A heavy mass is attached to a thin wire and is whirled in a ver-Turito

Physics-

General

Easy

Question

A heavy mass is attached to a thin wire and is whirled in a vertical circle. The wire is most likely to break

When the mass is at the highest point of the circle
When the mass is the lowest point of the circle
When the wire is horizontal
At an angle of $\cos^{- 1} (1 / 3)$ from the upward vertical

The correct answer is: When the mass is the lowest point of the circle

Because here tension is maximum

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Related Questions to study

General

physics-

Consider a vector $stack F with rightwards arrow on top equals 4 stack i with hat on top minus 3 stack j with hat on top$ . Another vector that is perpendicular to $stack F with rightwards arrow on top$ is

Since

stack F with rightwards arrow on top equals 4 stack i with hat on top minus 3 stack j with hat on top

is lying in

X minus Y

plane, hence the vector perpendicular to

stack F with rightwards arrow on top

must be lying perpendicular to

X minus Y

plane

i e comma

along

Z

-axis.

Consider a vector $stack F with rightwards arrow on top equals 4 stack i with hat on top minus 3 stack j with hat on top$ . Another vector that is perpendicular to $stack F with rightwards arrow on top$ is

physics-General

Since

stack F with rightwards arrow on top equals 4 stack i with hat on top minus 3 stack j with hat on top

is lying in

X minus Y

plane, hence the vector perpendicular to

stack F with rightwards arrow on top

must be lying perpendicular to

X minus Y

plane

i e comma

along

Z

-axis.

General

physics-

A circular road of radius $1000 blank m$ has banking angle $45 degree.$ The maximum safe speed of a car having mass $2000 blank k g$ will be, if the coefficient of friction between tyre and road is 0.5

The maximum velocity for a banked road with friction,

v to the power of 2 end exponent equals g r open parentheses fraction numerator u plus t a n theta over denominator 1 minus u t a n theta end fraction close parentheses

rightwards double arrow v to the power of 2 end exponent equals 9.8 cross times 1000 cross times open parentheses fraction numerator 0.5 plus 1 over denominator 1 minus 0.5 cross times 1 end fraction close parentheses rightwards double arrow v equals 172 m divided by s

A circular road of radius $1000 blank m$ has banking angle $45 degree.$ The maximum safe speed of a car having mass $2000 blank k g$ will be, if the coefficient of friction between tyre and road is 0.5

physics-General

The maximum velocity for a banked road with friction,

v to the power of 2 end exponent equals g r open parentheses fraction numerator u plus t a n theta over denominator 1 minus u t a n theta end fraction close parentheses

rightwards double arrow v to the power of 2 end exponent equals 9.8 cross times 1000 cross times open parentheses fraction numerator 0.5 plus 1 over denominator 1 minus 0.5 cross times 1 end fraction close parentheses rightwards double arrow v equals 172 m divided by s

General

chemistry-

NH₃ has a much higher boiling point than PH₃ because

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The maximum number of molecules that one water molecule can hold through hydrogen bonding is

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The wavelength of the wave function of a bonding molecular orbital formed by LCAO is

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In the formation of homo diatomic neutral molecule, if ‘N’ atomic orbitals combine, then the total number of bonding molecular orbitals formed is

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General

physics-

An object is moving in a circle of radius 100 m with a constant speed of $31.4 blank m s to the power of negative 1 end exponent$ . What is its average speed for one complete revolution?

The time taken by the particle for one complete revolution.

t equals fraction numerator 2 pi r over denominator s p e e d end fraction

equals fraction numerator 2 cross times 3.14 cross times 100 over denominator 31.4 end fraction equals 20 s

H e n c e comma blank a v e r g e blank s p e e d blank i s blank

v subscript a v end subscript equals fraction numerator 2 cross times 3.14 cross times 100 over denominator 20 end fraction equals 31.4 blank m s to the power of negative 1 end exponent

An object is moving in a circle of radius 100 m with a constant speed of $31.4 blank m s to the power of negative 1 end exponent$ . What is its average speed for one complete revolution?

physics-General

The time taken by the particle for one complete revolution.

t equals fraction numerator 2 pi r over denominator s p e e d end fraction

equals fraction numerator 2 cross times 3.14 cross times 100 over denominator 31.4 end fraction equals 20 s

H e n c e comma blank a v e r g e blank s p e e d blank i s blank

v subscript a v end subscript equals fraction numerator 2 cross times 3.14 cross times 100 over denominator 20 end fraction equals 31.4 blank m s to the power of negative 1 end exponent

