301. A force is \(1\,\text{N}\). Since \(1\,\text{N}=10^5\,\text{dyne}\), the force in dyne is
ⓐ. \(10^{-5}\,\text{dyne}\)
ⓑ. \(10^3\,\text{dyne}\)
ⓒ. \(10^5\,\text{dyne}\)
ⓓ. \(10^0\,\text{dyne}\)
Correct Answer: \(10^5\,\text{dyne}\)
Explanation: \( \textbf{Given conversion:} \)
\[
1\,\text{N}=10^5\,\text{dyne}
\]
The force is \(1\,\text{N}\), so no additional scaling factor is needed.
Therefore,
\[
1\,\text{N}=10^5\,\text{dyne}
\]
This conversion is consistent with force having dimension \([MLT^{-2}]\).
The dyne is a smaller unit of force than the newton, so the numerical value becomes larger in dyne.
\( \textbf{Final answer:} \) \(1\,\text{N}=10^5\,\text{dyne}\).
302. A work value is \(2\,\text{J}\). Using \(1\,\text{J}=10^7\,\text{erg}\), its value in \(\text{erg}\) is
ⓐ. \(5\times10^7\,\text{erg}\)
ⓑ. \(2\times10^{-7}\,\text{erg}\)
ⓒ. \(2\times10^7\,\text{erg}\)
ⓓ. \(2\times10^5\,\text{erg}\)
Correct Answer: \(2\times10^7\,\text{erg}\)
Explanation: \( \textbf{Given value:} \) Work \(=2\,\text{J}\).
\( \textbf{Conversion relation:} \)
\[
1\,\text{J}=10^7\,\text{erg}
\]
\( \textbf{Apply conversion:} \)
\[
2\,\text{J}=2\times10^7\,\text{erg}
\]
The erg is a smaller unit of work than the joule, so the numerical value becomes larger.
This conversion is also consistent with work having dimension \([ML^2T^{-2}]\).
\( \textbf{Final answer:} \) \(2\,\text{J}=2\times10^7\,\text{erg}\).
303. A pressure of \(1\,\text{Pa}\) is expressed in \(\text{dyne cm}^{-2}\). Given \(1\,\text{N}=10^5\,\text{dyne}\) and \(1\,\text{m}^2=10^4\,\text{cm}^2\), the value is
ⓐ. \(10^2\,\text{dyne cm}^{-2}\)
ⓑ. \(10\,\text{dyne cm}^{-2}\)
ⓒ. \(10^9\,\text{dyne cm}^{-2}\)
ⓓ. \(10^4\,\text{dyne cm}^{-2}\)
Correct Answer: \(10\,\text{dyne cm}^{-2}\)
Explanation: \( \textbf{Starting unit:} \)
\[
1\,\text{Pa}=1\,\text{N m}^{-2}
\]
Use
\[
1\,\text{N}=10^5\,\text{dyne}
\]
and
\[
1\,\text{m}^2=10^4\,\text{cm}^2
\]
So,
\[
1\,\text{N m}^{-2}=\frac{10^5\,\text{dyne}}{10^4\,\text{cm}^2}
\]
\[
=10\,\text{dyne cm}^{-2}
\]
The area conversion appears in the denominator because pressure is force divided by area.
\( \textbf{Final answer:} \) \(1\,\text{Pa}=10\,\text{dyne cm}^{-2}\).
304. A limitation of the dimensional method in deriving formulae is that it
ⓐ. cannot identify the dimensions of force
ⓑ. cannot find dimensionless constants
ⓒ. changes the physical quantity being studied
ⓓ. cannot be used to check any equation
Correct Answer: cannot find dimensionless constants
Explanation: Dimensional analysis compares powers of fundamental dimensions on both sides of a relation. Pure numbers such as \(2\), \(\frac{1}{2}\), or \(2\pi\) have no dimensions. Because of this, dimensional analysis cannot determine such constants. It can suggest that a period may be proportional to \(\sqrt{\frac{l}{g}}\), but it cannot produce the factor \(2\pi\) by dimensional reasoning alone. Exact numerical factors require deeper theory, derivation, or experiment.
