Two blocks of mass m and M are connected by amasseless thread which passes over a fixed massless pulley. The table under M is smooth. The force exerted by the pulley on the table is

Options

A. `T=(2mM)/((M-m))g,a(Mm)/((M+m))g`
B. `T=(2mM)/((M+m))g,a=((M-m)/(M+m))g`
C. `T=((M-m)/(M+m))g,a=((2mM)/((M+m)))g`
D. `T=((Mm)/(M+m)_g,a=((2mM)/((M+m)))g`

Answer

B

Explanation

For the particle of mass m,
T-mg=ma ......(i)
for the particle of mass M,
Mg-T=Ma ....(ii)

Add (i) and (ii) to eliminate T.
`Mg-mg=Ma+md`
`g(M-m)=a(M+m)`
or `a=((M-m)/(M+m))g`.....(iii)
Now `T-mg=mxx((M-m)/(M+m))g`
or `T=mg+mg((M-m)/(M+m))g`
or `T=(mg M+m^(2)g+mgM-m^(2)g)/((M+m))`
or `T=(2mM)/((M+m))g`.....(iv)