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9 months ago

Class 11th Oscillations

A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period. 2 m T A g π ρ = where m is mass of the body and ρ is density of the liquid.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Let the log be passed and the vertical displacement at the vertical displacement at the equilibrium position

So mg= buoyant forces = $ρ A x_{o} g$

When it is displaced by further displacement x, the buoyant force is A (x_o+x) $ρ g$

Net restoring force = buoyant forces -weight

=A (x_o+x) $ρ g$ -mg

=A $ρ g x$

As displacement x is downward and restoring force is upward

F_restoring₌-A $ρ g x$ =-kx

So motion is SHM

Acceleration a=F_restoring/m=-kx/m

a=-w²x

w²=k/m

w= $\sqrt[]{\frac{k}{m}}$

T= 2 $π \sqrt[]{\frac{m}{k}} = 2 π \sqrt[]{\frac{m}{A ρ g}}$

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9 months ago

Class 11th Oscillations

A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4cm below the point, where it was held in hand. (a) What is the amplitude of oscillation? (b) Find the frequency of oscillation?

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) When the support of the hand is removed the body oscillates about mean position

Suppose x is the maximum extension in the spring when it reaches the lowest point in oscillation.

Loss in PE of the block=mgx

Gain in elastic potential energy =1/2 kx²

By energy conservation we cam say that

Mgx=1/2kx²

Or x= 2mg/k

Now the mean position of oscillation will be when the block is balanced by spring

If x' is the extension in that case

F= kx'

F=mg

Mg=kx'

X'=mg/k

By dividing x by x'

x/x'= $\frac{2 m g / k}{m g / k} = 2$

so x=2x'

x'=4/2 =2cm

but the displacement of mass from the mean position when spring attains its natural l

...more

This is a long answer type question as classified in NCERT Exemplar

(a) When the support of the hand is removed the body oscillates about mean position

Suppose x is the maximum extension in the spring when it reaches the lowest point in oscillation.

Loss in PE of the block=mgx

Gain in elastic potential energy =1/2 kx²

By energy conservation we cam say that

Mgx=1/2kx²

Or x= 2mg/k

Now the mean position of oscillation will be when the block is balanced by spring

If x' is the extension in that case

F= kx'

F=mg

Mg=kx'

X'=mg/k

By dividing x by x'

x/x'= $\frac{2 m g / k}{m g / k} = 2$

so x=2x'

x'=4/2 =2cm

but the displacement of mass from the mean position when spring attains its natural length is equal to amplitude of the oscillation

A=x'=2cm

(b) Time period of the oscillating system depends on mass spring constant given by

T= $2 π \sqrt[]{\frac{m}{k}}$

2mg/k=x

2mg/k=4cm =4 $* 10^{- 2} m$

m/k =4 $* 10^{- 2}$ /2g

k/m=g/2 $* 10^{- 2}$

$v = \frac{1}{T} \sqrt[]{\frac{k}{m}}$

$v = \frac{1}{2 * 3.14} \sqrt[]{\frac{g}{2 * 10^{- 2}}}$

= $\frac{10}{628} * 2.21$ =3.51Hz

less

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9 months ago

Class 11th Oscillations

A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s^-1 and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time.

(a) Will there be any change in weight of the body, during the oscillation?

(b) If answer to part (a) is yes, what will be the maximum and minimum reading in the machine and at which position?

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) The weight of the body changes during oscillations.

(b) Considering the situations in two extreme positions

we can say mg-N= ma

so at the highest point the platform is accelerating downward.

N=mg-ma

a=w²A

N=mg-mw²A

A= amplitude of motion m=50kg v=2m/s

w=2 $π v = 4 π r a d / s$

A= 5cm = 5 $* 10^{- 2} m$

N= 50 $* 9.8 - 50 * {(4 π)}^{2} * 5 * 10^{- 2} = 95.5 N$

When it is accelerating towards mean position that is vertically upwards

N-mg=ma=Mw²A

N=mg+mw²A

N=m (g+w²A)

N= 50 [9.8+ ${(4 π)}^{2} * 5 * 10^{- 2}$ ]

N= 884N

Machine reads the normal reaction

Maximum weight =884N

Minimum weight=95.5N

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Class 11th Chemical Bonding and Molecular Structure

How do NCERT Exemplar questions improve understanding of bonding concepts?

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Pallavi Pathak

Contributor-Level 10

The NCERT Exemplar includes analytical, application-based, and reasoning-type questions. It goes beyond the NCERT textbooks and helps students apply theories like VSEPR, hybridization, and MOT. It helps students to prepare for various entrance exams like NEET and JEE by strengthening their conceptual clarity.

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9 months ago

Class 11th Chemical Bonding and Molecular Structure

In chemical bonding, what is the significance of the hybridization?

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Pallavi Pathak

Contributor-Level 10

Hybridization predicts the geometry and bond angles in complex molecules and it helps in the understanding of how four equivalent bonds in methane (CH? ) can be formed from atoms like carbon. For observed molecular geometries, it explains the formation of the equivalent orbitals (like sp³, sp², sp).

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Class 11th Chemical Bonding and Molecular Structure

For predicting the molecular shapes, why is the VSEPR theory important?

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Pallavi Pathak

Contributor-Level 10

In molecular shape prediction, VSEPR (Valence Shell Electron Pair Repulsion) theory is important because based on the repulsions between the electron pairs around the central atom, it helps in predicting the three-dimensional shape of molecules. The 3D shape influences the molecule's chemical and physical properties including reactivity, polarity, and intermolecular forces.

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9 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

A sparingly soluble salt having general formula A^p+_x B^q- _y and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt.

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alok kumar singh

Contributor-Level 10

The equation can be written as-

A^p+_x+ B^q- _y? xA^p+ (aq) + yB^q-

S moles of A and B dissolves to give x S moles of A^p+and y S moles of Bq-.

K_sp = [A^p+]^x [B^q-]^y = [x5]^x [y5]^y

= x^xy^y5^x+y

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9 months ago

Class 11th Redox Reactions

40. Best reducing agent is:
(a) He (b) Be (c) Li (d) Al

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Vishal Baghel

Contributor-Level 10

Kindly go through the solution

(c) Li

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Class 11th Redox Reactions

37. Which element is considered to have oxidation number of +3 in all its compounds?

(a) Be (b) Li (c) Al (d) Sc

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Vishal Baghel

Contributor-Level 10

Kindly go through the solution

(c) Al

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Class 11th Redox Reactions

36. Oxidation number ofS in S₂O₈^-2is

(a) -2 (b) -6 (c) +2 (d) +6

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Vishal Baghel

Contributor-Level 10

Kindly go through the solution

(d) +6

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