1

### JEE Main 2019 (Online) 9th January Evening Slot

The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.

The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :
A
5.755 mm
B
5.950 mm
C
5.725 mm
D
5.740 mm

## Explanation

We know,

Least count (LC) = ${{Pitch} \over {no.\,of\,divisions}}$

$\therefore$  LC = ${{0.5} \over {100}}$

= 0.5 $\times$ 10$-$2 mm

Reading = MSR + CSR $-$ positive error

Given, Main scale reading (MSR) = 5.5 mm

= 48 $\times$ 0.5 $\times$ 10$-$2 mm

= 0.24

As zero of its circular scale lines 3 division below the mean line, it means error is position error.

$\therefore$  positive error

= 3 $\times$ 0.5 $\times$ 10$-$2 mm

= 0.015 mm

$\therefore$  Reading = 5.5 + 0.24 $-$ 0.015

= 5.725 mm
2

### 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
3

### 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
4

### 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

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