A test-tube consists of a cylinder with a hemispherical base. The diameter of the hemisphere is 2a and the height of the cylinder is b (internal measurements). Three marks are made on the test tube to indicate when the test-tube is half full, one-quarter full and three quarters full. Find the heights of these three marks above the lowest point of the test-tube in order of half full, one quarter full and three quarters full?

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mi>a</mi> <mo>+</mo> <mfrac> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mfrac> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mi>b</mi> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mi>a</mi> <mo>+</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mi>b</mi> </mrow> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mi>a</mi> <mo>+</mo> <mfrac> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mfrac> <mrow> <mi>a</mi> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mi>a</mi> <mo>+</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mi>b</mi> </mrow> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mfrac> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mi>b</mi> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mi>a</mi> <mo>+</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mi>b</mi> </mrow> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>a</mi> <mo>+</mo> <mfrac> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mfrac> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mi>b</mi> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mi>a</mi> <mo>+</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mi>b</mi> </mrow> </math> </span></p>
2 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
R
7 months ago
Correct Option - 1
Detailed Solution:

Volume of tube = 2 3 π a 3 + π a 2 b

When it is half-full = 1 3 π a 3 + 1 2 π a 2 b

Volume of Cylinder

= 1 3 π a 3 + 1 2 π a 2 b 2 3 π a 3 = 1 2 π a 2 b 1 3 π a 3

Length in the cylinder

= ( 1 2 π a 2 b 1 3 π a 3 ) ÷ π a 2 = 1 2 b 1 3 a

The height above the lowest point

1 2 b 1 3 a + a = 2 3 a + 1 2 b

When it is 1 4  full, volume = 1 6 π a 3 + 1 4 π a b

Volume of cylinder

= 1 6 π a 3 + 1 4 π a 2 b 2 3 π a 3 = 1 4 π a 2 b 1 2 π a 3

Length in the cylinder

= ( 1 4 π a 2 b 1 2 π a 3 ) ÷ π a 2 = 1 4 b 1 2 a

The height above the lowest point

= 1 4 b 1 2 a + a = 1 2 a + 1 4 b

When it is full, volume 

= 3 4 ( 2 3 π a 3 + π a 2 b ) = 1 2 π a 3 + 3 4 π a 2 b

Volume in cylinder

= 1 2 π a 2 + 3 4 π a 2 b 2 3 π a 3 = 3 4 π a 2 b 1 6 π a 3

Length in the cylinder

= ( 3 4 π a 2 b 1 6 π a 3 ) ÷ π a 2 = 3 4 b 1 6 a

The

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