00:01
Hello and welcome to this video solution of numerate.
00:04
Here it's given that an acrylic beaker with coefficient of linear expansion, i'm writing alpha a, i'm writing okay, which is 6 .8 times of 10 to the power of minus 5 per degree celsius is fitted with mercury at 25 degrees celsius, right, thus having capacity of 500 ml.
00:26
So what we can do is v naught is 500 ml, i'm taking this is at t1 which is at 25 degrees celsius, right.
00:34
Now it is cooled down to 15 degrees celsius.
00:37
So t2 we have got that is equal to 15 degrees celsius.
00:46
Now you have to calculate the mass of mercury measured at 15 degrees celsius that must be added to make the beaker completely full, right, and it's given that the coefficient of linear expansion of mercury alpha m, i'm writing which is 1 .82 times of 10 to the power of minus 4 per degree celsius.
01:05
So initially what we have is this is filled up, right, this was completely filled up.
01:10
Now what we have is, now we can do one thing, we can calculate the change in volume delta v will be equal to v naught beta delta t, right.
01:22
So this is a formula, but here what happens there will be different expansions, right, for mercury and that of glass, right.
01:30
So what we can do is we can we have this alpha m that is the coefficient of linear expansion of mercury is more, right, than that of the glass acrylic, sorry acrylic beaker, right.
01:46
That means the mercury will get contract more faster, right.
01:50
So what i do is v naught vm delta t, this is a contraction in volume minus v naught beta a delta t, and beta is the coefficient of volumetric expansion, right.
02:06
Now we know as per empirical relationships, beta, the volumetric coefficient will be equal to thrice the alpha value, right.
02:18
So here we can do is v naught times of 3 of alpha m delta t, that is minus v naught times 3 alpha a delta t, right.
02:32
And what we have is equal to 3 v naught delta t alpha m minus of alpha a, right...