NCERT Exemplar

Assertion: Hydrometallurgy involves dissolving the ore in a suitable reagent followed by precipitation by a more electropositive metal. Reason: Copper is extracted by hydrometallurgy.

(i) Both assertion and reason are true and the reason is the correct explanation of assertion. (ii) Both assertion and reason are true but the reason is not the correct explanation of assertion....

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Wrought iron is the purest form of iron. Write a reaction used for the preparation of wrought iron from cast iron. How can the impurities of sulphur, silicon and phosphorus be removed from cast iron?

Limestone can be combined with flux to remove impurities of sulfur, silicon and phosphorus. They form an easily removable slag. The metal is removed from the slag by passing through the rollers....

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The mixture of compounds A and B is passed through a column of Al2O3 by using alcohol as eluant. Compound A is eluted in preference to compound B. Which of the compounds A or B, is more readily adsorbed on the column?

Compound A does not absorb well in the column and moves down the column with alcohol i.e. eluant, while compound B is well absorbed in the column and cannot go down the column.

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A 50 MHz sky wave takes 4.04 ms to reach a receiver via re-transmission from a satellite 600 km above earth’s surface. Assuming re-transmission time by satellite negligible, find the distance between source and receiver. If communication between the two was to be done by Line of Sight (LOS) method, what should size and placement of receiving and transmitting antenna be?

Answer: According to the question, the velocity of waves = 3 × 108 m/s The time to reach a receiver = 4.04 × 10-3 s we know that the height of satellite is: h = 600 km And the radius of earth = 6400...

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The maximum frequency for reflection of sky waves from a certain layer of the ionosphere is found to be f max = 9(Nmax) 1/2, where Nmax is the maximum electron density at that layer of the ionosphere. On a certain day, it is observed that signals of frequencies higher than 5MHz are not received by reflection from the F1 layer of the ionosphere while signals of frequencies higher than 8MHz are not received by reflection from the F2 layer of the ionosphere. Estimate the maximum electron densities of the F1 and F2 layers on that day.

Answer: Expression fpor the maximum frequency is: fmax = 9(Nmax)1/2 According to the question, for Layer F1, fmax = 5 MHz $ {{N}_{\max }}=\frac{F_{\max }^{2}}{9\times 9}=\frac{5\times...

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If the whole earth is to be connected by LOS communication using space waves (no restriction of antenna size or tower height), what is the minimum number of antennas required? Calculate the tower height of these antennas in terms of earths radius?

Answer: We know that the distance or range of transmission tower is given by the expression: $dT=\sqrt{2{{h}_{T}}R}$ R represents the radius of the earth (approximately 6400 km). hT​ denotes the...

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Two waves A and B of frequencies 2 MHz and 3 MHz, respectively are beamed in the same direction for communication via skywave. Which one of these is likely to travel a longer distance in the ionosphere before suffering total internal reflection?

Answer: The refractive index rises as the frequency rises, implying that the angle of refraction is smaller for higher frequency waves. In other words, bending is less. As a result, after covering a...

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A 1 KW signal is transmitted using a communication channel which provides attenuation at the rate of – 2dB per km. If the communication channel has a total length of 5 km, the power of the signal received is [gain in dB = 10 log P0/P1]

(a) 900 W (b) 100 W (c) 990 W (d) 1010 W Answer: The correct option is (b) 100 W Explanation: According to the question, Pi​ =1 kW = 1000W The rate of attenuation of the signal = -2dB/km Length of...

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Suppose a ‘n’-type wafer is created by doping Si crystal having 5 × 1028 atoms/m3 with 1ppm concentration of As. On the surface 200 ppm Boron is added to create the ‘P’ region in this wafer. Considering ni = 1.5 × 1016 m–3,

(i) Calculate the densities of the charge carriers in the n & p regions. (ii) Comment which charge carriers would contribute largely for the reverse saturation current when the diode is reverse...

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Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on the concentration of compound and KOH both. (i) Write down the structural formula of both compounds ‘A’ and ‘B’. (ii) Out of these two compounds, which one will be converted to the product with an inverted configuration.

Because the rate of reaction of compound ‘A' (C4H9Br) with aqueous KOH is solely determined by the concentration of compound ‘A,' the reaction happens via the SN1 mechanism, and product ‘A' is...

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Which of the following statements are correct about the kinetics of this reaction? (i) The rate of reaction depends on the concentration of only (b). (ii) The rate of reaction depends on the concentration of both (a) and (b). (iii) Molecularity of reaction is one. (iv) Molecularity of reaction is two.

Option (i) and (iii) are the answers. The SN1 mechanism is used in the given reaction. The production of carbocation is a gradual process in the SN1 mechanism. As a result, the pace of reaction is...

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A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs 300/- per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of Rs 1/- one subscriber will discontinue the service. Find what increase will bring maximum profit?

We should consider that the organization expands the yearly membership by \[\mathbf{Rs}\text{ }\mathbf{x}.\] Along these lines, x is the quantity of supporters who end the administrations.  ...

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Arrange the following compounds in increasing order of boiling point. Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol (i) Propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol (ii) Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol (iii) Pentan-1-ol, butan-2-ol, butan-1-ol, propan-1-ol (iv) Pentan-1-ol, butan-1-ol, butan-2-ol, propan-1-ol

Option (i) is the answer As the moleculer mass of the alcohol grows, the boiling point rises. Furthermore, $1^0$ alcohols have greater boiling points than $2^0$ alcohols among isomeric alcohols.

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