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A point light source shines equally in all directions. The light intensity measured at a sensor is inversely proportional to the square of the distance from the source.

When the distance is 6 meters, the intensity is 8 units.

Select for Distance a distance in meters and select for Intensity an intensity in units that would be jointly consistent with the given information, Make only two selections, one in each column.


light intensity is inversely proportional to square of distance from source

8 = k / ( 6)^2
k = 8 * 36 ; 288
I = 288/d^2
put values of I & D
if D is 12 then I we get is 2

Distance is 12 & Intensity is 2
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Let Intensity be I and distance be d
I proportional to 1/d^2
I * d^2 = k where k is a constant

k = 6^2 * 8 = 36 * 8 = 288

Of the answer choices d = 12 and I = 2 works as 288 = 2 * 12^2

d = 12
I = 2
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A point light source shines equally in all directions. The light intensity measured at a sensor is inversely proportional to the square of the distance from the source.

When the distance is 6 meters, the intensity is 8 units.

Select for Distance a distance in meters and select for Intensity an intensity in units that would be jointly consistent with the given information, Make only two selections, one in each column.
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A point light source shines equally in all directions.
The light intensity measured at a sensor is inversely proportional to the square of the distance from the source.
When the distance is 6 meters, the intensity is 8 units.

Intensity(I) = k * 1/d^2 ; where d = distance from the source in meters

8 = k/6^2 ; k = 8*6^2 = 288

I = 288/d^2

d = 2; I = 288/2^2 = 72; Not an option
d = 4; I = 288/4^2 = 18; Not an option
d = 12; I = 288/12^2 = 2; Possible

Distance12
Intensity2
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A point light source shines equally in all directions. The light intensity measured at a sensor is inversely proportional to the square of the distance from the source.

When the distance is 6 meters, the intensity is 8 units.

Select for Distance a distance in meters and select for Intensity an intensity in units that would be jointly consistent with the given information, Make only two selections, one in each column.

according to info
k = intneisty * distance * distance
k = 8 * 6 * 6 = 288
for distance = 12, intensty = 288/ (12*12)
= 2

answer is 12, 2
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A point light source shines equally in all directions. The light intensity measured at a sensor is inversely proportional to the square of the distance from the source.

When the distance is 6 meters, the intensity is 8 units.

Select for Distance a distance in meters and select for Intensity an intensity in units that would be jointly consistent with the given information, Make only two selections, one in each column.


I = k/D^2

8 * 36 = k

k = 288

So we have to find a value for which the k comes out to be 288

Let say intensity =2

D^2 = 288/2 = 144

D = 12

We do have this combination, hence we don't have to test out any other values.

Distance = 12
Intensity = 2
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intensity = k / (d)^2
8=k / 36
k= 288

now we can try value of intensity with given value of k and find distance

when i=2, d= 12
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A point light source shines equally in all directions. The light intensity measured at a sensor is inversely proportional to the square of the distance from the source.

When the distance is 6 meters, the intensity is 8 units.

Select for Distance a distance in meters and select for Intensity an intensity in units that would be jointly consistent with the given information, Make only two selections, one in each column.
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A point light source shines equally in all directions. The light intensity measured at a sensor is inversely proportional to the square of the distance from the source.

When the distance is 6 meters, the intensity is 8 units.

Select for Distance a distance in meters and select for Intensity an intensity in units that would be jointly consistent with the given information, Make only two selections, one in each column.
Light intensity is INVERSELY proportional to square of the distance.

8 is inversely proportional to 36

8 = k(1/36)

K = 8*36

K= 4*2*9*4

Light intensity = (4*2*9*4) * (1/d^2)

If d =2, then light intensity = 4*2*9 = 72 ( not in the options)

If d = 4, then light intensity = 2*9 = 18 ( not in the options).

If d = 12, then light intensity = 2. ( Matches with the options).

d=12 , LI = 2
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Let Intensity = K/distance^2 hence 8 = K/36 -------> K = 8*36 = 288

From options for distance 12 Intensity = 288/12^2 = 2
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A point light source shines equally in all directions. The light intensity measured at a sensor is inversely proportional to the square of the distance from the source.

When the distance is 6 meters, the intensity is 8 units.

