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6 months ago

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

A body is projected vertically upward from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be:

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alok kumar singh

Contributor-Level 10

$V_{C} = \sqrt[]{\frac{2 G M}{R}}$

Using conservation of Mechanical Energy

$\Rightarrow \frac{- G M m}{R} + \frac{1}{2} * m (\frac{V e^{2}}{9}) = \frac{- G M m}{(R + h)}$

$\Rightarrow \frac{1}{R + h} = \frac{8}{9 R} \Rightarrow h = \frac{R}{8} = \frac{6400}{8} = 800 k m$

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

The total charge on the system of capacitors $C_{1} = 1 μ F, C_{2} = 2 μ F, C_{3} = 4 μ F a n d C_{4} = 3 μ F$ connected in parallel is:

(Assume a battery of 20V is connected to the combination)

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Vishal Baghel

Contributor-Level 10

$C_{e q} = C_{1} + C_{2} + C_{3} + C_{4}$

= 1 + 2 + 4 + 3

$= 10 ? F$

Q = C_eq v

= (10 x 20) $? C$

Q = 200 $? C$

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6 months ago

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

Light travels in two media M₁ and M₂ with speeds 1.5 * 108 ms1 and 2.0 * 108 ms1 respectively. The critical angle between them is:

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alok kumar singh

Contributor-Level 10

$s i n i_{C} = \frac{μ_{r a r e r}}{μ_{d e n e r}} = \frac{v e l o c i t y o f r o r e r}{v e l o c i t y o f d e n s e r}$

$\Rightarrow s i n i_{c} = \frac{1.5 * 1 0^{8}}{2 * 1 0^{8}} = \frac{3}{4}$

$\Rightarrow t a n i_{c} = \frac{3}{\sqrt[]{7}}$

$\Rightarrow i_{c} = t a n^{- 1} (\frac{3}{\sqrt[]{7}})$

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6 months ago

Class 12th Electromagnetic Waves

The oscillating magnetic field in a plane electromagnetic wave is given by $B_{y} = 5 * 1 0^{- 6} s i n 100 π (5 x - 4 * 1 0^{8} t) T .$ The amplitude of electric field will be:

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alok kumar singh

Contributor-Level 10

$B_{0} = 5 * 1 0^{- 6}$

Speed of wave V = $\frac{4 * 1 0^{8}}{5} = 8 * 1 0^{7}$

$E ? = V * B ? = 4 ? * 10 = 4 * 1 0^{2} V / m$

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6 months ago

Class 12th Communication Systems

Draw the output signal Y in the given combination of gates.

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Payal Gupta

Contributor-Level 10

$y = \bar{\bar{A} + B} = A . \bar{B}$

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

The current I in the above given circuit will be:

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Vishal Baghel

Contributor-Level 10

$R = \frac{8}{2} Ω = 4 Ω$ (wheat stone bridge)

$I = \frac{V}{R} = \frac{40}{4} = 10 A$

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6 months ago

Class 12th Electric Charges and Fields

Given below are two statements:

Statement I : An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere.

Statement II : If R is the radius of a solid metallic sphere and Q be the total charge on it. The electric field at any point on the spherical surface of radius r (

In the light of the above statements, choose the correct answer from the option given below

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Payal Gupta

Contributor-Level 10

All the charge given to a conducting sphere resides on outer surface.

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6 months ago

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

B_X and B_Y are the magnetic fields at the centre of two coils X and Y respectively each carrying equal current. If coil X has 200 turns and 20 cm radius and coil Y has 400 turns and 20 cm radius, the ratio of B_X and B_Y is:

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Vishal Baghel

Contributor-Level 10

$B = N \frac{μ_{0} l}{2 μ}$

$B α \frac{N}{μ}$

$\frac{β_{x}}{β_{y}} = \frac{N_{x}}{r_{x}} * \frac{r_{y}}{N_{y}}$

$\frac{B_{x}}{B_{y}} = \frac{200}{20} * \frac{20}{400} = \frac{1}{2}$

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6 months ago

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

Two uniformly charged spherical conductors A and B of radii 5mm and 10mm are separated by a distance of 2cm. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere A and B will be:

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alok kumar singh

Contributor-Level 10

$V_{A} = V_{B}$

$\Rightarrow \frac{k Q_{A}}{R_{A}} = \frac{k Q_{B}}{R_{B}}$

$\Rightarrow \frac{Q_{A}}{Q_{B}} = \frac{R_{A}}{R_{B}} * {(\frac{R_{B}}{R_{A}})}^{2} = \frac{1}{2} * {(\frac{2}{1})}^{2} = \frac{2}{1}$

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Class 12th Atoms

The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be

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Payal Gupta

Contributor-Level 10

$Δ E = 13.6 (\frac{1}{1^{2}} - \frac{1}{5^{2}}) = 13.6 * \frac{24}{25} e V$

$\Rightarrow \frac{h c}{λ} = 13.6 * \frac{24}{25} e V . . . . . . . . . . (1)$

With the help of conservation of linear momentum, we can write

$\frac{h}{λ} = m_{H} v_{H} \Rightarrow \frac{h c}{λ} = c m_{H} v_{H} \Rightarrow v_{H} = \frac{\frac{h c}{λ}}{c m_{H}} = \frac{13.6 * \frac{24}{25} * 1.6 * 1 0^{- 19}}{3 * 1 0^{8} * 1.67 * 1 0^{- 27}} = 4.17 m / s$

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