Question

Suppose two students are trying to make a new measurement system so that they can use it like a code measurement system and others do not understand it. Instead of taking 1 kg, 1 m and 1 sec as basic units they took unit of mass as $\alpha kg$ the unit of length as $\beta m$ and unit of time as $\gamma$ second. They called power in new system as $ACME.$

List-I		List-I
I	1 N in new system	A	${\alpha }^{-1}{\beta }^{-2}{\gamma }^2$
II	1J in new system	B	${\alpha }^{-1}{\beta }^{-1}{\gamma }^2$
III	1 pascal (SI unit of pressure) in new system	C	${\alpha }^{-1}\beta {\gamma }^2$
IV	$\alpha ACME$ in watt	D	${\alpha }^2{\beta }^2{\gamma }^{-2}$

Now, match the List I with List II and mark the correct choice from given codes

• A. $I—B,II—A,III—C,IV—D$
• B. $I—C,II—B,III—D,IV—A$
• C. $I—A,II—B,III—D,IV—C$
• D. $I—C,II—D,III—A,IV—B$

Answer 1

Answer: (A)

$\left(I\right)1N=\frac{kg1m}{1s^2}=\frac{\left(\frac{1\text{unit mass}}{\alpha }\right)\left(\frac{1}{\beta }\text{unit length}\right)}{\left(\frac{1}{\gamma }\text{unit time}\right)}=\frac{{\gamma }^2}{\alpha \beta }$
unit force in new system
$\left(II\right)1J=\frac{1kg{\left(1m\right)}^2}{{\left(1s\right)}^2}=\frac{\frac{1}{\alpha }{\left(\frac{1}{\beta }\right)}^2}{{\left(\frac{1}{\gamma }\right)}^2}$ unit energy
$={\alpha }^{-1}{\beta }^{-2}{\gamma }^2$
$\left(III\right)1$ pascal $=\frac{1kg}{\left(1m\right){\left(1sec\right)}^2}=\frac{\frac{1}{\alpha }}{\frac{1}{\beta }{\left(\frac{1}{\gamma }\right)}^2}$ unit pressure
$={\alpha }^{-1}\beta {\gamma }^2$
$\left(IV\right)\alpha ACME=\alpha \times \frac{\left(1\text{unit mass}\right){\left(1\text{unit length}\right)}^2}{{\left(1\text{unit time}\right)}^3}$
$=\alpha \frac{\alpha kg{\left(\beta m\right)}^2}{{\left(\gamma sec\right)}^3}={\alpha }^2{\beta }^2{\gamma }^{-3}$ watt