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7 months ago

Class 11th Maths

$\underset{n \to \infty}{l i m} t a n {\sum_{r = 1}^{n} t a n^{- 1} (\frac{1}{1 + r + r^{2}})}$ is equal to___________.

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alok kumar singh

Contributor-Level 10

$\sum_{r = 1}^{n} t a n^{- 1} \frac{1}{1 + r + r^{2}} = \sum_{r = 1}^{n} t a n^{- 1} \frac{(r + 1)}{1 + r (r + 1)}$

$= t a n^{- 1} 2 - t a n^{- 1} + . . . . . . + t a n^{- 1} (n + 1) - t a n^{- 1} n$

For n $\infty$ value = $\frac{π}{2} - \frac{π}{4} = \frac{π}{4}$

$∴ t a n (\frac{π}{4}) = 1$

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7 months ago

Class 11th Physics

A large number of water drops, each of radius r, combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be

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Payal Gupta

Contributor-Level 10

From volume conservation

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

Decrease in surface area = $n * 4 π r^{2} - 4 π R^{2}$

$(Δ A) = 4 π [n r^{2} - R^{2}] = 4 π [\frac{n * r^{3}}{r} - R^{2}] = 4 π [\frac{R^{3}}{r} - R^{2}] = 4 π R^{3} [\frac{1}{r} - \frac{1}{R}]$

Energy released (W) = $T * Δ A = 4 π R^{3} T [\frac{1}{r} - \frac{1}{R}]$

Heat produced (Q) = $\frac{W}{J} = \frac{4 π T R^{3}}{J} [\frac{1}{r} - \frac{1}{R}]$

$N o w, Q = m s Δ θ$

$\Rightarrow \frac{4 π R^{3} T}{J} [\frac{1}{r} - \frac{1}{R}] = (\frac{4}{3} π R^{3}) * | x | * Δ θ = \frac{3 T}{J} [\frac{1}{r} - \frac{1}{R}]$

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7 months ago

Class 11th Physics

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R/2) from the earth's centre, where 'R' is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period

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Payal Gupta

Contributor-Level 10

F = m E cos θ

$= m (\frac{G M}{R^{3}} * r) * \frac{x}{r}$ $m a = \frac{m g}{R} * \Rightarrow a = \frac{g}{R} x$

$\Rightarrow T = 2 π \sqrt[]{\frac{R}{g}}$

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7 months ago

Class 11th Physics

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : Body 'P' having mass M moving with speed 'u' has head – on collision elastically with another body 'Q' having mass 'm' initially at rest. If m << M, Body 'Q' will have a maximum speed equal to '2u' after collision.

Reason R : During elastic collision, the momentum and kinetic energy are both conserved. In the light of the above statements, choose the most appropriate answer from the options given below

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Payal Gupta

Contributor-Level 10

Mass m will acquire velocity 2u. Total momentum of system will be conserved but total kinetic energy is not conserved during collision.

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7 months ago

Class 11th Physics Motion in Plane

A particle is moving with uniform speed along the circumference of a circle of radius of radius R under the action of a central fictitious force F which is inversely proportional to R³. Its time period of revolution will be given by

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Payal Gupta

Contributor-Level 10

$F = \frac{k}{R^{3}} = \frac{m v^{2}}{R} \Rightarrow v = \sqrt[]{\frac{k}{m R^{2}}} = \sqrt[]{\frac{k}{m}} * \frac{1}{R}$

$T = \frac{2 π R}{v} = \frac{2 π R}{\sqrt[]{\frac{k}{m} * (\frac{1}{R})}} \Rightarrow T α R^{2}$

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7 months ago

Class 11th Units and Measurement

In a typical combustion engine the work done by a gas molecule is given by W = $α^{2} β e^{- \frac{β x^{2}}{k T}},$ where x is the displacement, k is the Boltzmann constant and T is the temperature. If and are constants, dimensions of will be

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Payal Gupta

Contributor-Level 10

$W = α^{2} β e^{- \frac{- β x^{2}}{k T}}$

$[β x^{2}] = [k T] = [M L^{2} T^{- 2}]$

$β = M T^{- 2}$

$[W] = [α^{2} β] \Rightarrow [M L^{2} T^{- 2}] = [α^{2}] [M T^{- 2}] \Rightarrow [α^{2}] = [L^{2}] \Rightarrow [α] = [L]$

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7 months ago

Class 11th Physics

Find the gravitational force of attraction between the ring and sphere as shown in the diagram where the plane of the ring is perpendicular to the joining the centres. If $\sqrt[]{8} R$ is the distance between the centres of a ring (of mass 'm') and a sphere (mass 'M') where both have equal radius 'R'.

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Payal Gupta

Contributor-Level 10

Method – I:

$F = M_{s p h e r e} E_{r i n g} = M * \frac{G m x}{{(R^{2} + x^{2})}^{\frac{3}{2}}} = M * \frac{G m x \sqrt[]{8} R}{{(R^{2} + 8 R^{2})}^{\frac{3}{2}}} = \frac{\sqrt[]{8} G M m}{27 R^{2}}$

Method – II:

$F = \int d F c o s θ = c o s θ \int (d m) \frac{G M}{(R^{2} + 8 R^{2})} = \frac{\sqrt[]{8} G M m}{27 R^{2}}$

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7 months ago

Class 11th Chemistry

Is knowing just the atomic number or just the mass number enough to identify isotopes and isobars?

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Syed Aquib Ur Rahman

Contributor-Level 10

Correctly identifying isotopes and isobars requires knowing both the atomic and mass numbers. Relying on only one is a common error.

Isotopes: Same element (atomic number), different mass.
Isobars: Different elements (atomic number), same mass.

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7 months ago

Class 11th Chemistry

Why did Rutherford's Nuclear Model fail if it explained that atoms have a nucleus?

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Syed Aquib Ur Rahman

Contributor-Level 10

Rutherford's atomic model was a breakthrough, but it was flawed. It couldn't explain atomic stability, as orbiting electrons should lose energy and spiral into the nucleus. It also failed to account for the discrete line spectra observed from excited atoms.

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7 months ago

Class 11th Relations and Functions

The minimum values of a for which the equation $\frac{4}{s i n x} + \frac{1}{1 - s i n x} = α$ has at least one solution in $(0, \frac{π}{2}) i s_____.$

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alok kumar singh

Contributor-Level 10

$\frac{4}{s i n x} + \frac{1}{1 - s i n x} = a \forall x \in (0, \frac{π}{2})$

Let sin x = t, t $\in$ (0, 1)

$g (t) = \frac{4}{t} + \frac{1}{1 - t}$

$g' (t) = 0 \Rightarrow t = \frac{2}{3}$

$g'' (\frac{2}{3}) > 0$

$∴ g {(t)}_{m i n i m u m} = \frac{4}{2 / 3} + \frac{1}{1 - 2 / 3} = 9$

Minimum value of a for which solution exist = 9

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