Delhi Police Constable practice question
Raj had a cake in the shape of a sphere of radius 8 cm. He then dipped it in chocolate and kept it in the freezer. After some time, he observed that the chocolate formed a layer of thickness 2 cm over the cake. What is the percentage increase of the surface area of the cake after it is dipped in chocolate?
From a Delhi Police Constable General Awareness practice set · General Awareness
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Correct answer: (B) 56.25%
Original radius = 8 cm, new radius after coating = 10 cm. Surface areas are proportional to r², so % increase = (10²−8²)/8²×100 = (100−64)/64×100 = 56.25%.
हिंदी में प्रश्न
राज के पास 8 cm ित्र ा का एक गोल आकार का केक था। उसने उसे चॉकलेट म' डुबोकर ीजर म' रख िदया। कुछ समय बाद उसने देखा िक केक पर चॉकलेट की 2 cm मोटी परत जम जाती है। चॉकलेट म' डुबोने के बाद केक के पृ ीय क्षेत्रफल म' िकतने प्रितशत की वृ द्ध हुई ?
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उत्तर और व्याख्या देखें
सही उत्तर: (B) 56.25%
मूल त्रिज्या = 8 सेमी, लेप के बाद त्रिज्या = 10 सेमी। सतह क्षेत्रफल r² के अनुपातिक होता है, अत: % वृद्धि = (100−64)/64×100 = 56.25%।
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