WBJEE 2019 Physics Solved Paper

Given, cross sectional area of nozzle
A=‌

5

√21

×10−3m2
and rate of heat transfer Q=‌

1

10

=10−1m3/s
∵ Heat transfer Q = AV
When it hits a rigid wall then maximum possible increase in temperature of water can be expressed as
‌

1

2

mv2=ms∆T....(i)
where,
m = mass,
s = specific heat of water
=1cal/g=42×103J/kg
and ∆T= increase in temperature
From Eq⋅(i),∆T=‌

1

2

‌

v2

s

=‌

(2√21×101)2

2×4.2×103

=‌

84×102

8.4×103

=1∘C