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

Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density $d$ . The area of the base of both vessels is $S$ but the height of liquid in one vessel is $x 1$ and in the other, $x 2$ . When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is :

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alok kumar singh

Contributor-Level 10

Sol. $U_{initial} = \frac{k (4 q) (q)}{(d / 2)} + \frac{k (q) (- q)}{(d / 2)}$

$\begin{array}{r} \Rightarrow \frac{6 k q^{2}}{d} \\ \Rightarrow U_{final} = \frac{4 (4 q) (q)}{(\frac{3 d}{2})} + \frac{k (q) (- q)}{(d / 2)} \\ \Rightarrow \frac{2}{3} \frac{k q^{2}}{d} \\ \Rightarrow Δ U = (\frac{2}{3} - 6) \frac{k q^{2}}{d} \Rightarrow \frac{- 16}{3} \frac{{k q}^{2}}{d} \end{array}$

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

The value of current $i_{1}$ flowing from $A$ to $C$ in the circuit diagram is:

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alok kumar singh

Contributor-Level 10

Sol. $\overset{?}{τ} = \overset{?}{μ} * \overset{?}{B}$

$\begin{array}{l} \Rightarrow & 0.018 = μ (0.06) (s i n ? 30^{?}) \\ \Rightarrow & μ = 0.6 \\ \Rightarrow & Work = U_{f} - U_{i} \\ \Rightarrow & 2 μ B \\ \Rightarrow & 7.2 * 10^{- 2} J . \end{array}$

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Class 12th 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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Class 12th 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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Class 12th 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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Class 12th 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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Class 12th 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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Class 12th 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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Class 12th 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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Class 12th 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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