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# Assertion\$1{\mkern 1mu} m{m^2} = {10^{ - 6}}{\mkern 1mu} {m^2}\$ReasonThe numerical value of physical quantity is less for a smaller unit(A) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion(B) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion(C) Assertion is correct but Reason is incorrect(D) Assertion is incorrect but Reason is correct

Last updated date: 21st Apr 2024
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Hint In such problems, we will read the Assertion and the Reason separately, and after that evaluate each of them independently. In this case, the Assertion comes out to be true since the conversion given above, i.e., \$1{\mkern 1mu} m{m^2} = {10^{ - 6}}{\mkern 1mu} {m^2}\$ is true, where the Reason contradicts the Assertion itself.
Formula used: Conversion formula of \$1mm = {10^{ - 3}}m\$ is required.

Complete Step by step answer First we will consider the Assertion as given in the question above and evaluate whether it is true or not. We know that \$1mm = {10^{ - 3}}m\$.
Therefore, squaring both sides, we get
\$1m{m^2} = {(1mm)^2} = {({10^{ - 3}}m)^2} = {10^{ - 6}}{m^2}\$.
Therefore, it is evident that the Assertion is true.
Next, we will consider the Reason given in the question. The Reason in itself is false since it can be checked that the numerical value of the above given physical quantity which is \$m\$ (meters), is less for a larger unit of that same physical quantity. This is because of the general logical explanation that the lesser unit will need more of itself to represent a given quantity whereas the bigger unit will require less of itself when required to represent the same quantity.

Therefore, since the Assertion is True but the Reason is false, the correct answer is option (C) Assertion is correct but Reason is incorrect.

Note In Assertion and Reasoning, sometimes we have to make it clear why the Reasoning is false, and it would be necessary then to include certain examples. An example of such a problem would be \$1.5{\mkern 1mu} foot = 18{\mkern 1mu} inches\$ . Here \$foot\$ being the larger physical quantity \$(1{\mkern 1mu} foot = 12{\mkern 1mu} inches)\$ has a lesser numerical value than \$inches\$ , which is a smaller physical quantity.