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Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

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3 months ago

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

An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is

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Payal Gupta

Contributor-Level 10

Volume = constant

$A l = C$

$A d l + l d A = 0$

$\frac{d l}{l} + \frac{d A}{A} = 0$

$\frac{d R}{R} * 100 = (\frac{d l}{l} - \frac{d A}{A}) * 100$

= 0.4 (0.4)

= 0.8 %

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Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Semiconductor Electronics

The variation of acceleration due to gravity (g) with distance (r) from the centre of the earth is correctly represented by: (Given R = radius of earth)

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Payal Gupta

Contributor-Level 10

$g \propto r 0 \leq r \leq R$

$g \propto \frac{1}{r^{2}} r > R$

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Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Semiconductor Electronics

The variation of acceleration due to gravity (g) with distance (r) from the centre of the earth is correctly represented by: (Given R = radius of earth)

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Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Semiconductor Electronics

A transistor is used in an amplifier circuit in common emitter mode. If the best current changes by 100 $μ A,$ it brings a change of 10 mA is collector current. If the load resistance is 2 $k Ω$ and input resistance is 1 $k Ω,$ the value of power gain is x * 10⁴. The value of x is_____________.

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alok kumar singh

Contributor-Level 10

Power gain = ${(\frac{Δ i_{c}}{Δ i_{b}})}^{2} * \frac{R_{0}}{R_{i}}$

= ${(\frac{10 * 1 0^{- 3}}{100 * 1 0^{- 6}})}^{2} * \frac{2}{1}$

= 2 * 104 = x * 104

= 2

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Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Physics Oscillations

A body is performing simple harmonic with an amplitude of 10cm. The velocity if the body was tripled by air Jet when it is at 5cm from its mean position. The new amplitude vibration is $\sqrt[]{x} c m .$ The value of x is__________.

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alok kumar singh

Contributor-Level 10

v = $ω \sqrt[]{A^{2} - x^{2}}$ $^{}^{}$

at x = 5, A = 10

$v' = 3 v = 3 ω \sqrt[]{A^{2} - 5^{2}} = ω \sqrt[]{A'^{2} - 5^{2}}$

$= 3 \sqrt[]{A^{2} - 5^{2}} = \sqrt[]{A'^{2} - 5^{2}}$

$\sqrt[]{1 0^{2} - 5^{2}} = \sqrt[]{A'^{2} - 25}$

$A'^{2} = 25 + 9 * 75$ $A'^{2} = 700$

$A' = \sqrt[]{700} c m$

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Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Physics Oscillations

As per given figures, two spring constants k and 2k are connected to mass m. If period of oscillation in figure (a) is 3s, then the period of oscillation in figure (b) will be $\sqrt[]{x} s .$ The value of x is________.

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alok kumar singh

Contributor-Level 10

In figure a

$K_{e q} = \frac{k * 2 k}{3 k} = \frac{2 k}{3}$

$T = 2 Π \sqrt[]{\frac{3 M}{2 K}} = 3 s$

$\frac{I n f i g u r e b}{K_{e q} = 3 K}$

$T' = 2 Π \sqrt[]{\frac{M}{3 K}}$

$T' = \sqrt[]{2} \Rightarrow X = 2$

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Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Semiconductor Electronics

The typical transfer characteristics of a transistor in CE configuration is shown in figure. A load resistor of $2 k Ω$ is connected in the collector branch of the circuit used. The input resistance of the transistor is $0.50 k Ω .$ The voltage gain of the transistor is________.