Physics-
General
Easy

Question

A tube is open at both ends with the air oscillating in the 4th harmonic. How many displacement nodes are located within the tube?

  1. 2    
  2. 3    
  3. 4    
  4. 5    

The correct answer is: 4

Related Questions to study

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Heat of reaction at constant volume is measured in the apparatus:

Heat of reaction at constant volume is measured in the apparatus:

Chemistry-General
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The dissociation energy ofCH4 is 400 kcal mol and that of ethane is 670 kcal m or the C minus C bond energy is:

The dissociation energy ofCH4 is 400 kcal mol and that of ethane is 670 kcal m or the C minus C bond energy is:

Chemistry-General
General
Chemistry-

The bond dissociation energy of C-H in CH4 from the equation
C left parenthesis g right parenthesis plus 4 H left parenthesis g right parenthesis ⟶ C H subscript 4 end subscript left parenthesis g right parenthesis semicolon capital delta H equals negative 397.8Kcal is:

The bond dissociation energy of C-H in CH4 from the equation
C left parenthesis g right parenthesis plus 4 H left parenthesis g right parenthesis ⟶ C H subscript 4 end subscript left parenthesis g right parenthesis semicolon capital delta H equals negative 397.8Kcal is:

Chemistry-General
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General
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The heats of combustion of rhombic and monoclinic sulphur are -70960 and -71030 calorie respectively What will be the heat of conversion of rhombic sulphur to monoclinic sulphur?

The heats of combustion of rhombic and monoclinic sulphur are -70960 and -71030 calorie respectively What will be the heat of conversion of rhombic sulphur to monoclinic sulphur?

Chemistry-General
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Given, C left parenthesis s right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis ⟶ C O subscript 2 end subscript left parenthesis g right parenthesis semicolon capital delta H equals negative 395 KJ
S left parenthesis s right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis ⟶ S O subscript 2 end subscript left parenthesis g right parenthesis semicolon capital delta H equals negative 295 KJ
C S subscript 2 end subscript open parentheses l close parentheses plus 3 O subscript 2 end subscript open parentheses g close parentheses ⟶ C O subscript 2 end subscript open parentheses g close parentheses plus 2 S O subscript 2 end subscript open parentheses g close parentheses semicolon capital delta H equals negative 1110 KJ The heat of formation of C S subscript 2 end subscript left parenthesis l right parenthesis

Given, C left parenthesis s right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis ⟶ C O subscript 2 end subscript left parenthesis g right parenthesis semicolon capital delta H equals negative 395 KJ
S left parenthesis s right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis ⟶ S O subscript 2 end subscript left parenthesis g right parenthesis semicolon capital delta H equals negative 295 KJ
C S subscript 2 end subscript open parentheses l close parentheses plus 3 O subscript 2 end subscript open parentheses g close parentheses ⟶ C O subscript 2 end subscript open parentheses g close parentheses plus 2 S O subscript 2 end subscript open parentheses g close parentheses semicolon capital delta H equals negative 1110 KJ The heat of formation of C S subscript 2 end subscript left parenthesis l right parenthesis

Chemistry-General
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Chemistry-

Heat of combustion of CH4 , C2, C2H4 and C2H4 are -212.8, -373.0, -337.0 and -310.5 kcal respectively at the same temperature. The best fuel among these gases is:

Heat of combustion of CH4 , C2, C2H4 and C2H4 are -212.8, -373.0, -337.0 and -310.5 kcal respectively at the same temperature. The best fuel among these gases is:

Chemistry-General
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General
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Standard heat of formation for CCI4, H20, CO2 and HCI at 298K are - 25.5, - 57.8, 94.1 and 22.1 kJl mol respectively. For the reaction,
C C l subscript 4 end subscript plus 2 H subscript 2 end subscript O ⟶ C O subscript 2 end subscript plus 4 H C l what will be capital delta H

Standard heat of formation for CCI4, H20, CO2 and HCI at 298K are - 25.5, - 57.8, 94.1 and 22.1 kJl mol respectively. For the reaction,
C C l subscript 4 end subscript plus 2 H subscript 2 end subscript O ⟶ C O subscript 2 end subscript plus 4 H C l what will be capital delta H

Chemistry-General
General
Chemistry-

If capital delta H subscript f end subscript superscript ring operator end superscript for H subscript 2 end subscript O subscript 2 end subscript left parenthesis l right parenthesis and H subscript 2 end subscript O left parenthesis l right parenthesis subscript negative 188 end subscript -188 kJ mol-I and-286kJ mol , what will be the enthalpy change of the reaction2 H subscript 2 end subscript O subscript 2 end subscript left parenthesis l right parenthesis ⟶ 2 H subscript 2 end subscript O left parenthesis l right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis

If capital delta H subscript f end subscript superscript ring operator end superscript for H subscript 2 end subscript O subscript 2 end subscript left parenthesis l right parenthesis and H subscript 2 end subscript O left parenthesis l right parenthesis subscript negative 188 end subscript -188 kJ mol-I and-286kJ mol , what will be the enthalpy change of the reaction2 H subscript 2 end subscript O subscript 2 end subscript left parenthesis l right parenthesis ⟶ 2 H subscript 2 end subscript O left parenthesis l right parenthesis plus O subscript 2 end subscript left parenthesis g right parenthesis

Chemistry-General
General
Chemistry-

Which of the following units represents the largest amount of energy?

