Two concentric circles drawn with the radius of inner circle 6cm and outer circle radius 50% more than inner circle. What is the area of the annulus formed between two circles?
\( \frac{890}{7} \) cm²
\( \frac{900}{7} \) cm²
990 cm²
\( \frac{990}{7} \) cm²
To find the area of the annulus, we need to calculate the area of the outer circle and subtract the area of the inner circle. The area of a circle is given by the formula \( \pi r^2 \), where \( r \) is the radius. First, we'll determine the radii of both circles, then calculate their areas, and finally find the difference.
D) \( \frac{990}{7} \) cm² is the calculated area of the annulus.
A bus is travelling with \( \frac{7}{8} \) of its actual speed to cover 42 km in 1 h 16 min. Find the actual speed of the bus.
To find the actual speed of the bus, we first need to calculate the speed at which it is currently travelling. We are given the distance covered and the time taken. Once we have the current speed, we can use the given ratio of the current speed to the actual speed to find the actual speed.
The given time is 1 hour 16 minutes. We need to convert 16 minutes into hours.
\[ 16 \text{ minutes} = \frac{16}{60} \text{ hours} = \frac{4}{15} \text{ hours} \]Total time \[T = 1 + \frac{4}{15} = \frac{15+4}{15} = \frac{19}{15} \text{ hours}\]
The distance \( D = 42 \text{ km} \). The formula for speed is \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \).
Current speed \[S_{\text{current}} = \frac{42}{\frac{19}{15}} = 42 \times \frac{15}{19} \text{ km/h}\]
\[ S_{\text{current}} = \frac{630}{19} \text{ km/h} \approx 33.1579 \text{ km/h} \]We are given that the bus is travelling at \( \frac{7}{8} \) of its actual speed.
Let the actual speed be \( S_{\text{actual}} \).
\[ S_{\text{current}} = \frac{7}{8} \times S_{\text{actual}} \]Now, we can find \( S_{\text{actual}} \):
\[ S_{\text{actual}} = S_{\text{current}} \times \frac{8}{7} \] \[ S_{\text{actual}} = \frac{630}{19} \times \frac{8}{7} \]Simplify the expression:
\[ S_{\text{actual}} = \frac{90 \times 7}{19} \times \frac{8}{7} \] \[ S_{\text{actual}} = \frac{90 \times 8}{19} \] \[ S_{\text{actual}} = \frac{720}{19} \text{ km/h} \]Calculate the numerical value:
\[ S_{\text{actual}} \approx 37.8947 \text{ km/h} \]Rounding to one decimal place, the actual speed is approximately 37.9 km/h.
Correct Option: B) 37.9 km/h
The question asks to identify a plant whose body is not differentiated into true roots, stems, and leaves. This refers to plants with a thalloid body structure, characteristic of lower plant groups.
Correct Option: C) Chara — Chara is a type of green alga. Algae are characterized by a simple, undifferentiated plant body called a thallus, which lacks true roots, stems, and leaves. Their body is not organized into specialized vascular tissues or organs.
The question asks about the number of basic elements (Tatvas) described by early Indian philosophers to classify matter. This refers to the ancient Indian philosophical concept of the 'Panchamahabhutas' or 'Five Great Elements'.
A) Five. Early Indian philosophers, particularly in the Samkhya school and Ayurveda, described five basic elements (Panchamahabhutas) to classify matter: Earth, Water, Fire, Air, and Ether (Space).
The question asks us to identify valid conclusions based on a given statement. We need to analyze the statement carefully to determine what can be logically inferred from it, without making assumptions not supported by the text.
Based on this analysis, only conclusion (A) is a valid inference from the given statement.
Correct Option: C) A alone
The question asks to identify the individual who was NOT awarded the Padma Bhushan in 2021. To answer this, one needs to recall the list of Padma awardees for the year 2021, specifically focusing on the Padma Bhushan recipients.
C) S P Balasubramaniam was awarded the Padma Vibhushan, the second-highest civilian award, posthumously in 2021, not the Padma Bhushan. Therefore, he is the individual among the options who was NOT awarded the Padma Bhushan in 2021.
If cosecA – cotA = 3 then cosecA + cotA = ?
