TS EAMCET 11-Sep-2020 Shift 1 Solved Paper

The given situation is shown in the following figureEffective mass of the system of n masses placed on the table,
M=m+‌

m

2

+‌

m

6

+...+‌

m

n!

=m(1+‌

1

2

+‌

1

6

+...+‌

1

n!

)
=m(‌

1

1!

+‌

1

2!

+‌

1

3!

+...+‌

1

n!

)
M=m(e−1)
The given system of masses can be redrawn asIf a be the acceleration of system of masses, then according to Newton's second law of motion,
T=Ma⇒T=m(e−1)a....(i)
and mg−T=ma⇒T=mg−ma
T=m(g−a)...(ii)
From Eq. (i) and Eq. (ii), we get
m(e−1)a=m(g−a)⇒a=g∕e