What must all physical quantities have?
Correct Answer: \( \boxed{\mathrm{D}} \)
A physical quantity is any measurable quantity in physics.
Every physical quantity must have:
\( \bullet \) a numerical value (magnitude), and
\( \bullet \) an appropriate unit.
A direction is only required for vector quantities, not for all physical quantities.
Therefore, the correct answer is (D).
The time period \(T\) of a pendulum is given by
\(T=2\pi\left(\dfrac{L}{g}\right)^n\)
where \(L\) is the length of the pendulum and \(g\) is the acceleration of free fall.
The equation is homogeneous.
What is the value of \(n\)?
Correct Answer: \( \boxed{\mathrm{C}} \)
Since the equation is homogeneous, both sides must have the same dimensions.
The dimensions are
\( [L]=\mathrm{L} \)
\( [g]=\mathrm{LT^{-2}} \)
Therefore,
\( \left[\dfrac{L}{g}\right]=\dfrac{\mathrm{L}}{\mathrm{LT^{-2}}}=\mathrm{T^2} \)
Hence,
\( \left(\dfrac{L}{g}\right)^n=\mathrm{T^{2n}} \)
Since the left-hand side has dimensions of time,
\( \mathrm{T}=\mathrm{T^{2n}} \)
Equating powers of \(T\),
\(2n=1\)
\(n=\dfrac{1}{2}\)
Therefore, the correct answer is (C).
The number of atoms in a mobile phone handset may be estimated by dividing the approximate volume of the handset by the approximate volume of an atom.
What is a reasonable estimate of the number of atoms in a mobile phone handset?
Correct Answer: \( \boxed{\mathrm{B}} \)
A typical mobile phone has a volume of about \(10^{-4}\,\mathrm{m^{3}}\), while an atom occupies roughly \(10^{-30}\,\mathrm{m^{3}}\).
Hence,
\( \dfrac{10^{-4}}{10^{-30}} = 10^{26} \)
Therefore, the estimated number of atoms in a mobile phone handset is approximately \(10^{26}\).
Therefore, the correct answer is (B).
