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11 months ago

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

A ball is thrown vertically upwards with a velocity of 19.6 ms^-1 from the top of a tower. The ball strikes the ground after 6s. The height from the ground up to which the ball can rise will be $(\frac{k}{5}) m .$ The value of k is_______(use g = 9.8m/s²)

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Vishal Baghel

Contributor-Level 10

h = 39.2 + 19.6 = 58.8 m

Height above tower

$h' = \frac{u^{2}}{2 g} = \frac{19.6 * 19.6}{2 * 9.8}$

= 19.6

$A s, \frac{k}{5} = 78.4 \Rightarrow k = 392$

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11 months ago

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

Let y = y(x) be the solution of the differential equation $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0, x > 1, w i t h y (2) = - 2 .$ Then y(3) is equal to

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Payal Gupta

Contributor-Level 10

$x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0$

$\Rightarrow x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} - 1) y = 4 x^{3}$

$\Rightarrow y = 2 x - 1 (x^{3} - x) ∴ y (3) = - 18$

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11 months ago

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

A FM Broad cast transmitter, using modulating signal of frequency 20 kHz has a deviation ratio of 10. The Bandwidth required for transmission is:

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Vishal Baghel

Contributor-Level 10

f = 20 kHz

Deviation ratio = 10

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11 months ago

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

Let the solution curve y = y(x) of the differential equation $[\frac{x}{\sqrt[]{x^{2} - y^{2}}} + e^{\frac{y}{x}}] x \frac{d y}{d x} = x + [\frac{x}{\sqrt[]{x^{2} - y^{2}}} + e^{\frac{y}{x}}] y$ pass through the points (1, 0) and (2α, α), α > 0. Then a is equal to

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Payal Gupta

Contributor-Level 10

$[\frac{x}{\sqrt[]{x^{2} - y^{2}}} + e^{y / x}] x \frac{d y}{d x} = x + [\frac{x}{\sqrt[]{x^{2} - y^{2}}} + e^{y / x}] y$

$\Rightarrow e^{y / x} [x d y - y d x] = x d x + \frac{x}{\sqrt[]{x^{2} - y^{2}}} (y d x - x d y)$

$\Rightarrow e^{y / x} d (y / x) = \frac{d x}{x} - \frac{d (y / x)}{\sqrt[]{1 - {(y / x)}^{2}}}$

Integrating

$e^{y / x} = l n x - s i n^{- 1} (\frac{y}{x}) + c$

Passes (1, 0)

1 = c

$\Rightarrow α = \frac{1}{2} e x p (\sqrt[]{e} - 1 + \frac{π}{6})$

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11 months ago

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

Perpendicular to the plane $A B C$ is $I_{0}$ . If part $A D E$ is removed, the moment of inertia of the remaining part about the same axis is $\frac{{N I}_{0}}{16}$ where $N$ is an integer. Value of $N$ is

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alok kumar singh

Contributor-Level 10

$\frac{1}{λ_{min}} = R [\frac{1}{1} - \frac{1}{\infty}] = R [n = \infty \to n = 1]$

$\frac{1}{λ_{m a x}} = R [\frac{1}{1} - \frac{1}{4}] = \frac{3 R}{4} [n = 2 \to n = 1]$

$\Rightarrow Δ λ \Rightarrow \frac{4}{3 R} - \frac{1}{R} \Rightarrow \frac{1}{3 R} = 340$

For Paschan $\Rightarrow \frac{1}{λ_{min}} = R [\frac{1}{9}] [n = \infty \to h = 3]$

$\Rightarrow \frac{1}{λ_{m a x}} = R [\frac{1}{9} - \frac{1}{16}] = \frac{7 R}{144}$

$\Rightarrow Δ λ = \frac{81}{7 R}$

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11 months ago

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

An n.p.n transistor with current gain = 100in common emitter configuration is shown in figure. The output voltage of the amplifier will be

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Vishal Baghel

Contributor-Level 10

n.p.m. transistor current gain B = 100

As, B = $\frac{I_{C}}{I_{B}} V o l t a g e g a i n, \frac{V_{o u t}}{V_{i n}} = (\frac{I_{C}}{I_{B}}) \frac{R_{2}}{R_{i}}$

$V_{o u t} = \frac{B R_{2}}{R_{i n}} V i n$

$\frac{1 0^{3}}{1 0^{3}} V = 1 v$

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11 months ago

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

A radioactive sample decays $\frac{7}{8}$ times its original quantity in 15 minutes. The half-life of the sample is

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Vishal Baghel

Contributor-Level 10

$N_{0} \to_{}^{t = 15 m i n} (N_{0} - \frac{7}{8} N_{0}) = \frac{N_{0}}{8}$

So, $N = \frac{N_{0}}{2^{t / t_{1 / 2}}} \Rightarrow \frac{N_{0}}{8} = \frac{N_{0}}{2^{t / t_{1 / 2}}}$

$\Rightarrow 2^{3} = 2^{(\frac{15 m i n}{T})}$

$\Rightarrow T = \frac{15 m i n}{3} = 5 m i n$

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11 months ago

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

$A B C$ is a plane lamina of the shape of an equilateral triangle. $D, E$ are mid points of $A B, A C$ and $G$ is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through $G$ and (Rotational)

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alok kumar singh

Contributor-Level 10

$\begin{array}{l} A H \to \frac{H}{2} & H = \frac{\sqrt[]{3} a}{2} \\ A G \to \frac{H}{\sqrt[]{3}} & A l = \frac{H}{2 \sqrt[]{3}} \\ M_{A B C} = M & M_{A D E} = \frac{M}{4} \end{array}$

$\Rightarrow I_{G} = \frac{M a^{2}}{12} - [\frac{M}{4} \frac{{[\frac{a}{2}]}^{2}}{12} + \frac{M}{4} {[\frac{a}{2 \sqrt[]{3}}]}^{2}] = (\frac{11}{16}) \frac{{M a}^{2}}{12}$

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11 months ago

Class 12th Application of Integrals

The area of the region S = ${(x, y) : y^{2} \leq 8 x, y \geq \sqrt[]{2} x, x \geq 1}$ is

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Payal Gupta

Contributor-Level 10

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11 months ago

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

Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively, will be

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Vishal Baghel

Contributor-Level 10

$E_{1} = 5 ? - ? = 4 ? = \frac{1}{2} m v_{1}^{2}$

$E_{2} = 10 ? - ? = \frac{1}{2} m V_{2}^{2}$

$\Rightarrow \frac{V_{1}}{V_{2}} = \frac{\sqrt[]{\frac{8 ?}{m}}}{\sqrt[]{\frac{18 ?}{m}}} = \sqrt[]{\frac{8}{18}} = \frac{\sqrt[]{4}}{9} = \frac{2}{3}$

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