In figure, the blocks A, B and C each of mass m have acceleration a1, a2 and a3 respectively. F1 and F2 are external forces of magnitude 2 mg and mg respectively. Then

Question

In figure, the blocks A, B and C each of mass m have acceleration a1, a2 and a3 respectively. F1 and F2 are external forces of magnitude 2 mg and mg respectively. Then (A) a1=a2=a3 (B) a1 > a3 > a2 (C) a1=a2, a2=a3 (D) a1=a2, a1=a3

Tardigrade Admin · Accepted Answer

For A, T=f=2 m g 2 m g-m g =m a1 ∴ a1=g For B, From force diagram shown in figure,

2 mg - mg = 3 ma2 a2 = (g/3) For discussion of motion of the point mass, we assume that for C

∴ 2 m g-m g=2 m a3 ∴ a3=(g/2) So, a1 > a3 > a2

Q. In figure, the blocks $A, B$ and $C$ each of mass $m$ have acceleration $a_{1}, a_{2}$ and $a_{3}$ respectively. $F_{1}$ and $F_{2}$ are external forces of magnitude $2 m g$ and $m g$ respectively. Then

$a_{1} = a_{2} = a_{3}$

$a_{1} > a_{3} > a_{2}$

$a_{1} = a_{2}, a_{2} = a_{3}$

$a_{1} = a_{2}, a_{1} = a_{3}$

Q. In figure, the blocks A,B and C each of mass m have acceleration a1​,a2​ and a3​ respectively. F1​ and F2​ are external forces of magnitude 2mg and mg respectively. Then

a1​=a2​=a3​

a1​>a3​>a2​

a1​=a2​,a2​=a3​

a1​=a2​,a1​=a3​

For A,T=f=2mg 2mg−mg=ma1​ ∴a1​=g For B, From force diagram shown in figure, 2mg−mg=3ma2​ a2​=3g​ For discussion of motion of the point mass, we assume that for C ∴2mg−mg=2ma3​ ∴a3​=2g​ So, a1​>a3​>a2​

Q. In figure, the blocks $A, B$ and $C$ each of mass $m$ have acceleration $a_{1}, a_{2}$ and $a_{3}$ respectively. $F_{1}$ and $F_{2}$ are external forces of magnitude $2 m g$ and $m g$ respectively. Then

$a_{1} = a_{2} = a_{3}$

$a_{1} > a_{3} > a_{2}$

$a_{1} = a_{2}, a_{2} = a_{3}$

$a_{1} = a_{2}, a_{1} = a_{3}$