1

### JEE Main 2019 (Online) 10th January Morning Slot

The density of a material in SI units is 128 kg m–3 . In certain units in which the unit of length is 25 cm and the unit of mass is 50 g, the numerical value of density of the material is -
A
40
B
640
C
16
D
410

## Explanation

Here given that

density of a material in SI units is 128 kg m–3

And a new unit system is introduced where 1 unit of length = 25 cm and 1 unit of mass = 50 g

You should know that, physical quantity is same in any unit system. And to calculate a physical quantity yo should know two things

(1) numerical value of the physical quantity (n)

(2) unit of the physical quantity (u)

And n $\times$ u = constant in any unit system.

Here in SI unit system,

n1 = 128

u1 = kg/m3

And in new unit system,

n2 = ?

u2 = 50gm/(25cm)3

As n1u1 = n2u2

$\therefore$ 128 $\times$ (kg/m3) = n2 $\times$ 50gm/(25cm)3

$\Rightarrow$ 128 $\times$ ${{1000\,gm} \over {{{\left( {100\,cm} \right)}^3}}}$ = n2 $\times$ ${{50\,gm} \over {{{\left( {25\,cm} \right)}^3}}}$

$\Rightarrow$ n2 = 128 $\times$ ${{20} \over {{4^3}}}$ = 40
2

### JEE Main 2019 (Online) 10th January Evening Slot

The diameter and height of a cylinder are measured by a meter scale to be 12.6 $\pm$ 0.1 cm and 34.2 $\pm$ 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures ?
A
4264.4 $\pm$ 81.0 cm3
B
4264 $\pm$ 81 cm3
C
4300 $\pm$ 80 cm3
D
4260 $\pm$ 80 cm3

## Explanation

Volume of cylinder(V) = $\pi$r2h

= $\pi {{{d^2}} \over 4}h$

= $3.14 \times {{{{\left( {12.6} \right)}^2}} \over 4} \times 34.2$

= 4260

${{\Delta V} \over V} = 2{{\Delta d} \over d} + {{\Delta h} \over h} = 2\left( {{{0.1} \over {12.6}}} \right) + {{0.1} \over {34.2}}$ = 0.0188

$\therefore$ $\Delta$V = 0.0188 $\times$ 4260 = 80
3

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

The force of interaction between two atoms is given by F = $\alpha$$\beta$exp $\left( { - {{{x^2}} \over {\alpha kt}}} \right)$; where x is the distance, k is the Boltzmann constant and T is temperature and $\alpha$ and $\beta$ are two constants. The dimension of $\beta$ is :
A
M2L2T$-$2
B
M2LT$-$4
C
MLT$-$4
D
M0L2LT$-$4

## Explanation

$F = \alpha \beta {e^{\left( {{{ - {x^2}} \over {\alpha KT}}} \right)}}$

$\left[ {{{{x^2}} \over {\alpha KT}}} \right] = {M^o}{L^o}{T^o}$

${{{L^2}} \over {\left[ \alpha \right]M{L^2}{T^{ - 2}}}}$ $=$ ${M^o}{L^o}{T^o}$

$\Rightarrow$  $\left[ \alpha \right] = {M^{ - 1}}{T^2}$

$\left[ F \right] = \left[ \alpha \right]\left[ \beta \right]$

MLT$-$2 = M$-$1T2[$\beta$]

$\Rightarrow$  [$\beta$] = M2LT$-$4
4

### 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