\( \frac{1}{3} \)
\( \frac{1}{2} \)
\( \frac{1}{5} \)
\( \frac{3}{2} \)
This problem can be solved using a fundamental trigonometric identity relating cosecant and cotangent. The identity \( \text{cosec}^2 A - \text{cot}^2 A = 1 \) is key to finding the relationship between \( \text{cosecA} - \text{cotA} \) and \( \text{cosecA} + \text{cotA} \).
We know the identity:
\[ \text{cosec}^2 A - \text{cot}^2 A = 1 \]This identity is in the form of \( a^2 - b^2 = (a-b)(a+b) \). Applying this, we get:
\[ (\text{cosecA} - \text{cotA})(\text{cosecA} + \text{cotA}) = 1 \]We are given that \( \text{cosecA} - \text{cotA} = 3 \). Substitute this value into the factored identity:
\[ (3)(\text{cosecA} + \text{cotA}) = 1 \]Divide both sides by 3:
\[ \text{cosecA} + \text{cotA} = \frac{1}{3} \]A) \( \frac{1}{3} \) is the correct answer, as derived from the trigonometric identity.
The question asks about the location where India established the CIRUS reactor. This requires knowledge of India's nuclear research facilities and their historical development.
Correct Option: A) Trombay, Mumbai
The question asks for the venue of the Shanghai Cooperation Organisation's (SCO) Heads of State Council meeting in 2019. This requires knowledge of major international summits and their locations.
Correct Option: C) Bishkek
The 19th meeting of the Shanghai Cooperation Organisation (SCO) Heads of State Council was indeed held in Bishkek, Kyrgyzstan, from June 13-14, 2019. This was a significant summit where leaders discussed regional security, economic cooperation, and other key issues.
If \( \frac{a}{b} \) = \( \frac{b}{c} \) = \( \frac{c}{d} \), then \(\frac{b^3+c^3+d^3}{a^3+b^3+c^3}\) will be equal to:
\( \frac{a}{d} \)
\( \frac{d}{c} \)
\( \frac{d}{a} \)
\( \frac{b}{d} \)
The problem involves a continued proportion. We can solve this by introducing a constant \(k\) for the given ratios and expressing \(b, c,\) and \(d\) in terms of \(a\) and \(k\). Then, substitute these expressions into the given fraction and simplify.
Given the relation:
\[ \frac{a}{b} = \frac{b}{c} = \frac{c}{d} \]Let this common ratio be \(k\). So, we have:
\[ \frac{a}{b} = k \implies a = bk \quad \text{(1)} \] \[ \frac{b}{c} = k \implies b = ck \quad \text{(2)} \] \[ \frac{c}{d} = k \implies c = dk \quad \text{(3)} \]Now, express \(b\) and \(c\) in terms of \(d\) and \(k\):
From (3), \(c = dk\).
Substitute \(c\) in (2): \(b = (dk)k = dk^2\).
Substitute \(b\) in (1): \(a = (dk^2)k = dk^3\).
So, we have:
Substitute these values into the expression \(\frac{b^3+c^3+d^3}{a^3+b^3+c^3}\):
\[ \frac{(dk^2)^3 + (dk)^3 + d^3}{(dk^3)^3 + (dk^2)^3 + (dk)^3} \] \[ = \frac{d^3k^6 + d^3k^3 + d^3}{d^3k^9 + d^3k^6 + d^3k^3} \]Factor out \(d^3\) from both the numerator and the denominator:
\[ = \frac{d^3(k^6 + k^3 + 1)}{d^3(k^9 + k^6 + k^3)} \] \[ = \frac{k^6 + k^3 + 1}{k^9 + k^6 + k^3} \]Factor out \(k^3\) from the denominator:
\[ = \frac{k^6 + k^3 + 1}{k^3(k^6 + k^3 + 1)} \]Cancel out the common term \((k^6 + k^3 + 1)\) from the numerator and denominator:
\[ = \frac{1}{k^3} \]From step 2, we know that \(a = dk^3\). Therefore, \(k^3 = \frac{a}{d}\).
Substitute this back into the simplified expression:
\[ \frac{1}{k^3} = \frac{1}{\frac{a}{d}} = \frac{d}{a} \]Correct Option: C) \( \frac{d}{a} \)