A body weighing 10 N in air is partially immersed in water. It displaces water of mass 0.3 kg. What is the upthrust on the body? (g = 10 ms-2)
13.0 N
7.0 N
3.3N
3.0N
0.3 N.
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The answer is 'D'. The question is technical; from Archimedes principle it states that when a body is immersed in a fluid (gas/liquid), it experiences an UPTHRUST which is EQUAL to the WEIGHT of fluid DISPLACED i.e the weight of the fluid displaced is;from W=mg, W=0.3×10=3.0N. Thus, the correct answer is 3.0N, 'D'. Thanks
upthrust= pvg
but p=m/v
v=m/p
v=0.3/1000
=0.0003
Uw= pvg
= 1000 x 0.0003 x10
= 3N
the upthrust of an object is the weight of water it displaces. it displaces 0.3kg mass
so its weight is = mg =0.3*10 = 3
U=W-V, U=10-3, U=7. Therefore the answer is B NOTE* that 0.3 was converted to newton to make 3N
To find the upthrust on the body, we can use Archimedes' principle, which states that the buoyant force acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body.
Given:
Weight of the body in air,
W
=
10
N
W=10N
Mass of water displaced,
m
=
0.3
kg
m=0.3kg
Acceleration due to gravity,
g
=
10
m/s
2
g=10m/s
2
We know that the weight of the body in air is equal to the sum of its weight and the buoyant force acting on it when it is immersed in water.
So, buoyant force
=
Weight in air
−
Weight in water
=Weight in air−Weight in water
Buoyant force
=
W
−
m
g
=W−mg
Buoyant force
=
10
N
−
(
0.3
kg
×
10
m/s
2
)
=10N−(0.3kg×10m/s
2
)
Buoyant force
=
10
N
−
3
N
=10N−3N
Buoyant force
=
7
N
=7N
Therefore, the upthrust on the body is
7
N
7N.
A body weighing 10N in air is partially immersed in water. It displaces water of mass 0.3kg. What is the upthrust on the body?
