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(B) To find the maximum length that can measure the given lengths exactly,we need to calculate the $HCF$ (Highest Common Factor) of the lengths.First,

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$25$

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(B) To find the maximum length that can measure the given lengths exactly,we need to calculate the $HCF$ (Highest Common Factor) of the lengths.First,convert all lengths into centimeters $(cm)$:$6 \, m = 600 \, cm$$4 \, m \, 75 \, cm = 475 \, cm$$10 \, m \, 25 \, cm = 1025 \, cm$Now,find the $HCF$ of $600$,$475$,and $1025$:Prime factorization of $600 = 2^3 	imes 3 	imes 5^2$Prime factorization of $475 = 5^2 	imes 19$Prime factorization of $1025 = 5^2 	imes 41$The common factor with the lowest exponent is $5^2 = 25$.Therefore,the maximum possible length is $25 \, cm$.

Find the maximum possible length which can be used to measure exactly the lengths $6 \, m$,$4 \, m \, 75 \, cm$,and $10 \, m \, 25 \, cm$. (in $, cm$)

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