305. A derived unit is converted from one system to another using \(n_1u_1=n_2u_2\). In this relation, \(n_1\) and \(n_2\) represent
ⓐ. names of two different physical quantities
ⓑ. errors of the measuring instrument only
ⓒ. dimensions of the physical quantity
ⓓ. numerical values in the two unit systems
Correct Answer: numerical values in the two unit systems
Explanation: A physical quantity remains the same when it is expressed in two different unit systems. If it is written as \(n_1u_1\) in one system and \(n_2u_2\) in another, then \(n_1u_1=n_2u_2\). Here \(u_1\) and \(u_2\) are the unit sizes, while \(n_1\) and \(n_2\) are the corresponding numerical values. When the unit becomes smaller, the numerical value becomes larger for the same physical quantity. The relation is a quantity-conservation statement, not a statement about two different quantities.
306. A physical quantity has dimensional formula \([M^1L^2T^{-2}]\). If the units of mass, length, and time become \(10\) times, \(100\) times, and unchanged respectively, the new numerical value becomes
ⓐ. \(10^3\) times the old numerical value
ⓑ. \(10^5\) times the old numerical value
ⓒ. \(10^{-5}\) times the old numerical value
ⓓ. \(10^{-3}\) times the old numerical value
Correct Answer: \(10^{-5}\) times the old numerical value
Explanation: \( \textbf{Dimensional formula:} \) \([M^1L^2T^{-2}]\).
Let the new mass unit be \(10\) times the old mass unit, the new length unit be \(100\) times the old length unit, and the time unit remain unchanged.
The new derived unit becomes
\[
u_2=(10)^1(100)^2(1)^{-2}u_1
\]
\[
u_2=10\times10^4u_1=10^5u_1
\]
For the same physical quantity,
\[
n_1u_1=n_2u_2
\]
\[
n_1u_1=n_2(10^5u_1)
\]
\[
n_2=\frac{n_1}{10^5}
\]
A larger derived unit gives a smaller numerical value for the same quantity.
\( \textbf{Final answer:} \) The new numerical value is \(10^{-5}\) times the old value.
307. A quantity \(Q\) has dimensions \([MLT^{-2}]\). It is \(10\,\text{N}\) in SI. Since \(1\,\text{N}=10^5\,\text{dyne}\), the same quantity in \(\text{dyne}\) is
ⓐ. \(10^6\,\text{dyne}\)
ⓑ. \(10^5\,\text{dyne}\)
ⓒ. \(10^{-4}\,\text{dyne}\)
ⓓ. \(10^{10}\,\text{dyne}\)
Correct Answer: \(10^6\,\text{dyne}\)
Explanation: \( \textbf{Given value:} \) \(Q=10\,\text{N}\).
\( \textbf{Conversion relation:} \)
\[
1\,\text{N}=10^5\,\text{dyne}
\]
Multiply both sides by \(10\):
\[
10\,\text{N}=10\times10^5\,\text{dyne}
\]
\[
=10^6\,\text{dyne}
\]
The dyne is a smaller force unit than the newton, so the numerical value increases.
This is consistent with force having dimensions \([MLT^{-2}]\), where the smaller CGS base units produce a smaller derived force unit.
\( \textbf{Final answer:} \) \(10\,\text{N}=10^6\,\text{dyne}\).
308. A density is \(1000\,\text{kg m}^{-3}\). Its value in \(\text{g cm}^{-3}\) is
ⓐ. \(0.001\,\text{g cm}^{-3}\)
ⓑ. \(10\,\text{g cm}^{-3}\)
ⓒ. \(1\,\text{g cm}^{-3}\)
ⓓ. \(1000\,\text{g cm}^{-3}\)
Correct Answer: \(1\,\text{g cm}^{-3}\)
Explanation: \( \textbf{Given density:} \) \(1000\,\text{kg m}^{-3}\).
