Two masses M1 and M2 are accelerated uniformly on a frictionless surface as shown in figure. The ratio of the tensions (T1/T2) is

Question

Two masses M1 and M2 are accelerated uniformly on a frictionless surface as shown in figure. The ratio of the tensions (T1/T2) is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/86222cb9aafb2fc6bf17faba57cd4df3-.png" /> (A) (M1/M2) (B) (M2/M1) (C) ((M1+M2)/M2) (D) (M1/(M1+M2))

Tardigrade Admin · Accepted Answer

Let a be the common acceleration of the system. The equations of motion of masses M1 and M2 are T1=M1 a ... (i) T2-T1=M2 a ....(ii) Adding (i) and (ii), we get T2=(M1+M2) a ∴ (T1/T2)=(M1/M1+M2)

Q. Two masses $M_{1}$ and $M_{2}$ are accelerated uniformly on a frictionless surface as shown in figure. The ratio of the tensions $\frac{T _{1}}{T _{2}}$ is

$\frac{M _{1}}{M _{2}}$

$\frac{M _{2}}{M _{1}}$

$\frac{( M _{1} + M _{2} )}{M _{2}}$

$\frac{M _{1}}{( M _{1} + M _{2} )}$

Let $a$ be the common acceleration of the system. The equations of motion of masses $M_{1}$ and $M_{2}$ are $T_{1} = M_{1} a ... (i)$
$T_{2} - T_{1} = M_{2} a .... (ii)$
Adding $(i)$ and $(ii)$ , we get $T_{2} = (M_{1} + M_{2}) a$
$∴ \frac{T _{1}}{T _{2}} = \frac{M _{1}}{M _{1} + M _{2}}$

Q. Two masses M1​ and M2​ are accelerated uniformly on a frictionless surface as shown in figure. The ratio of the tensions T2​T1​​ is

M2​M1​​

M1​M2​​

M2​(M1​+M2​)​

(M1​+M2​)M1​​