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Match List - I with List - II.
List - I List - II
(a) Torque (i) MLT$$^{-1}$$
(b) Impulse (ii) MT$$^{-2}$$
(c) Tension (iii) ML$$^2$$ T$$^{-2}$$
(d) Surface Tension (iv) MLT$$^{-2}$$
Choose the most appropriate answer from the option given below:
We start by recalling that in dimensional analysis we express any physical quantity in terms of the fundamental dimensions: mass $$M$$, length $$L$$ and time $$T$$. The exponents of these symbols tell us how the quantity depends on each fundamental dimension.
For torque we use the definition “torque = force × perpendicular distance.” First we write the dimensional formula of force. From Newton’s second law, “force = mass × acceleration,” and acceleration is “length / time2.” Hence
$$[\text{Force}] = M^1L^1T^{-2}.$$
Multiplying by distance (another factor of $$L$$) gives
$$[\text{Torque}] = M^1L^1T^{-2}\; \times \; L^1 = M^1L^{1+1}T^{-2} = ML^{2}T^{-2}.$$
So torque corresponds to $$ML^{2}T^{-2},$$ which is entry (iii).
For impulse we recall the relation “impulse = force × time.” Stating the dimensional formula once more, force is $$MLT^{-2}$$ and time is $$T^1$$. Hence
$$[\text{Impulse}] = (MLT^{-2}) \times (T^1) = ML\,T^{-2+1} = MLT^{-1}.$$
Thus impulse matches $$MLT^{-1},$$ which is entry (i).
For tension we simply note that tension is a kind of force (a pulling force in a string). Therefore its dimensional formula is identical to that of force:
$$[\text{Tension}] = MLT^{-2}.$$
This equals entry (iv).
For surface tension we remember the definition “surface tension = force per unit length.” We again insert the force dimension $$MLT^{-2}$$ and divide by length $$L^1$$:
$$[\text{Surface Tension}] = \dfrac{MLT^{-2}}{L} = MT^{-2}.$$
This matches entry (ii).
Collecting all the matches, we have
$$(a) \longrightarrow (iii), \quad (b) \longrightarrow (i), \quad (c) \longrightarrow (iv), \quad (d) \longrightarrow (ii).$$
Comparing with the options supplied, we see that Option B lists exactly this set of pairings.
Hence, the correct answer is Option B.
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