Select for Distance a distance in meters and select for Intensity an intensity in units that would be jointly consistent with the given information, Make only two selections, one in each column.
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Intensity (I) is inversely proportional to the square of the distance (d)
I = x / d^2

Find x: 8 = x/36 -> x = 8*36 = 288

Check answer choices considering that as d increases, I decreases.

The only possible answers are distance = 12, intensity = 2
2 = 288/12^2 = 288/144
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I= K/(d^2)
8=K/(6^2)
K=8*36

I = (8*36)/d^2

if d=12, I =2 satisfy.

Ans 12 & 2
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I inversely proportional to square of distance

\(I1 * D1^{2} = I2 * D2^{2}\)

\(8 * 6^{2} = I2 * D2^{2}\)

\(2*2^{2} * 6^{2} = I2 * D2^{2}\)

\(2 * 12^{2} = I2 * D2^{2}\)

Distance = 12, Intensity = 2

Ans: 12, 2 or C, A
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A point light source shines equally in all directions. The light intensity measured at a sensor is inversely proportional to the square of the distance from the source.

When the distance is 6 meters, the intensity is 8 units.

Select for Distance a distance in meters and select for Intensity an intensity in units that would be jointly consistent with the given information, Make only two selections, one in each column.
Intensity I varies as [1][/d^2], so I = [k][/d^2]

Given d = 6, I = 8
8 = k/36
k = 288

so I = 288/d^2

Lets pick a distance from the left column that makes I match one of {2,4,12,16,24,32}

Lets take 12
I = 288/12^2 = 288/144 = 2

Distance = 12
Intensity = 2
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let intensity and distance be represented by I & D respectively.
given I is inversely proportional to the square of the D from the source.
I=k/(D^2) or k=I*D^2 k-constant
k=8*6^2=8*36
to find another set of the I and D we just need to rearrange above calculated value of k so that its the product of a perfect square denoting distance and an integer denoting I.
k=8*36=9*4*4*2=12^2*2
I=2 and D=12 are the correct choices
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A point light source shines equally in all directions. The light intensity measured at a sensor is inversely proportional to the square of the distance from the source.

When the distance is 6 meters, the intensity is 8 units.

Select for Distance a distance in meters and select for Intensity an intensity in units that would be jointly consistent with the given information, Make only two selections, one in each column.

If intensity is inversely proportional to distance squared, that means:

\(intensity=\frac{k}{distance^2}\), where k is a constant.

Subbing in the example we have, we get:
\(8=\frac{k}{6^2}\) which results in \(k=288\).

So now we're looking for:
\(intensity=\frac{288}{distance^2}\) which can be rearranged into \(intensity*distance^2=288\).

The prime factorization of \(288=2^5*3^2\)

Looking at the answer choices, only 12 and 24 would contribute a 3, so they are most likely to be the value for distance, because we need a \(3^2\).

\(24^2=3^2*2^6\). 24 gives too many factors for 2, so distance has to be 12.

\(12^2=3^2*2^4\), so we need one more 2 - thus intensity needs to be 2.
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The inverse relationship can be expressed as Intensity (I)=1/distance squared (d^2)x Constant (k)
Given I=8 and d=6, we can find k
8=1/6^2xk
k=288
So intensity=288/d^2. From here we can test cases using the answer choices starting with intensity of 2
2=288/d^2
d^2=288/2=144
d=12
Given we have both 2 and 12 among the answer choices, there is no need for further testing so
Intensity = 2
Distance= 12


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A point light source shines equally in all directions. The light intensity measured at a sensor is inversely proportional to the square of the distance from the source.

When the distance is 6 meters, the intensity is 8 units.

Select for Distance a distance in meters and select for Intensity an intensity in units that would be jointly consistent with the given information, Make only two selections, one in each column.
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A point light source shines equally in all directions. The light intensity measured at a sensor is inversely proportional to the square of the distance from the source.

When the distance is 6 meters, the intensity is 8 units.

Select for Distance a distance in meters and select for Intensity an intensity in units that would be jointly consistent with the given information, Make only two selections, one in each column.

As per the relation

Intensity = k / (distance)^2

find K and the answer will be easy to find

when intensity is 8 units = K / 6 x 6

k = 36 x 8
hence K = 288

plug in the options.


when distance is = 12, intensity = 288/12x12 ..hence 2 is your answer
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Intensity I =k/distance d^2

K=I×d^2=8×6^2=288

If d=12 then I =288÷144=2

Answer 12, 2
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