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Sia ? 6 years, 5 months ago
Temperature at which a real gas obeys idea gas behaviour (law) over an appreciable range of pressure.
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Yogita Ingle 6 years, 5 months ago
and Q have different dimensions... so P - Q is not proper..
Perhaps PQ has the same dimensions as that of R.
PQ/R quantity is alright.
PR could probably have the same dimension as Q2
R + Q is not possible... as they have different dimensions.
Posted by Netik Verma 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work done on the body by the resultant force acting on that body. Conversely, a decrease in kinetic energy is caused by an equal amount of negative work done by the resultant force.
Posted by Netik Verma 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let m and r be the respective masses of the hollow cylinder and the solid sphere.
moment of inertia of the hollow cylinder I1 = mr2
The moment of inertia of the solid sphere I2 {tex}= \frac { 2 } { 5 } m r ^ { 2 }{/tex}
We have the relation:
{tex}\tau = I a{/tex}
For the hollow cylinder, {tex}\tau _ { 1 } = I _ { 1 } \alpha _ { 1 }{/tex}
For the solid sphere, {tex}\tau _ { 2 } = I _ { 2 } \alpha _ { 2 }{/tex}
As an equal torque is applied to both the bodies, {tex}\tau _ { 1 } = \tau _ { 2 }{/tex}
{tex}\therefore \frac {\alpha_ { 2 } } { \alpha _ { 1 } } = \frac { I _ { 1} } { I _ { 2 } } = \frac { M r ^ { 2 } } { \frac { 2 } { 5 } M r ^ { 2 } } = \frac { 2 } { 5 }{/tex}
{tex}a _ { 2 } > a _ { 1 }{/tex} ….(i)
Now, using the relation:
{tex}\omega = \omega _ { 0 } + a t{/tex}
{tex}\omega \propto a{/tex} …(ii)
From equations (i) and (ii), we can write:
{tex}\omega _ { 2 } > \omega _ { 1 }{/tex}
Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.
Posted by Netik Verma 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let us consider the earth to be spherical mass of ‘M’ and radius ‘R’. A body of mass ‘m’ is placed initially on the surface and finally taken x distance deep into the earth.
We know acceleration due to gravity on the surface of earth is
This expression shows that acceleration due to gravity decreases as we go deep into the earth.
At the centre of earth x=R, so g’=0. Hence, the acceleration due to gravity at the centre of earth is zero (0).
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