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Class 12th Solutions

Of the following four aqueous solutions, total number of those solutions whose freezing point is lower than that of 0.10 M C₂H₅OH is___________

(Integer answer)

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Vishal Baghel

Contributor-Level 10

All the solution have higher ion production with respect to 0.1 M C₂H₅OH i.e. 0.1 M Ba₃ (PO₄)₂, 0.1 M Na₂SO₄, 0.1 M KCl and 0.1 M Li₃PO₄. Hence all have lowered freezing point than 0.1 M C₂H₅OH (which is non-ionisable in aqueous medium).

Ans. = 4

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8 months ago

Class 12th English

The table given below shows the annual expenditure (in crores) spend by 5 different countries on their vaccination program for children.

(in crores) Countries	2015	2016	2017	2018	2019	2020	2021
Nigeria	3387	4167	4500	5769	6354	8394	9300
UK	584793	747609	798294	976788	1314684	1779978	2285862
Egypt	107001	127137	130083	147750	171255	226686	285288
France	1252830	1437747	1210488	1250178	1415451	1697025	1784715
Italy	431085	344534	451317	487413	581451	761535	853257
Sri Lanka	3408	4938	7143	9138	7449	7140	8775
Austria	344040	413412	365253	375531	479706	595911	688947
Brazil	175320	207171	194904	204324	240972	288744	330330
Portugal	34620	43347	45087	49590	60528	76875	94875

According to the given data which of the following statements are not true?

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Payal Gupta

Contributor-Level 10

From statement (I), the table, we can observe that Myanmar has the highest annual growth

$\Rightarrow \frac{(\frac{4938 - 3408}{3408} * 100)}{1} = 44.89 %$

Statement (II), in interval 2019-20, Egyptn has second the highest growth rate, while Nigeria have the highest growth.

Statement (III), Nigeria register third the highest growth is wrong because it exhibits 28.2% which is the highest in 2017 to 2018.

Statement (IV),

Growth rate of 2016-2017

$= (\frac{194904 - 207171}{207171}) * 100 = - 5.9 %$

Growth rate of 2017-2018

$= (\frac{68108 - 64968}{64968}) * 100 = 4.8 %$

Difference = 4.8 – (–5.9) = 10.7%

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8 months ago

Class 12th English

Directions for questions: Answer the questions on the basis of the information given below.

The table given below shows the annual expenditure (in crores) spend by 5 different countries on their vaccination program for children.

(in crores) Countries	2015	2016	2017	2018	2019	2020	2021
Nigeria	3387	4167	4500	5769	6354	8394	9300
UK	584793	747609	798294	976788	1314684	1779978	2285862
Egypt	107001	127137	130083	147750	171255	226686	285288
France	1252830	1437747	1210488	1250178	1415451	1697025	1784715
Italy	431085	344534	451317	487413	581451	761535	853257
Sri Lanka	3408	4938	7143	9138	7449	7140	8775
Austria	344040	413412	365253	375531	479706	595911	688947
Brazil	175320	207171	194904	204324	240972	288744	330330
Portugal	34620	43347	45087	49590	60528	76875	94875

The third highest average annual expenditure growth on the vaccination program for children over the entire period (2015-2021) has been experienced by

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Payal Gupta

Contributor-Level 10

Average annual growth

Nigeria $= \frac{(\frac{9300 ? 3387}{3387} * 100)}{6} = 29.09 %$

UK $= \frac{(\frac{2285862 ? 584793}{584793} * 100)}{6} = 48.48 %$

Egypt $= \frac{(\frac{285288 ? 107001}{107001} * 100)}{6} = 27.77 %$

Portugal $= \frac{(\frac{94875 ? 34620}{34620} * 100)}{6} = 29 %$

Portugal experienced third the highest growth.

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8 months ago

Class 12th Chemistry

Given below are two statements.

Statement I: During electrolytic refining, blister copper deposits precious metals.

Statement II: In the process of obtaining pure copper by electrolysis method, copper blister is used to make the anode.

In the light of the above statements, choose the correct answer form the option given below:

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Raj Pandey

Contributor-Level 9

In electro-refining, impure metal (blister copper) is used as an anode while precious metal like Au, Pt gets deposited as anode mud.

