मराठी

Assertion (A): A hemispherical bowl is made of steel of thickness 1 cm. If the inner radius of the bowl is 5 cm then the volume of steel used is $$\left(\frac{182}{3}\pi\right)\ \mathrm{cm}^{3}$$.

Advertisements
Advertisements

प्रश्न

Assertion (A): A hemispherical bowl is made of steel of thickness 1 cm. If the inner radius of the bowl is 5 cm then the volume of steel used is $$\left(\frac{182}{3}\pi\right)\ \mathrm{cm}^{3}$$.

Reason (R): The volume of steel is calculated as the difference between the volumes of two hemispheres with radii 6 cm and 5 cm.

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
Advertisements

उत्तर

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 outer radius is $$5 + 1 = 6\ \mathrm{cm}$$. The volume of steel is: $$V = \frac{2}{3}\pi(6^{3} - 5^{3})$$ $$V = \frac{2}{3}\pi(216 - 125) = \frac{182}{3}\pi\ \mathrm{cm}^{3}$$

Step 2 – Reason: R is true. The steel occupies the region between the outer and inner hemispheres, so its volume is their volume difference.

Step 3 – Link: Applying the difference-of-hemispheres method with radii 6 cm and 5 cm gives exactly the assertion; R correctly explains A.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Additional Questions - Volume and Surface Areas of Solids [पृष्ठ १०४८]

APPEARS IN

आर. एस. अग्रवाल Mathematics [English] Class 10
पाठ 20 Additional Questions
Volume and Surface Areas of Solids | Q 12. | पृष्ठ १०४८
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×