Question

A cube of mass ‘m’ and side ‘x’ rests on the floor. An another cube of mass ‘2m’ and side ‘3x’is placed on the same floor near the first cube. The ratio of the pressures exerted by the firstcube and second cube on the floor is -
(1) 1 : 9 (2) 9 : 2 (3) 2 : 3 (4) 3 : 4

Answer 1

Answer:

Answer is option c buddy

Answer 2

Given: A cube of mass m and side x, another cube of mass 2m and aide 3x

To find: Ratio of the pressure exerted by first cube and second cube on the floor

Solution: Let the cube be A with mass m and let the other cube of mass 2m be B.

we know that only one side of both the cube will be touching the floor, and only that area will be exerting some pressure on it.

Therefore, area of A = side^2= x^2

area of B = (3x)^2= 9x^2

we know that, pressure exerted= force applied/Area of cross section

here in this case, force applied will be mass of the cube× acceleration due to gravity

let the pressure exerted by cube A be Pa and pressure exerted by cube B be Pb

we need to find pa/pb

Pa/pb= (Ma×g/x^2)/(Mb×g/9x^2)

Pa/Pb= 9/2

Therefore, ratio of the pressures exerted by the first cube and second cube on the floor is 2) 9 : 2