This question tests knowledge about the properties of washing soda (sodium carbonate decahydrate) and its behavior when exposed to the atmosphere, especially under hot conditions. The key concept here is efflorescence, where a hydrated salt loses its water of crystallization.
Correct Option: C) loses water molecules and becomes powdery
This option accurately describes the phenomenon of efflorescence that washing soda undergoes when exposed to a hot atmosphere. It loses its water of crystallization and its crystalline structure breaks down, turning into a powder.
Ram takes 2 hours more than Ravi to walk 20 km. If Ram doubles his speed, then he is ahead of Ravi by one hour 20 minutes. Find the speed of Ravi (in km/hr).
This problem involves relative speeds and time taken to cover a fixed distance. We will set up equations based on the given information about Ram's and Ravi's speeds and the time differences, and then solve for Ravi's speed.
Let the speed of Ram be \(S_R\) km/hr and the speed of Ravi be \(S_{Ra}\) km/hr.
The distance to be covered is 20 km.
Time taken by Ram to walk 20 km is \(T_R = \frac{20}{S_R}\) hours.
Time taken by Ravi to walk 20 km is \(T_{Ra} = \frac{20}{S_{Ra}}\) hours.
From the first condition, Ram takes 2 hours more than Ravi:
\[ T_R - T_{Ra} = 2 \]\[ \frac{20}{S_R} - \frac{20}{S_{Ra}} = 2 \quad \ldots(1) \]From the second condition, if Ram doubles his speed, his new speed is \(2S_R\). The time taken by Ram with doubled speed is \[T'_R = \frac{20}{2S_R} = \frac{10}{S_R}\] hours.
In this case, Ram is ahead of Ravi by 1 hour 20 minutes. Convert 1 hour 20 minutes to hours:
\[ 1 \text{ hour } 20 \text{ minutes} = 1 + \frac{20}{60} \text{ hours} = 1 + \frac{1}{3} \text{ hours} = \frac{4}{3} \text{ hours} \]Being ahead of Ravi by \(\frac{4}{3}\) hours means Ram takes \(\frac{4}{3}\) hours less than Ravi:
\[ T_{Ra} - T'_R = \frac{4}{3} \]\[ \frac{20}{S_{Ra}} - \frac{10}{S_R} = \frac{4}{3} \quad \ldots(2) \]Now we have a system of two equations:
Equation (1): \[\frac{20}{S_R} - \frac{20}{S_{Ra}} = 2\]
Equation (2): \[\frac{20}{S_{Ra}} - \frac{10}{S_R} = \frac{4}{3}\]
Let \(x = \frac{1}{S_R}\) and \(y = \frac{1}{S_{Ra}}\). The equations become:
\( 20x - 20y = 2 \quad \ldots(3) \)
\( 20y - 10x = \frac{4}{3} \quad \ldots(4) \)
From equation (3), divide by 2:
\( 10x - 10y = 1 \quad \ldots(5) \)
Add equation (4) and equation (5):
\[ (20y - 10x) + (10x - 10y) = \frac{4}{3} + 1 \]\[ 10y = \frac{4+3}{3} \]\[ 10y = \frac{7}{3} \]\[ y = \frac{7}{30} \]Since \(y = \frac{1}{S_{Ra}}\), we have:
\[ S_{Ra} = \frac{1}{y} = \frac{30}{7} \]Convert this to a mixed fraction:
\[ S_{Ra} = \frac{30}{7} = 4 \frac{2}{7} \text{ km/hr} \]C) \(4\frac{2}{7}\) is the correct speed of Ravi.
In a displacement reaction, if a more reactive metal displaces a less reactive metal from its salt solution, what is the result?
The less reactive metal is displaced and the more reactive metal becomes a part of the compound
The less reactive metal is displaced and solidifies
The more reactive metal remains unchanged
Formation of an insoluble salt
This question tests the understanding of displacement reactions, a fundamental concept in chemistry. In a displacement reaction, a more reactive element displaces a less reactive element from its compound. This principle is based on the reactivity series of metals.
Correct Option: A) The less reactive metal is displaced and the more reactive metal becomes a part of the compound. This accurately describes the outcome of a displacement reaction where the more reactive metal forms a new compound and the less reactive metal is set free.
The problem presents two pairs of letter-clusters and asks to find a third pair that exhibits a similar relationship. We need to identify the pattern of letter transformations (forward/backward shifts in alphabetical order) between the letters in the given pairs and then apply the same pattern to the options.
The pattern is: \((+5, -1, -2)\)
The pattern is consistent: \((+5, -1, -2)\)
A) NCL : SBJ
This pair follows the established pattern \((+5, -1, -2)\).
The problem asks us to identify the letter-cluster that does not follow the same pattern as the other three, based on the English alphabetical order. We need to find the numerical position of each letter in the alphabet and then look for a consistent mathematical or sequential pattern within each cluster.
Let's convert each letter to its numerical position:
Differences:
The pattern of differences is \((+6, +3, +5)\).
Differences:
The pattern of differences is \((+6, +3, +5)\).
Differences:
The pattern of differences is \((+6, +3, +4)\).
Differences:
The pattern of differences is \((+6, +3, +5)\).
Comparing the patterns:
The letter-cluster RXAE has a different pattern of differences \((+6, +3, +4)\) compared to the other three, which all follow the \((+6, +3, +5)\) pattern.
C) RXAE — This cluster has a difference of +4 between its third and fourth letters (A and E), whereas the other clusters have a difference of +5 for the corresponding letters.
The question asks about a unique feature of sex chromosomes in humans compared to other chromosomes (autosomes). We need to recall the basic structure and pairing behavior of human chromosomes during cell division and inheritance.
A) They are not always a perfect pair in males. This is correct. In males, the sex chromosomes are XY. The X and Y chromosomes are significantly different in size and gene content, meaning they do not form a perfect homologous pair like autosomes or the XX pair in females. They only share small pseudoautosomal regions for pairing during meiosis.
To solve this problem, we need to track Rakesh's movements in a 2D coordinate system or by separately calculating his net displacement in the East-West and North-South directions. We will assume Point A is the origin \((0,0)\) and then determine the final coordinates of Point P. The shortest distance and direction to return to Point A will be the negative of the displacement from A to P.
D) 5 km to the south
The final position of Point P is \((0, 5)\). To return to Point A \((0,0)\), Rakesh must travel 5 km in the negative Y-direction, which corresponds to 5 km south.
The given series is a number series where numbers are repeated, and there is a pattern in the unique numbers. We need to identify the pattern in the unique numbers and then apply it to find the missing term.
Correct Option: D) 82
The question asks why a gardener successfully grew a new potato plant from a potato with buds (eyes). This relates to the method of reproduction in plants, specifically asexual reproduction.
Correct Option: C) vegetative propagation is by buds (eyes)
This option accurately describes the biological process by which new potato plants grow from the buds (eyes) of a potato tuber. It is a classic example of vegetative propagation.
The question asks to identify the state with the highest density of railway network as of the 2019-20 period. This requires knowledge of geographical distribution and infrastructure development of railways in India, specifically focusing on the density metric (length of railway lines per unit area).
Correct Option: D) Uttar Pradesh had the highest density of railway network as of 2019-20. Its extensive flat terrain, high population density, and historical development as a major agricultural and administrative region have led to a very dense railway infrastructure.