A spherical ball of ice melts in such a way that the rate at which its

Question 27 A spherical ball of ice melts in such a way that the rate at which its volume decreases at any instant is directly proportional to its surface area. Prove that the radius of the ice ball decreases at a constant rate.Since Ball of ice is spherical Let r be the radius of bubble & V be the volume of bubble Given that Rate at which its volume decreases at any instant is directly proportional to its surface area. So, we write 𝒅𝑽/𝒅𝒕 ∝ S 𝒅𝑽/𝒅𝒕=−𝒌𝑺 Negative because Volume is decreasing And k is some constant Finding Surface Area and Rate of Change of Volume Surface Area of Sphere Surface Area of Sphere = S = 𝟒𝝅𝒓^𝟐 Rate of Change of Volume of Sphere We need to calculate the rate the Volume of the ice is decreasing i.e. we need to calculate 𝒅𝑽/𝒅𝒕 Volume of sphere = V = 𝟒/𝟑 πr3 Now, 𝑑𝑉/𝑑𝑡 = 𝑑(4/3 𝜋𝑟3)/𝑑𝑡 𝑑𝑉/𝑑𝑡 = 4/3 π (𝑑 (𝑟3))/𝑑𝑡 𝑑𝑉/𝑑𝑡 = 4/3 π (𝑑 (𝑟3))/𝑑𝑡 𝑑𝑉/𝑑𝑡 = 4/3 π . (𝑑(𝑟3))/𝑑𝑡 × 𝒅𝒓/𝒅𝒓 𝑑𝑉/𝑑𝑡 = 4/3 π . (𝑑(𝑟3))/𝑑𝑡 × 𝑑𝑟/𝑑𝑡 𝑑𝑉/𝑑𝑡 = 4/3 π .3r2 . 𝑑𝑟/𝑑𝑡 𝒅𝑽/𝒅𝒕 = 𝟒𝝅𝒓^𝟐 × 𝒅𝒓/𝒅𝒕 Putting 𝒅𝑽/𝒅𝒕=−𝒌𝑺 in (3) −𝒌𝑺 = 4𝜋𝑟^2 × 𝑑𝑟/𝑑𝑡 Putting S = 𝟒𝝅𝒓^𝟐 from (2) −𝒌 × 𝟒𝝅𝒓^𝟐 = 4𝜋𝑟^2 × 𝑑𝑟/𝑑𝑡 −𝑘 ×(4𝜋𝑟^2)/(4𝜋𝑟^2 ) = 𝑑𝑟/𝑑𝑡 −𝑘 = 𝑑𝑟/𝑑𝑡 𝒅𝒓/𝒅𝒕=−𝒌 Thus, radius of the ice-ball decreases at a constant rate Hence proved

Question 27 - CBSE Class 12 Sample Paper for 2026 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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