The area bounded by y=cos x, y=x+1 and y=0 in the second quadra-Turito

General

Easy

Maths-

The area bounded by y=cos x, y=x+1 and y=0 in the second quadrant is

Maths-General

3/2 sq units
2 sq units
1 sq unit
1/2 sq units

Answer:The correct answer is: 1/2 sq units

Book A Free Demo

Name*

Mobile*

+91

Email*

Grade*

I agree to get WhatsApp notifications & Marketing updates

Related Questions to study

General

maths-

Area of the region bounded by y= $open vertical bar x close vertical bar blank$ and y=2 is

Area

blank equals fraction numerator 1 over denominator 2 end fraction.4.2

equals 4 blank

sQ. units

Area of the region bounded by y= $open vertical bar x close vertical bar blank$ and y=2 is

maths-General

Area

blank equals fraction numerator 1 over denominator 2 end fraction.4.2

equals 4 blank

sQ. units

General

maths-

Area of the region bounded by y $equals open vertical bar x close vertical bar blank a n d blank y equals 1 minus open vertical bar x close vertical bar blank$ is

maths-General

General

maths-

The area of the region bounded by y= $1 plus open vertical bar x close vertical bar$ and the x-axis is

maths-General

General

physics-

A solid sphere of radius $R$ has moment of inertia $I$ about its geometrical axis. If it is melted into a disc of radius $r$ and thickness $t$ . If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$ , then the value of $r$ is equal to

fraction numerator 2 over denominator 5 end fraction M R to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction M r to the power of 2 end exponent plus M r to the power of 2 end exponent

fraction numerator 2 over denominator 5 end fraction M R to the power of 2 end exponent equals fraction numerator 3 over denominator 2 end fraction M r to the power of 2 end exponent

therefore r equals fraction numerator 2 over denominator square root of 15 end fraction R

A solid sphere of radius $R$ has moment of inertia $I$ about its geometrical axis. If it is melted into a disc of radius $r$ and thickness $t$ . If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$ , then the value of $r$ is equal to

physics-General

fraction numerator 2 over denominator 5 end fraction M R to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction M r to the power of 2 end exponent plus M r to the power of 2 end exponent

fraction numerator 2 over denominator 5 end fraction M R to the power of 2 end exponent equals fraction numerator 3 over denominator 2 end fraction M r to the power of 2 end exponent

therefore r equals fraction numerator 2 over denominator square root of 15 end fraction R

General

maths-

The area of the elliptic quadratic with the semi major axis and semi minor axis as 6 and 4 respectively

maths-General

General

maths-

The area of the region bounded by y=|x-1| and y=1 in sq. units is

maths-General

General

physics-

A force of- F $stack k with hat on top$ acts on $O$ , the origin of the coordinate system. The torque about the point 1, -(a) is

tau equals r cross times F

tau equals open parentheses stack i with hat on top minus stack j with hat on top close parentheses cross times left parenthesis negative F stack k with hat on top right parenthesis

equals F left square bracket open parentheses negative stack i with hat on top cross times stack k with hat on top close parentheses plus open parentheses stack j with hat on top cross times stack k with hat on top close parentheses right square bracket

equals F left square bracket stack i with hat on top plus stack j with hat on top right square bracket

A force of- F $stack k with hat on top$ acts on $O$ , the origin of the coordinate system. The torque about the point 1, -(a) is

physics-General

tau equals r cross times F

tau equals open parentheses stack i with hat on top minus stack j with hat on top close parentheses cross times left parenthesis negative F stack k with hat on top right parenthesis

equals F left square bracket open parentheses negative stack i with hat on top cross times stack k with hat on top close parentheses plus open parentheses stack j with hat on top cross times stack k with hat on top close parentheses right square bracket

equals F left square bracket stack i with hat on top plus stack j with hat on top right square bracket

General

physics-

A $T$ joint is formed by two identical rods $A$ and $B$ each of mass $m$ and length $L$ in the $X Y$ plane as shown. Its moment of inertia about axis coinciding with $A$ is

I equals I subscript x end subscript plus I subscript y end subscript equals fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction equals fraction numerator m L to the power of 2 end exponent over denominator 6 end fraction

