac{r}{2} = \frac{7}{2} \]Multiply both sides by 2 to solve for \(r\):
\[ r = 7 \]Thus, the radius of the cylindrical pillar is 7 meters.
The problem states that \(P\) is a prime number and \(P\) divides \(Q^2\). We need to determine which of the given expressions \(P\) will NOT necessarily divide. This involves applying the fundamental theorem of arithmetic (unique prime factorization) and properties of prime numbers related to divisibility.
Understand the given condition: We are given that \(P\) is a prime number and \(P\) divides \(Q^2\). According to Euclid's Lemma (a fundamental property of prime numbers), if a prime number \(P\) divides a product of two integers \(ab\), then \(P\) must divide \(a\) or \(P\) must divide \(b\). In our case, \(Q^2 = Q \times Q\). Since \(P\) divides \(Q^2\), it implies that \(P\) must divide \(Q\).
Analyze Option A: \(3Q\)
Since \(P\) divides \(Q\) (from step 1), and \(3\) is an integer, \(P\) will also divide \(3Q\). For example, if \(Q = kP\) for some integer \(k\), then \(3Q = 3kP\), which is clearly divisible by \(P\).
Analyze Option C: \(2Q^2\)
We are given that \(P\) divides \(Q^2\). If \(P\) divides \(Q^2\), then \(P\) will certainly divide any multiple of \(Q^2\), including \(2Q^2\). For example, if \(Q^2 = mP\) for some integer \(m\), then \(2Q^2 = 2mP\), which is divisible by \(P\).
Analyze Option D: \(Q\)
As established in step 1, if \(P\) is a prime number and \(P\) divides \(Q^2\), then \(P\) must divide \(Q\). This is a direct consequence of Euclid's Lemma.
Analyze Option B: \(Q + 1\)
We know that \(P\) divides \(Q\). Let's assume \(P\) also divides \(Q+1\). If \(P\) divides \(Q\) and \(P\) divides \(Q+1\), then \(P\) must divide their difference: \((Q+1) - Q = 1\). However, a prime number \(P\) cannot divide \(1\) unless \(P=1\), but \(1\) is not considered a prime number. Therefore, \(P\) cannot necessarily divide \(Q+1\).
Consider a counterexample: Let \(P = 3\). We know that \(P\) divides \(Q^2\). Let \(Q = 6\). Then \(Q^2 = 36\), and \(3\) divides \(36\). Here, \(P=3\) divides \(Q=6\). Now, let's check \(Q+1\). \(Q+1 = 6+1 = 7\). \(3\) does not divide \(7\). Thus, \(P\) does not necessarily divide \(Q+1\).
Correct Option: B) \(Q + 1\)
The question asks to identify the term for old alluvial soil. Alluvial soils are formed by the deposition of sediments by rivers. They are broadly classified into two types based on their age and characteristics: new alluvial soil and old alluvial soil. Understanding these classifications is key to answering the question.
D) Bangar is the correct term for old alluvial soil. These soils are found in the higher reaches of river plains, away from the annual flood limits, and contain calcareous deposits known as 'kankar'.
The United Nations International Day of Women and Girls in Science is annually observed on _____.
29th January
15th January
15th February
11th February
This question tests knowledge of important international days observed by the United Nations. Specifically, it asks for the date of the International Day of Women and Girls in Science.
Correct Option: D) 11 th February
The question asks for the common term used to refer to the main page of a website or a browser. This is a basic terminology question related to internet and web browsing.
Correct Option: B) Home page is the correct term for the main page of a website or a browser. It is the default or introductory page that users typically see first.
To find the sum of the measures of the interior angles of any polygon, we can use a standard formula that relates the number of sides of the polygon to the total sum of its interior angles. A hexagon is a polygon with 6 sides.
Thus, the sum of the measures of the interior angles of a hexagon is \(720^\circ\).
Correct Option: A) 720° is the correct sum of the interior angles of a hexagon.
The question asks to identify the first commercial and marketing arm of ISRO. This requires knowledge of the history and organizational structure of ISRO (Indian Space Research Organisation).
D) Antrix Corporation Limited — Antrix Corporation Limited was indeed the first commercial and marketing arm of ISRO, established in 1992 to promote and market ISRO's space products and services globally.
The question asks to identify the type of disease that Citrus canker is. This requires knowledge of common plant diseases and their causative agents.
Correct Option: A) bacterial disease
Citrus canker is caused by the bacterium Xanthomonas axonopodis pv. citri. It is a highly contagious disease that affects citrus plants, leading to significant economic losses in citrus-growing regions. Symptoms include characteristic canker-like lesions on leaves, stems, and fruits.
The problem requires simplifying a product of several algebraic expressions. We can use algebraic identities, specifically the difference of cubes identity \((x-y)(x^2+xy+y^2) = x^3-y^3\) and the difference of squares identity \((x-y)(x+y) = x^2-y^2\), repeatedly to simplify the given expression.
Rearrange the terms to group them strategically:
\[ (1 - a)(1 + a)(1 - a + a^{2})(1 + a + a^{2})(1 + a^{6}) \]First, apply the difference of squares identity to the first two terms: \((1 - a)(1 + a) = 1^2 - a^2 = 1 - a^2\).
Next, group the terms \((1 - a + a^{2})\) and \((1 + a + a^{2})\). These resemble the factors of a sum/difference of cubes. Let's rewrite them slightly:
Consider the product of \((1 - a^2)\) and \((1 + a^2)\). This is also a difference of squares: \((1 - a^2)(1 + a^2) = 1 - a^4\). However, we don't have \((1+a^2)\) directly.
Let's re-examine the original grouping. We have \((1-a)(1-a+a^2)\) and \((1+a)(1+a+a^2)\).
So, the expression becomes:
\[ [(1 - a)(1 + a + a^{2})] \cdot [(1 + a)(1 - a + a^{2})] \cdot (1 + a^{6}) \]\[ (1 - a^{3}) \cdot (1 + a^{3}) \cdot (1 + a^{6}) \]Now, apply the difference of squares identity to the first two terms: \((1 - a^{3})(1 + a^{3})\). Let \(x=1\) and \(y=a^3\).
\[ (1 - a^{3})(1 + a^{3}) = 1^2 - (a^{3})^2 = 1 - a^{6} \]Substitute this back into the expression:
\[ (1 - a^{6})(1 + a^{6}) \]Finally, apply the difference of squares identity one more time: \((1 - a^{6})(1 + a^{6})\). Let \(x=1\) and \(y=a^6\).
\[ 1^2 - (a^{6})^2 = 1 - a^{12} \]D) \(1 - a^{12}\) is the simplified value of the given expression, as derived using the algebraic identities.
The question asks us to find the relationship between the first two terms and apply the same relationship to the third term to find the fourth term. This is a classic analogy question where we need to identify the category or classification that the first term belongs to, which is represented by the second term.
D) Art — Dance is a form of art, specifically a performing art, which aligns with the relationship that Tennis is a form of sport.