Measure the length, breadth and thickness of a glass block using a metre rule (each reading correct to a mm), taking the mean of three readings in each case. Calculate the volume of the block in cm3 and m3 . Determine the mass (not weight) of the block using any convenient balance in g and kg. Calculate the density of glass in cgs and SI units using mass and volume in the respective units. Obtain the relation between the two density units
Answers
Answer:
0.48
Explanation:
Given that,
Length of the block l=(5.12+0.01)cm
Breadth of the block b=(10.15+0.01) cm
Thickness of the block t=(5.28+0.01) cm
We know that,
Relative error in volume of cuboid
V
dV
=
l
dl
+
b
db
+
t
dt
V
dV
=
5.12
0.01
+
10.15
0.01
+
5.28
0.01
V
dV
=
512
1
+
1015
1
+
528
1
Now, the percentage error in volume
V
ΔV
×100=0.48
%ΔV=0.48%
Hence, the percentage error in volume is 0.48.
Answer:
Given that,
Length of the block l=(5.12+0.01)cm
Breadth of the block b=(10.15+0.01) cm
Thickness of the block t=(5.28+0.01) cm
We know that,
Relative error in volume of cuboid
V
dV
=
l
dl
+
b
db
+
t
dt
V
dV
=
5.12
0.01
+
10.15
0.01
+
5.28
0.01
V
dV
=
512
1
+
1015
1
+
528
1
Now, the percentage error in volume
V
ΔV
×100=0.48
%ΔV=0.48%
Hence, the percentage error in volume is 0.48.