1

### JEE Main 2018 (Online) 16th April Morning Slot

A small soap bubble of radius 4 cm is trapped inside another bubble of radius 6 cm without any contact. Let P2 be the pressure inside the inner bubble and P0, the pressure outside the outer bubble. Radius of another bubble with pressure difference P2 $-$ P0 between its inside and outside would be :
A
12 cm
B
2.4 cm
C
6 cm
D
4.8 cm

## Explanation

Pressure difference inside the inner bubble,

p2 $-$ p1 = ${{4T} \over {{r_2}}}$b . . . . . (1)

And for outer bubble

p1 $-$ p0 = ${{4T} \over {{r_1}}}$ . . . . . . . (2)

$\therefore\,\,\,$ p2 $-$ p0 = 4T $\left( {{1 \over {{r_2}}} + {1 \over {{r_1}}}} \right)$

$\Rightarrow $$\,\,\, p2 - p0 = {{4T} \over r} Here r is the radius of the bubble. \therefore\,\,\, {1 \over r} = {1 \over {{r_2}}} + {1 \over {{r_1}}} \Rightarrow$$\,\,\,$ r = ${{{r_1}{r_2}} \over {{r_1} + {r_2}}}$

= ${{4 \times 6} \over {4 + 60}}$

= 2.4 cm
2

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

A rod, of length L at room temperature and uniform area of cross section A, is made of a metal having coefficient of linear expansion $\alpha$/oC. It is observed that an external compressive force F, is applied on each of its ends, prevents any change in the length of the rod, when its temperature rises by $\Delta$TK. Young's modulus, Y, for this metal is :
A
${F \over {A\alpha \Delta T}}$
B
${F \over {A\alpha (\Delta T - 273)}}$
C
${F \over {2A\alpha \Delta T}}$
D
${{2F} \over {A\alpha \Delta T}}$

## Explanation

We know,

Young's Modulus, Y = ${{Stress} \over {Strain}}$

Stress = ${F \over A}$

Strain = ${{\Delta l} \over l}$

$\therefore$  Y = ${{{F \over A}} \over {{{\Delta l} \over l}}}$

We also know,

$l$f = $l$i (1 + $\alpha $$\Delta T) \Rightarrow lf = li + li\alpha$$\Delta$T

$\Rightarrow$   ${{{l_f} - {l_i}} \over {{l_i}}}$ = $\alpha $$\Delta T \Rightarrow {{\Delta l} \over {{l_i}}} = \alpha$$\Delta$T

$\therefore$   y = ${{{F \over A}} \over {\alpha \Delta T}}$

= ${F \over {A\alpha \Delta T}}$
3

### 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
4

### 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