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Class 12th physics ncert exemplar solutions class 12th chapter two

A quantity $X$ is given by $({I F v}^{2} M L L^{4})$ in terms of moment of inertia $I$ , force $F$ , velocity $v$ , work $W$ and Length $L$ . The dimensional formula for $x$ is same as that of:

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alok kumar singh

Contributor-Level 10

Sol.

$B_{A} = \frac{μ_{0} I θ}{4 π R}$

$\Rightarrow \frac{B_{A}}{B_{B}} = \frac{I_{A} θ_{A} R_{B}}{I_{B} θ_{B} R_{A}}$

$\Rightarrow \frac{2 (\frac{3 π}{2}) (4)}{3 [\frac{5 π}{3}] [2]}$

$\Rightarrow \frac{6}{5}$ .

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

Find the binding energy per neucleon for $_{50}^{120} S n$ . Mass of proton $m_{p} = (A) 1.00783 U$ , mass of neutron $m_{n} = (A) 1.00867 U$ and mass of tin nucleus $m_{S n} = 119.902199 U$ . (take $1 U = 931 M e V$ )

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alok kumar singh

Contributor-Level 10

Sol. $\Rightarrow {\overset{?}{V}}_{1} = 2 \overset{?}{U_{c m}} - \overset{?}{U_{1}} \Rightarrow 2 [\frac{m / 2 V_{o}}{\frac{m}{2} + \frac{m}{3}}] - V_{o}$

$\Rightarrow \frac{6}{5} V_{o} - V_{o} \Rightarrow \frac{V_{0}}{5}$

$\Rightarrow λ_{o} = \frac{h c}{(\frac{m}{2} V_{o})} λ_{f} = \frac{h c}{(\frac{M}{2} \frac{V_{o}}{5})}$

$\Rightarrow Δ λ = \frac{8 h c}{{m V}_{o}}$

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

A circular coil has moment of inertia $0.8 k g m^{2}$ around any diameter and is carrying current to produce a magnetic moment of $20 {A m}^{2}$ . The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of $4 T$ is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by $60^{?}$ will be:

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alok kumar singh

Contributor-Level 10

Sol. $K = \frac{(\frac{Q}{r}) Δ x}{Δ A T}$

$\begin{array}{l} \Rightarrow & \frac{({M L}^{2} T^{- 2}) (L)}{(L^{2}) (θ) (T)} \\ \Rightarrow & M^{1} L^{1 -} T^{- 3} θ^{- 1} \end{array}$

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

Consider two uniform discs of the same thickness and different radii $R_{1} = R$ and $R_{2} = α R$ made of the same material. If the ratio of their moments of inertia $I_{1}$ and $I_{2}$ , respectively, about their axes is $I_{1} : I_{2} = 1 : 16$ then the value of $α$ is:

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alok kumar singh

Contributor-Level 10

$\begin{array}{r} \Rightarrow V^{2} = U^{2} + 2 g S \\ \Rightarrow V^{2} = 0 + 2 g (h - y) \\ \Rightarrow V^{2} = 2 g h - 2 g y \\ \Rightarrow V = \sqrt[]{2 g h - 2 g y} \end{array}$

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Class 12th Photoelectric Effect

In photoelectric effect experiment, the graph of stopping potential $\lor$ versus reciprocal of wavelength obtained is shown in the figure. As the intensity of incident radiation is increased:

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alok kumar singh

Contributor-Level 10

Sol. $e V s = h v - ?$

$\begin{array}{r} At V_{s} = 0 \Rightarrow h v = ? \\ \Rightarrow ? = [6.62 * 10^{- 34}] [10^{14}] [5.5] \\ \Rightarrow ? = \frac{ [6.62 * 10^{- 34}] [10^{14}] [5.5] e V}{(1.6 * 10^{- 19}]} \end{array}$

$= 2.27$

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Class 12th Physics Electrostatic Potential and Capacitance

A capacitor $C$ is fully charged with voltage $V_{0}$ . After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance. $\frac{C}{2}$ . The energy loss in the process after the charge is distributed between the two capacitor is:

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alok kumar singh

Contributor-Level 10

Sol. $g = \frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$

$\Rightarrow \int_{v}^{0} ? d V = - \int_{x}^{\infty} ? g d x$

$\Rightarrow O - V = - [\int \frac{A x}{{(a^{2} + x^{2})}^{3 / 2}}]$

Let, $a^{2} + x^{2} = t^{2}$

$\Rightarrow 2 x d x = 2 t d t$

$\Rightarrow x d x = t d t$

$\Rightarrow V = \int \frac{A t d t}{t^{3}} \Rightarrow - \frac{A}{t} \Rightarrow - {\frac{A}{\sqrt[]{a^{2} + x^{2}}}|}_{x}^{\infty}$

$\Rightarrow V = \frac{A}{\sqrt[]{a^{2} + x^{2}}}$

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Class 12th Alternating Current

If L, C and R are the self inductance, capacitance and resistance respectively, which of the following does not have the dimension of time?

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Payal Gupta

Contributor-Level 10

Dimension

RC → [T]

$\frac{L}{R} \to [T]$

$\sqrt[]{L c} \to [T]$

$\frac{L}{C} \to$ dimensionless

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Class 12th Moving Charges and Magnetism

Two long parallel conductors S₁ and S₂ are separated by a distance 10 cm and carrying currents of 4A and 2A respectively. The conductors are placed along x-axis in X-Y plane. There is a point P located between the conductors (as shown in figure).

A charge particle of 3p coulomb is passing through the point P with velocity $\vec{v} = (2 \hat{i} + 3 \hat{j})$ m/s; where $\hat{i} a n d \hat{j}$ represents unit vector along x & y axis respectively. The force acting on the charge particle is 4p * 10^-5 $(- x \hat{i} + 2 \hat{j}) N .$ The value of x is

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Payal Gupta

Contributor-Level 10

${\vec{B}}_{N e t} = \frac{μ_{0}}{4 π} * 2 [\frac{i_{1}}{d_{1}} - \frac{i_{2}}{d_{2}}] (- \hat{k})$

$= 1 0^{- 7} * 2 [\frac{4}{4} * 100 - \frac{2}{6} * 100] (- \hat{k})$

$= 2 * 1 0^{- 5} [1 - \frac{1}{3}] - \hat{k}$

${\vec{B}}_{N e t} = \frac{4}{3} * 1 0^{- 5} (- \hat{k})$

q = 3πc

$\vec{v} = 2 \hat{i} + 3 \hat{j}$

$\vec{F} = q (\vec{v} * \vec{B})$

= $3 π [(2 \hat{i} * 3 \hat{j}) * \frac{4}{3} * 1 0^{- 5} (- \hat{k})]$

$\vec{F} = 4 π * 1 0^{- 5} (- 3 \hat{i} + 2 \hat{j}) N$

x = 3

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Class 12th Electric Charges and Fields

Three identical charged balls each of charge 2 C are suspended from a common point P by silk threads of 2 m each (as shown in figure). They form an equilateral triangle of side 1 m. The ratio of net force on a charged ball to the force between any two charged balls will be