Question

1. Two boys A and B are asked to take 4 chairs from the ground floor to a room at the second floor. A takes 30 minutes to do this work while B takes 60 minutes for the same. (a) Name the physical quantity that is same for both boys. (b) Name the physical quantity that is different and find its ratio for the two boys.

Answer 1

(a). Work done is the physical quantity which is same for both boys.

(b).Power is the physical quantity which is different for both boys.

The ratio of power of both boys is 2:1

Step-by-step explanation:

Given that,

Time of boy A= 30 min

Time of boy B = 60 min

(a), We need to find the physical quantity that is same for both boys

Using given that,

We know that,

Work done depends on the force and displacement.

Here, force and displacement is same for both boys.

So, we can say that the work done will be same for both boys.

(b). We need to find the physical quantity that is different for both boys

Using given that,

We know that,

Power is equal to the work done divided by time.

Time is different for both boys so, the power will be different for both boys.

We need to calculate the ratio of power of both boys

Using formula of power

`P=(W)/(t)`

Put the value into the formula

For boy A,

`P_(A)=(W_(A))/(t_(A))`....(I)

For boy B,

`P_(B)=(W_(B))/(t_(B))`....(II)

From equation (I) and (II)

`(P_(A))/(P_(B))=((W)/(30))/((W)/(60))`

`(P_(A))/(P_(B))=(60)/(30)`

`(P_(A))/(P_(B))=(2)/(1)`

Hence, (a). Work done is the physical quantity which is same for both boys.

(b).Power is the physical quantity which is different for both boys.

The ratio of power of both boys is 2:1