MHT CET 2021 Physics Solved Paper

(b) The maximum tension in the string will be at lowest point i.e.,
Tmax=‌

mv12

I

+mg
and minimum tension in the string will be the highest point
i.e., Tmin=‌

mv22

I

−mg
Therefore, ‌

Tmax

Tmin

=‌

‌ mv12 I +mg

mv22 I −mg

=4
⇒‌‌‌

v12+gl

v22−gl

=4‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(i)
As we know, v12=v22+4gl‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(ii)
So from Eqs. (i) and (ii), we get
v22+4gl+gl=4v22−4gl
⇒‌‌3v22=9gl
v22=3gl=3×10×‌

10

3

⇒‌‌v22=100⇒v2=10m∕s