Question

An empty box is put on the pan of a physical balance and the scale is adjusted to zero by counterpoising. A stream of small identical beads each of mass m are then dropped into the box from a height h at a constant rate of n beads per second, if the collision between the beads and the box is completely inelastic, find the reading of the scale, t second after the beads begin to fill the box:
A. \[nm\sqrt{\dfrac{2h}{g}t}\]
B. \[\dfrac{nmt}{gh}\]
C. \[nmg(1+\sqrt{\dfrac{2h}{g}})\]
D. \[nmght\]

Answer 1

Hint:To solve the above question where we have to find the reading on the scale of the pan, we will use some formulas and equations like the newton’s second law of motion and law of conservation of momentum so as to find the force exerted by the beads on the pan.

Complete step-by-step solution:
In the above question we have to find the reading of the scale in physical balance produced by the force exerted by the beads on the pan. If the mass of the beads is m and we have n number of beads dropping per second
So, we will use the following equation to find the answer of our question:
Force on a pan is equal to force due to weight of beads plus force due to collision of beads.
Now, we will first find the force on the pan due to weight of beads,
We know that weight of one bead is equal to \[mg\]by using Newton’s second law of motion as the mass of bead is m
We have n number of beads so, the force on the pan due to weight of beads is equal to
\[n\times mg=nmg\]
Now, we will find the force on the pan due to the collision of the beads so, we will use the law of change in momentum for the beads.
And we know that change in momentum of one bead is equal to
\[=\dfrac{mv-0}{t}\]
Change in momentum of n number of beads is given as
\[=n\times (\dfrac{mv-0}{t})=n(\dfrac{mv-0}{t})\]
Now, we will further proceed towards our solution and find out that the force on the pan which is equal to the force due to collision \[+\]the force due to collision of beads
\[=nmg+n(\dfrac{mv-0}{t})\]
We know that \[v=\sqrt{2gh}\]
So, we can also say that,
\[=nmg+n(\dfrac{m\sqrt{2gh}}{t})\]
Also, we know that for \[t=0\]
\[=nmg+n(m\sqrt{2gh})\]
\[=nmg(1+\sqrt{\dfrac{2gh}{h}})\]
\[=nmg(1+\sqrt{\dfrac{2h}{g}})\]
With the above solution we can clearly say that the option (c) is the correct answer.

Note:In the above given question where we had to find the force on pan so as get the reading on the physical balance we have added the force of beads due to collision by using change in momentum because the momentum in beads is transferred to the pan and thus increasing the force on the pan.