The mouse and the lion

First we convert all measurements to grams: the mouse has mass $100 \; \text{g}$ and the lion has mass $200000 \; \text{g}.$

Let the lengths of the arms with the mouse and lion on them be $m$ and $l,$ respectively. Then $100m = 200000l,$ by the basic properties of balances. Dividing both sides by $100,$ we get $m = 2000l.$ Therefore, the balance arm with the mouse on it is $\boxed{2000}$ times longer than the balance arm with the lion on it.

If the balance is perfectly balanced, then the sum of all torques must be zero.

If for the mouse we call:

$m_1 = 0.1$ its mass in Kg

$l_1$ the length of the arm on the mouse's side.

and for the lion:

$m_2 = 200$ its mass in Kg

$l_2$ the length of the arm on the lion's side.

Then we have: $l_1 (0.1 \times g) - l_2 (200 \times g) = 0$ $l_1 = l_2 \frac{200 g}{0.1 g} = 2000 l_2$

Therefore $l_1$ must be 2000 times longer than $l_2$

* $Weight Of Lion\times x$ * =_ $Weight Of Mouse\times x.m$ so, $200000g \times x$ = $100g \times x.m$ now m= \frac{\( 200000g \times x }{100g \times x )} ) so m=2000

The moment of mouse and lion must be equal so that their will be balance in the pan.Moment is equal to force multiply by moment arm. 0.1x d1=200xd2 by algebraic solution you will get d1=2000d2.

M1 x L1 = M2 x L2 0.1kg L1 = 200kg L2 L1 = 2000L2

since the torque has to be balanced, and torque is proportional to arm length, the lion is 2000 times the weight of mouse, so the arm length of the mouse have to be 2000 times of the lion to make the balance balanced.

just put the number in.

• mouse's balance arm is 200kg/100g times longer than lion's balance arm. Mouse's balance arm is 200000/100 times longer than lion's balance arm. Mouse's balance arm is 2000 or 2E3 times longer than lion's balance arm

• The lion has a mass of $200 kg$
• The mouse has a mass of $100 g$ = $0.1 kg$

AFAIK gravity does not matter in this question because it is assumed that both the mouse, the lion and the pan balance are on the same surface; alternatively, the force of gravity can be taken as $1 m/s^{2}$ , making the lion's weight $200N$ and the mouse's weight $0.1N$ (unchanged)

If the lion's perpendicular distance from the center of the pan balance is 1 metre then the moment of the lion about the center of the pan balance is $200N * 1 m = 200N m$ .

The question states that the pan balance is perfectly balanced (is in equilibrium), so that must mean that the total moment about the center of the pan balance is $0N m$ .

Thus, the mouse's moment about the center of the pan balance must also be $200N m$ ; thus, $200N m = 0.1N * d m$ , where $d$ is the (perpendicular) distance from the center of the pan balance.

$200 / 0.1 = d = 2000$ ; units do not matter because $d$ is relative as far as the question is concerned (i.e. if the arm of the pan balance holding the lion is 1 cm from the center of the pan balance, the length of the arm of the balance holding the mouse is 2000 times that distance, or 2000 cm from the center of the pan balance.)

Thus, the balance arm with the mouse is $2000$ times longer than the balance arm with the lion.

Sine the balance is perfectly balanced,it implies that the torques produced by the two masses is the same,let arm length with mouse be x and that with lion be y. so, 0.1 x X = 200 x Y so, x = 2000y