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8 months ago

Class 12th Maths

The value of $2 s i n (\frac{π}{8}) s i n (\frac{2 π}{8}) s i n (\frac{3 π}{8}) s i n (\frac{5 π}{8}) s i n (\frac{6 π}{8}) s i n (\frac{7 π}{8}) i s :$

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alok kumar singh

Contributor-Level 10

$2 s i n \frac{π}{8} s i n \frac{2 π}{8} t a n \frac{3 π}{8} s i n (π - \frac{5 π}{8}) s i n (π - \frac{6 π}{8}) s i n (π - \frac{7 π}{8})$

$= {(\frac{1}{\sqrt[]{2}})}^{2} . 2 s i n^{2} \frac{π}{8} s i n^{2} \frac{3 π}{8}$

^{$= \frac{1}{4} . {(2 s i n \frac{π}{8} s i n \frac{3 π}{8})}^{2}$}

$= \frac{1}{4} . {(2 s i n \frac{π}{8} s i n \frac{3 π}{8})}^{2}$

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8 months ago

Class 12th English

The following bar chart gives the increase in internet user percentage of villages, towns and cities in the four states for the period between 2018-19 and 2020-21. The table provides information about the number of households who uses internet, average expenditure by a household and growth on expenditure on internet by a household.

	Number of households in 1987-88 (in thousands)	Average expenditure by house hold in internet on 2018-19	Growth in average household expenditure on internet (2018-19 over 2019-20)
Village	400	Rs. 30,00	50%
Town	100	Rs. 50,00	60%
City	50	Rs. 75,00	90%

The average income for the Tamil Nadu in 2020-21 was

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Payal Gupta

Contributor-Level 10

Average expenditure on internet for the Tamil Nadu

= $\frac{(400 * 30 + 100 * 50 + 50 * 75)}{400 + 100 + 50}$ = 37727.

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8 months ago

Class 12th Maths

If the value of the integral $\int_{0}^{5} \frac{x + [x]}{e^{x - [x]}} d x = α e^{- 1} + β,$ where $α, β ε R, 5 α + 6 β = 0,$ and [x] denotes the greatest integer less than or equal to x; then the value of ${(α + β)}^{2}$ is equal to

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alok kumar singh

Contributor-Level 10

$\int_{0}^{5} \frac{x + [x]}{e^{x - [x]}} d x$

= \sum_{r = 1}^{5} \int_{r - 1}^{r} \frac{(x + 1 - r)}{e^{x + 1 - r}} d x

Put x + 1 – r = t ->dx = dt

$= 5 {[- t e^{- t} - e^{- t}]}_{0}^{1}$

$= 5 [- 1 e^{- 1} - e^{- 1} + e^{0}] = 5 (1 - \frac{2}{e})$

$\Rightarrow α = - 10 β = 5 \Rightarrow {(α + β)}^{2} = 25$

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Class 12th English

	Number of households in 1987-88 (in thousands)	Average expenditure by house hold in internet on 2018-19	Growth in average household expenditure on internet (2018-19 over 2019-20)
Village	400	Rs. 30,00	50%
Town	100	Rs. 50,00	60%
City	50	Rs. 75,00	90%

The total expenditure on the internet in the cities than the town increased by how much percentage in year 2018-19?

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Payal Gupta

Contributor-Level 10

Total expenditure on internet by cities users = 7500 * 50000 = 375000000

Total expenditure on internet by town users = 5000 * 100000 = 50000000

= $\frac{125000000}{500000000} * 100$ = 25%

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New answer posted

8 months ago

Class 12th English

	Number of households in 1987-88 (in thousands)	Average expenditure by house hold in internet on 2018-19	Growth in average household expenditure on internet (2018-19 over 2019-20)
Village	400	Rs. 30,00	50%
Town	100	Rs. 50,00	60%
City	50	Rs. 75,00	90%

What is the percentage increase in total numbers of households that uses internet in towns of Tamil Nadu's over the given periods ?

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Payal Gupta

Contributor-Level 10

The percentage increase in total number of households that uses internet in Tamil Nadu is 200%.

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New answer posted

8 months ago

Class 12th English

Directions: Answer the questions based on the following Information.

	Number of households in 1987-88 (in thousands)	Average expenditure by house hold in internet on 2018-19	Growth in average household expenditure on internet (2018-19 over 2019-20)
Village	400	Rs. 30,00	50%
Town	100	Rs. 50,00	60%
City	50	Rs. 75,00	90%

What was the total households expenditure on internet in towns of Tamil Nadu's town who uses internet?

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Payal Gupta

Contributor-Level 10

Statewise breakup is not available. Hence, question cannot be answered.

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