JEE Main 27-Aug-2021 Shift 1 Solved Paper

Given, initial load, M1=185‌kg
Initial pressure, p1=76‌cm of Hg
Initial temperature, T1=27°C=(273+27)K=300K
Final pressure at height, p2=45‌cm of Hg
Final temperature, T2=−7°C=[273+(−7)]K=266K
Volume of balloon is constant, V1=V2=V
We know that, for air,
p=ρ‌RT
Now, for initial condition, we can write
ρ1=ρ1RT1 ...(i)
Similarly, for final condition, we can write
P2=ρ2RT2 ...(ii)
Dividing Eq. (i) by Eq. (ii), we get

p1

p2

=

ρ1T1

ρ2T2

Substituting the given values in above expression, we get

76‌cm‌of‌Hg

45‌cm‌of‌Hg

=

ρ1

ρ2

(

300K

266K

)
⇒

ρ1

ρ2

=

76×266

45×300

We know that, density of balloon can be written as
ρ

M

V

As, volume is constant ρ ∝ M
We can write,

ρ1

ρ2

=

M1

M2

=

76×266

45×300

⇒

M1

M2

=

20216

13500

Substituting the value of M1 in above expression, we get
M2=185×

13500

20216

=123.54‌kg
Thus, the load carried by balloon will be 123.54 kg.