The question asks for the launch year of ISRO's Chandrayaan-1 mission, which was instrumental in confirming the presence of solid ice on the Moon. This is a factual question related to India's space program history.
C) 2008 — ISRO launched Chandrayaan-1 on October 22, 2008, from Satish Dhawan Space Centre, Sriharikota. This mission was a landmark achievement for India's space program and provided crucial data that helped confirm the presence of water ice on the Moon.
If 5 tan α = 4, then find the value of \(\frac{5\sin\alpha-3\cos\alpha}{5\sin\alpha+2\cos\alpha}\).
The problem provides a relationship between \(\tan\alpha\) and asks for the value of an expression involving \(\sin\alpha\) and \(\cos\alpha\). The most straightforward approach is to express \(\tan\alpha\) in terms of \(\sin\alpha\) and \(\cos\alpha\), then manipulate the given expression to use this relationship. This can be done by dividing the numerator and denominator of the target expression by \(\cos\alpha\).
First, we are given the relation:
\[ 5 \tan \alpha = 4 \]From this, we can find the value of \(\tan\alpha\):
\[ \tan \alpha = \frac{4}{5} \]Now, consider the expression we need to evaluate:
\[ \frac{5\sin\alpha-3\cos\alpha}{5\sin\alpha+2\cos\alpha} \]To use the value of \(\tan\alpha\), we can divide both the numerator and the denominator by \(\cos\alpha\). This is valid as long as \(\cos\alpha \neq 0\). If \(\cos\alpha = 0\), then \(\tan\alpha\) would be undefined, which is not the case here since \(\tan\alpha = 4/5\).
\[ \frac{\frac{5\sin\alpha}{\cos\alpha}-\frac{3\cos\alpha}{\cos\alpha}}{\frac{5\sin\alpha}{\cos\alpha}+\frac{2\cos\alpha}{\cos\alpha}} \]Simplify the expression using the identity \(\tan\alpha = \frac{\sin\alpha}{\cos\alpha}\):
\[ \frac{5\tan\alpha-3}{5\tan\alpha+2} \]Substitute the value of \(\tan\alpha = \frac{4}{5}\) into this simplified expression:
\[ \frac{5\left(\frac{4}{5}\right)-3}{5\left(\frac{4}{5}\right)+2} \]Perform the multiplications:
\[ \frac{4-3}{4+2} \]Perform the subtractions and additions:
\[ \frac{1}{6} \]Correct Option: D) The value of the expression is \(\frac{1}{6}\).
The question asks for the specific year when Delhi was officially declared the capital of British India by Emperor George V. This is a direct factual recall question related to Indian history during the British Raj.
Correct Option: A) 1911
Delhi was officially announced as the Capital of British India by Emperor George V during the Delhi Durbar held in 1911. The transfer became effective in 1912, but the announcement was made in 1911.
The question asks to identify the type of goods that are used for producing other goods rather than for immediate consumption. This requires understanding the basic classification of goods in economics.
Correct Option: A) Capital Goods are goods that are used in the production of other goods and services, rather than being consumed directly by the end-user. Examples include machinery, tools, equipment, and buildings.
When multiple objects are melted and recast into a single new object, the total volume remains constant. Therefore, to find the edge of the new cube, we need to calculate the sum of the volumes of the three smaller cubes and then find the cube root of this total volume.
Correct Option: C) 6 cm
The question asks to identify the state with the lowest population density according to the 2011 Census of India. Population density is calculated as the total population divided by the total land area. To answer this, one needs to recall or know the population density figures for the given states from the 2011 Census data.
A) Arunachal Pradesh has a population density of 17 persons per square kilometer, which is the lowest among the given options according to the 2011 Census.
To find the mean speed (average speed), we need to calculate the total distance traveled and divide it by the total time taken. The problem provides information about different segments of the journey, each with a specific duration and average speed. We will calculate the distance for each segment and then sum them up.
P drives for 1 hour at an average speed of 40 km/h.
\[ \text{Distance}_P = \text{Speed}_P \times \text{Time}_P \]\[ \text{Distance}_P = 40 \text{ km/h} \times 1 \text{ h} = 40 \text{ km} \]Q drives for 2 hours at an average speed of 50 km/h.
\[ \text{Distance}_Q = \text{Speed}_Q \times \text{Time}_Q \]\[ \text{Distance}_Q = 50 \text{ km/h} \times 2 \text{ h} = 100 \text{ km} \]R drives for 3 hours at an average speed of 60 km/h.
\[ \text{Distance}_R = \text{Speed}_R \times \text{Time}_R \]\[ \text{Distance}_R = 60 \text{ km/h} \times 3 \text{ h} = 180 \text{ km} \]The total distance is the sum of the distances covered by P, Q, and R.
\[ \text{Total Distance} = \text{Distance}_P + \text{Distance}_Q + \text{Distance}_R \]\[ \text{Total Distance} = 40 \text{ km} + 100 \text{ km} + 180 \text{ km} = 320 \text{ km} \]The total time is the sum of the durations P, Q, and R drove.
\[ \text{Total Time} = \text{Time}_P + \text{Time}_Q + \text{Time}_R \]\[ \text{Total Time} = 1 \text{ h} + 2 \text{ h} + 3 \text{ h} = 6 \text{ h} \]Mean speed is total distance divided by total time.
\[ \text{Mean Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \]\[ \text{Mean Speed} = \frac{320 \text{ km}}{6 \text{ h}} \]\[ \text{Mean Speed} \approx 53.333 \text{ km/h} \]B) 53.33 km/h — The calculated mean speed is approximately 53.33 km/h.
The question asks to identify the specific event where Mahatma Gandhi made his first major public appearance in India after his return from South Africa. This is a factual question related to modern Indian history, specifically Gandhi's early activities in India.
D) Banaras Hindu University. Mahatma Gandhi's first major public appearance in India after returning from South Africa was indeed at the opening of the Banaras Hindu University in February 1916. His speech there was a significant moment, signaling his arrival on the Indian political scene.
The question asks to identify the Deputy Prime Minister of India during the period 1977 to 1979. This requires knowledge of the political history of India, specifically the Janata Party government era.
Correct Option: B) Jagjivan Ram served as the Deputy Prime Minister of India from 1977 to 1979 under Prime Minister Morarji Desai. He was a prominent leader of the Janata Party and held significant portfolios.
To find the smallest number to subtract from a given number to make it a perfect square, we need to find the largest perfect square less than or equal to the given number. The difference between the given number and this perfect square will be the required smallest number.
The given number is \(0.000327\). We can write this as:
\[ 0.000327 = \frac{327}{1000000} \]We need to find the largest perfect square less than or equal to \(327\). Let's check squares of numbers:
The largest perfect square less than or equal to \(327\) is \(324\), which is \(18^2\).
The denominator \(1000000\) is a perfect square: \(1000^2 = 1000000\).
The largest perfect square less than or equal to \(\frac{327}{1000000}\) will be \(\frac{324}{1000000}\).
The smallest number to be subtracted is the difference between the original number and the perfect square we found:
\[ \text{Number to subtract} = 0.000327 - \frac{324}{1000000} \]\[ \text{Number to subtract} = \frac{327}{1000000} - \frac{324}{1000000} \]\[ \text{Number to subtract} = \frac{327 - 324}{1000000} \]\[ \text{Number to subtract} = \frac{3}{1000000} \]Thus, \(0.000003\) should be subtracted from \(0.000327\) to make it a perfect square (\(0.000324 = (0.018)^2\)).
D) 0.000003 is the correct answer because subtracting this value from \(0.000327\) results in \(0.000324\), which is the square of \(0.018\).