A $T$ joint is formed by two identical rods $A$ and $B$ each of mass $m$ and length $L$ in the $X Y$ plane as shown. Its moment of inertia about axis coinciding with $A$ is

physics-General

I equals I subscript x end subscript plus I subscript y end subscript equals fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m L to the power of 2 end exponent over denominator 12 end fraction equals fraction numerator m L to the power of 2 end exponent over denominator 6 end fraction

General

physics-

A binary star consists of two stars $A$ (mass 2.2 $M subscript s end subscript$ ) and B (mass 11 $M subscript s end subscript$ ), where $M subscript s end subscript$ is the mass of the sun. They are separated by distance $d$ and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star $B$ about the centre of mass is

fraction numerator L subscript T o t a l end subscript over denominator L subscript B end subscript end fraction equals fraction numerator left parenthesis I subscript A end subscript plus I subscript B end subscript right parenthesis omega over denominator I subscript B end subscript. omega end fraction

(as

omega

will be same in both cases)

equals fraction numerator I subscript A end subscript over denominator I subscript B end subscript end fraction plus 1 equals fraction numerator m subscript A end subscript r subscript A end subscript superscript 2 end superscript over denominator m subscript B end subscript r subscript B end subscript superscript 2 end superscript end fraction plus 1

equals fraction numerator r subscript A end subscript over denominator r subscript B end subscript end fraction plus 1

(as

m subscript A end subscript r subscript A end subscript equals m subscript B end subscript r subscript B end subscript right parenthesis

equals fraction numerator 11 over denominator 2.2 end fraction plus 1 left parenthesis a s r proportional to fraction numerator 1 over denominator m end fraction right parenthesis

equals 6

therefore

The correct answer is 6.

A binary star consists of two stars $A$ (mass 2.2 $M subscript s end subscript$ ) and B (mass 11 $M subscript s end subscript$ ), where $M subscript s end subscript$ is the mass of the sun. They are separated by distance $d$ and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star $B$ about the centre of mass is

physics-General

fraction numerator L subscript T o t a l end subscript over denominator L subscript B end subscript end fraction equals fraction numerator left parenthesis I subscript A end subscript plus I subscript B end subscript right parenthesis omega over denominator I subscript B end subscript. omega end fraction

(as

omega

will be same in both cases)

equals fraction numerator I subscript A end subscript over denominator I subscript B end subscript end fraction plus 1 equals fraction numerator m subscript A end subscript r subscript A end subscript superscript 2 end superscript over denominator m subscript B end subscript r subscript B end subscript superscript 2 end superscript end fraction plus 1

equals fraction numerator r subscript A end subscript over denominator r subscript B end subscript end fraction plus 1

(as

m subscript A end subscript r subscript A end subscript equals m subscript B end subscript r subscript B end subscript right parenthesis

equals fraction numerator 11 over denominator 2.2 end fraction plus 1 left parenthesis a s r proportional to fraction numerator 1 over denominator m end fraction right parenthesis

equals 6

therefore

The correct answer is 6.

General

physics-

Consider a body, shown in figure, consisting of two identical balls, each of mass $M$ connected by a light rigid rod. If an impulse $J equals M v$ is impared to the body at one of its ends, what would be its angular velocity?

Let

omega

be the angular velocity of the rod. Applying ,
Angular impulse = change in angular momentum about centre of mass of the system

J. fraction numerator L over denominator 2 end fraction equals I subscript c end subscript omega

therefore open parentheses M v close parentheses open parentheses fraction numerator L over denominator 2 end fraction close parentheses equals open parentheses 2 close parentheses open parentheses fraction numerator M L to the power of 2 end exponent over denominator 4 end fraction close parentheses. omega

therefore omega equals fraction numerator v over denominator L end fraction

Consider a body, shown in figure, consisting of two identical balls, each of mass $M$ connected by a light rigid rod. If an impulse $J equals M v$ is impared to the body at one of its ends, what would be its angular velocity?

physics-General

Let

omega

be the angular velocity of the rod. Applying ,
Angular impulse = change in angular momentum about centre of mass of the system