General

physics-

The horizontal range of a projectile $4 square root of 3$ times the maximum height achieved by it, then the angle of projection is

fraction numerator u to the power of 2 end exponent sin invisible function application 2 theta over denominator g end fraction equals 4 square root of 3 cross times fraction numerator u to the power of 2 end exponent sin invisible function application theta over denominator 2 g end fraction

fraction numerator u to the power of 2 end exponent over denominator g end fraction 2 sin invisible function application theta cos invisible function application theta equals 2 square root of 3 fraction numerator u to the power of 2 end exponent over denominator g end fraction sin to the power of 2 end exponent invisible function application theta

The horizontal range of a projectile $4 square root of 3$ times the maximum height achieved by it, then the angle of projection is

physics-General

fraction numerator u to the power of 2 end exponent sin invisible function application 2 theta over denominator g end fraction equals 4 square root of 3 cross times fraction numerator u to the power of 2 end exponent sin invisible function application theta over denominator 2 g end fraction

fraction numerator u to the power of 2 end exponent over denominator g end fraction 2 sin invisible function application theta cos invisible function application theta equals 2 square root of 3 fraction numerator u to the power of 2 end exponent over denominator g end fraction sin to the power of 2 end exponent invisible function application theta

General

physics-

Two wires $A C$ and $B C$ are tied at $C$ of small sphere of mass 5 kg, which revolves at a constant speed $v$ in the horizontal circle of radius 1.6 m. The minimum value of $v$ is

From force diagram shown in figure

T subscript 1 end subscript cos invisible function application 30 degree plus T subscript 2 end subscript cos invisible function application 45 degree equals m g

(i)

T subscript 1 end subscript sin invisible function application 30 degree plus T subscript 2 end subscript sin invisible function application 45 degree equals fraction numerator m v to the power of 2 end exponent over denominator r end fraction

(ii)
After solving Eq. (i) and eq. (ii), we get

T subscript 1 end subscript equals fraction numerator m g minus fraction numerator m v to the power of 2 end exponent over denominator r end fraction over denominator open parentheses fraction numerator square root of 3 minus 1 over denominator 2 end fraction close parentheses end fraction

But

T subscript 1 end subscript greater than 0

therefore fraction numerator m g minus fraction numerator m v to the power of 2 end exponent over denominator r end fraction over denominator fraction numerator square root of 3 minus 1 over denominator 2 end fraction end fraction greater than 0

m g greater than fraction numerator m v to the power of 2 end exponent over denominator r end fraction

v less than square root of r g end root

therefore v subscript m a x end subscript equals square root of r g end root equals square root of 1.6 cross times 9.8 end root equals 3.96 blank m s to the power of negative 1 end exponent

Two wires $A C$ and $B C$ are tied at $C$ of small sphere of mass 5 kg, which revolves at a constant speed $v$ in the horizontal circle of radius 1.6 m. The minimum value of $v$ is

physics-General

From force diagram shown in figure

T subscript 1 end subscript cos invisible function application 30 degree plus T subscript 2 end subscript cos invisible function application 45 degree equals m g

(i)

T subscript 1 end subscript sin invisible function application 30 degree plus T subscript 2 end subscript sin invisible function application 45 degree equals fraction numerator m v to the power of 2 end exponent over denominator r end fraction

(ii)
After solving Eq. (i) and eq. (ii), we get

T subscript 1 end subscript equals fraction numerator m g minus fraction numerator m v to the power of 2 end exponent over denominator r end fraction over denominator open parentheses fraction numerator square root of 3 minus 1 over denominator 2 end fraction close parentheses end fraction

But

T subscript 1 end subscript greater than 0

therefore fraction numerator m g minus fraction numerator m v to the power of 2 end exponent over denominator r end fraction over denominator fraction numerator square root of 3 minus 1 over denominator 2 end fraction end fraction greater than 0

m g greater than fraction numerator m v to the power of 2 end exponent over denominator r end fraction

v less than square root of r g end root

therefore v subscript m a x end subscript equals square root of r g end root equals square root of 1.6 cross times 9.8 end root equals 3.96 blank m s to the power of negative 1 end exponent

General

physics-

Two stones are projected with the same velocity in magnitude but making different angles with the horizontal. Their ranges are equal. If the angel of projection of one is $pi divided by 3$ and its maximum height is $y subscript 1 end subscript$ , the maximum height of the other will be

Given

theta subscript 1 end subscript equals pi divided by 3 equals 30 degree

Horizontal range is same if

theta subscript 1 end subscript plus theta subscript 2 end subscript equals 90 degree

therefore theta subscript 2 end subscript equals 90 degree minus 30 degree equals 60 degree

y subscript 1 end subscript equals fraction numerator u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application 30 degree over denominator 2 g end fraction

and

y subscript 2 end subscript fraction numerator u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application 60 degree over denominator 2 g end fraction

therefore fraction numerator y subscript 2 end subscript over denominator y subscript 1 end subscript end fraction equals fraction numerator sin to the power of 2 end exponent invisible function application 30 degree over denominator sin to the power of 2 end exponent invisible function application 60 degree end fraction equals open parentheses fraction numerator 1 divided by 4 over denominator square root of 3 divided by 4 end fraction close parentheses to the power of 2 end exponent blank equals fraction numerator 1 over denominator 2 end fraction blank o r blank y subscript 2 end subscript equals fraction numerator y subscript 1 end subscript over denominator 3 end fraction

