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11 months ago

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

The decreasing order of reactivity of the following organic molecules towards ${A g N O}_{3}$ solution is:

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alok kumar singh

Contributor-Level 10

Reactivity towards ${A g N O}_{3}$ depend on stability of carbocation formed

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11 months ago

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

For the given circuit the current through battery of 6 V just after closing the switch 'S' will be_______A.

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Vishal Baghel

Contributor-Level 10

Kindly consider the following answer

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11 months ago

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

For the given circuit the current through battery of 6 V just after closing the switch ‘S’ will be_______A.

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11 months ago

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

An electrical bulb rated 220 V, 100 W, is connected in series with another bulb rated 220 V, 60 W. If the voltage across combination is 220 V, the power consumed by the 100 W bulb will be about ______W.

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Vishal Baghel

Contributor-Level 10

$\frac{v_{1}}{v_{2}} = \frac{R_{1}}{R_{2}} = \frac{P_{2}}{P_{1}} = \frac{60}{100} = \frac{3}{5}$

$v_{1} = \frac{3}{8} * 220 = 3 * 27.5 = 82.5 v$

$∴ P_{1} = {(\frac{v_{1}}{v})}^{2} P$

= 14 w

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11 months ago

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

Among statements (a) - (d), the correct ones are:

(a) Lime stone is decomposed to $C a O$ during the extraction of iron from its oxides.

(b) In the extraction of silver, silver is extracted as an anionic complex.

(c) Nickel is purified by Mond's process.

(d) Zr and Ti are purified by Van Arkel method.

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alok kumar singh

Contributor-Level 10

(a) In extraction of iron lime stone is added on a flux

$\begin{array}{r} {C a C O}_{3} ? C a O + {C O}_{2} \\ {C O}_{2} + {S i O}_{2} ? {C a S i O}_{3} \end{array}$

(b) Extraction of silver

$A g + N a C N ? {[\begin{array}{l} anionic complex \\ A g (C N)_{2} \end{array}]}^{-}$

(c) Nickel is purified by Mond's process

(d) $Z r$ and $T i$ are purified by Van Arkel method by converting to volatile iodide.

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let the eccentricity of the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 b e \sqrt[]{\frac{5}{2}}$ and length of its latus rectum be $6 \sqrt[]{2}, i f y = 2 x + c$ is a tangent to the hyperbola H, then the value of c²is equal to

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Payal Gupta

Contributor-Level 10

$a^{2} (e^{2} - 1) = b^{2}$

$e = \sqrt[]{\frac{5}{2}} \Rightarrow b^{2} = \frac{3 a^{2}}{2}$

Length of latus rectum $\frac{2 b^{2}}{a} = 6 \sqrt[]{2}$

$3 a = 6 \sqrt[]{2} \Rightarrow a = 2 \sqrt[]{2}$

$b = 2 \sqrt[]{3}$

y = 2x + c is tangent to hyperbola

$∴ c^{2} = a^{2} m^{2} - b^{2} = 20$

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11 months ago

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

The potential energy of a particle of mass 4 kg in motion along the x-axis is given by $U = 4 (1 - c o s 4 x) J .$ The time period of the particle for small oscillation (sin = ) is $(\frac{π}{K}) s .$ The value of K is_______.

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Vishal Baghel

Contributor-Level 10

m = 4kg

U = 4 (1 – cos 4x) J

$F = - \frac{\partial υ}{\partial x} = - 4 * 4 s i n 4 x$

$∴ a = - \frac{16}{4} s i n 4 x = - 4 s i n 4 x = - 16 x$

$\Rightarrow a = - 16 x$

$ω^{2} = 16 \Rightarrow ω = 4 = \frac{2 π}{T} \Rightarrow T = \frac{2 π}{4} = \frac{π}{2} s e c$

k = 2

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11 months ago

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

At a certain temperature, the degrees of freedom per molecule for gas is 8. The gas performs 150 J at work when it expands under constant pressure. The amount of heat absorbed by the gas will be________J.

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Vishal Baghel

Contributor-Level 10

Degree of freedom, f = 8

$∴ C_{V} = \frac{8}{2} R = 4 R a n d C_{P} = 4 R + R = 5 R$

So, H = $Δ U + W$

at constant pressure, W = $P Δ v$

150 = $n R Δ T$

So, $H = n C_{V} Δ T = n 5 R Δ T = 5 (n R Δ T)$

= 5 * 150

= 750 J

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11 months ago

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

A string of area of cross – section 4 mm² and length 0.5 m is connected with a rigid body of mass 2 kg. The body is rotated in a vertical circular path of radius 0.5m. The body acquires a speed of 5 m/s at the bottom of the circular path. Strain produce in the string when the body is at bottom of the circle is________* 10^-5.

(use young's modulus 10¹¹ N/m² and g = 10 m/s²)

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Vishal Baghel

Contributor-Level 10

For circular motion:-

$T - m g = \frac{m v^{2}}{r}$

$\Rightarrow T = m (g + \frac{v^{2}}{r})$

$= 2 (10 + \frac{25}{1 / 2}) = 120 N$

$\Rightarrow S t r a i n = \frac{s t r e s}{Y}$

= 30 * 10^-5

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11 months ago

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

The distance of centre of mass from end A of a one dimensional rod (AB) having mass density ? = ?0 $(1 - \frac{x^{2}}{L^{2}}) k g / m$ and length L (in meter) is $\frac{3 L}{α} m .$ The value of is _________. (where x is the distance from end A)

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Vishal Baghel

Contributor-Level 10

Linear mass density, $λ = λ_{0} (1 - \frac{x^{2}}{L^{2}}) k g / m$

$M = \int λ d x = \int_{0}^{L} λ_{0} (1 - \frac{x^{2}}{L^{2}}) d x$

$= λ_{0} (L - \frac{L^{3}}{3 L^{2}}) = \frac{2 L λ_{0}}{3}$

$A s, x_{C} = \frac{3 L}{α} = \frac{9}{8} L . \Rightarrow α = \frac{3 * 8}{9} = 8 / 3$

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