Tags Class 11th

Class 11th

Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th

Follow Ask Question

Questions

Discussions

Active Users

Followers

New answer posted

4 months ago

Class 11th Maths

The sum of the infinite series $1 + \frac{5}{6} + \frac{12}{6^{2}} + \frac{22}{6^{3}} + \frac{35}{6^{4}} + \frac{51}{6^{5}} + \frac{70}{6^{6}} + . . . .$ is equal to

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

Let S = 1 $+ \frac{5}{6} + \frac{12}{6^{2}} + \frac{22}{6^{3}} + . . . .$

$\underline{- \frac{S}{6} = \frac{1}{6} + \frac{5}{6^{2}} + . . . .}$

$\frac{5 S}{6} = 1 + \frac{4}{6} + \frac{7}{6^{2}} + \frac{10}{6^{3}} + . . . .$

$\underline{- \frac{5 S}{36} = \frac{1}{6} + \frac{4}{6^{2}} + \frac{7}{6^{3}} + . . . . .}$

$\frac{25 S}{36} = 1 + \frac{3}{6} + \frac{3}{6^{2}} + \frac{3}{6^{3}} + . . . . . .$

$\frac{25 S}{36} = 1 + \frac{3 / 6}{1 - 1 / 6}$

$\frac{25 S}{36} = 1 + \frac{3 / 5}{1}$

$S = \frac{288}{125}$

0 0

New answer posted

4 months ago

Class 11th Maths

Let arg(z) represent the principal argument of the complex number z. Then $| z | = 3$ and art(z – 1) – arg(z + 1) = $\frac{π}{4}$ intersect

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

$| z | = 3 \Rightarrow$ circle with radius = 3

arg $(\frac{z - 1}{z + 1}) = \frac{π}{4},$ part of a circle (with radius $\sqrt[]{2}$ ). no common points

0 0

New answer posted

4 months ago

Class 11th Maths

Let α be root of the equation 1 + x² + x⁴ = 0. Then the value of α¹⁰¹¹ + α²⁰²² - α³⁰³³ is equal to

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

$\Rightarrow (x^{2} + 1 + x) (x^{2} - x + 1) = 0$

$x = \pm ω, ? ω^{2}$

Now, = $α^{1011} + α^{2022} - α^{3033}$

= $ω^{1011} + ω^{2022} - ω^{3033}$

= 1 + 1 – 1 = 1

0 0

New answer posted

4 months ago

Class 11th Physics

The momentum of inertia of a uniform thin rod about a perpendicular axis passing through on end is I₁. The same rod is into a ring and its moment of inertia about a diameter is I₂. If $\frac{I_{1}}{I_{2}} i s \frac{X π^{2}}{3},$ then the value of x will be__________.

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

According to question, we can write

$l = 2 π r \Rightarrow r = \frac{l}{2 π}$

$I_{1} = \frac{m l^{2}}{3}, a n d I_{2} = \frac{m r^{2}}{2} = \frac{m l^{2}}{8 π^{2}}$

$\Rightarrow \frac{I_{1}}{I_{2}} = \frac{m l^{2}}{3} * \frac{8 π^{2}}{m l^{2}} = \frac{8 π^{2}}{3}$

0 0

New answer posted

4 months ago

Class 11th Physics Waves

In an experiment to determine the velocity of sound in air at room temperature using a resonance tube, the first resonance is observed when the air column has a length of 20.0 cm for a tuning form of frequency 400 Hz is used. The velocity of the sound at room temperature is 336 ms^-¹. The third resonance is observed when the air column has a length of ____________cm.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

According to Concept of resonance tube, we can write

$\frac{λ}{4} + e = l_{1} a n d \frac{3 λ}{4} + e = l_{2} \Rightarrow \frac{λ}{2} = l_{2} - l_{1} \Rightarrow \frac{V}{2 ν} = l_{2} - l_{1}$

$\Rightarrow l_{2} = \frac{v}{2 ν} + l_{1} = \frac{336}{400} + 0.20 = 0.84 + 0.20 = 1.04 m = 104 c m$

0 0

New answer posted

4 months ago

Class 11th Physics Mechanical Properties of Fluids

A small spherical ball of radius 0.1mm and density 10⁴kgm^-³ falls freely under gravity through a distance h before entering a tank of water. If, after entering the water the velocity inside water, then the value of h will be_________ m.

(Given g = 10 ms^-², viscosity of water = 1.0 * 10^-⁵ N-sm^-²).

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

mg = F_B + F_v $\Rightarrow \frac{4 π}{3} r^{3} ρ g = \frac{4 π}{3} r^{3} σ g + 6 π η r v$

$\Rightarrow v = \frac{2 r^{2} (ρ - σ) g}{9 η} = \frac{2 * 0.1 * 0.1 * 1 0^{- 6} * (1 0^{4} - 1 0^{3}) * 10}{9 * 1.0 * 1 0^{- 5}}$

$\Rightarrow h = \frac{400}{2 g} = 20 m$

0 0

New answer posted

4 months ago

Class 11th Units and Measurement

The Vernier constant of Vernier calipers is 0.1 mm and it has zero error of (-0.05) cm. White measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The correct diameter will be__________ * 10^-2 cm.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

Least count of Vernier = 0.1mm

$\Rightarrow$ Reading of Vernier Scale = 5 * 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 * 102 cm.

0 0

New answer posted

4 months ago

Class 11th Physics Motion in Plane

A person can throw a ball upto a maximum range of 100m. How high above the ground he can the same ball?

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

According to question, we can write

$R = \frac{u^{2} s i n 2 θ}{g} \Rightarrow u^{2} = g R_{m a x}, w h e r e θ = 45 °$

$\Rightarrow H_{m a x} = \frac{u^{2}}{2 g} = \frac{g R_{m a x}}{2 g} = \frac{R_{m a x}}{2} = \frac{100}{2} = 50 m$

0 0

New answer posted

4 months ago

Class 11th Physics Laws of Motion

A block of mass M placed inside a box descends vertically with acceleration 'a'. The block exerts a force equal to one-fourth of its weight on the floor of the box. The value of 'a' will be

0 Follower 6 Views