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.

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

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}

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.

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.

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
   = KE_rot + KE_trans
   =   {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}
   The time taken to return to the bottom is:
   {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.