A particle moves along the sides ab,b,cd of a square of side 25-Turito

Physics-

General

Easy

Question

A particle moves along the sides $A B comma B C comma C D$ of a square of side $25 blank m$ wuth a velocity of $15 blank m s to the power of negative 1 end exponent$ . Its average velocity is

$15 m s^{- 1}$
$10 m s^{- 1}$
$7.5 m s^{- 1}$
$5 m s^{- 1}$

The correct answer is: $5 m s to the power of negative 1 end exponent$

Average velocity = $blank fraction numerator T o t a l blank D i s p l a c e m e n t blank over denominator T i m e blank t a k e n blank end fraction equals fraction numerator 25 over denominator 75 divided by 15 end fraction equals 5 m divided by s$

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Related Questions to study

General

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The velocity-time graph of a body moving in a straight line is shown in the figure. The displacement and distance travelled by the body in $6 blank s e c$ are respectively

Displacement = Summation of all the area with sign

equals open parentheses A subscript 1 end subscript close parentheses plus open parentheses negative A subscript 2 end subscript close parentheses plus open parentheses A subscript 3 end subscript close parentheses equals open parentheses 2 cross times 4 close parentheses plus open parentheses negative 2 cross times 2 close parentheses plus open parentheses 2 cross times 2 close parentheses

therefore D i s p l a c e m e n t equals 8 blank m

Distance = Summation of all the areas without sign

equals open vertical bar A subscript 1 end subscript close vertical bar plus open vertical bar negative A subscript 2 end subscript close vertical bar plus open vertical bar A subscript 3 end subscript close vertical bar equals open vertical bar 8 close vertical bar plus open vertical bar negative 4 close vertical bar plus open vertical bar 4 close vertical bar equals 8 plus 4 plus 4

therefore D i s t a n c e equals 16 blank m

The velocity-time graph of a body moving in a straight line is shown in the figure. The displacement and distance travelled by the body in $6 blank s e c$ are respectively

physics-General

Displacement = Summation of all the area with sign

equals open parentheses A subscript 1 end subscript close parentheses plus open parentheses negative A subscript 2 end subscript close parentheses plus open parentheses A subscript 3 end subscript close parentheses equals open parentheses 2 cross times 4 close parentheses plus open parentheses negative 2 cross times 2 close parentheses plus open parentheses 2 cross times 2 close parentheses

therefore D i s p l a c e m e n t equals 8 blank m

Distance = Summation of all the areas without sign

equals open vertical bar A subscript 1 end subscript close vertical bar plus open vertical bar negative A subscript 2 end subscript close vertical bar plus open vertical bar A subscript 3 end subscript close vertical bar equals open vertical bar 8 close vertical bar plus open vertical bar negative 4 close vertical bar plus open vertical bar 4 close vertical bar equals 8 plus 4 plus 4

therefore D i s t a n c e equals 16 blank m

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The area bounded by $y equals e to the power of x end exponent$ , Y-axis and the line y=e is

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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

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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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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 $theta$ = ---

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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].

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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)

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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

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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 magnitude $a subscript 0 end subscript$ is 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

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

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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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

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

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therefore l equals 2 r cos invisible function application alpha

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A cyclist starts from the centre $O$ 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 centre $O$ 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

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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

Distance travelled by motor bike at

t equals 18

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 18

s subscript c a r end subscript equals s subscript 2 end subscript

=(18)(60)=720 m
Therefore, separation between them at

t equals 18

s 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 9

s beyond 18s or
Hence,

t equals open parentheses 18 plus 9 close parentheses

s=27s from start and distant travelled by both is

s subscript b i k e end subscript

s subscript c a r end subscript equals 1080

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 18

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 18

s subscript c a r end subscript equals s subscript 2 end subscript

=(18)(60)=720 m
Therefore, separation between them at

t equals 18

s 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 9

s beyond 18s or
Hence,

t equals open parentheses 18 plus 9 close parentheses

s=27s from start and distant travelled by both is

s subscript b i k e end subscript

s subscript c a r end subscript equals 1080

General

physics-

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.

Assertion : Owls can move freely during night.
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.

General

physics-

A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point

Instantaneous velocity is given by the slope of the curve at that instant

v equals fraction numerator d s over denominator d t end fraction equals tan invisible function application theta

from the figure it is clear that slope of the curve is maximum at point ‘

C

’

A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point

physics-General

Instantaneous velocity is given by the slope of the curve at that instant

v equals fraction numerator d s over denominator d t end fraction equals tan invisible function application theta

from the figure it is clear that slope of the curve is maximum at point ‘

C

’

General

Maths-

The area bounded by y=3x and $y equals x to the power of 2 end exponent$ is

Maths-General

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Question

A particle moves along the sides of a square of side wuth a velocity of . Its average velocity is

Answer

The correct answer is:

Solution

Average velocity =

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A particle moves along the sides $A B comma B C comma C D$ of a square of side $25 blank m$ wuth a velocity of $15 blank m s to the power of negative 1 end exponent$ . Its average velocity is

The correct answer is: $5 m s to the power of negative 1 end exponent$

Average velocity = $blank fraction numerator T o t a l blank D i s p l a c e m e n t blank over denominator T i m e blank t a k e n blank end fraction equals fraction numerator 25 over denominator 75 divided by 15 end fraction equals 5 m divided by s$

The velocity-time graph of a body moving in a straight line is shown in the figure. The displacement and distance travelled by the body in $6 blank s e c$ are respectively

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The area bounded by $y equals e to the power of x end exponent$ , Y-axis and the line y=e is

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If $32 Tan to the power of 8 theta equals 2 Cos squared space alpha minus 3 cos space alpha text and end text 3 cos space 2 theta equals 1$ then the general value of 'α ' is

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