JEE Main 25 Feb 2021 Shift 1 Solved Paper

Given, atomic mass of monoatomic gas,
m=4u
Velocity of container, u=30ms−1
Final velocity of container, v=0ms−1
Let ∆T be the change in temperature
and ∆U be the internal energy change.
Therefore,
∆KE=∆U
‌‌=(1∕2)m(v2−u2)=nCv∆T
∆T=‌

1

2

‌

m(v2−u2)

n⋅3∕2R

⇒‌‌‌‌‌‌‌‌‌‌‌‌(∵Cv=‌

3

2

R)
‌‌‌

1

2

×m[v2−u2]=n⋅‌

3

2

⋅R∆T
‌ Now, ‌‌

1

2

×4|02−302|=1×‌

3

2

R∆T
∆T=‌

4×900

3R

=‌

x

3R

‌‌x=3600