1

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

Water flows into a large tank with flat bottom at the rate of 10–4m3s–1. Water is also leaking out of a hole ofarea 1 cm2 at its bottom. If the height of the water in the tank remains steady, then this height is -
A
2.9 cm
B
5.1 cm
C
4 cm
D
1.7 cm

## Explanation Since height of water column is constant therefore, water inflow rate (Qin)

= water outflow rate

Qin = 10$-$4 m3s$-$1

Qout = Au = 10$-$4 $\times$ $\sqrt {2gh}$

10$-$4 = 10$-$4 $\sqrt {20 \times h}$

h = ${1 \over {20}}m$

h = 5cm
2

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

A liquid of density $\rho$ is coming out of a hose pipe of radius a with horizontal speed $\upsilon$ and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% looses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be :
A
${3 \over 4}\rho {v^2}$
B
${1 \over 4}\rho {v^2}$
C
${1 \over 2}\rho {v^2}$
D
$\rho {v^2}$

## Explanation

Momentum per second carried by liquid per second is $\rho$av2

net force due to reflected liquid = 2$\times$$\left[ {{1 \over 4}\rho a{v^2}} \right]$

net force due to stopped liquid = ${{1 \over 4}\rho a{v^2}}$

Total force = ${{3 \over 4}\rho a{v^2}}$

net pressure = ${{3 \over 4}\rho {v^2}}$
3

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

A load of mass M kg is suspended from a steel wire of length 2m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8.

The new value of increase in length of the steel wire is:
A
5.0 mm
B
zero
C
3.0 mm
D
4.0 mm

## Explanation ${F \over A} = y.{{\Delta \ell } \over \ell }$

$\Delta \ell \propto F$         . . .. (i)

T $=$ mg

T $=$ mg $-$ fB $=$ mg $-$ ${m \over {{\rho _b}}}.{\rho _\ell }.$g

$= \left( {1 - {{{\rho _\ell }} \over {{\rho _b}}}} \right)$ mg

$= \left( {1 - {2 \over 8}} \right)$ mg

T' $=$ ${3 \over 4}$ mg

From (i)

${{\Delta \ell '} \over {\Delta \ell }} = {{T'} \over T} = {3 \over 4}$

$\Delta \ell ' = {3 \over 4}.\Delta \ell = 3$ mm
4

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

A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above the piston is $\ell$1, and that below the piston is $\ell$2, such that $\ell$1 > $\ell$2. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass, m, will be given by :
(R is universal gas constant and g is the acceleration due to gravity)
A
${{nRT} \over g}\left[ {{{{\ell _1} - {\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]$
B
${{RT} \over g}\left[ {{{2{\ell _1} + {\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]$
C
${{nRT} \over g}\left[ {{1 \over {{\ell _2}}} + {1 \over {{\ell _1}}}} \right]$
D
${{RT} \over {ng}}\left[ {{{{\ell _1} - 3{\ell _2}} \over {{\ell _1}{\ell _2}}}} \right]$

## Explanation P2A = P1A + mg

${{nRT.A} \over {A{\ell _2}}}$ = ${{nRT.A} \over {A{\ell _1}}}$ + mg

nRT$\left( {{1 \over {{\ell _2}}} - {1 \over {{\ell _1}}}} \right)$ = mg

m = ${{nRT} \over g}\left( {{{{\ell _1} - {\ell _2}} \over {{\ell _1}.{\ell _2}}}} \right)$