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The parabolas y to the power of 2 end exponent equals 4 x comma x to the power of 2 end exponent equals 4 y divide the square region bounded by the lines x=4, y=4 and the co-ordinate axes. If S subscript 1 end subscript comma S subscript 2 end subscript comma S subscript 3 end subscript are respectively the areas of these parts numbered from top to bottom then S subscript 1 end subscript colon S subscript 2 end subscript colon S subscript 3 end subscript is

  1. 2:1:1    
  2. 1:1:1    
  3. 1:2:1    
  4. 1:2:3    

The correct answer is: 1:1:1

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Related Questions to study

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The figure shows a double slit experiment P and Q are the slits. The path lengths PX and QX are n lambda and left parenthesis n plus 2 right parenthesis lambda respectively, where n is a whole number and lambda is the wavelength. Taking the central fringe as zero, what is formed at X

For brightness, path difference equals n lambda equals 2 lambda
So second is bright.

The figure shows a double slit experiment P and Q are the slits. The path lengths PX and QX are n lambda and left parenthesis n plus 2 right parenthesis lambda respectively, where n is a whole number and lambda is the wavelength. Taking the central fringe as zero, what is formed at X

physics-General
For brightness, path difference equals n lambda equals 2 lambda
So second is bright.
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maths-

If gamma Sin space theta equals 3 comma gamma equals 4 left parenthesis 1 plus Sin space theta right parenthesis comma 0 less or equal than theta less or equal than 2 pi then theta = ---

If gamma Sin space theta equals 3 comma gamma equals 4 left parenthesis 1 plus Sin space theta right parenthesis comma 0 less or equal than theta less or equal than 2 pi then theta = ---

maths-General
General
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A particle starts from rest. Its acceleration open parentheses a close parentheses blank v e r s u s time open parentheses t close parentheses is as shown in the figure. The maximum speed of the particle will be

Area under acceleration-time graph gives the change in velocity. Hence,
v subscript m a x end subscript equals fraction numerator 1 over denominator 2 end fraction cross times 10 cross times 11 equals 55 blank m s to the power of negative 1 end exponent
Therefore, the correct option is [b].

A particle starts from rest. Its acceleration open parentheses a close parentheses blank v e r s u s time open parentheses t close parentheses is as shown in the figure. The maximum speed of the particle will be

physics-General
Area under acceleration-time graph gives the change in velocity. Hence,
v subscript m a x end subscript equals fraction numerator 1 over denominator 2 end fraction cross times 10 cross times 11 equals 55 blank m s to the power of negative 1 end exponent
Therefore, the correct option is [b].
General
physics-

In figure, one car at rest and velocity of the light from head light is c, tehn velocity of light from head light for the moving car at velocity v, would be

Since c greater than greater than v (negligible)

In figure, one car at rest and velocity of the light from head light is c, tehn velocity of light from head light for the moving car at velocity v, would be

physics-General
Since c greater than greater than v (negligible)
General
maths-

If α and β are two different solutions lying between fraction numerator negative pi over denominator 2 end fraction and  of the equation 2 Tan space theta plus Sec space theta equals 2 then Tan α + Tan β is

If α and β are two different solutions lying between fraction numerator negative pi over denominator 2 end fraction and  of the equation 2 Tan space theta plus Sec space theta equals 2 then Tan α + Tan β is

maths-General
General
physics-

A particle of mass m is initially situated at the point P inside a hemispherical surface of radius r as shown in figure. A horizontal acceleration of magnitudea subscript 0 end subscriptis suddenly produced on the particle in the horizontal direction. If gravitational acceleration is neglected, the time taken by particle to touch the sphere again is

