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    General
    General
    Easy

    Question

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

    1. True
    2. False

    Hint:

    Blue is scattered more than other colors

    The correct answer is 'False'.

      Sky appears blue because of Scattering.Blue light is scattered in all directions by the tiny molecules of air in Earth's atmosphere. Blue is scattered more than other colors because it travels as shorter, smaller waves. This is why we see a blue sky most of the time.

      So False

      Related Questions to study

      General
      Physics-

      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

      1. 367 J    
      2. 3.7 J    
      3. 3.67 MJ    
      4. 0.367 MJ    

      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
      1. 367 J    
      2. 3.7 J    
      3. 3.67 MJ    
      4. 0.367 MJ    
      General
      Chemistry-

      Acetic acid is a weak electrolyte because:

      1. ltls11ignly unstable
            
      2. it does not dissociate much, or its ionisation is very small    
      3. its molecular mass is high    
      4. it in covalent compound    

      Acetic acid is a weak electrolyte because:

      Chemistry-General
      1. ltls11ignly unstable
            
      2. it does not dissociate much, or its ionisation is very small    
      3. its molecular mass is high    
      4. it in covalent compound    
      General
      Physics-

      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)

      1. 40 m    
      2. 80square root of 2m    
      3. 80m    
      4. 80/square root of 2 m    

      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
      1. 40 m    
      2. 80square root of 2m    
      3. 80m    
      4. 80/square root of 2 m    
      parallel
      General
      Chemistry-

      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 ?

      1. A only    
      2. neither A nor B    
      3. both A and B    
      4. B only    

      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
      1. A only    
      2. neither A nor B    
      3. both A and B    
      4. B only    
      General
      Physics-

      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)

      1. 2 m    
      2. 1    
      3. 3 m    
      4. 4 m    

      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
      1. 2 m    
      2. 1    
      3. 3 m    
      4. 4 m    
      General
      Chemistry-

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

      1.      
      2.      
      3.      
      4.      

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

      Chemistry-General
      1.      
      2.      
      3.      
      4.      
      parallel
      General
      Chemistry-

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

      1.      
      2.      
      3.      
      4.      

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

      Chemistry-General
      1.      
      2.      
      3.      
      4.      
      General
      Chemistry-

      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:

      1. 0.125 mol L to the power of negative 1 end exponent m i n to the power of negative 1 end exponent    
      2. 0.5 mol L to the power of negative 1 end exponent m i n to the power of negative 1 end exponent      
      3. 1.25 L m o l to the power of negative 1 end exponent m i n to the power of negative 1 end exponent    
      4. 1.25 mol L to the power of negative 1 end exponent m i n to the power of negative 1 end exponent    

      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
      1. 0.125 mol L to the power of negative 1 end exponent m i n to the power of negative 1 end exponent    
      2. 0.5 mol L to the power of negative 1 end exponent m i n to the power of negative 1 end exponent      
      3. 1.25 L m o l to the power of negative 1 end exponent m i n to the power of negative 1 end exponent    
      4. 1.25 mol L to the power of negative 1 end exponent m i n to the power of negative 1 end exponent    
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      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:

      1. 3    
      2. 0    
      3. 2    
      4. 1    

      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
      1. 3    
      2. 0    
      3. 2    
      4. 1    
      parallel
      General
      Chemistry-

      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:

      1. rate = k open square brackets N H subscript 4 end subscript O C N close square brackets    
      2. rate =k open square brackets N H subscript 2 end subscript C O N H subscript 2 end subscript close    
      3. none of these    
      4. rate = k open square brackets N H subscript 4 end subscript close square brackets to the power of plus end exponent left square bracket O C N right square bracket to the power of minus end exponent    

      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
      1. rate = k open square brackets N H subscript 4 end subscript O C N close square brackets    
      2. rate =k open square brackets N H subscript 2 end subscript C O N H subscript 2 end subscript close    
      3. none of these    
      4. rate = k open square brackets N H subscript 4 end subscript close square brackets to the power of plus end exponent left square bracket O C N right square bracket to the power of minus end exponent    
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      Chemistry-

      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 moltext  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?

      1. 6 cross times 10 to the power of negative 10 end exponentsec    
      2. 6 cross times 10 to the power of negative 12 end exponentsec    
      3. 6 cross times 10 to the power of negative 9 end exponentsec    
      4. 6 cross times 10 to the power of negative 8 end exponentsec    

      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 moltext  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
      1. 6 cross times 10 to the power of negative 10 end exponentsec    
      2. 6 cross times 10 to the power of negative 12 end exponentsec    
      3. 6 cross times 10 to the power of negative 9 end exponentsec    
      4. 6 cross times 10 to the power of negative 8 end exponentsec    
      General
      Chemistry-

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

      1. P2O5    
      2. H4P2O7    
      3. H3PO3    
      4. H3PO4    

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

      Chemistry-General
      1. P2O5    
      2. H4P2O7    
      3. H3PO3    
      4. H3PO4    
      parallel
      General
      Chemistry-

      Cu2+ + 2e– ¾® Cu; log[Cu2+] vs Ered graph is of the type shown in figure where OA = 0.34V, then electrode potential of the half cell of Cu/Cu2+ (0.1 M) will be

      1. 0.34 V    
      2. None    
      3. 0.34 + 0.059V    
      4. – 0.34 + fraction numerator 0.059 over denominator 2 end fractionV    

      Cu2+ + 2e– ¾® Cu; log[Cu2+] vs Ered graph is of the type shown in figure where OA = 0.34V, then electrode potential of the half cell of Cu/Cu2+ (0.1 M) will be

      Chemistry-General
      1. 0.34 V    
      2. None    
      3. 0.34 + 0.059V    
      4. – 0.34 + fraction numerator 0.059 over denominator 2 end fractionV    
      General
      Chemistry-

      Which statement is correct for alkaline earth metals

      1. They are diatomic and form ions of the type M2+    
      2. They are monoatomic and form ions of the type M2+    
      3. They are highly electronegative elements    
      4. They are diatomic and form ions of the type M2–    

      Which statement is correct for alkaline earth metals

      Chemistry-General
      1. They are diatomic and form ions of the type M2+    
      2. They are monoatomic and form ions of the type M2+    
      3. They are highly electronegative elements    
      4. They are diatomic and form ions of the type M2–    
      General
      Chemistry-

      The incorrect statement is

      1. Zn blende and iron pyrites are sulphates    
      2. Malachite and azurite are ores of Cu    
      3. Calamine and siderite are carbonates    
      4. Argentite and cuprites are oxides    

      The incorrect statement is

      Chemistry-General
      1. Zn blende and iron pyrites are sulphates    
      2. Malachite and azurite are ores of Cu    
      3. Calamine and siderite are carbonates    
      4. Argentite and cuprites are oxides    
      parallel

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