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Class 12th Inverse Trigonometric Functions

Let A = ${Z \in C : | \frac{z + 1}{z - 1} | < 1}$ and B = ${z \in C : a r g (\frac{z - 1}{z + 1}) = \frac{2 π}{3}} .$ Then $A \cup B$ is

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Class 12th Inverse Trigonometric Functions

Let A = ${Z \in C : | \frac{z + 1}{z - 1} | < 1}$ and B = ${z \in C : a r g (\frac{z - 1}{z + 1}) = \frac{2 π}{3}} .$ Then $A \cup B$ is

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New answer posted

a year ago

Class 12th Maths Matrices

Let f(x) = $\frac{x - 1}{x + 1}, x \in R - {0, - 1, 1} .$ If $f^{n + 1} (x) = f (f^{n} (x)) f o r a l l n \in N,$ then f⁶ (6) + f⁷ (7) is equal a to

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Payal Gupta

Contributor-Level 10

$f (x) = \frac{x - 1}{x + 1} \Rightarrow f (f (x)) = \frac{\frac{x - 1}{x + 1} - 1}{\frac{x - 1}{x + 1} + 1} = - \frac{1}{x}$

$f^{3} (x) = - \frac{x + 1}{x - 1} \Rightarrow f^{4} (x) = \frac{\frac{x - 1}{x + 1} + 1}{\frac{x - 1}{x + 1} - 1} = x$

$S o, f^{6} (6) + f^{7} (7) = f^{2} (6) + f^{3} (7)$

= $- \frac{1}{6} - \frac{7 + 1}{7 - 1} = - \frac{9}{6} = - \frac{3}{2}$

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a year ago

Class 12th Physics Wave Optics

A liquid of density 750 kgm^-3 flows smoothly through a horizontal pipe that tapers in cross-sectional area from A₁ = 1.2 * 10^-2m² to A₂ = $\frac{A_{1}}{2} .$ The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is_________ * 10^-3m³s^-1.

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alok kumar singh

Contributor-Level 10

Given A₂ = $\frac{A_{1}}{2}$

Bernoulli's b/w (1) & (2)

$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + z_{1} = \frac{ρ_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + z_{2}$

z₁ = z₂]

$\frac{P_{1} - P_{2}}{ρ g} = \frac{v_{2}^{2} - v_{1}^{2}}{2 g}$

[Q = A₁V₁ = A₂V₂]

$\frac{4500}{750} = \frac{{(\frac{Q}{A_{2}})}^{2}}{2} - {(\frac{Q}{A_{1}})}^{2}$

$12 = Q^{2} [\frac{1}{A_{2}^{2}} - \frac{1}{A_{1}^{2}}]$

$Q = 2 A_{1} = 2 * 1.2 * 1 0^{- 2}$

= 2.4 * 10^-2 m³ /sec

= 24 * 10^-3 m³ /sec

= 24

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a year ago

Class 12th Units and Measurement

A uniform disc with mass M = 4 kg and radius R = 10 cm is mounted on a fixed horizontal axle as shown in figure. A block with mass m = 2kg hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is_________ N.

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alok kumar singh

Contributor-Level 10

NLM² for (1)

mg – T = ma

Þ 2 * 10 – T = 2a

Þ T + 2a = 20 - (1)

Rotation equestion for (2)

T * R = I_cma

Þ T R = $I_{C M} \frac{a}{R}$

$[α = \frac{a}{R}, N o s l i p p i n g c o n d i t i o n]$

T = $\frac{M a}{2}$

$T = \frac{4}{2} a$

T = 2a - (2)

from (1) & (2)

2T = 20

T = 10N

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a year ago

Class 12th Nuclei

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alok kumar singh

Contributor-Level 10

Given A₂ = $\frac{_{}}{}$ $\frac{A_{1}}{2}$

Bernoulli's b/w (1) & (2)

$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + z_{1} = \frac{ρ_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + z_{2}$

[z₁= z₂]

$\frac{P_{1} - P_{2}}{ρ g} = \frac{v_{2}^{2} - v_{1}^{2}}{2 g}$

[Q = A₁V₁ = A₂V₂]

$\frac{4500}{750} = \frac{{(\frac{Q}{A_{2}})}^{2}}{2} - {(\frac{Q}{A_{1}})}^{2}$

$12 = Q^{2} [\frac{1}{A_{2}^{2}} - \frac{1}{A_{1}^{2}}]$

$Q = 2 A_{1} = 2 * 1.2 * 1 0^{- 2}$

= 2.4 * 10^-2 m³ /sec

= 24 * 10^-3 m³ /sec

= 24

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New answer posted

a year ago

Class 12th Physics

A running fork of frequency 340 Hz resonates in the fundamental mode with a air column of length 125cm in a cylindrical tube closed at one end. When water is slowly poured in it, the minimum height of water required for observing resonance once again is __________cm.

(Velocity of sound in air is 340 ms^-1)

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alok kumar singh

Contributor-Level 10

First resonating length = $λ = \frac{V}{f} = \frac{340}{340} = 1 m$ $\frac{}{}$

Second resonating length = $\frac{3 λ}{4} = 75 c m$ $\frac{}{}$

Third resonating length = $\frac{5 λ}{4} = 125 c m$ $\frac{}{}$

Height of water required = 125 – 75 = 50cm

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Class 12th Engineering

I completed class 12th with Arts + ITI with engineering trade. Which courses am I eligible for?

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Class 12th Deenbandhu Mahavidyalay

Kya mujhe B.Sc me Deenbandhu Mahavidyalaya me admission milega agr mere class 12th me 61.2% hai?

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a year ago

Class 12th Physics

A capacitor C₁ of capacitance $5 μ F$ is charged to a potential of 30V using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor C₂ of capacitance $10 μ F$ as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor C₂ is__________ $μ C .$

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alok kumar singh

Contributor-Level 10

Q = C₁v

...more

Q = C₁v

= 5 * 30

Q = 150 $μ c$

Now it is connected to an unchanged capacitor C₂

Let change on C₁ ® q₁

& 4 C₂ ® q₂

Apply kvL

$\frac{q_{1}}{q} - \frac{q_{2}}{C_{2}} = 0$

Þ q₂ = 2q₁ – (1)

& conservation of change

q₁ + q₂ = 150 – (2)

Þ from (1) & (2)

3q₁ = 150

q₁ = 50 $μ c$

$q_{2} = 100 μ c$

$q_{2} = 100 μ c o n c_{2}$

less

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