Two stones are projected with the same velocity in magnitude but making different angles with the horizontal. Their ranges are equal. If the angel of projection of one is $pi divided by 3$ and its maximum height is $y subscript 1 end subscript$ , the maximum height of the other will be

physics-General

Given

theta subscript 1 end subscript equals pi divided by 3 equals 30 degree

Horizontal range is same if

theta subscript 1 end subscript plus theta subscript 2 end subscript equals 90 degree

therefore theta subscript 2 end subscript equals 90 degree minus 30 degree equals 60 degree

y subscript 1 end subscript equals fraction numerator u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application 30 degree over denominator 2 g end fraction

and

y subscript 2 end subscript fraction numerator u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application 60 degree over denominator 2 g end fraction

therefore fraction numerator y subscript 2 end subscript over denominator y subscript 1 end subscript end fraction equals fraction numerator sin to the power of 2 end exponent invisible function application 30 degree over denominator sin to the power of 2 end exponent invisible function application 60 degree end fraction equals open parentheses fraction numerator 1 divided by 4 over denominator square root of 3 divided by 4 end fraction close parentheses to the power of 2 end exponent blank equals fraction numerator 1 over denominator 2 end fraction blank o r blank y subscript 2 end subscript equals fraction numerator y subscript 1 end subscript over denominator 3 end fraction

General

physics

Angle of projection, maximum height and time to reach the maximum height of a particle are $theta$ ,H and respectivley. Find the true relation.

physicsGeneral

General

physics-

A particle $A$ is projected from the ground with an initial velocity of $10 blank m s to the power of negative 1 end exponent$ at an angle of 60 $degree$ with horizontal. From what height $h$ should an another particle $B$ be projected horizontally with velocity $5 m s to the power of negative 1 end exponent$ so that both the particles collide in ground at point $C$ if both are projected simultaneously? $left parenthesis g equals 10 blank m s to the power of negative 2 end exponent right parenthesis$

Horizontal component of velocity of

A

is 10

c o s blank 6 0 degree

5 blank m s to the power of negative 1 end exponent

which is equal to the velocity of

B

in horizontal direction. They will collide at

C

if time of flight of the particles are equal or

t subscript A end subscript equals t subscript B end subscript

fraction numerator 2 u sin invisible function application theta over denominator g end fraction equals square root of fraction numerator 2 h over denominator g end fraction end root open parentheses therefore h equals fraction numerator 1 over denominator 2 end fraction g t subscript B end subscript superscript 2 end superscript close parentheses

h equals fraction numerator 2 u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application theta over denominator g end fraction

equals fraction numerator 2 open parentheses 10 close parentheses to the power of 2 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses to the power of 2 end exponent over denominator 10 end fraction equals 15 blank m

A particle $A$ is projected from the ground with an initial velocity of $10 blank m s to the power of negative 1 end exponent$ at an angle of 60 $degree$ with horizontal. From what height $h$ should an another particle $B$ be projected horizontally with velocity $5 m s to the power of negative 1 end exponent$ so that both the particles collide in ground at point $C$ if both are projected simultaneously? $left parenthesis g equals 10 blank m s to the power of negative 2 end exponent right parenthesis$

physics-General

Horizontal component of velocity of

A

is 10

c o s blank 6 0 degree

5 blank m s to the power of negative 1 end exponent

which is equal to the velocity of

B

in horizontal direction. They will collide at

C

if time of flight of the particles are equal or

t subscript A end subscript equals t subscript B end subscript

fraction numerator 2 u sin invisible function application theta over denominator g end fraction equals square root of fraction numerator 2 h over denominator g end fraction end root open parentheses therefore h equals fraction numerator 1 over denominator 2 end fraction g t subscript B end subscript superscript 2 end superscript close parentheses

h equals fraction numerator 2 u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application theta over denominator g end fraction

equals fraction numerator 2 open parentheses 10 close parentheses to the power of 2 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses to the power of 2 end exponent over denominator 10 end fraction equals 15 blank m

General

physics-

A particle is projected with a speed $v$ at $45 degree$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

Velocity of particle at maximum height

h

v to the power of ´ end exponent equals v cos invisible function application theta

where

v equals

initial velocity of particle at which it is projected,

theta equals

angle of projection
Angular momentum,

L equals m v to the power of ´ end exponent h equals m v cos invisible function application theta h

equals m v h cos invisible function application 45 degree equals fraction numerator m v h over denominator square root of 2 end fraction

