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6 months ago

Physics 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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6 months ago

Physics 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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6 months ago

Physics 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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6 months ago

Physics 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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6 months ago

Physics 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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6 months ago

Physics 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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6 months ago

Physics 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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6 months ago

Physics 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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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The power of lens (biconvex) is 1.25 m^-¹ in particular medium. Refractive index of the lens is 1.5 and radii of curvature are 20 cm and 40 cm respectively. The refractive index of surrounding medium:

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Vishal Baghel

Contributor-Level 10

Power of a lens, P = 1.25 m^-1

$μ_{l e n s} = 1.5, r_{1} = 20 c m, r_{2} = 40 c m$

As, P = $\frac{1}{f} = 1.25$

$\Rightarrow \frac{1.5}{μ_{1}} = \frac{1}{6} + 1 = \frac{7}{6} \Rightarrow μ_{1} = \frac{6}{7} * 1.5 = \frac{9}{7}$

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

In a compound microscope, the magnified virtual image is formed at a distance of $25 c m$ from the eye-piece. The focal length of its objective lens is $1 c m$ . If the magnification is 100 and the tube length of the microscope is $20 c m$ , then the focal length of the eyepiece lens (in $c m$ ) is

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alok kumar singh

Contributor-Level 10

$M = \frac{1}{f_{o}} [1 + \frac{D}{F e}]$

$\Rightarrow 100 = \frac{20}{1} [1 + \frac{25}{F e}]$

$\Rightarrow 1 + \frac{25}{F e} = 5$

$\Rightarrow F e = \frac{25}{4} = 6.25 c m$

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