[tex]\pink{\underline{\large{\underline{\mathfrak{\boxed{Question}}}}}}[/tex]
A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
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Answer:
Actual length of bacteria is 1 × 10^(-4) cm.
and, the enlarged length of bacteria = 2 cm.
Step-by-step explanation:
Given,
Length of bacteria is 5 cm if its photograph is enlarged 50,000 times
Finding the actual length of bacteria
Let the actual length of bacteria = x cm
Since bacteria is enlarged 50,000 times to 5 cm
50,000 × Actual length = 5 cm
50,000 × x = 5
x = (5 / 50,000) cm
x = (1 / 10,000) cm
x = 1 / 10⁴ cm
x = 1 × 10^(-4)
Actual length of bacteria is 1 × 10^(-4) cm.
Now, our question is
If the photograph is enlarged 20,000 times only, what would be its enlarged length?
Finding the enlarged length
Now, given that
Length of bacteria is 5 cm if its photograph is enlarged 50,000 times
Let y cm be the length of the bacteria if the bacteria is enlarged 20,000 times
As photograph is enlarged,
The length of bacteria also increases,
Therefore, the length and enlarged photograph are in direct proportion
(50,000 / 5) = (20,000 / y)
10,000 = (20,000 / y)
10,000 × y = 20,000
y = 20,000 / 10,000
y = 2 cm
Therefore, the enlarged length of bacteria = 2 cm.
[tex]\sf\underline{Answer}[/tex]
→ Given: 50,000 times enlarged, the photograph of a bacteria attains a length of 5 cm.
→ To find: The enlarged length of the bacteria when the photograph is enlarged 20,000 times.
→ Let us assume the actual length of the bacteria as 'x' cm.
We get the relation:
[tex]\frac{\sf50,000}{\sf1}=\frac{\sf5\:cm}{\sf x}[/tex]
Solving the equation:
[tex]\sf x=\frac{\sf5\:cm}{\sf50,000} = \sf1\times\sf10^-^4\:cm[/tex]
→ We get the actual length of the bacteria as [tex]\sf10^-^4\:cm[/tex] which is too small to be observed through naked eyes.
→ Let us assume the length of bacteria when the photograph is enlarged 20,000 times as 'y.'
We get the relation:
[tex]\frac{\sf20,000}{\sf50,000} = \frac{\sf y}{\sf5\:cm}[/tex]
Solving for y:
[tex]\sf y=\frac{\sf20,000\:\times\:\sf5\:cm}{\sf50,000} = \sf2\:cm[/tex]
So, if the photograph is enlarged 20,000 times, the enlarged length would be 2 cm.
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