मराठी

Assertion (A): The total surface area of a solid hemisphere of diameter $$d$$ is $$\frac{3}{4}\pi d^{2}$$. Reason (R): If the diameter of a solid hemisphere is halved

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प्रश्न

Assertion (A): The total surface area of a solid hemisphere of diameter $$d$$ is $$\frac{3}{4}\pi d^{2}$$.

Reason (R): If the diameter of a solid hemisphere is halved then its total surface area reduces to 25% of the original value.

पर्याय

  • 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
विधान आणि तर्क
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उत्तर

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

Explanation:

Step 1 – Assertion: A is true. Since the total surface area of a solid hemisphere is $$3\pi r^{2}$$ and $$r = \frac{d}{2}$$ : $$3\pi r^{2} = 3\pi\left(\frac{d}{2}\right)^{2} = \frac{3}{4}\pi d^{2}$$

Step 2 – Reason: R is true. Halving the diameter also halves the radius, and surface area scales with the square of the radius; hence the new area is $$\left(\frac{1}{2}\right)^{2} = \frac{1}{4}$$ of the original, or 25%.

Step 3 – Link: R describes how surface area changes when the diameter is halved; it does not establish the formula stated in A.

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पाठ 20: Additional Questions - Volume and Surface Areas of Solids [पृष्ठ १०४६]

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आर. एस. अग्रवाल Mathematics [English] Class 10
पाठ 20 Additional Questions
Volume and Surface Areas of Solids | Q 3. | पृष्ठ १०४६
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