00:02
In the problem number 22, there is an object which is attached with a pulley through a string.
00:18
Now, let's assume that the moment of inertia of this pulley is i.
00:27
Now, coming to this mass m, the forces acting on this mass m will be m -g in downwards.
00:36
And let's suppose t is the tension in the string so now we can write for this mass m the forces acting on the mass m and balancing it we can write m g minus t equals to m a where a is the net acceleration of this object for pulley we can write i alpha equals to t into r where t is the tension in the string and r is the radius of this pulley so this is basically torque now this torque is equals to i alpha where i is the moment of inertia and alpha is the angular acceleration let's call this equation one and this equation two we can write here t equals to i alpha divide by r let's say this is equation one and this is equation two we can write here t equals to i alpha divide by r let's say this is number 3 now adding equation 1 and 3 we can write m g minus t plus t equals to m a plus i alpha divide by r so this this will cancel out it will be m g equals to m a plus i alpha divided by r now here alpha can angular resolution alpha can also be written as a that is a linear acceleration divided by radius r so we can write m g equals to m a plus i a divide by r square so from here we can write a equals to mg divided by m plus i divided by r square so we can get the value of acceleration after putting the values of mass m which is now here mass m is 0 .64 kg, that is the mass of the object and i is the moment of inertia which is equals to half mr squared.
02:55
After putting the values, we'll get 0 .11 kg, sorry, 0 .11 newton meter square or sorry, kg meter square.
03:05
That is, i will be equals to 0 .11 kg meter square.
03:13
Putting these all values will get a equals to 3 .6 meter per second square.
03:20
Now velocity at h equals to 6 .82 meter can be calculated as d square equals to u square plus 2 a h.
03:29
Now initial velocity is 0, so d will be equals to 2 into a into h.
03:35
Now putting the values will get 2 into 3 .6 into 6 .82.
03:41
The velocity at h equals to 6 .82 meter down will be equals to 7 meter per second and the angular velocity omega will be equals to b divided by r so that will be 7 divided by 0 .1 meter that will be equals to 70 radian per second so this will be the angular velocity and this will be the linear velocity so this is the answer for problem number 22.
04:18
For question number 23, let's assume that q is the amount of energy required to melt 2 .62 kg of ice at minus 10 degrees celsius to water at 0 degrees celsius.
04:35
So we can write q equals to m, where m is the mass of ice, into cp, that is the specific heat of the ice.
04:49
Now that is equals to 2 0 or simply 2 kilojoule per kj.
04:57
Per kelvin.
05:02
That is the specific heat of for ice into temperature difference that is equals to 10 degrees celsius.
05:12
Plus, now this is the amount of energy which will bring the ice from minus 10 degrees celsius.
05:20
To 0 degrees celsius into ice only.
05:25
Now that ice which is at 0 degrees celsius it will require latent heat of fusion or melting heat which is equal to 3 .35 into 10 to the power 5 joule per kg as given in the question.
05:43
So we can write m into l that l is equals to 3 .3...