J. fraction numerator L over denominator 2 end fraction equals I subscript c end subscript omega

therefore open parentheses M v close parentheses open parentheses fraction numerator L over denominator 2 end fraction close parentheses equals open parentheses 2 close parentheses open parentheses fraction numerator M L to the power of 2 end exponent over denominator 4 end fraction close parentheses. omega

therefore omega equals fraction numerator v over denominator L end fraction

General

maths-

Assertion (A): If the system of equations $3 x minus 2 y plus z equals 0 comma lambda x minus 14 y plus 15 z equals 0 comma x plus 2 y minus 3 z equals 0$ have non zero solution then $lambda equals 5$
Reason (R): If the system of equations $A X equals 0$ has a non zero solution then $bold italic A$ is singular

maths-General

General

physics-

A particle of mass $m$ moves in the $X Y$ plane with a velocity $v$ along the straight line $A B$ . If the angular momentum of the particle with respect to origin $O$ is $L subscript A end subscript$ when it is at $A$ and $L subscript B end subscript$ when it is at $B$ , then

From the definition of angular momentum,

L equals r cross times p equals r m v sin invisible function application empty set left parenthesis negative stack k with hat on top right parenthesis

Therefore, the magnitude of

L

L equals m v r sin invisible function application empty set equals m v d

where

d equals r sin invisible function application empty set

is the distance of closest approach of the particle to the origin. As

d

is same for both the particles, hence

L subscript A end subscript equals L subscript B end subscript

A particle of mass $m$ moves in the $X Y$ plane with a velocity $v$ along the straight line $A B$ . If the angular momentum of the particle with respect to origin $O$ is $L subscript A end subscript$ when it is at $A$ and $L subscript B end subscript$ when it is at $B$ , then

physics-General

From the definition of angular momentum,

L equals r cross times p equals r m v sin invisible function application empty set left parenthesis negative stack k with hat on top right parenthesis

Therefore, the magnitude of

L

L equals m v r sin invisible function application empty set equals m v d

where

d equals r sin invisible function application empty set

is the distance of closest approach of the particle to the origin. As

d

is same for both the particles, hence

L subscript A end subscript equals L subscript B end subscript

General

physics-

Four balls each of radius 10 cm and mass 1 kg, 2kg, 3 kg and 4 kg are attached to the periphery of massless plate of radius 1 m. What is moment of inertia of the system about the centre of plate?

Moment of inertia of the system about the centre of plane is given by

I equals open square brackets fraction numerator 2 over denominator 5 end fraction cross times 1 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 1 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 2 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 2 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 3 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 3 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 4 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 4 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

equals 1.004 plus 2.008 plus 3.012 plus 4.016

equals 10.04 blank k g minus m to the power of 2 end exponent

Four balls each of radius 10 cm and mass 1 kg, 2kg, 3 kg and 4 kg are attached to the periphery of massless plate of radius 1 m. What is moment of inertia of the system about the centre of plate?

physics-General

Moment of inertia of the system about the centre of plane is given by

I equals open square brackets fraction numerator 2 over denominator 5 end fraction cross times 1 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 1 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 2 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 2 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 3 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 3 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

plus open square brackets fraction numerator 2 over denominator 5 end fraction cross times 4 cross times open parentheses 0.1 close parentheses to the power of 2 end exponent plus 4 cross times open parentheses 1 close parentheses to the power of 2 end exponent close square brackets

equals 1.004 plus 2.008 plus 3.012 plus 4.016

equals 10.04 blank k g minus m to the power of 2 end exponent

General

physics-

For the given uniform square lamina $A B C D$ , whose centre is $O$

Let the each side of square lamina is

d

.
So,

I subscript E F end subscript equals I subscript G H end subscript

(due to symmetry)
And

I subscript A C end subscript equals I subscript B D end subscript

(due to symmetry)
Now, according to theorem of perpendicular axis,

I subscript A C end subscript plus I subscript B D end subscript equals I subscript 0 end subscript

rightwards double arrow 2 I subscript A C end subscript equals I subscript 0 end subscript

(i)
and

I subscript E F end subscript plus I subscript G H end subscript equals I subscript 0 end subscript

rightwards double arrow 2 I subscript E F end subscript equals I subscript 0 end subscript

(ii)
From Eqs. (i) and (ii), we get

I subscript A C end subscript equals I subscript E F end subscript

therefore I subscript A D end subscript equals I subscript E F end subscript plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction

equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction open parentheses a s I subscript E F end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction close parentheses

