JEE Challenger
More from Units and Measurements
PhyMedAdv 2007 • Paper 1

Match Physical Quantities with Their Corresponding SI Units

Some physical quantities are given in Column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physical quantities in Column I with the units in Column II and indicate your answer by darkening appropriate bubbles in the 4×44 \times 4 matrix given in the ORS.

Column IColumn II(A) GMeMs(p) (volt) (coulomb) (metre)G−universal gravitational constant,Me−mass of the earth,Ms−mass of the Sun(B) 3RTM(q) (kilogram) (metre)3 (second)−2R−universal gas constant,T−absolute temperature,M−molar mass(C) F2q2B2(r) (metre)2 (second)−2F−force,q−charge,B−magnetic field(D) GMeRe(s) (farad) (volt)2 (kg)−1G−universal gravitational constant,Me−mass of the earth,Re−radius of the earth\begin{array}{ll} \textbf{Column I} & \textbf{Column II} \\[6pt] \text{(A) } GM_e M_s & \text{(p) (volt) (coulomb) (metre)} \\ \quad G - \text{universal gravitational constant,} & \\ \quad M_e - \text{mass of the earth,} & \\ \quad M_s - \text{mass of the Sun} & \\[6pt] \text{(B) } \dfrac{3RT}{M} & \text{(q) (kilogram) (metre)}^3\text{ (second)}^{-2} \\ \quad R - \text{universal gas constant,} & \\ \quad T - \text{absolute temperature,} & \\ \quad M - \text{molar mass} & \\[6pt] \text{(C) } \dfrac{F^2}{q^2 B^2} & \text{(r) (metre)}^2\text{ (second)}^{-2} \\ \quad F - \text{force,} & \\ \quad q - \text{charge,} & \\ \quad B - \text{magnetic field} & \\[6pt] \text{(D) } \dfrac{GM_e}{R_e} & \text{(s) (farad) (volt)}^2\text{ (kg)}^{-1} \\ \quad G - \text{universal gravitational constant,} & \\ \quad M_e - \text{mass of the earth,} & \\ \quad R_e - \text{radius of the earth} & \end{array}

Options

A

(A)→(P),  (B)→(R,S),  (C)→(R),  (D)→(R,S)(\text{A}) \rightarrow (\text{P}),\; (\text{B}) \rightarrow (\text{R}, \text{S}),\; (\text{C}) \rightarrow (\text{R}),\; (\text{D}) \rightarrow (\text{R}, \text{S})

B

(A)→(P,Q),  (B)→(R,S),  (C)→(R,S),  (D)→(R,S)(\text{A}) \rightarrow (\text{P}, \text{Q}),\; (\text{B}) \rightarrow (\text{R}, \text{S}),\; (\text{C}) \rightarrow (\text{R}, \text{S}),\; (\text{D}) \rightarrow (\text{R}, \text{S})

Correct
C

(A)→(P,Q),  (B)→(R),  (C)→(S),  (D)→(R,S)(\text{A}) \rightarrow (\text{P}, \text{Q}),\; (\text{B}) \rightarrow (\text{R}),\; (\text{C}) \rightarrow (\text{S}),\; (\text{D}) \rightarrow (\text{R}, \text{S})

D

(A)→(Q),  (B)→(R,S),  (C)→(R,S),  (D)→(S)(\text{A}) \rightarrow (\text{Q}),\; (\text{B}) \rightarrow (\text{R}, \text{S}),\; (\text{C}) \rightarrow (\text{R}, \text{S}),\; (\text{D}) \rightarrow (\text{S})

Topics & Concepts

Units and MeasurementsGravitationKinetic Theory of GasesMagnetic Force

Step-by-Step Solution

To match the physical quantities in Column I with their corresponding units in Column II, let us determine the dimensions and SI units of each quantity.

Analysis of Column II:

  • (p) (volt) (coulomb) (metre)\text{(volt) (coulomb) (metre)}:
    Recall that the product of electric potential (volt\text{volt}) and charge (coulomb\text{coulomb}) represents energy (joule\text{joule}):
    1 volt×1 coulomb=1 joule=1 kg⋅m2⋅s−2\text{1 volt} \times \text{1 coulomb} = \text{1 joule} = \text{1 kg}\cdot\text{m}^2\cdot\text{s}^{-2}
    Therefore:
    (volt) (coulomb) (metre)=J⋅m=(kg⋅m2⋅s−2)⋅m=kg⋅m3⋅s−2\text{(volt) (coulomb) (metre)} = \text{J}\cdot\text{m} = \left(\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}\right)\cdot\text{m} = \text{kg}\cdot\text{m}^3\cdot\text{s}^{-2}
    The dimensional formula is [ML3T−2][\text{M}\text{L}^3\text{T}^{-2}].

