Match List I with List II.

Option 1 - (a)→(iii), (b)→(ii), (c)→(iv), (d)→(i)

Option 2 - (a)→(iii), (b)→(iv), (c)→(ii), (d)→(i)

Option 3 - (a)→(iv), (b)→(ii), (c)→(iii), (d)→(i)

Option 4 - (a)→(iii), (b)→(iv), (c)→(ii), (d)→(i)

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Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

Correct Option - 4

Detailed Solution:

[C] = Q/V = Q/ (W/Q) = Q²/W = (A²T²)/ (M¹L²T? ²) = M? ¹L? ²T? A²
[ε? ] = C/ (4πr²) * (Fr²/q²) = M? ¹L? ³T? A²
[µ? ] = 4πrF/I²l = M¹L¹T? ²A? ²
[E] = F/q = (MLT? ²)/ (AT) = MLT? ³A? ¹

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Similar Questions for you

Given below are two statements : One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A : Product of Pressure (P) and time (t) has the same dimension as that of coefficient of viscosity.

Reason R : Coefficient of viscosity = $\frac{F o r c e}{V e l o c i t y g r a d i e n t}$

Choose the correct answer from the options given below:

Raj Pandey

Pressure × time $= \frac{F t}{A} = [M L T^{- 2}] [T] [L^{- 2}] = [M L^{- 1} T^{- 1}]$

Coefficient of viscosity, $η = [M L^{- 1} T^{- 1}]$

$F_{v i s c o u s} = - η A \frac{Δ v}{Δ y}$

If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.


Alok Kumar Singh

E ∝ F? A? T?
[M¹L²T? ²] ∝ [M¹L¹T? ²]? [LT? ²]? [T]?
a = 1
a + b = 2 ⇒ b = 1
-2a - 2b + c = -2
⇒ c = 2
a = 1 b = 1 c = 2
E ∝ [F] [A] [T²]

Match List-I with List-II.

                  List-I                                             List-II

           (A) Torque                          (i) MLT^-1

           (B) Impulse                         (ii) MT^-2

           (C) Tension                         (iii) ML²T^-2

           (D) Surface Tension           (iv) MLT^-2

           Choose the most appropriate answer from the option given below:

Raj Pandey

(a) Torque ® ML²T^-2 ↑ (iii)

(b) Impulse ® MlT^-1 ↑ (i)

(d) Surface Tension ↑ $\frac{M L T^{- 2}}{L} = M T^{- 2}$ (ii)

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