English

Assertion (A): A solid is in the form of a cone of radius $$r$$ surmounted on a hemisphere of the same radius. If the height of the cone is the same as the diameter of its base

Advertisements
Advertisements

Question

Assertion (A): A solid is in the form of a cone of radius $$r$$ surmounted on a hemisphere of the same radius. If the height of the cone is the same as the diameter of its base then the volume of the solid is $$\frac{4}{3}\pi r^{3}$$.

Reason (R): Total volume of the solid $$=\text{volume of the cone}+\text{volume of the hemisphere}$$

Options

  • Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

  • Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

  • Assertion (A) is true and Reason (R) is false.

  • Assertion (A) is false and Reason (R) is true.

MCQ
Assertion and Reasoning
Advertisements

Solution

Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

Explanation:

Step 1 – Assertion: A is true. The cone height is $$2r$$, so: $$V = \frac{1}{3}\pi r^{2}(2r) + \frac{2}{3}\pi r^{3}$$ $$V = \frac{4}{3}\pi r^{3}$$

Step 2 – Reason: R is true. The total volume of the composite solid is the sum of the volumes of the cone and hemisphere.

Step 3 – Link: Adding those two component volumes, using the stated cone height, gives the volume in A; therefore, R correctly explains A.

shaalaa.com
  Is there an error in this question or solution?
Chapter 20: Additional Questions - Volume and Surface Areas of Solids [Page 1047]

APPEARS IN

R.S. Aggarwal Mathematics [English] Class 10
Chapter 20 Additional Questions
Volume and Surface Areas of Solids | Q 8. | Page 1047
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×