Let the particle touches the sphere t the point A.
Let P A equals 1
therefore P B equals fraction numerator l over denominator 2 end fraction
In increment O P B comma cos invisible function application alpha equals fraction numerator P B over denominator r end fraction

therefore P B equals r cos invisible function application a
or fraction numerator l over denominator 2 end fraction equals r cos invisible function application a
therefore l equals 2 r cos invisible function application alpha
B u t blank l equals fraction numerator 1 over denominator 2 end fraction a subscript 0 end subscript t to the power of 2 end exponent
therefore blank t equals square root of open parentheses fraction numerator 2 l over denominator a subscript 0 end subscript end fraction close parentheses end root equals square root of open parentheses fraction numerator 2 cross times 2 r cos invisible function application a over denominator a subscript 0 end subscript end fraction close parentheses end root equals square root of open parentheses fraction numerator 4 blank r cos invisible function application a over denominator a subscript 0 end subscript end fraction close parentheses end root

A particle of mass m is initially situated at the point P inside a hemispherical surface of radius r as shown in figure. A horizontal acceleration of magnitudea subscript 0 end subscriptis suddenly produced on the particle in the horizontal direction. If gravitational acceleration is neglected, the time taken by particle to touch the sphere again is

physics-General
Let the particle touches the sphere t the point A.
Let P A equals 1
therefore P B equals fraction numerator l over denominator 2 end fraction
In increment O P B comma cos invisible function application alpha equals fraction numerator P B over denominator r end fraction

therefore P B equals r cos invisible function application a
or fraction numerator l over denominator 2 end fraction equals r cos invisible function application a
therefore l equals 2 r cos invisible function application alpha
B u t blank l equals fraction numerator 1 over denominator 2 end fraction a subscript 0 end subscript t to the power of 2 end exponent
therefore blank t equals square root of open parentheses fraction numerator 2 l over denominator a subscript 0 end subscript end fraction close parentheses end root equals square root of open parentheses fraction numerator 2 cross times 2 r cos invisible function application a over denominator a subscript 0 end subscript end fraction close parentheses end root equals square root of open parentheses fraction numerator 4 blank r cos invisible function application a over denominator a subscript 0 end subscript end fraction close parentheses end root
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physics-

A cyclist starts from the centreO of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference and returns to the point O as shown in figure. If the round trip takes 10 min, the net displacement and average speed of the cyclist (in metre and kilometer per hour) are

Since, the initial position of cyclist coincides with final position, so his net displacement is zero.
A v e r a g e blank s p e e d equals fraction numerator t o t a l blank d i s t a n c e blank t r a v e l l e d over denominator t o t a l blank t i m e blank t a k e n end fraction
equals fraction numerator O P plus P Q plus Q O over denominator 10 end fraction k m blank m i n to the power of negative 1 end exponent
equals fraction numerator 1 plus fraction numerator pi over denominator 2 end fraction cross times 1 plus 1 over denominator 10 end fraction k m blank m i n to the power of negative 1 end exponent
equals fraction numerator pi plus 4 over denominator 20 end fraction cross times 60 blank k m h to the power of negative 1 end exponent equals 21.4 blank k m h to the power of negative 1 end exponent

A cyclist starts from the centreO of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference and returns to the point O as shown in figure. If the round trip takes 10 min, the net displacement and average speed of the cyclist (in metre and kilometer per hour) are

physics-General
Since, the initial position of cyclist coincides with final position, so his net displacement is zero.
A v e r a g e blank s p e e d equals fraction numerator t o t a l blank d i s t a n c e blank t r a v e l l e d over denominator t o t a l blank t i m e blank t a k e n end fraction
equals fraction numerator O P plus P Q plus Q O over denominator 10 end fraction k m blank m i n to the power of negative 1 end exponent
equals fraction numerator 1 plus fraction numerator pi over denominator 2 end fraction cross times 1 plus 1 over denominator 10 end fraction k m blank m i n to the power of negative 1 end exponent
equals fraction numerator pi plus 4 over denominator 20 end fraction cross times 60 blank k m h to the power of negative 1 end exponent equals 21.4 blank k m h to the power of negative 1 end exponent
General
physics-

A the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed-time graphs for motor bike and car are represented by O A B and C D respectively Then

