A light spring balance hangs from the hook of the other light spring b...
Explanation:
When a block of mass M kg hangs from the spring balance, it exerts a force equal to its weight on the spring balance.
Let's assume the spring constant of each spring balance to be k1 and k2 respectively.
The force exerted by the block on the upper spring balance is equal to Mg, where g is the acceleration due to gravity.
Similarly, the force exerted by the upper spring balance on the lower spring balance is also Mg.
Therefore, the lower spring balance also reads Mg.
Now, let's assume the extension in the spring of the upper balance to be x1 and that of the lower balance to be x2.
From Hooke's law, we know that the force exerted by a spring is directly proportional to its extension.
Therefore, the force exerted by the upper spring balance on the block is k1x1, and the force exerted by the lower spring balance on the upper spring balance is k2(x1 + x2).
Since the block is in equilibrium, the net force acting on it is zero.
Therefore, we can write:
k1x1 = Mg - k2(x1 + x2)
Solving this equation, we get:
x1 = Mg/(k1 + k2)
x2 = k1M/(k1 + k2)
The reading of the upper spring balance is given by:
Mg/k1 = k2x1/k1 = k2M/(k1 + k2)
The reading of the lower spring balance is given by:
Mg/k2 = k1x2/k2 = k1M/(k1 + k2)
Adding these two readings, we get:
Mg/k1 + Mg/k2 = M(k1 + k2)/(k1 + k2) = M
Therefore, both the scales read M/2 kg each.
A light spring balance hangs from the hook of the other light spring b...
I think option A is correct.