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A student obtained the value of Young's...

A student obtained the value of Young's modulus of steel as `1.9xx10^(11)N//m^(2)` while the standard value of Y is `2.1xx10^(11)N//m^(2)`. What is the percentage error in his measurement?

A

`8.5%`

B

`9.5%`

C

`9%`

D

`10%`

Text Solution

Verified by Experts

The correct Answer is:
b
Percentage error in the measurement of Y
`=("Standard value - Experimental value")/("Standard Value")xx100`
`=[((2.1-1.9)xx10^(11))/(2.1xx10^(11))]xx100`
`=((0.2)/(2.1))xx100=9.5%`
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