In statement (b), the percentage error in case of 1m length is l1Δl1×100=10.01×100=1%
while in case of 0.5m length, it is l2Δl2×100=0.50.01×100=2%
As percentage error in case of 0.5m length is greater. So, the accuracy is less as compared to that of 1m length. Hence this statements is incorrect.
In statement (d), according to the result of rounding off, if the number to be rounded off is 5, then the preceding digit remains unchanged if it is even or increased by 1 , if it is odd. So, the correct result is 2.44. Hence, this is also an incorrect statement.
However, In statement (a), as the second measured value i.e., 5.51km is more closed to the result i.e., 5.678km so, it is more precise. Hence, this statement is correct.
Similarly, in the statement (c), the significant digits in first number is 2 and in second number is also 2 as the trailing zero after non-zero number, having decimal point are significant. Thus, this statement is also correct.