So,

I subscript A D end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 3 end fraction equals 4 I subscript E F end subscript

For the given uniform square lamina $A B C D$ , whose centre is $O$

physics-General

Let the each side of square lamina is

d

.
So,

I subscript E F end subscript equals I subscript G H end subscript

(due to symmetry)
And

I subscript A C end subscript equals I subscript B D end subscript

(due to symmetry)
Now, according to theorem of perpendicular axis,

I subscript A C end subscript plus I subscript B D end subscript equals I subscript 0 end subscript

rightwards double arrow 2 I subscript A C end subscript equals I subscript 0 end subscript

(i)
and

I subscript E F end subscript plus I subscript G H end subscript equals I subscript 0 end subscript

rightwards double arrow 2 I subscript E F end subscript equals I subscript 0 end subscript

(ii)
From Eqs. (i) and (ii), we get

I subscript A C end subscript equals I subscript E F end subscript

therefore I subscript A D end subscript equals I subscript E F end subscript plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction

equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction open parentheses a s I subscript E F end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction close parentheses

So,

I subscript A D end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 3 end fraction equals 4 I subscript E F end subscript

General

physics-

A uniform rod $A B$ of length $l$ and mass $m$ is free to rotate about point $A$ . The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $fraction numerator m l to the power of 2 end exponent over denominator 3 end fraction$ , the initial angular acceleration of the rod will be

The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is

I equals fraction numerator m l to the power of 2 end exponent over denominator 3 end fraction

Where

m

is mass of rod and

l

is length.
Torque

left parenthesis tau equals I alpha right parenthesis

acting on centre of gravity of rod is given by

tau equals m g fraction numerator 1 over denominator 2 end fraction

I alpha equals m g fraction numerator 1 over denominator 2 end fraction

fraction numerator m l to the power of 2 end exponent over denominator 3 end fraction alpha equals m g fraction numerator 1 over denominator 2 end fraction

alpha equals fraction numerator 3 g over denominator 2 l end fraction

A uniform rod $A B$ of length $l$ and mass $m$ is free to rotate about point $A$ . The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $fraction numerator m l to the power of 2 end exponent over denominator 3 end fraction$ , the initial angular acceleration of the rod will be

physics-General

The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is

I equals fraction numerator m l to the power of 2 end exponent over denominator 3 end fraction

Where

m

is mass of rod and

l

is length.
Torque

left parenthesis tau equals I alpha right parenthesis

acting on centre of gravity of rod is given by

tau equals m g fraction numerator 1 over denominator 2 end fraction

I alpha equals m g fraction numerator 1 over denominator 2 end fraction

fraction numerator m l to the power of 2 end exponent over denominator 3 end fraction alpha equals m g fraction numerator 1 over denominator 2 end fraction

alpha equals fraction numerator 3 g over denominator 2 l end fraction

Have a query? Ask a Tutor!!

Can’t find what you’re looking for?

The area bounded by y=cos x, y=x+1 and y=0 in the second quadrant is

3/2 sq units
2 sq units
1 sq unit
1/2 sq units

Answer:The correct answer is: 1/2 sq units

Book A Free Demo

Related Questions to study

Area of the region bounded by y= $open vertical bar x close vertical bar blank$ and y=2 is

Area of the region bounded by y= $open vertical bar x close vertical bar blank$ and y=2 is

Area of the region bounded by y $equals open vertical bar x close vertical bar blank a n d blank y equals 1 minus open vertical bar x close vertical bar blank$ is

Area of the region bounded by y $equals open vertical bar x close vertical bar blank a n d blank y equals 1 minus open vertical bar x close vertical bar blank$ is

The area of the region bounded by y= $1 plus open vertical bar x close vertical bar$ and the x-axis is

The area of the region bounded by y= $1 plus open vertical bar x close vertical bar$ and the x-axis is

A solid sphere of radius $R$ has moment of inertia $I$ about its geometrical axis. If it is melted into a disc of radius $r$ and thickness $t$ . If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$ , then the value of $r$ is equal to

A solid sphere of radius $R$ has moment of inertia $I$ about its geometrical axis. If it is melted into a disc of radius $r$ and thickness $t$ . If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$ , then the value of $r$ is equal to

