Class 12th

103. Using differentials, find the approximate value of each of the following

(a) ${(\frac{17}{81})}^{\frac{1}{4}}$

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(a) Consider $y = x^{\frac{1}{4}} . L e t, x = \frac{16}{81} & ? x = \frac{1}{81} .$

$\begin{array}{l} T h e n, ? y = {(x + ? x)}^{\frac{1}{4}} - x^{\frac{1}{4}} \\ = {(\frac{17}{81})}^{\frac{1}{4}} - {(\frac{16}{81})}^{\frac{1}{4}} \\ = {(\frac{17}{81})}^{\frac{1}{4}} - \frac{2}{3} \\ ∴ {(\frac{17}{81})}^{\frac{1}{4}} = \frac{2}{3} + ? y \end{array}$

Now, $d y$ is approximately equal to $? y$ and is given by,

$\begin{array}{l} d y = (\frac{d y}{d x}) ? x = \frac{1}{4 {(x)}^{\frac{3}{4}}} (? x) (a s, y = x^{\frac{1}{4}}) \\ = \frac{1}{4 {(\frac{16}{81})}^{\frac{3}{4}}} (\frac{1}{81}) = \frac{27}{4 * 8} * \frac{1}{81} = \frac{1}{32 * 3} = \frac{1}{96} = 0.010 \end{array}$

Hence, the approximate value of ${(\frac{17}{81})}^{\frac{1}{4}}$ is $\frac{2}{3} + 0.010 = 0.667 + 0.010$

=0.677

(b) ${(33)}^{\frac{- 1}{5}}$

(b) Consider $y = x^{\frac{1}{5}} . L e t, x = 32 & ? x = 1$

Now, $d y$ is approximately equal to $? y$ and is given by,

$\begin{array}{l} d y = (\frac{d y}{d x}) (? x) = \frac{- 1}{5 {(x)}^{\frac{6}{5}}} (? x) (a s, y = x^{\frac{1}{5}}) \\ = - \frac{- 1}{5 {(x)}^{6}} (1) = - \frac{1}{320} = - 0.003 \end{array}$

Hence, the approximate value of $33^{\frac{1}{5}}$
is $\frac{1}{2} + (- 0.003 = 0.497)$

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Why is the mass determined by measuring a colligative property in case of some solutes abnormal? Discuss it with the help of Van't Hoff factor.

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This is a long answer type question as classified in NCERT Exemplar

When dissolved in suitable solvents, certain solutes/compounds either dissociate or associate. For instance, ethanoic acid dimerises in benzene due to hydrogen bonding, but dissociates and produces ions in water.

As a result, the number of chemical species in solution rises or falls in relation to the number of chemical species of solute used to create the solution. Because the magnitude of the colligative property is dependent on the number of solute particles in relation to the total number of particles in solution, it is expected that the molar mass calculated using c

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102. The maximum value of ${[x (x - 1) + 1]}^{\frac{1}{3}}$ , 0

(A) ${\frac{1}{3}}^{\frac{1}{3}}$ (B) $\frac{1}{2}$ (C) 1 (D) 0

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We have,

$f^{'} (x) = [X(x - 1) +$
${1]}^{\frac{1}{3}}, 0 \leq x \leq 1$

$f (x) = \frac{1}{3} {[x^{2} - x + 1]}^{\frac{- 2}{3}} \frac{d}{d x} (x^{2} - x + 1)$

$= \frac{2 x - 1}{3 {(x^{2} - x + 1)}^{\frac{2}{3}}}$

At f(x) = 0.

2x – 1 = 0

$x = \frac{1}{2} \in [0, 1]$

$f (\frac{1}{2}) = {[\frac{1}{2} (\frac{1}{2} - 1) + 1]}^{\frac{1}{3}} = {[\frac{1}{4} - \frac{1}{2} + 1]}^{\frac{1}{3}} .$

$= {[\frac{1 - 2 + 4}{4}]}^{\frac{1}{3}}$

= ${[\frac{3}{4}]}^{\frac{1}{3}} ? 0.90$

$f (0) = [0(0 - 1)$
$+ {1]}^{\frac{1}{3}} = 1 .$

$f (1) = [1 (1 - 1) + 1]$
$^{\frac{1}{3}} = 1 .$

Option (B) is

Hence maximum value of f(x) = at x = 1 and x = 0.

Option (c) is correct.

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How can you remove the hard calcium carbonate layer of the egg without damaging its semipermeable membrane? Can this egg be inserted into a bottle with a narrow neck without distorting its shape? Explain the process involved.

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This is a long answer type question as classified in NCERT Exemplar

(i) When egg is placed in a dilute mineral acid solution (preferably dilute HCl solution), the hard external CaCO₃ layer of the egg dissolves out /removed without damaging its semipermeable membrane.

