Question

In a bank principal increases at the rate of r% per year find the value of r of Rs.100 double itself in 10 years `(log e^2 = 0.6931)`

पर्याय

• 6.931

• 7.931

• 8.931

• 9.931

Answer 1

6.931

Explanation:

Let 'P' be the principal rate of interest = r%

∴ `(dp)/(dt) = r/100` P ⇒ `(dp)/p = r/100  dt`

Intergrating `int  (dp)/p = r/100 int  dt`

or `log p = r/100 t + log c`

or `log p - log c = r/100 t` or `log = p/c = r/100 t`

⇒ `p/c = e^(r/100)` or P = C `e^(r/100 t)`  ......(1)

Initially, t = 0, p = 100

∴ 100 = `ce^0`

∴ `c` = 100

Equation (1) becames P = 100 =  `e^(r/100 t)`

Now t = 10, p = 20

200 = 100 `e^(r/100 t)` or 2 = `e^(r/10)`

∴ `r/10 = log e^2` = 0.6931

∴ `r = 10 xx 0.6931` = 6.931

Thus the value of r = 6.931