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Sakshi Masal 1 day, 19 hours ago

the reflecting and transmitting power of a perfectly black body is 1,as it reflects the same amount of energy it absorbed.

Sakshi Masal 1 day, 18 hours ago

as the body slow downs to zero at maximum point , the velocity at max point/height is 0.

Galaxy Threpy 2 days, 17 hours ago

30

Anshi Bidhuri 3 days, 17 hours ago

Joule(J)

Gopi Kishan 5 days, 5 hours ago

that energy can neither be created or not be destroyed

Mitali Sharma 6 days, 18 hours ago

that energy can neither be created or not be destroyed..

Shabya Shaikh 5 days, 9 hours ago

Suppose, body is initially at rest and force 'F' is applied on the body to displace it through 'ds' along it's own direction. Then small workdene, dw = f.ds dw = fds Acc. to Newton's 2nd law of motion f = ma Therefore, dw = ma ds dw = m dv/dt ds dw = m ds/dt dv dw = mvdv Integrating both sides In order to increase it's velocity from 0-v workdone is given by -- w = m sign of integration upper limit v and lower limit 0 vdv w = m[v square/2] upper limit v and lower limit 0 w = m[v square/2-0] w = 1/2mv square workdone is stored in form of energy Therefore, K.E. = 1/2mv square

Preeti Dabral 1 week, 6 days ago

Similarities: Both opposes relative motion. ... Viscous force depends on the velocity gradient and area of contact and frictional force independent of area of contact and relative velocity. Viscosity of liquid decrease with increase in temperature, Where as friction independent of temperature.

Preeti Dabral 1 week, 5 days ago

benzene contracts in winter. So 5 litre of benzene will weigh more in winter than in summer.

........ Gd.. 2 weeks, 1 day ago

20m height

Sanjay Gangwar 2 weeks, 1 day ago

No

Aryan Singh 3 weeks, 1 day ago

Sol or ans.

Ayaana Riyas 3 weeks, 4 days ago

Angle= Length of arc/ Radius of arc ​ π/6=x/31 x =(31×π​)\6 ​ =31×3.14/6 =16.22cm ​

Vansh Gangwar 1 week, 1 day ago

Velocity: Rate of change of displacement

Ravi Choudhary 3 weeks, 5 days ago

Velocity is the rate of change of displacement. Acceleration is the rate of change of velocity.velocity Is a vector physical quantity because it consist of both magnitude and direction.accerleration is also a vector physical quantity as it is just the rate of change of velocity

Preeti Dabral 2 weeks ago

1. As AO = BO = CO = 1 m, hence we have
{tex}\left| \overrightarrow { F _ { A } } \right| = \left| \overrightarrow { F _ { B } } \right| = \left| \overrightarrow { F _ { C } } \right| = \frac { G M \cdot 2 M } { ( 1 ) ^ { 2 } } = 2 G M ^ { 2 }{/tex}
If we consider direction parallel to BC as x-axis and perpendicular direction as y-axis, then as shown in figure, we have
{tex}\vec { F } _ { A } = 2 G M ^ { 2 } \hat { j }{/tex}
{tex}\vec { F } _ { B } = \left( - 2 G M ^ { 2 } \cos 30 ^ { \circ } \hat { i } - 2 G M ^ { 2 } \sin 30 ^ { \circ } \hat { j } \right){/tex}
and {tex}\overrightarrow { F _ { C } } = \left( 2 G M ^ { 2 } \cos 30 ^ { \circ } \hat { i } - 2 G M ^ { 2 } \sin 30 ^ { \circ } \hat { j } \right){/tex}
Therefore, the net force on mass 2M placed at the centroid O is given by,
{tex}\vec { F } = \vec { F } _ { A } + \vec { F } _ { B } + \vec { F } _ { C } {/tex}
{tex}= 2 G M ^ { 2 } \left[ j + \left( - \frac { \sqrt { 3 } } { 2 } \hat { i } - \frac { 1 } { 2 } \hat { i } \right) + \left( \frac { \sqrt { 3 } } { 2 } \hat { i } - \frac { 1 } { 2 } \hat { j } \right) \right]{/tex}
{tex}=0{/tex}

Preeti Dabral 2 weeks ago

The work obtained in bringing a body from infinity to a point in gravitational field is called gravitational potential energy. Force of attraction between the earth and the object when an object is at the distance a from the center of the earth. F=x2GmM.

