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Physics-

General

Easy

Question

A ball is suspended by a thread of length l at the point O on the wall, forming a small angle $proportional to$ with the vertical (figure) Then the thread with the ball was deviated through 2 $proportional to$ and set free. Assuming the collision of the ball against the wall to be perfectly elastic, find the oscillation period of such a pendulum.

$\frac{4}{3} π \sqrt{\frac{l}{g}}$
$\frac{2}{3} π \sqrt{\frac{l}{g}}$
$2 π \sqrt{\frac{l}{g}}$
None

The correct answer is: $fraction numerator 4 over denominator 3 end fraction pi square root of fraction numerator l over denominator g end fraction end root$

Time taken from equilibrium to $fraction numerator A over denominator 2 end fraction$ is $fraction numerator T over denominator 12 end fraction$
Total time = $fraction numerator T over denominator 12 end fraction plus fraction numerator T over denominator 12 end fraction plus fraction numerator T over denominator 4 end fraction plus fraction numerator T over denominator 4 end fraction equals fraction numerator 4 over denominator 6 end fraction T$
$therefore$ Time period = $fraction numerator 4 pi over denominator 3 end fraction square root of fraction numerator l over denominator g end fraction end root$

Related Questions to study

A mass m attached to a spring A has a frequency of 3 Hz and a spring B has a frequency of 4 Hz. When both the springs are connected as shown, find the frequency (in Hz) of oscillation.

A mass m attached to a spring A has a frequency of 3 Hz and a spring B has a frequency of 4 Hz. When both the springs are connected as shown, find the frequency (in Hz) of oscillation.

Physics-General

If we go deep into earth, time period of simple pendulum will:

If we go deep into earth, time period of simple pendulum will:

Physics-General

"The sky appears blue. This is due to atmospheric refraction.”
Identify whether the above statement is true or false.

"The sky appears blue. This is due to atmospheric refraction.”
Identify whether the above statement is true or false.

parallel

A car is moving with a speed of 100 kmh^-1 . if the mass of the car is 950 kg, then its kinetic energy is

A car is moving with a speed of 100 kmh^-1 . if the mass of the car is 950 kg, then its kinetic energy is

Physics-General

Acetic acid is a weak electrolyte because:

Acetic acid is a weak electrolyte because:

Chemistry-General

A body is projected horizontally from the top of a high tower with a speed of 20 m/s. After 4 sec the displacement of the body is (take g=10m/s²)

A body is projected horizontally from the top of a high tower with a speed of 20 m/s. After 4 sec the displacement of the body is (take g=10m/s²)

Physics-General

parallel

Consider the reaction: $blank C l subscript 2 end subscript left parenthesis a q right parenthesis plus H subscript 2 end subscript S left parenthesis a q right parenthesis ⟶ S left parenthesis s right parenthesis plus 2 H to the power of plus end exponent left parenthesis a q right parenthesis plus 2 C l to the power of minus end exponent left parenthesis a q right parenthesis$ The rate equation for this reaction is : $blank R a t e blank equals k open square brackets C l subscript 2 end subscript close square brackets open square brackets H subscript 2 end subscript S close square brackets$ Which of these mechanisms is/are consistent with this rate equation ?

Consider the reaction: $blank C l subscript 2 end subscript left parenthesis a q right parenthesis plus H subscript 2 end subscript S left parenthesis a q right parenthesis ⟶ S left parenthesis s right parenthesis plus 2 H to the power of plus end exponent left parenthesis a q right parenthesis plus 2 C l to the power of minus end exponent left parenthesis a q right parenthesis$ The rate equation for this reaction is : $blank R a t e blank equals k open square brackets C l subscript 2 end subscript close square brackets open square brackets H subscript 2 end subscript S close square brackets$ Which of these mechanisms is/are consistent with this rate equation ?

