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6 months ago

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

In an experiment with a convex lens, The plot of the image distance (v') against the object distance $(μ')$ measured from the focus gives a curve v $' μ'$ = 225. If all the distances are measured in cm. The magnitude of the focal length of the lens is__________cm.

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Vishal Baghel

Contributor-Level 10

As, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v' + f} - \frac{1}{μ' - f} = \frac{1}{f}$

Also, $μ' v' = 225$

Using Newton's formula - $μ' v' = f^{2}$

So, f² = 225

f = 15 cm

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6 months ago

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

An object 'O' is placed at a distance of 100 cm in front of a concave mirror of radius of curvature 200 cm as shown in the figure. The object starts moving towards the mirror at a speed 2 cm/s. The position of the image from the mirror after 10 s will be at________cm.

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Vishal Baghel

Contributor-Level 10

u = 100 cm

$f = \frac{R}{2} = - 100 c m$

after 10 sec $i = \frac{6}{6} = 1 A$

u = 80 m

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 100} - \frac{1}{- 80} = \frac{1}{80} - \frac{1}{100} = \frac{5 - 4}{400} = \frac{1}{400}$

$∴ v = 400 c m$

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6 months ago

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

If the tangents drawn at the points O(0, 0) and $P (1 + \sqrt[]{5}, 2)$ on the circle x² + y² – 2x – 4y = 0 intersect at the point Q, then the area of the triangle OPQ is equal to

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Payal Gupta

Contributor-Level 10

x^{2} + y^{2} - 2 x - 4 y = 0

Centre (1, 2) r = $\sqrt[]{5}$

Equation of OQ is x . 0 + y . 0 – (x + 0) – 2 (y + 0) = 0

⇒ x + 2y = 0 ……… (i)

Equation of PQ is $x (1 + \sqrt[]{5}) + 2 y - (| x + 1 + \sqrt[]{5}) - 2 (y + 2) = 0$

Solving (i) & (ii), $Q (\sqrt[]{5} + 1, - \frac{\sqrt[]{5} + 1}{2})$

$= \frac{1}{2} | \frac{6 + 2 \sqrt[]{5} + 4 \sqrt[]{5} = 4}{2} | = \frac{6 \sqrt[]{5} + 10}{4} = \frac{3 \sqrt[]{5} + 5}{2}$

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6 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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6 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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6 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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6 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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6 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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6 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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6 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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