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6 months ago

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

The polymer, which can be stretched and retains its original status on releasing the force is

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alok kumar singh

Contributor-Level 10

Buna – N is synthetic rubber which can be stretched and retains its original status on releasing the force.

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6 months ago

Class 12th Oscillations

In a vernier calipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be__________* 10^-2mm.

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Payal Gupta

Contributor-Level 10

Let 'x' he the value of one division of main scale

$x = \frac{1}{20} c m = 0.05 c m$

Let y be value of one division on venire scale given

10 y = 9 x

$y = \frac{9 x}{10}$

Least count = $x - \frac{9 x}{10} = \frac{x}{10} = \frac{0.05}{10}$

= 0.005 cm

= 5 * 10^-2 mm

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6 months ago

Class 12th Ray Optics and Optical Instruments

A light ray is incident, at an incident angle q₁, on the system of two plane mirrors M₁ and M₂ having an inclination angle 75° between them (as shown in figure). After reflecting from mirror M₁ it gets reflected back by the mirror M₂ with an angle of reflection 30°. The total deviation of the ray will be ___________ degree.

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Payal Gupta

Contributor-Level 10

$δ_{t o t a l} = 360 ° - 2 θ$

= 360 – 2 * 75° = 210°

Total deviation = $δ = 150 °$

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6 months ago

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

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 $\overset{o}{A}$ is 0.42 V. If the threshold frequency is x * 10¹³ /s, where x is ________(nearest integer).

(Given, speed light = 3 * 10⁸ m/s, Planck's constant = 6.63 * 10^-34 Js)

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Vishal Baghel

Contributor-Level 10

$e V_{0} + h υ_{0} = \frac{h C}{λ}$

$\Rightarrow 1.6 * 1 0^{- 19} * 0.42 + 6.63 * 1 0^{- 34} υ_{0} = \frac{6.63 * 1 0^{- 34} * 3 * 1 0^{8}}{66 3^{0} * 1 0^{- 10}}$

$\Rightarrow 6.63 * 1 0^{- 34} υ_{0} = 1 0^{- 19} (3 - 1.6 * 0.42)$

$6.63 * 1 0^{- 34} υ_{0} = 2.328 * 1 0^{- 19}$

$υ_{0} = 35.11 * 1 0^{13} = 35$

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6 months ago

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

A travelling microscope is used to determine the refractive index of a glass slab. If 40 divisions are there in 1cm on main scale and 50 Vernier scale divisions are equal to 49 main scale division, then least count of the travelling microscope is __________* 10^-6 m.

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Vishal Baghel

Contributor-Level 10

Let 'x' be the value of one division on main scale

$x = \frac{1}{40} c m$

Let 'y' be the value of one division on vernier

Now given

50y = 49 x

$y = \frac{49 x}{50}$

$\begin{array}{l} = 5 * 1 0^{- 4} c m \\ = 5 * 1 0^{- 6} m \\ 5 \end{array}$

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6 months ago

Class 12th Electric Charges and Fields

A $10 Ω$ 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 $μ s .$ The average e.m.f. induced in the coil is__________V.

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Payal Gupta

Contributor-Level 10

Decay of current in Inductor is given by,

$i = i_{0} e^{- t / τ} [W h e r e τ = \frac{L}{R}; i_{0} = \frac{v}{R} = \frac{20}{10} = 2]$

At t = 100 $μ s$

i = 0

i.e. i = i0 $e^{- 100 / τ} = 0$ -(1)

e.m.f induced

$e = \frac{- L d i}{d t} = - L \frac{d}{d t} [i_{0} e^{- t / τ}]$

= $= - L i_{0} \frac{- 1}{2} e^{- t / τ}$

$e = \frac{L i_{0}}{τ} e^{- t / τ}$

$e_{a v g} = \frac{\int_{0}^{200 μ s} e d t}{\int_{0}^{200 μ s} d t} = \frac{\int_{0}^{200 μ s} \frac{L i_{0}}{τ} e^{- t / τ} d t}{\int_{0}^{200 μ s} d t}$

$= \frac{L i_{0}}{200 * 1 0^{- 6} s e c} = \frac{20 * 1 0^{- 3} * 2}{200 * 1 0^{- 6}} = 400$

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6 months ago

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

A small bulb is placed at the bottom of a tank containing water to a depth of $\sqrt[]{7} m .$ The refractive index of water is $\frac{4}{3} .$ The area of the surface of water through which light from the bulb can emerge out is xπm². The value of x is_________.

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Vishal Baghel

Contributor-Level 10

by snell's law

$\frac{4}{3} s i n θ_{C} = μ_{1} s i n 90 °$

$s i n θ_{C} = \frac{3}{4} = \frac{r}{\sqrt[]{r^{2} + {(\sqrt[]{7})}^{2}}}$

$\frac{3}{4} = \frac{r}{\sqrt[]{r^{2} + 7}}$

Squaring both side

$\frac{9}{16} = \frac{r^{2}}{r^{2} + 7}$

$A = r^{2} = π * 9 = 9 π$

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6 months ago

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

Two 10cm log, straight wires, each carrying a current of 5A are kept parallel to each other. If each wire experienced a force of 10^-5N, then separation between the wires is__________cm.

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Vishal Baghel

Contributor-Level 10

$F = \frac{μ_{0} * 2}{4 π} \frac{i_{1} i_{2}}{d} l$

$1 0^{- 5} = \frac{1 0^{- 7} * 2 * 5 * 5}{d} = \frac{10}{100}$

$d = 5 * 1 0^{- 2} m$

d = 5 cm

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6 months ago

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

Given below are two statements:

Statement I: The esterification of carboxylic acid with an alcohol is a nucleophilc acyl substitution.

Statement II: Electron withdrawing groups in the carboxylic acid will increase the rate of esterification reaction

Choose the most appropriate option:

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alok kumar singh

Contributor-Level 10

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New answer posted

6 months ago

Class 12th Nuclei

The elongation of a wire on the surface of the earth is 10^-⁴m. The same wire of same dimensions is elongated by 6 * 10^-⁵ m on another planet. The acceleration due to gravity on the planet will be__________ms^-². (Take acceleration due to gravity on the surface of earth = 10 ms^-²)

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Payal Gupta

Contributor-Level 10

$Δ L$ (change in length) $= \frac{F l}{A E} = \frac{m g l}{A E}$

$Δ l \propto g$

$\frac{Δ l_{1}}{g_{1}} = \frac{Δ l_{2}}{g_{2}} \Rightarrow \frac{1 0^{- 4}}{10} = \frac{6 * 1 0^{- 5}}{g_{1}}$

$g_{1} = \frac{6 * 1 0^{- 4}}{1 0^{- 4}} =$ 6 m/sec²

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