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)

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