The area of the elliptic quadratic with the semi major axis and semi minor axis as 6 and 4 respectively

The area of the elliptic quadratic with the semi major axis and semi minor axis as 6 and 4 respectively

The area of the region bounded by y=|x-1| and y=1 in sq. units is

The area of the region bounded by y=|x-1| and y=1 in sq. units is

A force of- F $stack k with hat on top$ acts on $O$ , the origin of the coordinate system. The torque about the point 1, -(a) is

A force of- F $stack k with hat on top$ acts on $O$ , the origin of the coordinate system. The torque about the point 1, -(a) is

A $T$ joint is formed by two identical rods $A$ and $B$ each of mass $m$ and length $L$ in the $X Y$ plane as shown. Its moment of inertia about axis coinciding with $A$ is

A $T$ joint is formed by two identical rods $A$ and $B$ each of mass $m$ and length $L$ in the $X Y$ plane as shown. Its moment of inertia about axis coinciding with $A$ is

Consider a body, shown in figure, consisting of two identical balls, each of mass $M$ connected by a light rigid rod. If an impulse $J equals M v$ is impared to the body at one of its ends, what would be its angular velocity?

Consider a body, shown in figure, consisting of two identical balls, each of mass $M$ connected by a light rigid rod. If an impulse $J equals M v$ is impared to the body at one of its ends, what would be its angular velocity?

A particle of mass $m$ moves in the $X Y$ plane with a velocity $v$ along the straight line $A B$ . If the angular momentum of the particle with respect to origin $O$ is $L subscript A end subscript$ when it is at $A$ and $L subscript B end subscript$ when it is at $B$ , then

A particle of mass $m$ moves in the $X Y$ plane with a velocity $v$ along the straight line $A B$ . If the angular momentum of the particle with respect to origin $O$ is $L subscript A end subscript$ when it is at $A$ and $L subscript B end subscript$ when it is at $B$ , then

Four balls each of radius 10 cm and mass 1 kg, 2kg, 3 kg and 4 kg are attached to the periphery of massless plate of radius 1 m. What is moment of inertia of the system about the centre of plate?

Four balls each of radius 10 cm and mass 1 kg, 2kg, 3 kg and 4 kg are attached to the periphery of massless plate of radius 1 m. What is moment of inertia of the system about the centre of plate?

For the given uniform square lamina $A B C D$ , whose centre is $O$

For the given uniform square lamina $A B C D$ , whose centre is $O$

Courses

Quick Links

Follow Us at

Support

Download App

Courses

Quick Links

Have a query? Ask a Tutor!!

Can’t find what you’re looking for?

The area bounded by y=cos x, y=x+1 and y=0 in the second quadrant is

3/2 sq units 2 sq units 1 sq unit 1/2 sq units

Answer:The correct answer is: 1/2 sq units

Book A Free Demo

Related Questions to study

Area of the region bounded by y=and y=2 is

Area of the region bounded by y=and y=2 is

Area of the region bounded by y is

Area of the region bounded by y is

The area of the region bounded by y= and the x-axis is

The area of the region bounded by y= and the x-axis is

A solid sphere of radius has moment of inertia about its geometrical axis. If it is melted into a disc of radius and thickness . If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to , then the value of is equal to

A solid sphere of radius has moment of inertia about its geometrical axis. If it is melted into a disc of radius and thickness . If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to , then the value of is equal to

The area of the elliptic quadratic with the semi major axis and semi minor axis as 6 and 4 respectively

The area of the elliptic quadratic with the semi major axis and semi minor axis as 6 and 4 respectively

The area of the region bounded by y=|x-1| and y=1 in sq. units is

The area of the region bounded by y=|x-1| and y=1 in sq. units is

A force of- F acts on , the origin of the coordinate system. The torque about the point 1, -(a) is

A force of- F acts on , the origin of the coordinate system. The torque about the point 1, -(a) is

A joint is formed by two identical rods and each of mass and length in the plane as shown. Its moment of inertia about axis coinciding with is

A joint is formed by two identical rods and each of mass and length in the plane as shown. Its moment of inertia about axis coinciding with is

Consider a body, shown in figure, consisting of two identical balls, each of mass connected by a light rigid rod. If an impulse is impared to the body at one of its ends, what would be its angular velocity?

