A solid iron sphere of diameter 42 cm is dropped into a cylindrical drum partly filled with water. If the radius of drum is 1.4 m, by how much will the surface of water be raised?
Answers
Step-by-step explanation:
We can solve this by taking logarithms of both sides. So,
log 4x = log 15
Now using the laws of logarithms, and in particular log An = n log A, the left hand side can be
re-written to give
x log 4 = log 15
This is more straightforward. The unknown is no longer in the power. Straightaway, dividing both
sides by log 4,
x =
log 15
log 4
This value can be found from a calculator. Check that this equals 1.953 (to 3 decimal places).
Example
Solve the equation 6
x = 2x−3
.
Solution
Take logarithms of both sides.
log 6x = log 2x−3
Now use the laws of logarithms.
x log 6 = (x − 3) log2
Notice now that the x we are trying to find is no longer in a power. Multiplying out the brackets
x log 6 = x log 2 − 3 log 2
Rearrange this equation to get the two terms involving x on the right hand side:
3 log 2 = x log 2 − x log 6
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Factorise the right hand side by extracting the common factor of x.
3 log 2 = x(log 2 − log 6)
= x log
1
3
using the laws of logarithms. And finally x =
3 log 2
log
1
3
.
This value can be found from a calculator. Check that this equals −1.893 (to 3 decimal places).
Example
Solve the equation ex = 17.
Solution
We could proceed as in the examples above. However note that the logarithmic form of this expression is loge 17 = x from which, with the use of a calculator, we can obtain x directly as 2.833.
Example
Solve the equation 102x−1 = 4.
Solution
The logarithmic form of this equation is log10 4 = 2x − 1 from which
Answer:
Step-by-step explanation:
Let r & R be the radius of the sphere and cylindrical vessel respectively.
Now, 2r=18cm⇒r=9cm
2R=36cm⇒R=18cm
Let the rise in water level in the cylindrical vessel be
′
h
′
cm.
Volume of sphere =
3
4
πr
3
Volume of liquid displaced in the cylindrical vessel =πR
2
h
If the sphere is completely submerged in the vessel, then volume of liquid displace in the cylindrical vessel = Volume of the sphere
∴πR
2
h=
3
4
πr
3
⇒(18)
2
×h=
3
4
×(9)
3
⇒h=
3×(18)
2
4×(9)
3
=3cm
Thus, the rise in water level in the cylindrical vessel is 3cm