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s subscript b i k e end subscript equals s subscript 1 end subscript equals fraction numerator 1 over denominator 2 end fraction(18)(60)=540 m
Distance travelled by car at t equals 18s
s subscript c a r end subscript equals s subscript 2 end subscript=(18)(60)=720 m
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s subscript c a r end subscript minus s subscript b i k e end subscript equals 0
rightwards double arrow blank 720 plus 40 t equals 540 plus 60 t
rightwards double arrow blank t equals 9s beyond 18s or
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A the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed-time graphs for motor bike and car are represented by O A B and C D respectively Then

physics-General
Distance travelled by motor bike at t equals 18s
s subscript b i k e end subscript equals s subscript 1 end subscript equals fraction numerator 1 over denominator 2 end fraction(18)(60)=540 m
Distance travelled by car at t equals 18s
s subscript c a r end subscript equals s subscript 2 end subscript=(18)(60)=720 m
Therefore, separation between them at t equals 18s is 180m. Let, separation between them decreases to zero at time t beyond 18s.
Hence, s subscript b i k e end subscript equals 540 plus 60 t and s subscript c a r end subscript equals 720 plus 40 t
s subscript c a r end subscript minus s subscript b i k e end subscript equals 0
rightwards double arrow blank 720 plus 40 t equals 540 plus 60 t
rightwards double arrow blank t equals 9s beyond 18s or
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Assertion : Owls can move freely during night.
Reason : They have large number of rods on their retina.

Owls can move freely during night, because they have large number of cones on their retina which help them to see in night.

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Reason : They have large number of rods on their retina.

physics-General
Owls can move freely during night, because they have large number of cones on their retina which help them to see in night.
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physics-

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physics-General
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Maths-General
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Maths-

The area bounded by y equals x squared plus 2 comma X- axis, x=1 and x=2 is

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Maths-General
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maths-

If Cos space 2 theta times Cos space 3 theta times cos space theta equals 1 fourth text  for  end text 0 less than theta less than pi then theta =

If Cos space 2 theta times Cos space 3 theta times cos space theta equals 1 fourth text  for  end text 0 less than theta less than pi then theta =

maths-General
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physics-

The x minus t graph shown in the figure represents

Up to time t subscript 1 end subscript slope of the graph is constant and after t subscript 1 end subscriptslope is zero i. e. the body travel with constant speed up to time t subscript 1 end subscript and then stops

The x minus t graph shown in the figure represents

physics-General
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General
physics-

For the velocity-time graph shown in figure below the distance covered by the body in last two seconds of its motion is what fraction of the total distance covered by it in all the seven seconds

fraction numerator open parentheses S close parentheses subscript open parentheses l a s t blank 2 s close parentheses end subscript over denominator open parentheses S close parentheses subscript 7 s end subscript end fraction equals fraction numerator table row cell fraction numerator 1 over denominator 2 end fraction cross times 2 cross times 10 end cell row cell end cell end table over denominator table row cell fraction numerator 1 over denominator 2 end fraction cross times 2 cross times 10 plus 2 cross times 10 plus fraction numerator 1 over denominator 2 end fraction cross times 2 cross times 10 end cell row cell end cell row cell end cell end table end fraction equals fraction numerator 1 over denominator 4 end fraction

For the velocity-time graph shown in figure below the distance covered by the body in last two seconds of its motion is what fraction of the total distance covered by it in all the seven seconds

physics-General
fraction numerator open parentheses S close parentheses subscript open parentheses l a s t blank 2 s close parentheses end subscript over denominator open parentheses S close parentheses subscript 7 s end subscript end fraction equals fraction numerator table row cell fraction numerator 1 over denominator 2 end fraction cross times 2 cross times 10 end cell row cell end cell end table over denominator table row cell fraction numerator 1 over denominator 2 end fraction cross times 2 cross times 10 plus 2 cross times 10 plus fraction numerator 1 over denominator 2 end fraction cross times 2 cross times 10 end cell row cell end cell row cell end cell end table end fraction equals fraction numerator 1 over denominator 4 end fraction