Question

The weight of a rider driving scooter is assumed to be evenly distributed on both the tyres. The area of contact of each tyre with the ground is $10c{m^2}$ . If the pressure inside the tyres is $2$ bar, find the mass of the rider. ( $g = 10m{s^{ - 2}}$ )

Answer 1

Hint : Here we will first observe the given data, convert all the given data in the same type of system. Here we will use the formula of force relating to pressure and area and also the formula of force relating to mass and acceleration to find the mass of the rider.

Complete Step By Step Answer:
Here, we are given the contact area of the tyre in centimetres so will convert it in the form of the meters.
Contact area of each tyre $= 10{\text{ c}}{{\text{m}}^2} = {10^{ - 3}}{m^2}$
Therefore, the net area of the contact of tyres $= 2 \times {10^{ - 3}}{m^2}$
Also, pressure $P = 3{\text{ bar = 3}} \times {\text{1}}{{\text{0}}^5}Pa$
Now, the force can be given as the product of the pressure and the area.
Force $=$ Pressure $\times$ Area
Place the given values in the above equation –
$F = 3 \times {10^5} \times 2 \times {10^{ - 3}}N$
Simplify the above equation –
$F = 6 \times {10^2}N$
Force on both the tyres $= 600\;{\text{N}}$
Now, using the formula of the force which states that the force is the product of mass and area of contact.
Hence, the mass of the rider can be given by $= \dfrac{{600}}{{2 \times 10}}$
Common factors from the numerator and the denominator cancels each other.
Mass of the rider $= 30{\text{ kg}}$
This is the required solution.

Note :
Always remember the units, derived units and SI units of the physical quantities to get the relation between the two or more physical quantities. Always check the given units and the units asked in the solutions. All the quantities should have the same system of units. Simplification using mathematical operations to be done carefully. Rest goes well in these types of problems.