Question

During Searle’s experiment, initially, zero of vernier scale lies between  and  of the main scale. The 20th division of the vernier scale coincides with one of the main scale divisions. When an additional load of 2kg is applied to the wire, the zero of the vernier scale still lies between and  of the main scale but now, 45thdivision of vernier scale coincides with one of the main scale divisions. Least count of vernier scale is . Maximum percentage error in the measurement of deformation of the wire is:​

Answer 1

Answer:

Y=lAFL since the experiment measures only change in the length of wire

∴YΔY×100=lΔl×100

From the obervation: 

l1=MSR+20(LC) (MSR-Main Scale Reading)

l2=MSR+45(LC)

⇒ change in lengths =25(LC)

And the maximum permissible error in elongation is one LC.

∴YΔY×100=25(LC)(LC)×100=4