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

A velocity selector consists of electric field $\vec{E} = E \hat{k}$ and magnetic field $\vec{B} = B \hat{j}$ with B = 12 mT. The value of E required for and electron of energy 728 eV moving along the positive x-axis to pass undeflected is:

(Given. Mass of electron = 9.1 * 1031kg)

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alok kumar singh

Contributor-Level 10

$K E = 728 * 1.6 * 1 0^{- 19} = \frac{1}{2} * 9.1 * 1 0^{- 31} * V^{2}$

V = 16 * 106 m/s

For Lorentz force to be zero

eE = eVB

$\Rightarrow E = v B = 16 * 1 0^{6} * 12 * 1 0^{- 3}$

= 192 * 103 V/m

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

Two concentric circular loops of radii r₁ = 30cm and r₂ = 50cm are placed in X – Y plane as shown in the figure. A current I = 7A is flowing through them in the direction as shown in figure. The net magnetic momentum of this system of two circular loops is approximately:

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alok kumar singh

Contributor-Level 10

$\vec{M} = I A_{1} - I A_{2}$

$\Rightarrow M = - 7 * \frac{22}{7} * [\frac{25}{100} - \frac{9}{100}] \hat{k}$

$\Rightarrow \vec{M} = \frac{- 7}{2} \hat{k} A m^{2}$

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

A source of potential difference V is connected to the combination of two identical capacitors as shown in the figure. When key 'K' is closed, the total energy stored across the combination is E₁. Now key 'K' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now E₂. The ratio E₁/E₂ will be:

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alok kumar singh

Contributor-Level 10

I > When switch is closed

Ceq = 2C

Energy E1 = $\frac{1}{2} * C_{e q} * v^{2} = \frac{1}{2} * 2 C * v^{2}$

E1 = CV2

II > When switch is opened

$E_{2} = \frac{1}{2} * (5 c) * V^{2} + \frac{{(C v)}^{2}}{2 * 5 C}$

$= \frac{13}{5} C V^{2}$

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{5}{13}$

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a year ago

Class 12th Semiconductor Electronics

The zener diode has a V_z = 30V. The current passing through the diode for the following circuit is…………mA.

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Payal Gupta

Contributor-Level 10

$l = \frac{90 ? 30}{4000} = 15 m A$

$I_{1} = \frac{30}{5000} = 6 m A & l_{2} = 9 m A$

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

A battery of 6V is connected to the circuit as shown below. The current I drawn from the battery is:

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alok kumar singh

Contributor-Level 10

The circuit is balanced whetstone bridge.

$\Rightarrow R_{e q} = \frac{6 * 12}{6 + 12} + 2 = 6 Ω$

$\Rightarrow I = \frac{V}{R_{e q}} = \frac{6}{6} = 1 A$

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a year ago

Class 12th Gravitation

In the reported figure of earth, the value of acceleration due to gravity is same at point A and C but it is smaller than that of its value at point B (surface of the earth). The value of OA: AB will be x : y. The value of x is

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Payal Gupta

Contributor-Level 10

$\frac{G M}{{(3 R / 2)}^{2}} = \frac{G M}{R^{3}} * r$

$\Rightarrow O A = \frac{4 R}{9} = r$

$A B = R - \frac{4 R}{9} = \frac{5 R}{9} \Rightarrow O A : A B = 4 : 5 = x : y \Rightarrow x = 4$

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a year ago

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

A screw gauge of pitch 0.5mm is used to measure the diameter of uniform wire of length 6.8cm, the main scale reading is 1.5mm and circular scale reading is 7. The calculated curved surface area wire to appropriate significant figures is:

[Screw gauge has 50 divisions on its circular scale]

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Vishal Baghel

Contributor-Level 10

Least count = $\frac{0.5}{50} m m$ = 0.01 mm

Diameter, d = 1.5 + 7 Ã— 0.01

= 1.57

$∴$ Surface Area = (2r) $l$

= 3.4 cm²

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a year ago

Class 12th Electrostatic Potential and Capacitance

27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is……… times that of a smaller drop.

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Payal Gupta

Contributor-Level 10

With the help of conservation of volume, we can write

$27 * \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 3 r . . . . . . . (1)$

With the help of conservation of charge, we can write

Q = 27 q. (2)

Potential energy of single drop = U1 = $\frac{q^{2}}{8 π ε_{0} r}$

Potential energy of bigger drop = $U_{2} = \frac{Q^{2}}{8 π ε_{0} R} = \frac{27 * 27 * q^{2}}{8 π ε_{0} (3 r)} = 243 (\frac{q^{2}}{8 π ε_{0} r}) = 243 U_{1}$

$\Rightarrow \frac{U_{2}}{U_{1}} = 243$

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a year ago

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

Two projectiles thrown at $30 ° a n d 45 °$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is:

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Vishal Baghel

Contributor-Level 10

H₁ = H₂

$\frac{u_{1}^{2} s i n^{2} θ_{1}}{2 g} = \frac{u_{2}^{2} s i n^{2} θ_{2}}{2 g}$

$\Rightarrow u_{1}^{2} {(s i n 30 °)}^{2} = {u_{2}}^{2} {(s i n 45 °)}^{2}$

$\Rightarrow \frac{u_{1}}{u_{2}} = \sqrt[]{2}$

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a year ago

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

A transverse wave s represented by y = $2 s i n (ω t - k x) c m .$ The value of wavelength (in cm) for which the wave velocity becomes equal to the maximum particle velocity. Will be:

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alok kumar singh

Contributor-Level 10

Maximum particle velocity = $A ω$

Wave velocity = $\frac{ω}{R}$

$\Rightarrow \frac{ω}{k} = A ω$

$\Rightarrow k = \frac{1}{A} = \frac{1}{2} = c m$

$λ = \frac{2 π}{k} = 4 π c m$

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