A particle is projected with a speed $v$ at $45 degree$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

physics-General

Velocity of particle at maximum height

h

v to the power of ´ end exponent equals v cos invisible function application theta

where

v equals

initial velocity of particle at which it is projected,

theta equals

angle of projection
Angular momentum,

L equals m v to the power of ´ end exponent h equals m v cos invisible function application theta h

equals m v h cos invisible function application 45 degree equals fraction numerator m v h over denominator square root of 2 end fraction

General

physics-

A body of mass 1kg is rotating in a vertical circle of radius 1m. What will be the difference in its kinetic energy at the top and bottom of the circle? (Take $g equals 10 blank m s to the power of negative 2 end exponent$ )

Difference in

K E equals fraction numerator 1 over denominator 2 end fraction m open square brackets open parentheses square root of 5 g r end root close parentheses to the power of 2 end exponent minus square root of g r end root close square brackets to the power of 2 end exponent

equals 2 m g r equals 2 cross times 1 cross times 10 cross times 1 equals 20 blank

A body of mass 1kg is rotating in a vertical circle of radius 1m. What will be the difference in its kinetic energy at the top and bottom of the circle? (Take $g equals 10 blank m s to the power of negative 2 end exponent$ )

physics-General

Difference in

K E equals fraction numerator 1 over denominator 2 end fraction m open square brackets open parentheses square root of 5 g r end root close parentheses to the power of 2 end exponent minus square root of g r end root close square brackets to the power of 2 end exponent

equals 2 m g r equals 2 cross times 1 cross times 10 cross times 1 equals 20 blank

General

physics-

A particle of mass $m$ moves with constant speed along a circular path of radius $r$ under the action of force $F$ . Its speed is

F equals fraction numerator m v to the power of 2 end exponent over denominator r end fraction

v equals square root of fraction numerator F r over denominator m end fraction end root

A particle of mass $m$ moves with constant speed along a circular path of radius $r$ under the action of force $F$ . Its speed is

physics-General

F equals fraction numerator m v to the power of 2 end exponent over denominator r end fraction

v equals square root of fraction numerator F r over denominator m end fraction end root

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Question

A heavy mass is attached to a thin wire and is whirled in a vertical circle. The wire is most likely to break

When the mass is at the highest point of the circle When the mass is the lowest point of the circle When the wire is horizontal At an angle of from the upward vertical

Answer

The correct answer is: When the mass is the lowest point of the circle

Solution

Because here tension is maximum

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When the mass is at the highest point of the circle
When the mass is the lowest point of the circle
When the wire is horizontal
At an angle of $\cos^{- 1} (1 / 3)$ from the upward vertical

Consider a vector $stack F with rightwards arrow on top equals 4 stack i with hat on top minus 3 stack j with hat on top$ . Another vector that is perpendicular to $stack F with rightwards arrow on top$ is

Consider a vector $stack F with rightwards arrow on top equals 4 stack i with hat on top minus 3 stack j with hat on top$ . Another vector that is perpendicular to $stack F with rightwards arrow on top$ is

A circular road of radius $1000 blank m$ has banking angle $45 degree.$ The maximum safe speed of a car having mass $2000 blank k g$ will be, if the coefficient of friction between tyre and road is 0.5

A circular road of radius $1000 blank m$ has banking angle $45 degree.$ The maximum safe speed of a car having mass $2000 blank k g$ will be, if the coefficient of friction between tyre and road is 0.5

NH₃ has a much higher boiling point than PH₃ because

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An object is moving in a circle of radius 100 m with a constant speed of $31.4 blank m s to the power of negative 1 end exponent$ . What is its average speed for one complete revolution?

An object is moving in a circle of radius 100 m with a constant speed of $31.4 blank m s to the power of negative 1 end exponent$ . What is its average speed for one complete revolution?

The horizontal range of a projectile $4 square root of 3$ times the maximum height achieved by it, then the angle of projection is

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Two wires $A C$ and $B C$ are tied at $C$ of small sphere of mass 5 kg, which revolves at a constant speed $v$ in the horizontal circle of radius 1.6 m. The minimum value of $v$ is

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Angle of projection, maximum height and time to reach the maximum height of a particle are $theta$ ,H and respectivley. Find the true relation.

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A particle is projected with a speed $v$ at $45 degree$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

A particle is projected with a speed $v$ at $45 degree$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

A body of mass 1kg is rotating in a vertical circle of radius 1m. What will be the difference in its kinetic energy at the top and bottom of the circle? (Take $g equals 10 blank m s to the power of negative 2 end exponent$ )

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A particle of mass $m$ moves with constant speed along a circular path of radius $r$ under the action of force $F$ . Its speed is

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