Convert the mass unit:
\[
1\,\text{kg}=1000\,\text{g}
\]
Convert the volume unit:
\[
1\,\text{m}=100\,\text{cm}
\]
\[
1\,\text{m}^3=(100\,\text{cm})^3=10^6\,\text{cm}^3
\]
So,
\[
1000\,\text{kg m}^{-3}=1000\times\frac{1000\,\text{g}}{10^6\,\text{cm}^3}
\]
\[
=\frac{10^6\,\text{g}}{10^6\,\text{cm}^3}
\]
\[
=1\,\text{g cm}^{-3}
\]
The volume conversion is cubic, so using only \(100\,\text{cm}\) instead of \(10^6\,\text{cm}^3\) would give a wrong result.
\( \textbf{Final answer:} \) \(1000\,\text{kg m}^{-3}=1\,\text{g cm}^{-3}\).
309. A speed is \(72\,\text{km h}^{-1}\). Its value in \(\text{m s}^{-1}\) is
ⓐ. \(10\,\text{m s}^{-1}\)
ⓑ. \(72\,\text{m s}^{-1}\)
ⓒ. \(25\,\text{m s}^{-1}\)
ⓓ. \(20\,\text{m s}^{-1}\)
Correct Answer: \(20\,\text{m s}^{-1}\)
Explanation: \( \textbf{Given speed:} \) \(72\,\text{km h}^{-1}\).
Use
\[
1\,\text{km}=1000\,\text{m}
\]
and
\[
1\,\text{h}=3600\,\text{s}
\]
So,
\[
72\,\text{km h}^{-1}=72\times\frac{1000\,\text{m}}{3600\,\text{s}}
\]
\[
=72\times\frac{5}{18}\,\text{m s}^{-1}
\]
\[
=20\,\text{m s}^{-1}
\]
The hour is in the denominator, so converting it to seconds divides by \(3600\), not multiplies the final speed by \(3600\).
\( \textbf{Final answer:} \) \(72\,\text{km h}^{-1}=20\,\text{m s}^{-1}\).
310. A graph is plotted with work \(W\) on the vertical axis and displacement \(s\) on the horizontal axis. If the relation is \(W=Fs\), the dimension of the slope is
ⓐ. \([ML^{-1}T^{-2}]\)
ⓑ. \([ML^2T^{-2}]\)
ⓒ. \([LT^{-1}]\)
ⓓ. \([MLT^{-2}]\)
Correct Answer: \([MLT^{-2}]\)
Explanation: \( \textbf{Slope meaning:} \) Slope is the ratio of vertical-axis quantity to horizontal-axis quantity.
Here,
\[
\text{slope}=\frac{W}{s}
\]
Work has dimensional formula \([ML^2T^{-2}]\).
Displacement has dimension \([L]\).
Therefore,
\[
[\text{slope}]=\frac{[ML^2T^{-2}]}{[L]}
\]
\[
=[MLT^{-2}]
\]
This is the dimensional formula of force.
The result matches the relation \(W=Fs\), where force is the multiplier of displacement.
\( \textbf{Final answer:} \) The slope has the dimension of force.
311. The area under a force-displacement graph has the dimension of
ⓐ. force
ⓑ. work
ⓒ. acceleration
ⓓ. pressure
Correct Answer: work
Explanation: Area under a graph is found by multiplying the vertical-axis quantity by the horizontal-axis quantity. For a force-displacement graph, this gives force multiplied by displacement. Dimensionally,
\[
[F][s]=[MLT^{-2}][L]
\]
\[
=[ML^2T^{-2}]
\]
This is the dimensional formula of work or energy. The graphical area is therefore associated with work done when force acts along displacement.
312. A dimensionally valid equation may still be physically wrong because dimensional analysis
ⓐ. it checks dimensions, not the full physical law
ⓑ. always gives the exact coefficient in the equation
ⓒ. ignores all dimensions of physical quantities
ⓓ. changes variables into unitless numbers
Correct Answer: it checks dimensions, not the full physical law
Explanation: Dimensional analysis tests whether the dimensions on both sides of an equation agree. Passing this test is necessary for a physical equation, but it is not sufficient to prove the equation. Many expressions with the same dimensions can differ by numerical constants, signs, vector directions, or physical conditions. For example, dimensional analysis cannot decide whether a coefficient is \(\frac{1}{2}\), \(2\), or \(2\pi\). A correct physical law also needs reasoning, experiment, or derivation beyond dimensional matching.
