# Surface Areas & Volumes | Exercise 13.4 | Q4

###### Class 9 Maths | NCERT Solution to Exercise | Surface Area of Sphere

Question 4: The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

## NCERT Solution to Class 10 Maths

### Explanatory Answer | Exrecise 13.4 Question 4

#### Step 1: Compute Surface Area of Balloon Before Filling Air

Radius of the spherical balloon initially, r1 = 7 cm
Surface area of the balloon initially = 4 × π × r12
= 4 × π × 72
= 4 × π × 49

#### Step 2: Compute Surface Area of Balloon After Filling Air

Radius of the spherical balloon after air is pumped, r2 = 14 cm
Surface area of the balloon after air is pumped = 4 × π × r22
= 4 × π × 142
= 4 × π × 196

#### Step 3: Compute Ratio Before & After Filling Air

Ratio of the two scenarios = $\frac{\text{Surface area of the balloon before air is pumped}}{\text{Surface area of the balloon after air is pumped}}$
= $\frac{4 × π × 49}{4 × π × 196}$
= $\frac{1}{4}$
Ratio of surface areas of the balloon in the two scenarios = 1 : 4

###### Free CBSE Online CoachingClass 9 Maths

Register in 2 easy steps and
Start learning in 5 minutes!