  • (q) (kilogram) (metre)3 (second)−2\text{(kilogram) (metre)}^3\text{ (second)}^{-2}:
    This unit is directly expressed in SI base units:
    kg⋅m3⋅s−2\text{kg}\cdot\text{m}^3\cdot\text{s}^{-2}
    The dimensional formula is [ML3T−2][\text{M}\text{L}^3\text{T}^{-2}].

  • (r) (metre)2 (second)−2\text{(metre)}^2\text{ (second)}^{-2}:
    This represents the square of velocity:
    m2⋅s−2\text{m}^2\cdot\text{s}^{-2}
    The dimensional formula is [L2T−2][\text{L}^2\text{T}^{-2}].

  • (s) (farad) (volt)2 (kg)−1\text{(farad) (volt)}^2\text{ (kg)}^{-1}:
    The energy stored in a capacitor is given by U=12CV2U = \frac{1}{2} C V^2, so:
    1 farad×1 volt2=1 joule=1 kg⋅m2⋅s−2\text{1 farad} \times \text{1 volt}^2 = \text{1 joule} = \text{1 kg}\cdot\text{m}^2\cdot\text{s}^{-2}
    Dividing by mass (kg\text{kg}):
    (farad) (volt)2 (kg)−1=Jkg=kg⋅m2⋅s−2kg=m2⋅s−2\text{(farad) (volt)}^2\text{ (kg)}^{-1} = \frac{\text{J}}{\text{kg}} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{kg}} = \text{m}^2\cdot\text{s}^{-2}
    The dimensional formula is [L2T−2][\text{L}^2\text{T}^{-2}].

Analysis of Column I:

  • (A) GMeMsG M_e M_s:
    From Newton's law of universal gravitation, the gravitational force between two masses is:
    F=GMeMsr2  ⟹  GMeMs=Fr2F = \frac{G M_e M_s}{r^2} \implies G M_e M_s = F r^2
    The dimensions are:
    [GMeMs]=[F][r2]=(MLT−2)(L2)=[ML3T−2][G M_e M_s] = [F][r^2] = (\text{M}\text{L}\text{T}^{-2})(\text{L}^2) = [\text{M}\text{L}^3\text{T}^{-2}]
    This corresponds to the units in (p) and (q).
    (A)→(p), (q)\text{(A)} \rightarrow \text{(p), (q)}

  • (B) 3RTM\dfrac{3RT}{M}:
    From the kinetic theory of gases, the root-mean-square speed of molecules is given by:
    vrms=3RTM  ⟹  3RTM=vrms2v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \implies \frac{3RT}{M} = v_{\text{rms}}^2
    Since this is the square of velocity, its dimensions are:
    [3RTM]=[v2]=[L2T−2]\left[\frac{3RT}{M}\right] = [v^2] = [\text{L}^2\text{T}^{-2}]
    This corresponds to the units in (r) and (s).
    (B)→(r), (s)\text{(B)} \rightarrow \text{(r), (s)}

  • (C) F2q2B2\dfrac{F^2}{q^2 B^2}:
    The magnetic Lorentz force on a charge moving with speed vv perpendicular to a magnetic field is:
    F=qvB  ⟹  v=FqBF = q v B \implies v = \frac{F}{qB}
    Squaring both sides gives:
    F2q2B2=v2\frac{F^2}{q^2 B^2} = v^2
    The dimensions are:
    [F2q2B2]=[v2]=[L2T−2]\left[\frac{F^2}{q^2 B^2}\right] = [v^2] = [\text{L}^2\text{T}^{-2}]
    This corresponds to the units in (r) and (s).
    (C)→(r), (s)\text{(C)} \rightarrow \text{(r), (s)}

  • (D) GMeRe\dfrac{GM_e}{R_e}:
    The orbital speed of a satellite near the surface of the Earth is given by v0=GMeRev_0 = \sqrt{\frac{GM_e}{R_e}}, or equivalently, the gravitational potential (energy per unit mass) is given by V=−GMeReV = -\frac{GM_e}{R_e}. Thus:
    GMeRe=v02\frac{GM_e}{R_e} = v_0^2
    The dimensions are:
    [GMeRe]=[v2]=[L2T−2]\left[\frac{GM_e}{R_e}\right] = [v^2] = [\text{L}^2\text{T}^{-2}]
    This corresponds to the units in (r) and (s).
    (D)→(r), (s)\text{(D)} \rightarrow \text{(r), (s)}

Conclusion:

  • (A) →\rightarrow (p), (q)
  • (B) →\rightarrow (r), (s)
  • (C) →\rightarrow (r), (s)
  • (D) →\rightarrow (r), (s)
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