313. The equation \(v^2=u^2+2as\) is dimensionally valid because every term has the dimension
ⓐ. \([L^1T^{-1}]\)
ⓑ. \([L^2T^{-2}]\)
ⓒ. \([L^1T^{-2}]\)
ⓓ. \([L^0T^{-2}]\)
Correct Answer: \([L^2T^{-2}]\)
Explanation: Velocity has dimension \([LT^{-1}]\), so \(v^2\) and \(u^2\) each have dimension \([L^2T^{-2}]\). Acceleration has dimension \([LT^{-2}]\), and displacement has dimension \([L]\). Therefore,
\[
[as]=[LT^{-2}][L]=[L^2T^{-2}]
\]
The numerical factor \(2\) is dimensionless.
All terms in \(v^2=u^2+2as\) therefore have the same dimensions.
A sum is meaningful here because the terms being added are dimensionally alike.
314. The proposed relation \(v=u+as\), where \(v\) and \(u\) are velocities, \(a\) is acceleration, and \(s\) is displacement, is dimensionally invalid because \(as\) has dimensions
ⓐ. \([L^{-1}T^{1}]\)
ⓑ. \([L^1T^{-1}]\)
ⓒ. \([L^1T^{-2}]\)
ⓓ. \([L^2T^{-2}]\)
Correct Answer: \([L^2T^{-2}]\)
Explanation: The terms \(v\) and \(u\) have dimension \([LT^{-1}]\). The product \(as\) has dimension
\[
[as]=[LT^{-2}][L]
\]
\[
=[L^2T^{-2}]
\]
This is the dimension of velocity squared, not velocity. Since a term with dimension \([L^2T^{-2}]\) cannot be added to a velocity term, the proposed relation fails dimensional homogeneity. A dimensionally similar expression would involve \(v^2\) and \(u^2\), not \(v\) and \(u\) alone.
315. A relation is proposed as \(E=Ax^2\), where \(E\) is energy and \(x\) is displacement. The dimensional formula of \(A\) is
ⓐ. \([MLT^{-2}]\)
ⓑ. \([MT^{-2}]\)
ⓒ. \([M^{-1}T^2]\)
ⓓ. \([ML^2T^{-2}]\)
Correct Answer: \([MT^{-2}]\)
Explanation: \( \textbf{Given relation:} \)
\[
E=Ax^2
\]
Rearrange for \(A\):
\[
A=\frac{E}{x^2}
\]
Energy has dimensional formula \([ML^2T^{-2}]\).
Displacement squared has dimension \([L^2]\).
So,
\[
[A]=\frac{[ML^2T^{-2}]}{[L^2]}
\]
\[
[A]=[MT^{-2}]
\]
This is the same dimensional formula as a force constant in a relation such as \(F=kx\).
\( \textbf{Final answer:} \) The dimensional formula of \(A\) is \([MT^{-2}]\).
316. In \(R=\alpha L+\beta t\), \(R\) is length, \(L\) is length, and \(t\) is time. The dimensions of \(\alpha\) and \(\beta\), respectively, are
ⓐ. \([L]\) and \([T]\)
ⓑ. \([LT^{-1}]\) and \([M^0L^0T^0]\)
ⓒ. \([M^0L^0T^0]\) and \([LT^{-1}]\)
ⓓ. \([L^{-1}]\) and \([T^{-1}]\)
Correct Answer: \([M^0L^0T^0]\) and \([LT^{-1}]\)
Explanation: \( \textbf{Given relation:} \)
\[
R=\alpha L+\beta t
\]
Since \(R\) is a length, each term on the right side must have dimension \([L]\).
For \(\alpha L\):
\[
[\alpha][L]=[L]
\]
\[
[\alpha]=[M^0L^0T^0]
\]
For \(\beta t\):
\[
[\beta][T]=[L]
\]
\[
[\beta]=[LT^{-1}]
\]
Thus \(\alpha\) is dimensionless, while \(\beta\) has the dimension of velocity.
\( \textbf{Final answer:} \) \([\alpha]=[M^0L^0T^0]\) and \([\beta]=[LT^{-1}]\).