(ii) Yes, this egg can be inserted into a bottle with a narrow neck without distorting in shape. The process involved utilising phenomenon of osmosis is explained as below -

Egg is placed in a mineral acid solution – after some time the egg is removed and placed in a hypertonic solution- size of the egg gradually decreases after some time and it shrivels due to osmos

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101. For all real values of x, the minimum value of $\frac{1 - x + x^{2}}{1 + x + x^{2}}$ is

(A) 0 (B) 1 (C) 3 (D) $\frac{1}{3}$

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We have,

$f (x) = \frac{1 - x + x^{2}}{1 + x + x^{2}} .$

$f (x) = \frac{(1 + x + x^{2}) \frac{d}{d x} (1 - x + x^{2}) - (1 - x + x^{2}) \frac{d}{d x} (1 + x + 2)}{{(1 + x + x^{2})}^{2}}$

$\Rightarrow f^{'} (x) = \frac{(1 + x + x^{2}) (- 1 + 2 x) - (1 - x + x^{2}) (1 + 2 x)}{{(1 + x + x^{2})}^{2}}$

$= \frac{- 1 + 2 x - x + 2 x^{2} - x^{2} + 2 x^{3} - 1 - 2 x + x + 2 x^{2} - x^{2} - 2 x^{3}}{{(1 + x + x^{2})}^{2}}$

$= \frac{- 2 + 2 x^{2}}{{(1 + x + x^{2})}^{2}} = \frac{- 2 (1 - x^{2})}{{(1 + x + x^{2})}^{2}} = - \frac{2 (1 + x) (1 - x)}{{(1 + x + x^{2})}^{2}}$

At f(x) = 0.

$\Rightarrow \frac{- 2 (1 + x) (1 - x)}{{(1 + x + x^{2})}^{2}} = 0$

x = 1 and x = -1.

At $x = 1, f (1) = \frac{1 - 1 + 1}{1 + 1 + 1} = \frac{1}{3}$ $x = 1, f (1) = \frac{1 - 1 + 1}{1 + 1 + 1} = \frac{1}{3}$

At x = -1, $f (- 1) = \frac{1 + 1 + 1}{1 - 1 + 1} = 3$

The maximum value of f(x) $= \frac{1}{3} x = 1$

Hence, option (D) is correct.

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100. The point on the curve x2 = 2y which is nearest to the point (0, 5) is

(A) (2√2 ,4) (B) (2√2 ,0) (C) (0, 0) (D) (2, 2)

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The equation of the given curve is

x² = 2y.

Let p (x, y) be a point on the curve.

The distance of p (x, y) from (0, 5) is say S is given by

$s^{2} = x^{2} + {(y - 5)}^{2}$

Let z = s² = x²+ y² + 25 – 10y = 2y + y²-10y + 25

z = y² – 8y + 25

So, $\frac{d z}{d y} = 2 y - 8$

$\frac{d^{2} z}{d y^{2}} = 2$

At $\frac{d z}{d y} = 0 \Rightarrow 2 y - 8 = 0 \Rightarrow y = \frac{8}{2} = 4$

At y = 4, $\frac{d^{2} z}{d y^{2}} = 2 > 0$

y = 4, is point of minimum distance.

So, x² = 2y->x2 = 2 * 4-> x² = 8 $x = \pm 2$

Hence, the point of the nearest distance are $(2√2, 4)$ and

$(- 2√2, 4) .$

Option (A) is correct.

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Discuss biological and industrial importance of osmosis.

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This is a long answer type question as classified in NCERT Exemplar

The process of osmosis is of immense biological and industrial importance as is evident from the following examples:

A. Biological Importance :

(i) Osmosis is responsible for some of the water movement from the soil into the plant roots and then into the top parts of the plant.

(ii) Adding salt to meat to protect it against bacterial action (ie. salting).

(iii) Sugar is used to protect fruits from bacterial attack. Bacterium in canned fruit shrivels and dies as a result of the osmosis process.

(iv) Salt blood cells break due to osmosis when placed in water containing les

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99. Show that semi-vertical angle of right circular cone of given surface area and maximum volume is $s i n^{- 1} \frac{1}{3}$ .

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Kindly go through the solution

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When kept in water, raisin swells in size. Name and explain the phenomenon involved with the help of a diagram. Give three applications of the phenomenon.

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Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Raised increases in size when submerged in water. It's due to a phenomenon known as "Osmosis." The process is depicted graphically in figure. A semipermeable membrane separates a solution from its solvent in this procedure, allowing solvent molecules to pass through but preventing solute particles from passing through.

The passage of solvent molecules from a pure solvent to a solution via a semipermeable membrane is known as osmosis. The following are three osmosis applications:

(i) Osmosis is responsible for some of the water movement from the soil into the plant roots

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Why is it not possible to obtain pure ethanol by fractional distillation? What general name is given to binary mixtures which show deviation from Raoult's law and whose components cannot be separated by fractional distillation. How many types of such mixtures are there?

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