This question tests the understanding of refraction, specifically how light behaves when moving from a denser medium to a rarer medium. The key principle here is Snell's Law and the concept of refractive index.
Correct Option: A) bend away from the normal. As explained by Snell's Law, when light moves from a denser medium to a rarer medium, its speed increases, causing it to bend away from the normal.
This problem involves determining relative directions between towns. We can solve this by drawing a diagram based on the given information, placing each town relative to others. We will use standard cardinal directions (North, South, East, West).
Let's place F first. Then C is directly above F.
C
|
F
O is to the right of F.
C
|
F --- O
This means Z is to the left of O. Since F is also to the west of O, Z can be at the same position as F, or to the west of F, or between F and O. The problem doesn't specify the exact distance, so Z is somewhere on the line to the west of O. For now, let's assume Z is at F's position or to its west.
C
|
Z/F --- O (Z is at F or to the left of F)
Y is directly below Z.
C
|
Z/F --- O
|
Y
This is a crucial piece of information. If C is north-east of Y, it means C is both to the north and to the east of Y. Since Y is south of Z, and C is north of F, for C to be north-east of Y, Z must be to the west of F. Let's refine our diagram:
C
| \
F --- O
| \
Z ---
| /
Y
More accurately, if C is north-east of Y, then Y must be south-west of C. Since Y is south of Z, and Z is west of F (because C is north of F and north-east of Y), the relative positions are:
C
|
F --- O
/|
Z |
| |
Y
From the statement "Town C is to the north-east of Town Y", we know that C is to the east of the vertical line passing through Y (and Z). Also, C is to the north of the horizontal line passing through Y.
Let's align them:
North
^
|
West <--- . ---> East
|
v
South
1. F is the reference point.
2. C is North of F.
3. O is East of F.
4. Z is West of O. (This means Z can be at F or to the West of F).
5. Y is South of Z.
6. C is North-East of Y. This implies that C is to the East of Z (because Y is South of Z, and C is North-East of Y, so C must be East of the vertical line passing through Y and Z).
Combining these:
So the arrangement is:
Z ---- F ---- O
| |
Y C
Wait, this diagram is incorrect. Let's re-evaluate step 6 carefully.
Let's denote coordinates. Let F be at \((0,0)\).
So, Z is to the West of F (since \(x_Z < 0\) and F is at \(x=0\)).
Let's draw again with Z being West of F:
C
|
F --- O
/|
Z |
| |
Y
This implies Z is to the west of F. Y is south of Z. C is north of F.
Now, check "C is to the north-east of Y".
If Y is at \((x_Z, y_Y)\) where \(x_Z < 0\) and \(y_Y < y_Z\).
If C is at \((0, y_C)\) where \(y_C > 0\).
For C to be North-East of Y, we need \(x_C > x_Y\) and \(y_C > y_Y\).
\(0 > x_Z\) (True, as \(x_Z\) is negative).
\(y_C > y_Y\) (True, as C is north of F (at \(y=0\)) and Y is south of Z, which can be south of F or north of F, but Y is definitely below C).
So the relative arrangement is:
C
|
F -- O
/
Z
|
Y
Now, we need to find the position of Town C with respect to Town Z.
From Z, to reach C, we need to go North and then East.
Therefore, C is to the North-East of Z.
Correct Option: D) North-East
The question asks to identify the substance with a pH lower than that of pure water. Pure water has a neutral pH of 7. Substances with a pH lower than 7 are acidic, while those with a pH higher than 7 are basic (alkaline).
B) Gastric juice has a pH value typically between 1.5 and 3.5, which is significantly lower than the pH of pure water (7). It is highly acidic due to the presence of hydrochloric acid.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters from the English alphabets in the surnames was obtained as follows:
| Number of letters | 1 - 4 | 4 - 7 | 7 - 10 | 10 - 13 | 13 - 16 | 16 - 19 |
| Number of surnames | 6 | 30 | 40 | 16 | 4 | 4 |
The median number of letters in the surnames is:
To find the median of a grouped frequency distribution, we first need to determine the median class. The median class is the class interval where the cumulative frequency first exceeds or equals half of the total number of observations. Once the median class is identified, we use the formula for the median of grouped data.