Preeti Dabral 2 weeks ago

As the depth increased the mass of the earth decreases. At the surface of earth this value will be maximum because radius will be maximum. When radius becomes less this value also decreases. Hence acceleration due to gravity decreases with increase in the depth.

Preeti Dabral 2 weeks ago

A solid cylinder rolling up an inclination is shown in the following figure.

Initial velocity of the solid cylinder, v = 5 m/s
Angle of inclination, {tex}\theta = 30 ^ { \circ }{/tex}
Height reached by the cylinder = h

1. Energy of the cylinder at point A will be purely kinetic due to the rotation and translational motion. Hence, total energy at A
= KErot + KEtrans
=   {tex}\frac { 1 } { 2 } I \omega ^ { 2 } + \frac { 1 } { 2 } m v ^ { 2 }{/tex}

The energy of the cylinder at point B will be purely in the form of gravitational potential energy  = mgh
Using the law of conservation of energy, we can write:
{tex}\frac { 1 } { 2 } I \omega ^ { 2 } +\frac { 1 } { 2 } m v ^ { 2 } = m g h{/tex}
Moment of inertia of the solid cylinder, {tex}I = \frac { 1 } { 2 } m r ^ { 2 }{/tex}
{tex}\therefore \frac { 1 } { 2 } \left( \frac { 1 } { 2 } m r ^ { 2 } \right) \omega ^ { 2 } + \frac { 1 } { 2 } m v ^ { 2 } = m g h{/tex}
{tex}\frac { 1 } { 4 } I \omega ^ { 2 } + \frac { 1 } { 2 } m v ^ { 2 } = m g h{/tex}
But we have the relation, {tex}v = r \omega{/tex}
{tex}\therefore \frac { 1 } { 4 } v ^ { 2 } + \frac { 1 } { 2 } v ^ { 2 } = g h{/tex}
{tex}\frac { 3 } { 4 } v ^ { 2 } = g h{/tex}
{tex}\therefore h = \frac { 3 } { 4 } \frac { v ^ { 2 } } { g }{/tex}
{tex}= \frac { 3 } { 4 } \times \frac { 5 \times 5 } { 9.8 } = 1.91 \mathrm { m }{/tex}

To find the distance covered along the inclined plane
In {tex}\Delta A B C{/tex}:
{tex}\sin \theta = \frac { B C } { A B }{/tex}
{tex}\sin 30 ^ { \circ } = \frac { h } { A B }{/tex}
{tex}A B = \frac { 1.91 } { 0.5 } = 3.82 \mathrm { m }{/tex}
Hence, the cylinder will travel 3.82 m up the inclined plane.

2. {tex}v = \left( \frac { 2 g h } { 1 + \frac { K ^ { 2 } } { R ^ { 2 } } } \right) ^ { \frac { 1 } { 2 } }{/tex}
{tex}\therefore v = \left( \frac { 2 g A B \sin \theta } { 1 + \frac { K ^ { 2 } } { R ^ { 2 } } } \right) ^ { \frac { 1 } { 2 } }{/tex}
For the solid cylinder, {tex}K ^ { 2 } = \frac { R ^ { 2 } } { 2 }{/tex}
{tex}\therefore v = \left( \frac { 2 g A B \sin \theta } { 1 + \frac { 1 } { 2 } } \right) ^ { \frac { 1 } { 2 } }{/tex}
{tex}= \left( \frac { 4 } { 3 } g A B \sin \theta \right) ^ { \frac { 1 } { 2 } }{/tex}
{tex}t = \frac { A B } { v }{/tex}
{tex}= \frac { A B } { \left( \frac { 4 } { 3 } g A B \sin \theta \right) ^ { \frac { 1 } { 2 } } } = \left( \frac { 3 A B } { 4 g \sin \theta } \right) ^ { \frac { 1 } { 2 } }{/tex}
{tex}= \left( \frac { 11.46 } { 19.6 } \right) ^ { \frac { 1 } { 2 } } = 0.7645{/tex}
So the total time taken by the cylinder to return to the bottom is (2 {tex}\times{/tex} 0.764)= 1.53 s.as time of ascend is equal to time of descend for the following problem.

ML^2T^-2
15 m s-1

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