A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22 mm , 1.23 mm , 1.19 mm and 1.20 mm. The percentage error is ( x /121) %. The value of x is

Question

A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22 mm , 1.23 mm , 1.19 mm and 1.20 mm. The percentage error is ( x /121) %. The value of x is  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

X =(1.22 mm +1.23 mm +1.19 mm +1.20 mm /4) X =1.21 mm Δ x =(0.01+0.02+0.02+0.01/4)=(0.06/4)=0.015 Percentage error =(0.015/1.21) × 100 X =150

Q. A student in the laboratory measures thickness of a wire using screw gauge. The readings are $1.22 mm, 1.23 mm, 1.19 mm$ and $1.20 mm$ . The percentage error is $\frac{x}{121} %$ . The value of $x$ is _____

$X = \frac{1.22 mm + 1.23 mm + 1.19 mm + 1.20 mm}{4}$
$X = 1.21 mm$
$Δ x = \frac{0.01 + 0.02 + 0.02 + 0.01}{4} = \frac{0.06}{4} = 0.015$
Percentage error $= \frac{0.015}{1.21} \times 100$
$X = 150$

Q. A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22mm,1.23mm,1.19mm and 1.20mm. The percentage error is 121x​%. The value of x is _____

X=41.22mm+1.23mm+1.19mm+1.20mm​ X=1.21mm Δx=40.01+0.02+0.02+0.01​=40.06​=0.015 Percentage error =1.210.015​×100 X=150

Q. A student in the laboratory measures thickness of a wire using screw gauge. The readings are $1.22 mm, 1.23 mm, 1.19 mm$ and $1.20 mm$ . The percentage error is $\frac{x}{121} %$ . The value of $x$ is _____

$X = \frac{1.22 mm + 1.23 mm + 1.19 mm + 1.20 mm}{4}$
$X = 1.21 mm$
$Δ x = \frac{0.01 + 0.02 + 0.02 + 0.01}{4} = \frac{0.06}{4} = 0.015$
Percentage error $= \frac{0.015}{1.21} \times 100$
$X = 150$