हिंदी

Assertion (A): The total surface area (internal and external) of a hemispherical bowl having radii $$r_{1}$$ and $$r_{2}\ (r_{2} > r_{1})$$ is $$3\pi r_{1}^{2} + \pi r_{2}^{2}$$.

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प्रश्न

Assertion (A): The total surface area (internal and external) of a hemispherical bowl having radii $$r_{1}$$ and $$r_{2}\ (r_{2} > r_{1})$$ is $$3\pi r_{1}^{2} + \pi r_{2}^{2}$$.

Reason (R): The total surface area of a solid hemisphere of radius $$r$$ is $$3\pi r^{2}$$.

विकल्प

  • 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
अभिकथन और तर्क
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उत्तर

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

Explanation:

Step 1 – Assertion: A is false. The bowl’s total area includes the outer curved hemisphere, inner curved hemisphere and the top annular rim: $$2\pi r_{2}^{2} + 2\pi r_{1}^{2} + \pi(r_{2}^{2} - r_{1}^{2}) = 3\pi r_{2}^{2} + \pi r_{1}^{2}$$. This differs from the expression in A.

Step 2 – Reason: R is true. The total surface area of a solid hemisphere of radius $$r$$ is $$3\pi r^{2}$$.

Therefore, the correct option is: Assertion (A) is false and Reason (R) is true.

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अध्याय 20: Additional Questions - Volume and Surface Areas of Solids [पृष्ठ १०४९]

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आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 20 Additional Questions
Volume and Surface Areas of Solids | Q 14. | पृष्ठ १०४९
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