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Physics Ncert Solutions Class 12th

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Physics Ncert Solutions Class 12th Class 12th

A circuit element X when connected to an a.c. supply of peak voltage 100V given a peak current of 5A which is in phase with the voltage. A second element Y when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by $\frac{π}{2} .$ If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

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Payal Gupta

Contributor-Level 10

v = 100 sin $ω t$

$i_{0} = \frac{100}{R_{x}} = 5 \Rightarrow R_{x} = 20 Ω$

i = $5 s i n (ω t - \frac{π}{2})$

i = 100 5sin $ω t$

$5 = \frac{100}{R_{y}} \Rightarrow R_{y} = \frac{100}{5} = 20 Ω$

When x and y both are connected in series :-

v = 100sin $ω t$

tan = 1 = 45°

R₀ = $\sqrt[]{R_{x}^{2} + R_{y}^{2}} = 20 \sqrt[]{2} Ω$

$∴ \overset{o}{l_{0}} = \frac{v_{0}}{R_{0}} = \frac{100}{20 \sqrt[]{2}} = \frac{5}{\sqrt[]{2}} A .$

$∴ l_{m s} = \frac{l_{0}}{\sqrt[]{2}}$

$= \frac{5}{\sqrt[]{2} \cdot \sqrt[]{2}}$

$= \frac{5}{2} A = \frac{5}{2} A$

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Physics Ncert Solutions Class 12th Class 12th

Two identical thin metal plates has charge q₁ and q₂ respectively such that q₁ > q₂. The plates were brought close to each other to from a parallel plate capacitor of capacitance C. The potential difference between them is

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Payal Gupta

Contributor-Level 10

V = $\frac{q_{(c a p a c i t a n c e)}}{C} = \frac{q_{1} - q_{2}}{2 c}$

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Physics Ncert Solutions Class 12th Class 12th

Two identical metallic spheres A and B when placed at certain distance in air repel each other with force of F. Another identical uncharged sphere C is first place in contact with A and then in contact with B and finally placed at midpoint between sphere A and B. The force experienced by sphere C will be

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Payal Gupta

Contributor-Level 10

For sphere 'C' after contacting with 'A'. q_A = q_C = $\frac{q}{2} .$

offer contacting with 'B'. q_B = q_C = $\frac{3 q}{4}$

$F_{N e t} = | F_{1} - F_{2} |$

F₂ = $\frac{K \cdot 3 q^{2} * 4}{r^{2} 8} = \frac{3}{2} \frac{k q^{2}}{r^{2}}$

$F_{1} = \frac{9 k q^{2} * 4}{16 r^{2}} = \frac{9 k q^{2}}{4 r^{2}} = \frac{9}{4} F$

$∴ F_{N e t} = | \frac{9}{4} F - \frac{3}{2} F |$

= $\frac{9 - 6}{4} F = \frac{3}{4} F$

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2 months ago

Physics Ncert Solutions Class 12th Communication Systems

The height of a transmitting antenna at the top of a tower is 25m and that of receiving antenna is, 49m. The maximum distance between them, for satisfactory communication is LOS (Line-Of-Sight) is $K \sqrt[]{5} * 1 0^{2} m .$ The value of K is___________.

(Assume radius of Earth is 64 * 10⁺⁵m) [Calculate upto nearest integer value]

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Vishal Baghel

Contributor-Level 10

Given, hT = 25m, R = 6400 km

hR = 49m

dm (Maximum distance for satisfactory communication)

dm = $\sqrt[]{2 R h_{T}} + \sqrt[]{2 R h_{R}}$ $_{}_{}$

dm = $\sqrt[]{2 * 6400 * 1 0^{3} * 25} + \sqrt[]{2 * 6400 * 1 0^{3} * 49}$ $^{}^{}$

= $\sqrt[]{5 * 1 0^{4} * 6400} + \sqrt[]{6400 * 1 0^{3} * 49 * 2}$ $^{}^{}$

= $192 \sqrt[]{5} * 1 0^{2}$ $^{}$

= $k \sqrt[]{5} * 1 0^{2}$ $^{}$

k= 192

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2 months ago

Physics Ncert Solutions Class 12th Class 12th

A capacitor of capacitance 50 pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is _________nJ.