Chemistry-General

Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located at one point and moved with velocities V₁ = 1 m/s and V₂ = 4 m/s horizontally in opposite directions. The distance between the particles at the moment when their velocity vectors become mutually perpendicular is (g=10m/s²)

Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located at one point and moved with velocities V₁ = 1 m/s and V₂ = 4 m/s horizontally in opposite directions. The distance between the particles at the moment when their velocity vectors become mutually perpendicular is (g=10m/s²)

Physics-General

An endothermic reaction with high activation energy for the forward reaction is given by the diagram:

An endothermic reaction with high activation energy for the forward reaction is given by the diagram:

Chemistry-General

parallel

Graph between concentration of the product' x' and time' t' for $A rightwards arrow B$ is given ahead:
The graph between $negative fraction numerator d left square bracket A right square bracket over denominator d t end fraction$ and time wil1 be of the type:

Graph between concentration of the product' x' and time' t' for $A rightwards arrow B$ is given ahead:
The graph between $negative fraction numerator d left square bracket A right square bracket over denominator d t end fraction$ and time wil1 be of the type:

Chemistry-General

Following is the graph between $left parenthesis a minus x right parenthesis to the power of negative 1 end exponent$ and time t for second order reaction. $theta equals t a n to the power of negative 1 end exponent invisible function application left parenthesis 1 divided by 2 right parenthesis semicolon O A equals 2 L m o l to the power of negative 1 end exponent$ hence rate at the start of reaction will be:

Following is the graph between $left parenthesis a minus x right parenthesis to the power of negative 1 end exponent$ and time t for second order reaction. $theta equals t a n to the power of negative 1 end exponent invisible function application left parenthesis 1 divided by 2 right parenthesis semicolon O A equals 2 L m o l to the power of negative 1 end exponent$ hence rate at the start of reaction will be:

Chemistry-General

Following is the graph between log tll2 and log a (a = initial concentration) for a given
reaction at $27 to the power of ring operator end exponent$ e. Hence, order is:

Following is the graph between log tll2 and log a (a = initial concentration) for a given
reaction at $27 to the power of ring operator end exponent$ e. Hence, order is:

Chemistry-General

parallel

For the reaction $N H subscript 4 end subscript superscript plus end superscript plus O C N ⟶ N H subscript 2 end subscript C O N H subscript 2 end subscript$ the probable mechanism is,
$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell NH subscript 4 superscript plus plus OCN to the power of minus not stretchy rightwards harpoon over leftwards harpoon NH subscript 4 OCN end cell row cell text and end text space of 1em NH subscript 4 OCN not stretchy ⟶ NH subscript 2 CONH subscript 2 space of 1em text (slow) end text end cell end table$ The rate law will be:

For the reaction $N H subscript 4 end subscript superscript plus end superscript plus O C N ⟶ N H subscript 2 end subscript C O N H subscript 2 end subscript$ the probable mechanism is,
$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell NH subscript 4 superscript plus plus OCN to the power of minus not stretchy rightwards harpoon over leftwards harpoon NH subscript 4 OCN end cell row cell text and end text space of 1em NH subscript 4 OCN not stretchy ⟶ NH subscript 2 CONH subscript 2 space of 1em text (slow) end text end cell end table$ The rate law will be:

Chemistry-General

A drop of solution (volume 0.05 mL) contains $3 cross times 10 to the power of negative 6 end exponent$ mole of W. If the rate constant of disappearance of H+ is 107 mol $text litre end text to the power of negative 1 end exponent sec to the power of negative 1 end exponent$ , how long would it take for W in the drop tell disappear?

A drop of solution (volume 0.05 mL) contains $3 cross times 10 to the power of negative 6 end exponent$ mole of W. If the rate constant of disappearance of H+ is 107 mol $text litre end text to the power of negative 1 end exponent sec to the power of negative 1 end exponent$ , how long would it take for W in the drop tell disappear?

Chemistry-General

P₄ + 5O₂ ¾ → X $Error converting from MathML to accessible text.$ Y, Y is

P₄ + 5O₂ ¾ → X $Error converting from MathML to accessible text.$ Y, Y is

Chemistry-General

parallel

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