If the diameter and the height of a cylinder are 12 cm and 10.5 cm, respectively, then using π = \( \frac{22}{7} \), find the volume of the cylinder.
To find the volume of a cylinder, we use the formula \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height. First, we need to calculate the radius from the given diameter, and then substitute the values into the formula.
The radius is half of the diameter.
\[ r = \frac{D}{2} = \frac{12 \text{ cm}}{2} = 6 \text{ cm} \]The volume (\(V\)) of a cylinder is given by:
\[ V = \pi r^2 h \]We can simplify the calculation by noting that \(10.5 = \frac{21}{2}\).
\[ V = \frac{22}{7} \times 36 \times \frac{21}{2} \]\[ V = 22 \times \frac{36}{2} \times \frac{21}{7} \]\[ V = 22 \times 18 \times 3 \]\[ V = 396 \times 3 \]\[ V = 1188 \text{ cm}^3 \]Correct Option: C) 1188 cm³
The value of
\(\left(-\frac{5}{3}\right)\times\left(\frac{9}{25}\right)+\left(\frac{1}{5}-2\right)\)
\( \frac{-12}{5} \)
0
\( \frac{-4}{5} \)
\( \frac{9}{5} \)
To solve this problem, we need to evaluate the given expression by following the order of operations (PEMDAS/BODMAS). This involves performing multiplication first, then simplifying the terms inside the parenthesis, and finally performing addition/subtraction.
The value of the expression is \(-\frac{12}{5}\).
A) \(\frac{-12}{5}\) is the correct answer, as derived from the step-by-step calculation.
To find the Least Common Multiple (LCM) of a set of numbers, we can use the prime factorization method. This involves finding the prime factors of each number and then taking the highest power of each common and uncommon prime factor.
Multiply the highest powers of all prime factors:
\[ \text{LCM}(24, 42, 56) = 2^3 \times 3^1 \times 7^1 = 8 \times 3 \times 7 \] \[ \text{LCM}(24, 42, 56) = 24 \times 7 = 168 \]Correct Option: C) 168
P, Q, R, S, T and U are six siblings who passed their class tenth exam in different years viz. 2001, 2003, 2005, 2007, 2010 and 2012 but not necessarily in the same order.
P passed in one of the years before S and after R. Only Q passed before R. T passed in the year immediately after S. U passed in one of the years before P. Who passed their class tenth exam in the year 2007?
This is a logical reasoning problem involving ordering and sequencing. We need to determine the passing year for each sibling based on the given clues and then identify who passed in 2007. We will use a table or a linear arrangement to keep track of the information.
List the years: The years are 2001, 2003, 2005, 2007, 2010, 2012.
Analyze the clues:
Fill in the knowns:
| Year | Sibling |
|---|---|
| 2001 | Q |
| 2003 | R |
| 2005 | |
| 2007 | |
| 2010 | |
| 2012 |
Deduce further positions:
Assign years to the remaining siblings:
Final arrangement:
| Year | Sibling |
|---|---|
| 2001 | Q |
| 2003 | R |
| 2005 | U |
| 2007 | P |
| 2010 | S |
| 2012 | T |
Answer the question: The person who passed in 2007 is P.
Correct Option: C) P
Which of the following is RHS of the equation
\(\frac{\sin^2 A}{1-\cos A}\)
To find the RHS of the given equation, we need to simplify the expression \(\frac{\sin^2 A}{1-\cos A}\) using fundamental trigonometric identities. The key identity here is \(\sin^2 A + \cos^2 A = 1\), which can be rearranged to express \(\sin^2 A\) in terms of \(\cos A\).
Recall the fundamental trigonometric identity:
\[ \sin^2 A + \cos^2 A = 1 \]From this identity, we can express \(\sin^2 A\) as:
\[ \sin^2 A = 1 - \cos^2 A \]Substitute this expression for \(\sin^2 A\) into the given equation:
\[ \frac{\sin^2 A}{1 - \cos A} = \frac{1 - \cos^2 A}{1 - \cos A} \]Recognize that the numerator \(1 - \cos^2 A\) is in the form of \(a^2 - b^2\), which can be factored as \((a - b)(a + b)\). Here, \(a = 1\) and \(b = \cos A\).
\[ 1 - \cos^2 A = (1 - \cos A)(1 + \cos A) \]Substitute the factored form back into the expression:
\[ \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A} \]Cancel out the common term \((1 - \cos A)\) from the numerator and the denominator, assuming \(1 - \cos A \neq 0\) (i.e., \(\cos A \neq 1\), which means \(A \neq 2n\pi\) for integer \(n\)).
\[ 1 + \cos A \]Thus, the simplified form of the given expression is \(1 + \cos A\).
C) cos A + 1 — As derived in the step-by-step explanation, the expression \(\frac{\sin^2 A}{1-\cos A}\) simplifies to \(1 + \cos A\), which is the same as \(\cos A + 1\).
A) cosec A — This is \(\frac{1}{\sin A}\). Our simplified expression is \(1 + \cos A\), which is generally not equal to \(\text{cosec } A\).
B) tan A — This is \(\frac{\sin A}{\cos A}\). Our simplified expression is \(1 + \cos A\), which is generally not equal to \(\tan A\).
D) sec A + tan A — This is \[\frac{1}{\cos A} + \frac{\sin A}{\cos A} = \frac{1 + \sin A}{\cos A}\]. Our simplified expression is \(1 + \cos A\), which is generally not equal to \(\text{sec } A + \tan A\).
The question asks to identify the gas with an unpleasant smell produced when sodium sulfide reacts with dilute hydrochloric acid. This is a classic acid-base reaction involving a salt of a weak acid and a strong acid, leading to the formation of a new salt and a weak acid, which often decomposes into a gas.
A) \(\text{H}_2\text{S}\) is hydrogen sulfide, which is produced when sodium sulfide reacts with hydrochloric acid. It is a gas with a characteristic unpleasant smell, often described as rotten eggs.
The question asks to identify the basic requirements that led scientists to classify elements. We need to evaluate each statement to determine its relevance to the historical development of element classification.
B) only (i)
Statement (i) accurately describes a fundamental motivation for the classification of elements. Scientists sought to find order and predictability among the vast number of known elements by identifying recurring patterns in their properties.
According to mirror formula, the focal length of a spherical mirror is equal to:
\(\frac{uv}{u-v}\)
\(\frac{uv}{u+v}\)
\(\frac{u-v}{uv}\)
\(\frac{u+v}{uv}\)
The question asks for the expression for the focal length of a spherical mirror based on the mirror formula. We need to recall the standard mirror formula and then rearrange it to solve for the focal length \(f\).
B) \(\frac{uv}{u+v}\)
This is the correct expression for the focal length \(f\) derived directly from the mirror formula.
CuSO4​ + Mg ⇒ MgSO4​ + Cu
In the above reaction Mg is the:
oxidising agent
dissolving agent
reducing agent
catalyst
This question asks to identify the role of Magnesium (Mg) in the given chemical reaction. To determine this, we need to analyze the oxidation states of the elements involved before and after the reaction and understand the definitions of oxidizing and reducing agents.
\( \text{CuSO}_4 + \text{Mg} \rightarrow \text{MgSO}_4 + \text{Cu} \)
C) reducing agent — As determined in the step-by-step analysis, Magnesium (Mg) goes from an oxidation state of 0 to +2, meaning it loses electrons and gets oxidized. A substance that gets oxidized acts as a reducing agent for another substance.