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Physics physics ncert exemplar solution class 12th chapter one

A particle of charge $q$ and mass $m$ is subjected to an electric field $E = E_{0} (1 - a x^{2})$ in the $x$ -direction, where $a$ and $E_{0}$ are constants. Initially the particle was at rest at $x = 0$ . Other then the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is:

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alok kumar singh

Contributor-Level 10

Sol. $E = (I) (t) (A) ?{c o s}^{2} ? θ?$
$\Rightarrow (3.3) [\frac{2 π}{31.4}] [3 * 10^{- 4}] * \frac{1}{2}$

$\Rightarrow 0.99 * 10^{- 4}$

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New answer posted

a year ago

Physics Semiconductor Electronics

The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6 V. The current through the resister of 40 $Ω$ is………….mA

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Payal Gupta

Contributor-Level 10

D₁ → forward biased

D₂ → Reversed biased

So, current will flow through only D₁

$l = \frac{(1 - 0.6) v}{(60 + 40) Ω} = \frac{0.4}{100}$

= 4 * 10^-³ A

= 4mA

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New answer posted

a year ago

Physics Electromagnetic Induction

A metallic rod of length 20 cm is placed in North-South direction and is moved at a constant speed of 20 m/s towards East. The horizontal component of the Earth's magnetic field at that place is 4 * 10^-³ T and the angle of dip is 45°. The emf induced in the rod is………….mV.

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Payal Gupta

Contributor-Level 10

$t a n δ = \frac{B_{v}}{B_{h}}$

$B v = B_{h} t a n 45$

= 4 * 10^-³

emf induced in the rod is = $B_{v} v l$

$= 4 * 1 0^{- 3} * 20 * \frac{20}{100}$

$\begin{array}{l} = 16 * 1 0^{- 3} \\ = 16 m v \end{array}$

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New answer posted

a year ago

Physics Moving Charges and Magnetism

A deuteron and a proton moving with equal kinetic energy enter into to a uniform magnetic field at right angle to the field. If r_d and r_p are the radii of their circular paths respectively, then the ratio $\frac{r_{d}}{r_{p}}$ will be $\sqrt[]{x} : 1$ where x is……….

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Payal Gupta

Contributor-Level 10

r_d = (Radius of deuteron)

$= \frac{m_{d} v_{d}}{q_{d} B}$

$= \frac{\sqrt[]{2 m_{d} k . E}}{q_{d} B}$

$= \frac{\sqrt[]{2 m_{P}}}{\sqrt[]{m_{P}}} * \frac{q_{P}}{q_{d}}$ $[q_{P} = q_{d} = 1.6 * 1 0_{C}^{- 19}]$

$= \sqrt[]{2} : 1$

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a year ago

Physics Semiconductors

Identify the operation performed by the circuit given below:

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

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a year ago

Physics Electrostatic Potential and Capacitance

A parallel plate capacitor is made up of stair like structure with a plate area A of each stair and that is connected with a wire of length b, as shown in the figure. The capacitance of the arrangement is $\frac{x}{15} \frac{\in_{0} A}{b},$ the value of x is……………

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Payal Gupta

Contributor-Level 10

$C_{1} = \frac{\in_{0} A}{5 b}$

$C_{2} = \frac{\in_{0} A}{3 b}$

$C_{3} = \frac{\in_{0} A}{b}$

$C_{e q} = C_{1} + C_{2} + C_{3}$

$= \frac{\in_{0} A}{b} [\frac{1}{5} + \frac{1}{3} + 1]$

$C_{e q} = \frac{23}{15} \frac{\in_{0} A}{b}$

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New answer posted

a year ago

Physics Oscillations

A particle executives simple harmonic motion. Its amplitude is 8cm and time period is 6s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is………… s.

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Payal Gupta

Contributor-Level 10

A = 8 cm

T = 6 sec

q = 60° from A to B

During reaching the point its maximum amplitude from point (A)

$θ = 60 ° = \frac{π^{C}}{3}$

$θ = ω t$

$\frac{π}{3} = \frac{2 π}{6} t$

t = 1 sec

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New answer posted

a year ago

Physics Thermodynamics

A diatomic gas $(γ = 1.4)$ does 400 J of work when it is expanded isobarically. The heat given to the gas in the process is…………….J.

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Payal Gupta

Contributor-Level 10

$w = P Δ v = n R Δ T = 400 J$ - (1)

$Q = n C_{P} Δ T = \frac{n R y}{y - 1} Δ T$

$Q = \frac{14}{4} * 400$

Q = 1400 J

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New answer posted

a year ago

Physics System of Particles and Rotational Motion

A rolling wheel of 12kg is on an inclined plane at position P and connected to a mass of 3kg through a string of fixed length and pulley as shown in figure.

Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be $\frac{1}{2} \sqrt[]{x g h} m / s .$ The value of x is…………

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Payal Gupta

Contributor-Level 10

By conservation of Energy

$m g h = \frac{1}{2} l_{P} ω^{2}$

3gh $= \frac{1}{2} (M R^{2} + M R^{2}) ω^{2}$

$3 g h = m v^{2}$

$v^{2} = \frac{3 g h}{12}$

$v = \frac{\sqrt[]{g h}}{2} = \frac{\sqrt[]{x g h}}{2}$

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New answer posted

a year ago

Physics Laws of Motion

A mass of 10kg is suspended vertically by a rope of length 5m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is q = tan^-¹ (x * 10^-¹). The value of x is…………….

(Given, g = 10 m/s²)

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Payal Gupta

Contributor-Level 10

F.B.D. of point 'P'

$T s i n θ = 30$ - (1)

$T c o s θ = 100$ - (2)

$(1) \div (2)$

$t a n θ = \frac{3}{10} \Rightarrow θ = t a n^{- 1} (3 * 1 0^{- 1})$

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