JEE Main Units and Measurements Previous year Question

31.(JEE Main 2020 (Online) 3rd September Evening Slot )

Amount of solar energy received on the earth’s surface per unit area per unit time is defined a solar constant. Dimension of solar constant is :

(A) MLT^–2

(B) ML⁰T^–3

(C) M²L⁰T^–1

(D) ML²T^–2

Correct answer is (B)

Solar constant = $\frac{E}{A T}$

= $\frac{[M^{1} L^{2} T^{- 2}]}{[L^{2} T]}$

= [ ML⁰T^–3 ]

32.(JEE Main 2020 (Online) 9th January Morning Slot )

A quantity f is given by $f = \sqrt{\frac{h c^{5}}{G}}$ where c is speed of light, G universal gravitational constant and h is the Planck's constant. Dimension of f is that of :

(A) Energy

(B) Momentum

(C) Area

(D) Volume

Correct answer is (A)

[h] = M¹L²T^–1
[C] = L¹T^–1
[G] = M^–1L³T^–2

[f] = $\sqrt{\frac{M L^{2} T^{- 1} \times L^{5} T^{- 5}}{M^{- 1} L^{3} T^{- 2}}}$ = M¹L²T^–2

33.(JEE Main 2020 (Online) 7th January Evening Slot )

The dimension of $\frac{B^{2}}{2 μ_{0}}$ , where B is magnetic field and $μ_{0}$ is the magnetic permeability of vacuum, is :

(A) ML²T^–2

(B) MLT^–2

(C) ML^-1T^–2

(D) ML²T^–1

Correct answer is (C)

As $\frac{B^{2}}{2 μ_{0}}$ = Energy per unit volume

$∴$ Dimension = $\frac{M L^{2} T^{- 2}}{L^{3}}$ = ML^-1T^–2

34.(JEE Main 2019 (Online) 12th April Morning Slot )

Which of the following combinations has the dimension of electrical resistance ( $\in$ ₀is the permittivity of vacuum and $μ$ ₀ is the permeability of vacuum)?

(A) $\sqrt{\frac{\in_{0}}{μ_{0}}}$

(B) $\frac{\in_{0}}{μ_{0}}$

(C) $\sqrt{\frac{μ_{0}}{\in_{0}}}$

(D) $\frac{μ_{0}}{\in_{0}}$

Correct answer is (C)

According to Coulomb's law

F = $\frac{1}{4 π \in_{0}} \frac{q^{2}}{r^{2}}$

$∴$ $\in_{0} = \frac{1}{4 π} \frac{q^{2}}{F r^{2}}$

$[\in_{0}] = \frac{{[A T]}^{2}}{[M L T^{- 2}] [L^{2}]}$ = $[M^{- 1} L^{- 3} T^{4} A^{2}]$

Force between two parallel current carrying wires,

$\frac{F}{L} = \frac{μ_{0}}{2 π} \frac{i^{2}}{r}$

$∴$ $μ_{0}$ = $\frac{2 π r F}{i^{2} L}$

$[μ_{0}] = \frac{[L] [M L T^{- 2}]}{[A^{2}] [L]}$ = $[M L T^{- 2} A^{- 2}]$

From Ohm's law,

V = IR

$∴$ $R = \frac{V}{I}$ $= \frac{U}{I t} \times \frac{1}{I}$

[R] = $\frac{[U]}{{[I]}^{2} [t]}$ = $\frac{[M L^{2} T^{- 2}]}{[A^{2}] [T]}$ = $[M L^{2} T^{- 3} A^{- 2}]$

Let, R $\propto$ ${[\in_{0}]}^{a} {[μ_{0}]}^{b}$

$\Rightarrow$ $[M L^{2} T^{- 3} A^{- 2}]$ =

$[M^{- 1} L^{- 3} T^{4} A^{2}]$ ^$a$ $[M L T^{- 2} A^{- 2}]$ ^b

$\Rightarrow$ $[M L^{2} T^{- 3} A^{- 2}]$ =

[ M^{- $a$ + b} L^{-3 $a$ + b} T^{4 $a$ - 2b} A^{2 $a$ - 2b} ]

By comparing both sides we get,

- $a$ + b = 1

- 3 $a$ + b = 2

By solving we get,

$a$ = $- \frac{1}{2}$ and b = $\frac{1}{2}$

$∴$ R = ${[\in_{0}]}^{- \frac{1}{2}} {[μ_{0}]}^{\frac{1}{2}}$ = $\sqrt{\frac{μ_{0}}{\in_{0}}}$

35.(JEE Main 2019 (Online) 10th April Evening Slot )

In the formula X = 5YZ² , X and Z have dimensions of capacitance and magnetic field, respectively. What are the dimensions of Y in SI units?

(A) [M^–3L^–2T⁸A⁴]

(B) [M^–2L^–2T⁶A³]

(C) [M^–1L^–2T⁴A²]

(D) [M^–2L⁰ T^–4A^–2]

Correct answer is (A)

Capacitance (C) = $\frac{Q^{2}}{2 E}$ = $\frac{[A^{2} T^{2}]}{[M L^{2} T^{- 2}]}$ = $[M^{- 1} L^{- 2} T^{4} A^{2}]$ X

Magnetic field (B) = $\frac{F}{I L}$ = $\frac{[M L T^{- 2}]}{[A] [L]}$ = $[M T^{- 2} A^{- 1}]$ = Z

Given,

X = 5YZ²

$∴$ Y = $\frac{X}{5 Z^{2}}$

[Y] = $\frac{[M^{- 1} L^{- 2} T^{4} A^{2}]}{{[M T^{- 2} A^{- 1}]}^{2}}$

$∴$ [Y] = [M^–3L^–2T⁸A⁴]