1

### JEE Main 2019 (Online) 11th January Evening Slot

If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young,s modulus will be:
A
V$-$2A2F2
B
V$-$4A$-$2F
C
V$-$4A2F
D
V$-$2A2F$-$2

## Explanation

We know,

Young's modulus (Y) = ${{{F \over A}} \over {{{\Delta l} \over l}}}$

$\therefore$ [Y] = ${{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ {{L^2}} \right]}}$ = [ ML-1T-2]

Let [Y] = [V]x [A]y [F]z

$\therefore$ [ ML-1T-2] =

[LT-1]x [LT-2]y [MLT-2]z

$\Rightarrow$ [ ML-1T-2] =

[ Mz Lx + y + z T-x -2y - 2z

For dimensional balance, the dimension on both sides should be same.

So, z = 1

x + y + z = -1

$\Rightarrow$ x + y = -2 ........(1)

and -x -2y - 2z = -2

$\Rightarrow$ x + 2y = 0 ...........(2)

By solving those two equations we get,

x = -4 and y = 2

$\therefore$ [Y] = V$-$4A2F1
2

### JEE Main 2019 (Online) 12th January Morning Slot

The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 $\mu$m diameter of a wire is :
A
500
B
100
C
200
D
50

## Explanation

Least count = ${{Pitch} \over {Number\,\,of\,\,division\,\,on\,\,circular\,scale}}$

5 $\times$ 10$-$6 = ${{{{10}^{ - 3}}} \over N}$

N = 200
3

### JEE Main 2019 (Online) 12th January Evening Slot

Let $\ell$, r, C and V represent inductance, resistance, capacitance and voltage, respectively. The dimension of ${\ell \over {rCV}}$ in SI units will be :
A
[A–1]
B
[LTA]
C
[LA–2]
D
[LT2]

## Explanation

$\left[ {{\ell \over r}} \right] =$ T

[CV] $=$ AT

So,   $\left[ {{\ell \over {rCV}}} \right]$ = ${T \over {AT}}$ = [A$-$1]
4

### JEE Main 2019 (Online) 8th April Morning Slot

In SI units, the dimesions of $\sqrt {{{{ \in _0}} \over {{\mu _0}}}}$ is :
A
A–1 TML3
B
A2T3M–1L–2
C
AT–3ML3/2
D
AT2M–1L–1

## Explanation

$\sqrt {{{{ \in _0}} \over {{\mu _0}}}}$ = ${{{ \in _0}} \over {\sqrt {{\mu _0}{ \in _0}} }}$ = c $\times$ ${{ \in _0}}$

$\therefore$ $\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } \right]$ = $\left[ {L{T^{ - 1}}} \right] \times \left[ {{ \in _0}} \right]$

We know, F = ${1 \over {4\pi { \in _0}}}{{{q^2}} \over {{r^2}}}$

$\therefore$ ${ \in _0} = {{{q^2}} \over {4\pi {r^2}F}}$

$\Rightarrow$ $\left[ {{ \in _0}} \right] = {{{{\left[ {AT} \right]}^2}} \over {\left[ {ML{T^{ - 2}}} \right] \times \left[ {{L^2}} \right]}}$ = $\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^4}} \right]$

$\therefore$ $\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } \right]$ = $\left[ {L{T^{ - 1}}} \right]$ $\times$ $\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^4}} \right]$

= [A2T3M–1L–2]