Question

In a new system of units, the unit of mass is 10 kg, unit of length is 1 km and unit of time is 1 minute. The value of one joule in this new hypothetical system is:
(A) $3.6 \times {10^{ - 4}}$ new units
(B) $6 \times {10^7}$ new units
(C) ${10^{11}}$ new units
(D) $1.67 \times {10^4}$ new units

Answer 1

Find the unit and dimension of joule in current S.I system. Express the new units and dimension for joule as per given in the question. Take the ratio of dimension of current unit system and new unit system for 1 joule.

Complete step by step answer:
We know 1 joule is equivalent to $1\,kg\,{m^2}{s^{ - 2}}$. We can express the dimensions of 1 joule as follows,
${n_1} = \left[ {{\text{M}}_1^1\,{\text{L}}_1^2\,{\text{T}}_1^{ - 2}} \right]$
Here, ${M_1}$ is the mass, ${L_1}$ is the length and ${T_1}$ is the time in the former unit system.
Now, the dimension of joule in the new unit system can be expressed as,
${n_2} = \left[ {{\text{M}}_2^1\,{\text{L}}_2^2\,{\text{T}}_2^{ - 2}} \right]$
Here, ${M_2}$ is the mass, ${L_2}$ is the length and ${T_2}$ is the time in the new unit system.
Dividing ${n_1}$ by ${n_2}$, we get,
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{\left[ {{\text{M}}_1^1\,{\text{L}}_1^2\,{\text{T}}_1^{ - 2}} \right]}}{{\left[ {{\text{M}}_2^1\,{\text{L}}_2^2\,{\text{T}}_2^{ - 2}} \right]}}$
$\Rightarrow {n_1} = {n_2}{\left[ {\dfrac{{{{\text{M}}_{\text{1}}}}}{{{{\text{M}}_{\text{2}}}}}} \right]^1}{\left[ {\dfrac{{{{\text{L}}_{\text{1}}}}}{{{{\text{L}}_{\text{2}}}}}} \right]^2}{\left[ {\dfrac{{{{\text{T}}_{\text{1}}}}}{{{{\text{T}}_{\text{2}}}}}} \right]^{ - 2}}$
We know in the current S.I unit system,1 kg is the S.I unit of mass, 1 m is S.I unit of length and 1 s is the S.I unit of time. We have given that in the new unit system unit of mass is 10 kg, unit of length is 1 km and unit of time is 1 minute. Therefore, we can write the above equation as,
$\Rightarrow {n_1} = {n_2}{\left[ {\dfrac{{{\text{1}}\,{\text{kg}}}}{{{\text{10}}\,{\text{kg}}}}} \right]^1}{\left[ {\dfrac{{{\text{1}}\,{\text{m}}}}{{{\text{1}}\,{\text{km}}}}} \right]^2}{\left[ {\dfrac{{{\text{1}}\,{\text{s}}}}{{1\min }}} \right]^{ - 2}}$
$\Rightarrow {n_1} = {n_2}{\left[ {\dfrac{{{\text{1}}\,{\text{kg}}}}{{{\text{10}}\,{\text{kg}}}}} \right]^1}{\left[ {\dfrac{{{\text{1}}\,{\text{m}}}}{{{\text{1}}{{\text{0}}^{\text{3}}}\,{\text{m}}}}} \right]^2}{\left[ {\dfrac{{{\text{1}}\,{\text{s}}}}{{60\,\operatorname{s} }}} \right]^{ - 2}}$
$\Rightarrow {n_1} = {n_2}\left( {\dfrac{1}{{10}}} \right)\left( {\dfrac{1}{{{{10}^6}}}} \right)\left( {3600} \right)$
$\therefore {n_1} = {n_2}\left( {3.6 \times {{10}^{ - 4}}} \right)$

Therefore, 1 joule would be equal to $3.6 \times {10^{ - 4}}$ new units.So, the correct answer is option (A).

Note: If you don’t know the equivalent of 1 joule, then you can use the formula for potential energy or kinetic energy to determine the unit of energy in terms of kg, m, and s. To answer this type of question, students should remember the units and dimensions of velocity, acceleration and force. Most of the other physical parameters depend upon these terms.