A particle starts from rest. its acceleration (a) versus time (-Turito

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Question

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

$110 {m s}^{- 1}$
$55 {m s}^{- 1}$
$550 {m s}^{- 1}$
$660 {m s}^{- 1}$

The correct answer is: $55 blank m s to the power of negative 1 end exponent$

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

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

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

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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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

’

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

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

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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 $theta$ =

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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 subscript

slope 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

Up to time

t subscript 1 end subscript

slope of the graph is constant and after

t subscript 1 end subscript

slope is zero

i. e.

the body travel with constant speed up to time

t subscript 1 end subscript

and then stops

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

General

physics-

A particle starts from rest at $t equals 0$ and undergoes an acceleration $a$ in $m s to the power of negative 2 end exponent$ with time $t$ in second which is as shownWhich one of the following plot represents velocity $v$ in $m s to the power of negative 1 end exponent blank v e r s u s$ time $t$ in second?

A particle starts from rest at

t equals 0

The equation of motion

v equals u plus a t equals 0 plus 3 cross times 2 equals 6 blank m s to the power of negative 1 end exponent

The velocity for next 2 s

v to the power of ´ end exponent equals v plus a t

blank equals 6 minus 3 cross times 2 equals 0

v to the power of ´ end exponent equals 0

Hence,

v minus t

graph will be as shown.

A particle starts from rest at $t equals 0$ and undergoes an acceleration $a$ in $m s to the power of negative 2 end exponent$ with time $t$ in second which is as shownWhich one of the following plot represents velocity $v$ in $m s to the power of negative 1 end exponent blank v e r s u s$ time $t$ in second?

physics-General

A particle starts from rest at

t equals 0

The equation of motion

v equals u plus a t equals 0 plus 3 cross times 2 equals 6 blank m s to the power of negative 1 end exponent

The velocity for next 2 s

v to the power of ´ end exponent equals v plus a t

blank equals 6 minus 3 cross times 2 equals 0

v to the power of ´ end exponent equals 0

Hence,

v minus t

graph will be as shown.

General

physics-

A body is at rest at $x equals 0$ . At $t equals 0$ , it starts moving in the positive $x$ -direction with a constant acceleration. At the same instant another body passes through $x equals 0$ moving in the positive $x$ -direction with a constant speed. The position of the first body is given by $x subscript 1 end subscript left parenthesis t right parenthesis$ after time ‘ $t$ ’ and that of the second body by $x subscript 2 end subscript left parenthesis t right parenthesis$ after the same time interval. Which of the following graphs correctly describes $left parenthesis x subscript 1 end subscript minus x subscript 2 end subscript right parenthesis$ as a function of time ‘ $t$ ’

x subscript 1 end subscript equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent

and

x subscript 2 end subscript equals u t blank therefore x subscript 1 end subscript minus x subscript 2 end subscript equals fraction numerator 1 over denominator 2 end fraction blank a t to the power of 2 end exponent minus u t

y equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent minus u t

.This equation is of parabola

fraction numerator d y over denominator d t end fraction equals a t minus u

and

fraction numerator d to the power of 2 end exponent y over denominator d t to the power of 2 end exponent end fraction equals a

fraction numerator d to the power of 2 end exponent y over denominator d t to the power of 2 end exponent end fraction greater than 0 i. e. comma

graph shows possess minima at

t equals fraction numerator u over denominator a end fraction

A body is at rest at $x equals 0$ . At $t equals 0$ , it starts moving in the positive $x$ -direction with a constant acceleration. At the same instant another body passes through $x equals 0$ moving in the positive $x$ -direction with a constant speed. The position of the first body is given by $x subscript 1 end subscript left parenthesis t right parenthesis$ after time ‘ $t$ ’ and that of the second body by $x subscript 2 end subscript left parenthesis t right parenthesis$ after the same time interval. Which of the following graphs correctly describes $left parenthesis x subscript 1 end subscript minus x subscript 2 end subscript right parenthesis$ as a function of time ‘ $t$ ’

physics-General

x subscript 1 end subscript equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent

and

x subscript 2 end subscript equals u t blank therefore x subscript 1 end subscript minus x subscript 2 end subscript equals fraction numerator 1 over denominator 2 end fraction blank a t to the power of 2 end exponent minus u t

y equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent minus u t

.This equation is of parabola

fraction numerator d y over denominator d t end fraction equals a t minus u

and

fraction numerator d to the power of 2 end exponent y over denominator d t to the power of 2 end exponent end fraction equals a

fraction numerator d to the power of 2 end exponent y over denominator d t to the power of 2 end exponent end fraction greater than 0 i. e. comma

graph shows possess minima at

t equals fraction numerator u over denominator a end fraction

General

maths-

General solution of $2 to the power of x equals n x end exponent plus 2 to the power of cos space x end exponent equals 2 to the power of 1 minus fraction numerator 1 over denominator square root of 2 end fraction end exponent$ is

maths-General

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Question

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

$110 {m s}^{- 1}$
$55 {m s}^{- 1}$
$550 {m s}^{- 1}$
$660 {m s}^{- 1}$

Answer

The correct answer is: $55 blank m s to the power of negative 1 end exponent$

Solution

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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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 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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Assertion : Owls can move freely during night.
Reason : They have large number of rods on their retina.

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

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The $x minus t$ graph shown in the figure represents

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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

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General solution of $2 to the power of x equals n x end exponent plus 2 to the power of cos space x end exponent equals 2 to the power of 1 minus fraction numerator 1 over denominator square root of 2 end fraction end exponent$ is

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Question

A particle starts from rest. Its acceleration time is as shown in the figure. The maximum speed of the particle will be

Answer

The correct answer is:

Solution

Area under acceleration-time graph gives the change in velocity. Hence,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 , tehn velocity of light from head light for the moving car at velocity , would be

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If α and β are two different solutions lying between and of the equation then Tan α + Tan β is

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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 and respectively Then

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A particle starts from rest at and undergoes an acceleration in with time in second which is as shownWhich one of the following plot represents velocity in time in second?

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General solution of is

General solution of is

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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

The correct answer is: $55 blank m s to the power of negative 1 end exponent$

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].

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

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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 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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Assertion : Owls can move freely during night.
Reason : They have large number of rods on their retina.

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

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

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

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The $x minus t$ graph shown in the figure represents

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General solution of $2 to the power of x equals n x end exponent plus 2 to the power of cos space x end exponent equals 2 to the power of 1 minus fraction numerator 1 over denominator square root of 2 end fraction end exponent$ is