Which of the following units represents the largest amount of energy?

Chemistry-General
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General
Chemistry-

If S plus O subscript 2 end subscript ⟶ S O subscript 2 end subscript semicolon capital delta H equals negative 298.2 KJ
S O subscript 2 end subscript plus fraction numerator 1 over denominator 2 end fraction O subscript 2 end subscript ⟶ S O subscript 3 end subscript semicolon capital delta H equals negative 98.7 KJ
S O subscript 3 end subscript plus H subscript 2 end subscript O ⟶ H subscript 2 end subscript S O subscript 4 end subscript semicolon capital delta H equals negative 130.2 KJ
H subscript 2 end subscript plus fraction numerator 1 over denominator 2 end fraction O subscript 2 end subscript ⟶ H subscript 2 end subscript O semicolon capital delta H equals negative 227.3 KJ
the heat of foundation o f blank H subscript 2 end subscript S O subscript 4 end subscriptwill be:

If S plus O subscript 2 end subscript ⟶ S O subscript 2 end subscript semicolon capital delta H equals negative 298.2 KJ
S O subscript 2 end subscript plus fraction numerator 1 over denominator 2 end fraction O subscript 2 end subscript ⟶ S O subscript 3 end subscript semicolon capital delta H equals negative 98.7 KJ
S O subscript 3 end subscript plus H subscript 2 end subscript O ⟶ H subscript 2 end subscript S O subscript 4 end subscript semicolon capital delta H equals negative 130.2 KJ
H subscript 2 end subscript plus fraction numerator 1 over denominator 2 end fraction O subscript 2 end subscript ⟶ H subscript 2 end subscript O semicolon capital delta H equals negative 227.3 KJ
the heat of foundation o f blank H subscript 2 end subscript S O subscript 4 end subscriptwill be:

Chemistry-General
General
Maths-

ABCD is a parallelogram and P is the intersection of the diagonals. If O is any point then OA+OB+OC+OD =

ABCD is a parallelogram and P is the intersection of the diagonals. If O is any point then OA+OB+OC+OD =

Maths-General
General
Maths-

If the vectors 4 i with not stretchy bar on top minus 7 j with not stretchy bar on top minus 2 k with not stretchy bar on top comma i with not stretchy bar on top plus 5 j with not stretchy bar on top minus 3 k with not stretchy bar on top and 3 i with not stretchy bar on top minus lambda j with not stretchy bar on top plus k with not stretchy bar on top form a triangle then λ=

If the vectors 4 i with not stretchy bar on top minus 7 j with not stretchy bar on top minus 2 k with not stretchy bar on top comma i with not stretchy bar on top plus 5 j with not stretchy bar on top minus 3 k with not stretchy bar on top and 3 i with not stretchy bar on top minus lambda j with not stretchy bar on top plus k with not stretchy bar on top form a triangle then λ=

Maths-General
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General
Maths-

Let f left parenthesis t with not stretchy bar on top right parenthesis equals left square bracket t right square bracket i with not stretchy bar on top minus left parenthesis t minus left square bracket t right square bracket right parenthesis j with not stretchy bar on top plus left square bracket t plus 1 right square bracket k with not stretchy bar on top comma left square bracket. right square bracket is greatest integer function. If f open parentheses 5 over 4 close parentheses and i with not stretchy bar on top plus lambda j with not stretchy bar on top plus mu k with not stretchy bar on top are parallel vectors thenleft parenthesis lambda comma mu right parenthesis =

For such questions, we should know the concept of greatest integer number. We should also know the properties of parallel vectors.

Let f left parenthesis t with not stretchy bar on top right parenthesis equals left square bracket t right square bracket i with not stretchy bar on top minus left parenthesis t minus left square bracket t right square bracket right parenthesis j with not stretchy bar on top plus left square bracket t plus 1 right square bracket k with not stretchy bar on top comma left square bracket. right square bracket is greatest integer function. If f open parentheses 5 over 4 close parentheses and i with not stretchy bar on top plus lambda j with not stretchy bar on top plus mu k with not stretchy bar on top are parallel vectors thenleft parenthesis lambda comma mu right parenthesis =

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For such questions, we should know the concept of greatest integer number. We should also know the properties of parallel vectors.

General
Maths-

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For such questions, we should know how to find the vector joining two points. Also, we should know the condition for two vectors to be collinear.

The points with position vectors bar a plus bar b comma space bar a minus bar b and a with not stretchy bar on top plus lambda b with not stretchy bar on top are collinear for

Maths-General

For such questions, we should know how to find the vector joining two points. Also, we should know the condition for two vectors to be collinear.

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A= (1,1,1) B=(1,2,3) C=(2,-1,1) then the length of the internal bisector of angle A is

A= (1,1,1) B=(1,2,3) C=(2,-1,1) then the length of the internal bisector of angle A is

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