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Vishal Baghel

Contributor-Level 10

Q = CV

= 50 * 10^-12 * 100

= 5 * 10^-12 * 10³

Q = 5 * 10^-9C

U_i (Initial Energy)

= $\frac{Q^{2}}{2 C} = \frac{25 * 1 0^{- 18}}{2 * 50 * 1 0^{- 12}}$

= $\frac{1}{4} * 1 0^{- 6}$

Now capacitor is connected to an identical uncharged capacitor

kvL

$\frac{Q_{1}}{C} = \frac{Q_{2}}{C} = 0$

q₁ = q₂

Initial change Q₁ + Q₂ = Q

$\Rightarrow 2 Q_{1} = Q$

$Q_{1} = \frac{Q}{2}$

$u_{F} = \frac{Q_{1}^{2}}{2 C} + \frac{Q_{1}^{2}}{2 C}$

$= \frac{{(\frac{Q}{2})}^{2}}{2 C} + \frac{{(Q / 2)}^{2}}{2 C}$

$= 2 * \frac{Q^{2}}{4 * 2 C}$

...more

Q = CV

= 50 * 10^-12 * 100

= 5 * 10^-12 * 10³

Q = 5 * 10^-9C

U_i (Initial Energy)

= $\frac{Q^{2}}{2 C} = \frac{25 * 1 0^{- 18}}{2 * 50 * 1 0^{- 12}}$

= $\frac{1}{4} * 1 0^{- 6}$

Now capacitor is connected to an identical uncharged capacitor

kvL

$\frac{Q_{1}}{C} = \frac{Q_{2}}{C} = 0$

q₁ = q₂

Initial change Q₁ + Q₂ = Q

$\Rightarrow 2 Q_{1} = Q$

$Q_{1} = \frac{Q}{2}$

$u_{F} = \frac{Q_{1}^{2}}{2 C} + \frac{Q_{1}^{2}}{2 C}$

$= \frac{{(\frac{Q}{2})}^{2}}{2 C} + \frac{{(Q / 2)}^{2}}{2 C}$

$= 2 * \frac{Q^{2}}{4 * 2 C}$

$U_{f} = \frac{Q_{2}}{4 * 2 C}$

Loss in energy = U_i = U_t

= $\frac{Q^{2}}{2 C} - \frac{Q^{2}}{4 C}$

= $\frac{Q^{2}}{4 C}$

= $\frac{1}{2} [\frac{Q^{2}}{2 C}]$

$\frac{1}{2} * 0.25 * 1 0^{- 6}$

= 125 * 10^-9

= 125 nJ

less

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2 months ago

Physics Ncert Solutions Class 12th Class 12th

Identify the correct Logic Gate for the following output (Y) of two inputs A and B.

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Vishal Baghel

Contributor-Level 10

Truth table for Input & Output

A	B	y
1	1	0
0	0	1
0	1	1
1	0	1
1	1	0
0	0	1
0	1	1
1	0	1

Now, Truth Table for option B.

A	B	y
1	1	0
1	0	1
0	1	1
0	0	1

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2 months ago

Physics Ncert Solutions Class 12th Class 12th

Consider a light ray travelling in air is incident into a medium of refractive index $\sqrt[]{2 n} .$ The incident angle is twice that of refracting angle. Then, the angle of incidence will be:

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Vishal Baghel

Contributor-Level 10

Given I = 2r

$μ_{1} = 1 (A i r)$

$μ_{2} = \sqrt[]{2 n}$

by, Snell's law

$μ s i n i = μ_{2} s i n r$

$i = 2 c o s^{- 1} [\sqrt[]{\frac{n}{2}}]$

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2 months ago

Physics Ncert Solutions Class 12th Class 12th

A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be.

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Vishal Baghel

Contributor-Level 10

$F = q E [E = \frac{σ}{ε_{0}}]$

$F = \frac{q σ}{\in_{0}} = 10 N$

Final case – Now one plate is removed, Electric field will becomes half; So, force will become half.

$F' = \frac{10}{2} = 5 N$

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2 months ago

Physics Ncert Solutions Class 12th Class 12th

Match List – I with List – II :

List – I List – II

(A) Ultravoilet &nbs

...more

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Vishal Baghel

Contributor-Level 10

Ultraviolet rays – sterilizing surgical instrument

Microwaves – Radar system

Infrared waves – Greenhouse effect

X-rays – Study crystal structure

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2 months ago

Physics Ncert Solutions Class 12th Class 12th

The susceptibility of a paramagnetic material is 99. The permeability of the material in Wb/A-m, is:

[Permeability of free space $μ_{0} = 4 π * 1 0^{- 7} W b / A - m]$

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Vishal Baghel

Contributor-Level 10

Given, x = 99

$μ_{0} = 4 π * 1 0^{- 7} W b / A - m$

$μ_{m} = ?$ (Permeability of the material)

$μ_{m} = 4 π * 1 0^{- 5}$

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