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12 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Ten

A cube of metal is subjected to a hydrostatic pressure 4GPa. The percentage change in the length of the side of the cube is close to: (Given bulk modulus of metal, $B = 8 * 10^{10}$ $P a)$

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alok kumar singh

Contributor-Level 10

Sol.

$\begin{array}{r} \end{array}$ $\begin{array}{r} ↑_{1}^{?} (1) ε = 3 \\ (2) ε - I r = 2.5 V \\ \Rightarrow I r = 0.5 \\ Now, IR = 2.5 \\ \Rightarrow \frac{R}{r} = 5 . \\ \Rightarrow \frac{P_{R}}{R_{r}} = \frac{I^{2} R}{I^{2} r} = \frac{R}{r} = 5 \\ \Rightarrow P_{r} = \frac{0.5}{5} = 0.1 \end{array}$

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12 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Four

A smooth circular groove has a smooth vertical wall as shown in figure. A block of mass m moves against the wall with a speed v. Which of the following curve represents the correct relation between the normal reaction on the block by the wall (N) and speed of the block (v)?

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Vishal Baghel

Contributor-Level 10

$N = \frac{m v^{2}}{R}$

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12 months ago

Class 12th Conic Sections

The number of points of intersection of $| z - (4 + 3 i) | = 2 a n d | z | + | z - 4 | = 6, z \in C,$ is

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Payal Gupta

Contributor-Level 10

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12 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

If t = $\sqrt[]{x} + 4, t h e n {(\frac{d x}{d t})}_{t = 4} i s :$

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Vishal Baghel

Contributor-Level 10

$t = \sqrt[]{x} + 4$

then $\frac{d t}{d x} = \frac{1}{2 \sqrt[]{x}} ∴ {(\frac{d x}{d t})}_{t = 4} = {(2 \sqrt[]{x})}_{t = 4} = {[2 (t - 4)]}_{a t t = 4}$

= 0

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12 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Six

Match the thermodynamics processes taking place in a system with the correct conditions. In the table: $Δ Q$ is the heat supplied, $Δ W$ is the work done and $Δ U$ is change in internal energy of the system. Match the following:

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alok kumar singh

Contributor-Level 10

Sol. $? v$

$\Rightarrow m v = 16 {m v}_{1}$

$\Rightarrow V_{1} = \frac{V}{16}$

$\Rightarrow Δ k$ loss $= \frac{1}{2} {m v}^{2} - \frac{1}{2} (16 m) {(\frac{V}{16})}^{2}$

$\Rightarrow \frac{1}{2} {m v}^{2} (\frac{15}{16})$

$%$ loss $= \frac{15}{16} * 100 = 93.75 %$

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12 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Three

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height $\frac{h}{3}$ while going up and coming down respectively.

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Vishal Baghel

Contributor-Level 10

$h = \frac{u^{2}}{2 g} \Rightarrow u = \sqrt[]{2 g h}$

$\frac{h}{3} = \sqrt[]{2 g h} t - 5 t^{2}$

$5 t^{2} - \sqrt[]{20 h} t + \frac{h}{3} = 0$

$t = \frac{\sqrt[]{20 h} \pm \sqrt[]{20 h - 4 * 5 * \frac{h}{3}}}{10}$

$∴ \frac{t_{1}}{t_{2}} = \frac{\sqrt[]{20 h} - \sqrt[]{\frac{40}{3} h}}{\sqrt[]{20 h} + \sqrt[]{\frac{40}{3} h}} = \frac{1 - \sqrt[]{\frac{2}{3}}}{1 + \sqrt[]{\frac{2}{3}}} = \frac{\sqrt[]{3} - \sqrt[]{2}}{\sqrt[]{3} + \sqrt[]{2}}$

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12 months ago

Class 12th Organic Chemistry - Some Basic Principles and Tech

0.25g of an organic compound containing chlorine gave 0.40g of silver chloride in Carius estimation. The percentage of chlorine present in the compound is___________. [ in nearest integer]

(Given: Molar mass of Ag is 108 g mol^-1 and that of Cl is 35.5 g mol^-1)

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Payal Gupta

Contributor-Level 10

Moles of chlorine in the given compound = Moles of chlorine in AgCl

= moles of AgCl

$n_{C l} = \frac{0.4}{143.5} m o l$

Mass of chlorine = $\frac{0.4}{143.5} * 35.5 g$

= 0.098 g

$% o f C l = \frac{0.098}{0.25} * 100 = 39.6 %$

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12 months ago

Class 12th Surface Chemistry

If the initial pressure of a gas is 0.03 atm, the mass of the gas adsorbed per gram of the adsorbent is____________* 10^-2g.

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Payal Gupta

Contributor-Level 10

Using, Freundlich adsorption isotherm ;

$\frac{x}{m} = k . p^{1 / n}$ ………………. (i)

$l o g \frac{x}{m} = \frac{1}{n} l o g p + l o g k$

Comparing with y = mx + C

Slope = $\frac{1}{n}$ = 1

Intercept, log k = 0.602

log k = log 4

k = 4

from equation (i)

$\frac{x}{m} = 4 * {(0.03)}^{1}$

= 0.12

= 12 * 10^-²

So, 12 * 10^-² g of gas is adsorbed per gram of adsorbent,

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12 months ago

Class 12th physics ncert exemplar solutions class 12th chapter two

Given below are two statements: One is labeled as Assertion (A) and other is labeler as Reason (R).

Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is $ρ$ and radius of the drop is r, then

$T = K \sqrt[]{\frac{ρ r^{3}}{S^{\frac{3}{2}}}}$ is dimensionally correct, where K is dimensionless.

Reason (R) : Using dimensional analysis we get R.H.S. having different dimension than that of time period.

In the light of above statements, choose the correct answer from the options given below.

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Vishal Baghel

Contributor-Level 10

$T = k \sqrt[]{\frac{ρ r^{3}}{S^{3 / 2}}}$

$[T]$ ${[M L^{- 3}]}^{1 / 2} {[L^{3}]}^{1 / 2} {[M L T^{- 2} L^{- 1}]}^{\frac{- 3}{4}}$

$[T]$ $= [M^{1 / 2} L^{- \frac{3}{2}} L^{\frac{3}{2}} M^{- \frac{3}{4}} T^{\frac{3}{2}}]$

$= [M^{\frac{1}{4}} T^{3 / 2}]$

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