JEE Main 26-Aug-2021 Shift 1 Solved Paper

The given situation can be shown as below. Points A and B are at same horizontal level.
So, the pressure at A and B will be equal
i.e. pA=pB.... (i)
If p0 is the atmospheric pressure, then from Bernoulli's theorem,
pA=p0−

2T

rA

+(h+y)ρg
pB=p0−

2T

rB

+ρgy
Substituting the value of pA and pB in Eq. (i), we get
p0−

2T

rA

+(h+y)ρg=p0−

2T

rB

+ρgy
⇒2T(

1

rB

−

1

rA

)=ρgy−(h+y)ρg
⇒2T(

1

rB

−

1

rA

)=−ρgh
or2T(

1

rA

−

1

rB

)=cg‌h ...(ii)
Here, rA=2.5‌mm=2.5×10−3m,
and T=7.3×10−2N∕m,
rB=4‌mm=4×10−3m,g=10m∕s2 and ρ=1×103kgm−3
Substituting the given value in Eq. (ii), we get
h=

2×7.3×10−2

1×103×10

(

1

2.5×10−3

−

1

4×10−3

)
h=

2×7.3×10−2

10−3×104

(

1

2.5

−

1

4

)
h=2.19×10−3m=2.19‌mm