1

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

The top of a water tank is open to air and its water level is mainted. It is giving out 0.74 m3 water per minute through a circular opening of 2 cm radius in its wall. The depth of the center of the opening from the level of water in the tank is close to :
A
6.0 m
B
4.8 m
C
9.6 m
D
2.9 m

## Explanation Here water level is kept same all the time. So, the amount of water remove from the hole put in the tank the top to keep the water level same.

$\therefore$   Inflow volume role = outflow volume

$\Rightarrow$   ${{0.74} \over {60}} = Av$

$\Rightarrow$   ${{0.74} \over {60}} = \pi {r^2}\left( {\sqrt {2gh} } \right)$

$\Rightarrow$   ${{0.74} \over {60}} = \pi \left( {4 \times {{10}^{ - 4}}} \right) \times \sqrt {2gh}$

$\Rightarrow$   $\sqrt {2gh}$ = ${{740} \over {24\pi }}$

$\Rightarrow$   2gh = ${{740 \times 740} \over {24 \times 24 \times {\pi ^2}}}$

$\Rightarrow$   h = 4.8 m
2

### 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
3

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

### 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