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Question
একটি গোলকের ব্যাসার্ধ পরিমাপে 1.2% ভুল করলে, ঐ গোলকের আয়তনে শতকরা কত ভুল হবে?
2.72%
3.60%
4.12%
5.56%
ANSWER : 2
Descrption
<p style="margin-left:0px;"><strong>3.60%</strong></p><p style="margin-left:0px;">একটি গোলকের আয়তন V হল,</p><p>V = 4/3πr³ </p><p style="margin-left:0px;">যেখানে,</p><ul><li>V = আয়তন (cm³)</li><li>π = 3.14</li><li>r = ব্যাসার্ধ (cm)</li></ul><p style="margin-left:0px;">ব্যাসের পরিমাপে 1.2% ভুল হলে, ব্যাসার্ধের পরিমাপে ভুল হবে,</p><p>Δr = 1.2/100 * r </p><p style="margin-left:0px;">আয়তনের পরিমাপে ভুল হবে,</p><p>ΔV = 4/3π(Δr)³ </p><p>ΔV = 4/3π(1.2/100 * r)³ </p><p>ΔV = 0.036r³ </p><p style="margin-left:0px;">আয়তনের শতকরা ভুল হবে,</p><p>%ΔV = ΔV/V * 100 </p><p>%ΔV = 0.036r³/V * 100 </p><p>%ΔV = 0.036r³/(4/3πr³) * 100 </p><p>%ΔV = 0.036 * 3/4 * 100 </p><p>%ΔV = 3.60% </p><p style="margin-left:0px;">অতএব, উত্তর হল <strong>3.60%</strong>।</p>
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