317. A quantity \(Q\) is given by \(Q=\frac{P}{\rho g}\), where \(P\) is pressure, \(\rho\) is density, and \(g\) is acceleration. The dimensional formula of \(Q\) is
ⓐ. \([L]\)
ⓑ. \([T]\)
ⓒ. \([M]\)
ⓓ. \([L^2]\)
Correct Answer: \([L]\)
Explanation: \( \textbf{Given relation:} \)
\[
Q=\frac{P}{\rho g}
\]
Pressure has dimension \([ML^{-1}T^{-2}]\).
Density has dimension \([ML^{-3}]\), and acceleration has dimension \([LT^{-2}]\).
So,
\[
[\rho g]=[ML^{-3}][LT^{-2}]
\]
\[
=[ML^{-2}T^{-2}]
\]
Now,
\[
[Q]=\frac{[ML^{-1}T^{-2}]}{[ML^{-2}T^{-2}]}
\]
Cancel common factors:
\[
[Q]=[L]
\]
The expression has the dimension of length, which is why such combinations often represent a height or head in fluids.
\( \textbf{Final answer:} \) The dimensional formula of \(Q\) is \([L]\).
318. A relation has the form \(X=A\cos(\omega t)\), where \(X\) is displacement and \(t\) is time. The dimensions of \(A\) and \(\omega\), respectively, are
ⓐ. \([M]\) and \([M^0L^0T^0]\)
ⓑ. \([L]\) and \([T^{-1}]\)
ⓒ. \([T]\) and \([L]\)
ⓓ. \([LT^{-1}]\) and \([T]\)
Correct Answer: \([L]\) and \([T^{-1}]\)
Explanation: \( \textbf{Given relation:} \)
\[
X=A\cos(\omega t)
\]
The cosine factor is dimensionless because the argument \(\omega t\) must be dimensionless.
Since \(X\) is displacement, the multiplier \(A\) must also have dimension \([L]\).
For the argument:
\[
[\omega][T]=[M^0L^0T^0]
\]
Therefore,
\[
[\omega]=[T^{-1}]
\]
The angular unit may be written as \(\text{rad s}^{-1}\), but the radian itself is dimensionless in dimensional analysis.
\( \textbf{Final answer:} \) \([A]=[L]\) and \([\omega]=[T^{-1}]\).
319. A dimensionally consistent expression for a time using only \(l\) and \(g\), where \(l\) is length and \(g\) is acceleration, must be proportional to
ⓐ. \(\sqrt{\frac{l}{g}}\)
ⓑ. \(\sqrt{\frac{g}{l}}\)
ⓒ. \(\frac{l}{g}\)
ⓓ. \(\frac{1}{\sqrt{lg}}\)
Correct Answer: \(\sqrt{\frac{l}{g}}\)
Explanation: Length has dimension \([L]\), and acceleration has dimension \([LT^{-2}]\). The ratio \(\frac{l}{g}\) has dimension
\[
\frac{[L]}{[LT^{-2}]}=[T^2]
\]
Taking its square root gives
\[
\sqrt{[T^2]}=[T]
\]
So \(\sqrt{\frac{l}{g}}\) has the dimension of time. The expression \(\frac{l}{g}\) has the dimension of time squared, not time.
320. A physical equation contains the term \(\ln\left(\frac{x}{x_0}\right)\), where \(x\) and \(x_0\) are lengths. This logarithmic term is dimensionally acceptable because
ⓐ. \(\frac{x}{x_0}\) is dimensionless
ⓑ. logarithms can take dimensional quantities directly
ⓒ. \(x\) alone is dimensionless
ⓓ. \(x_0\) has dimension \([T]\)
Correct Answer: \(\frac{x}{x_0}\) is dimensionless
Explanation: The argument of a logarithmic function must be dimensionless. In \(\ln\left(\frac{x}{x_0}\right)\), both \(x\) and \(x_0\) are lengths, so their ratio has no dimension. This makes the logarithmic expression dimensionally meaningful. Writing \(\ln x\) directly, with \(x\) as a length, would not be dimensionally proper. Ratios are often used inside logarithmic, exponential, and trigonometric functions for this reason.
