# If the radius of a sphere is measured as 7m with an error of 0.02 m, then find the approximate error in calculating its volume

**Solution:**

We can use differentials to calculate small changes in the dependent variable of a function corresponding to small changes in the independent variable

Let r be the radius of the sphere and Δr be the error in measuring the radius

Then, r = 7 m and Δr = 0.0 2m

Now, the volume V of the sphere is given by,

V = 4/3π r^{3}

Therefore,

dV/dr = 4πr^{2}

Hence,

dV= (dV//dr) Δr

= (4π r^{2})(0.02)

= 4π (7)^{2} (0.02)

= 3.92 π

Thus, the approximate error in calculating its volume is 3.92π m^{3}

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.4 Question 6

## If the radius of a sphere is measured as 7m with an error of 0.02 m, then find the approximate error in calculating its volume

**Summary:**

Given that the radius of a sphere is measured as 7m with an error of 0.02 m. Thus, the approximate error in calculating its volume is 3.92π m^{3}