Consider a body, shown in figure, consisting of two identical balls, each of mass connected by a light rigid rod. If an impulse is impared to the body at one of its ends, what would be its angular velocity?

Assertion (A): If the system of equations have non zero solution then Reason (R): If the system of equations has a non zero solution then is singular

Assertion (A): If the system of equations have non zero solution then Reason (R): If the system of equations has a non zero solution then is singular

A particle of mass moves in the plane with a velocity along the straight line . If the angular momentum of the particle with respect to origin is when it is at and when it is at , then

A particle of mass moves in the plane with a velocity along the straight line . If the angular momentum of the particle with respect to origin is when it is at and when it is at , then

Four balls each of radius 10 cm and mass 1 kg, 2kg, 3 kg and 4 kg are attached to the periphery of massless plate of radius 1 m. What is moment of inertia of the system about the centre of plate?

Four balls each of radius 10 cm and mass 1 kg, 2kg, 3 kg and 4 kg are attached to the periphery of massless plate of radius 1 m. What is moment of inertia of the system about the centre of plate?

For the given uniform square lamina , whose centre is

For the given uniform square lamina , whose centre is

A uniform rod of length and mass is free to rotate about point . The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about is, the initial angular acceleration of the rod will be

A uniform rod of length and mass is free to rotate about point . The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about is, the initial angular acceleration of the rod will be

Courses

Quick Links

3/2 sq units
2 sq units
1 sq unit
1/2 sq units

Area of the region bounded by y= $open vertical bar x close vertical bar blank$ and y=2 is

Area of the region bounded by y= $open vertical bar x close vertical bar blank$ and y=2 is

Area of the region bounded by y $equals open vertical bar x close vertical bar blank a n d blank y equals 1 minus open vertical bar x close vertical bar blank$ is

Area of the region bounded by y $equals open vertical bar x close vertical bar blank a n d blank y equals 1 minus open vertical bar x close vertical bar blank$ is

The area of the region bounded by y= $1 plus open vertical bar x close vertical bar$ and the x-axis is

The area of the region bounded by y= $1 plus open vertical bar x close vertical bar$ and the x-axis is

A solid sphere of radius $R$ has moment of inertia $I$ about its geometrical axis. If it is melted into a disc of radius $r$ and thickness $t$ . If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$ , then the value of $r$ is equal to

A solid sphere of radius $R$ has moment of inertia $I$ about its geometrical axis. If it is melted into a disc of radius $r$ and thickness $t$ . If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$ , then the value of $r$ is equal to

A force of- F $stack k with hat on top$ acts on $O$ , the origin of the coordinate system. The torque about the point 1, -(a) is

A force of- F $stack k with hat on top$ acts on $O$ , the origin of the coordinate system. The torque about the point 1, -(a) is

A $T$ joint is formed by two identical rods $A$ and $B$ each of mass $m$ and length $L$ in the $X Y$ plane as shown. Its moment of inertia about axis coinciding with $A$ is

A $T$ joint is formed by two identical rods $A$ and $B$ each of mass $m$ and length $L$ in the $X Y$ plane as shown. Its moment of inertia about axis coinciding with $A$ is

Consider a body, shown in figure, consisting of two identical balls, each of mass $M$ connected by a light rigid rod. If an impulse $J equals M v$ is impared to the body at one of its ends, what would be its angular velocity?

Consider a body, shown in figure, consisting of two identical balls, each of mass $M$ connected by a light rigid rod. If an impulse $J equals M v$ is impared to the body at one of its ends, what would be its angular velocity?

A particle of mass $m$ moves in the $X Y$ plane with a velocity $v$ along the straight line $A B$ . If the angular momentum of the particle with respect to origin $O$ is $L subscript A end subscript$ when it is at $A$ and $L subscript B end subscript$ when it is at $B$ , then

A particle of mass $m$ moves in the $X Y$ plane with a velocity $v$ along the straight line $A B$ . If the angular momentum of the particle with respect to origin $O$ is $L subscript A end subscript$ when it is at $A$ and $L subscript B end subscript$ when it is at $B$ , then

For the given uniform square lamina $A B C D$ , whose centre is $O$

For the given uniform square lamina $A B C D$ , whose centre is $O$