find the relative error and percentage error of Length = 2± 0.04metre Mass = 4± 0.1 kg Radius= 3±0.03 m of a cylinder
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Answer:
( 50 ± 3.5) N
Force(f)=
r
mv
2
=
8
4×(10)
2
=50
∴
F
ΔF
=
m
Δm
+2(
V
ΔV
)+(
r
Δr
)
=
4
0.1
+2
10
(0.1)
+
8
0.8
∴
F
ΔF
=0.07
Percentage error in F=25%+2+2.5%
∴ Percentage error in F=7%
Eror in ΔF in F=F×0.07
=0×0.07
∴ Error in ΔF in F=3.5N
∴F=(50±3.5)N
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