In the arrangement $M_{1}$ and $M_{2}$ , masses of two blocks $A$ and $B$ , respectively such that $M_{1} > M_{2}$ . A weightless spring balance $S$ is connected by ideal spring passing over ideal pulleys. The reading of the spring balance (in $k g . w t$ ) is:

(M_{1} - M_{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{M_{1} + M_{2}}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 M_{1} M_{2}}{M_{1} + M_{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{4 M_{1} M_{2}}{M_{1} + M_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{2 M_{1} M_{2}}{M_{1} + M_{2}}$
Here, the reading of the spring shows the tension in the string.
For mass $M_{1} : M_{1} a = M_{1} g - T . . . (1)$
For mass $M_{2} : M_{2} a = T - M_{2} g . . . (2)$
$(1) + (2), (M_{1} + M_{2}) a = (M_{1} - M_{2}) g \Rightarrow a = \frac{(M_{1} - M_{2})}{(M_{1} + M_{2})} g$
putting value in $(1)$ , we get
$T = \frac{2 M_{1} M_{2}}{(M_{1} + M_{2})} g N$
as $9.8 N = 1 k g . w t$ , so $T = \frac{2 M_{1} M_{2}}{(M_{1} + M_{2})} k g . w t$ (as $g = 9.8 m / s^{2}$ )

Suggest Corrections

Similar questions

Q. Two masses

m_{1}

and

m_{2}

are attached to a spring balance S as shown in Figure.if

m_{1}

m_{2}

then the reading of spring balance will be

Two spheres of masses m1 and m2 (m1>m2) respectively are tied to the ends of a light, inextensible string which passes over a light frictionless pulley. When the masses are released from their initial state of rest, the acceleration of their centre of mass is: