Relative density problems | Examples - Kisembo Academy

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In this video, we get to calculate relative density problems. For those who keep funding me via
Relative density calculation examples | solved problems - Kisembo Academy

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transcript;

we have quite a long question here I

think s we will have to first summarize

it it says that the block of mass 0.1

0:00:10.139,0:00:13.590
kilograms is suspended from a spring

0:00:11.849,0:00:15.059
balance when the block is immersed in

0:00:13.590,0:00:16.980
the water of density a thousand

0:00:15.059,0:00:18.480
kilograms per meter cubed the spring

balance reads zero point six three

0:00:18.480,0:00:22.890
Newtons when the block is immersed in a

0:00:20.850,0:00:25.230
liquid of unknown density the spring

0:00:22.890,0:00:28.470
balance reads zero point seven neutrons

0:00:25.230,0:00:32.219
the question is find that density of

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block the density of the liquid so we'll

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go ahead and first summarize this

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question so we are talking about a block

0:00:35.550,0:00:42.739
it is having a mass of 0.1 kilograms so

0:00:39.239,0:00:47.100
meaning mass of the block is going to be

0:00:42.739,0:00:49.020
0.1 kilograms and it is suspended from a

0:00:47.100,0:00:51.570
spring when the block is immersed in

0:00:49.020,0:00:55.610
water of density that so meaning that

0:00:51.570,0:01:00.210
the density of the water is a thousand

0:00:55.610,0:01:02.250
kilograms per meter cubed now

0:01:00.210,0:01:04.530
approaching is to say when the block is

0:01:02.250,0:01:07.979
immersed in water of density a thousand

0:01:04.530,0:01:08.939
kilograms the spring balance reads 0.63

0:01:07.979,0:01:12.720
Newtons

0:01:08.939,0:01:16.430
so when this window weight in water when

0:01:12.720,0:01:19.710
this is the weight when with this solid

0:01:16.430,0:01:22.439
or the block is immersed in water it's

0:01:19.710,0:01:24.570
going to weigh 0.63 newtons for stop

0:01:22.439,0:01:26.189
continue and say when the block is

0:01:24.570,0:01:28.590
immersed in the liquid of unknown

0:01:26.189,0:01:31.439
density so meaning that when it is in

0:01:28.590,0:01:33.720
within its weight in liquid whose

0:01:31.439,0:01:36.600
density is not known so meaning that the

0:01:33.720,0:01:38.579
density of the liquid is not known the

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weight of the liquid the question says

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the spring balance reads zero point

0:01:40.650,0:01:47.210
seven Newton so it's going to breed 0.7

0:01:43.369,0:01:53.250
Newton's find that density of the block

0:01:47.210,0:01:55.680
so we need the density of the block so

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the density of the block is also unknown

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and then the density of the liquid you

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so do not know the density of the liquid

0:02:00.229,0:02:04.110
from what the question has given us you

0:02:02.549,0:02:07.229
can see that we are having this is mass

0:02:04.110,0:02:08.849
of the block we have weight of when that

0:02:07.229,0:02:09.869
block is immersed in water we have the

0:02:08.849,0:02:11.849
weight of that blow

0:02:09.869,0:02:13.950
it's a must in the liquid and we have

0:02:11.849,0:02:15.690
the density of the liquid but we do best

0:02:13.950,0:02:17.879
your water we do not know the density of

0:02:15.690,0:02:19.620
the liquid and this ordinate may be this

0:02:17.879,0:02:22.560
chapter no you know what we need to find

0:02:19.620,0:02:24.300
these two out so what do we do here is

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simple first of all let's find the

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weight of the block we know that the

0:02:26.910,0:02:32.610
mass of the block is 0.1 kilograms so

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the weight of the block or call it the

0:02:32.610,0:02:37.580
weight of the block in air is going to

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be 0.1 which is the mass of the block

0:02:37.580,0:02:41.160
given here multiply that by the

0:02:39.720,0:02:44.220
acceleration due to gravity which is

0:02:41.160,0:02:53.880
9.81 and so the weight of the block is

0:02:44.220,0:02:55.590
going to become 0.98 1 Newton's help the

0:02:53.880,0:02:57.989
weight of the block we have the weight

0:02:55.590,0:02:59.640
when the block is immersed in water we

0:02:57.989,0:03:02.280
have the weight of the block when it's

0:02:59.640,0:03:04.739
immersed in the liquid whose density we

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do not know so let's go ahead and find

0:03:04.739,0:03:11.340
the relative density of the block so the

0:03:08.910,0:03:13.470
relative density of the block is going

0:03:11.340,0:03:15.120
to be given by from a previous session

0:03:13.470,0:03:17.940
we say that the relative density is

0:03:15.120,0:03:19.590
given by weight of a substance in air
0:03:17.940,0:03:21.269
divided by the apparent loss of weight in water and that is exactly what we are
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