Let's list the given data and calculate the cumulative frequency (cf).
| Number of letters (Class Interval) | Number of surnames (Frequency, \(f\)) | Cumulative Frequency (\(cf\)) |
|---|---|---|
| 1 - 4 | 6 | 6 |
| 4 - 7 | 30 | \(6 + 30 = 36\) |
| 7 - 10 | 40 | \(36 + 40 = 76\) |
| 10 - 13 | 16 | \(76 + 16 = 92\) |
| 13 - 16 | 4 | \(92 + 4 = 96\) |
| 16 - 19 | 4 | \(96 + 4 = 100\) |
The total number of surnames is \(N = 100\).
The median position is \[\frac{N}{2} = \frac{100}{2} = 50\].
The cumulative frequency just greater than or equal to 50 is 76, which corresponds to the class interval 7 - 10. So, the median class is 7 - 10.
The formula for the median is:
\[\text{Median} = L + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h\]Substitute the values:
\[\text{Median} = 7 + \left(\frac{50 - 36}{40}\right) \times 3\] \[\text{Median} = 7 + \left(\frac{14}{40}\right) \times 3\] \[\text{Median} = 7 + \left(\frac{7}{20}\right) \times 3\] \[\text{Median} = 7 + \frac{21}{20}\] \[\text{Median} = 7 + 1.05\] \[\text{Median} = 8.05\]A) 8.05 is the calculated median number of letters.
Tanvi saves Rs 5,000 from her salary every month after spending 80% of her monthly salary. Tanvi’s salary is:
This problem can be solved by understanding the relationship between spending, saving, and total salary. If Tanvi spends 80% of her salary, the remaining percentage must be what she saves. We can then equate this percentage of her salary to the actual amount she saves to find her total salary.
Tanvi spends 80% of her monthly salary. The total salary represents 100%.
\[ \text{Percentage Saved} = \text{Total Percentage} - \text{Percentage Spent} \] \[ \text{Percentage Saved} = 100\% - 80\% = 20\% \]We are given that Tanvi saves Rs 5,000 every month. This amount corresponds to the 20% of her salary that she saves.
\[ 20\% \text{ of Salary} = \text{Rs } 5,000 \]Let Tanvi's monthly salary be \(S\). We can write the equation as:
\[ 0.20 \times S = 5,000 \]To find \(S\), divide both sides by 0.20:
\[ S = \frac{5,000}{0.20} \] \[ S = \frac{5,000}{\frac{20}{100}} \] \[ S = 5,000 \times \frac{100}{20} \] \[ S = 5,000 \times 5 \] \[ S = 25,000 \]So, Tanvi's monthly salary is Rs 25,000.
Correct Option: C) 25,000
The question asks about the concentration of commercially exploitable wind resources in India, specifically how many states account for more than 95% of this potential, according to the Ministry of New and Renewable Energy (MNRE). This requires knowledge of the distribution of wind energy potential in India as reported by official sources.
A) Seven. The states of Gujarat, Maharashtra, Tamil Nadu, Karnataka, Andhra Pradesh, Rajasthan, and Madhya Pradesh together hold more than 95% of India's commercially exploitable wind resources, as per the Ministry of New and Renewable Energy.
The question asks for the location of India's first e-waste clinic. This is a factual question related to current affairs and environmental initiatives in India. To answer correctly, one needs to recall the specific city where this clinic was established.
The e-waste clinic aims to enable segregation, processing, and disposal of waste from both household and commercial units. It also promotes the practice of reducing, reusing, and recycling electronic waste.
A) Bhopal — India's first e-waste clinic was indeed set up in Bhopal, Madhya Pradesh, in 2019.
The problem presents a proportion in the format \(a : b :: c : d\), which means that the ratio \(a/b\) is equal to the ratio \(c/d\). We need to find the value of the unknown variable \(x\) by setting up the proportion as an equation and solving for \(x\).
Therefore, the value of \(x\) is 65.
Correct Option: B) 65
The question asks to identify the specific Government's Department of Women and Child Development (WCD) that formulated a policy in July 2021 to define street children and tasked its agencies with their identification and rehabilitation. This is a factual question requiring knowledge of recent government initiatives.
Correct Option: C) In July 2021, the Delhi Government's Department of Women and Child Development (WCD) indeed came up with a policy to define street children and asked 10 of its agencies to identify and rehabilitate them. This initiative was part of a larger effort to ensure the welfare and protection of vulnerable children in the capital.
This is a syllogism problem. We will use Venn diagrams to represent the given statements and then check which conclusions logically follow from these diagrams. We assume the statements are true, even if they contradict common knowledge.
D) Only conclusions I and II follow. Both Conclusion (I) and Conclusion (II) are logically derivable from the given statements using Venn diagrams.