A body takes 4 minutes to cool from 100- Cto 70- C- To cool from 70- C to 40- C it will take room temperature is 15- C A- 5 minutesB- 4 minutesC- 7 minutesD- 6 minutes

The correct option is D 6 minutes θ_1 θ_2/t = K ( θ_1 + θ_2/2 θ_0 ) ∴100 70 /4 = K ( 100 + 70 /2 15 ) = 60 K ⇒ K = 1/8 Again 70 40 /t = 1/8( 70 + 4 ...

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Standard XII

Mathematics

Focii of Ellipse

A body takes ...

Question

A body takes 4 minutes to cool from $100^{\circ} C$

to $70^{\circ} C$ . To cool from $70^{\circ} C$ to $40^{\circ} C$ it will take

( room temperature is $15^{\circ} C$ )

7 minutes

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6 minutes

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5 minutes

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4 minutes

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Solution

The correct option is B
6 minutes

$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$

$∴ \frac{100 - 70}{4} = K (\frac{100 + 70}{2} - 15) = 60 K \Rightarrow K = \frac{1}{8}$

Again $\frac{70 - 40}{t} = \frac{1}{8} (\frac{70 + 40}{2} - 15) = 5 \Rightarrow t = 6 m i n .$

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A body takes 4 minutes to cool from 100∘C to 70∘C . To cool from 70∘C to 40∘C it will take ( room temperature is 15∘C )

The correct option is B 6 minutes θ1−θ2t=K(θ1+θ22−θ0) ∴100−704=K(100+702−15)=60K⇒K=18 Again 70−40t=18(70+402−15)=5⇒t=6min.

A body takes 4 minutes to cool from $100^{\circ} C$

to $70^{\circ} C$ . To cool from $70^{\circ} C$ to $40^{\circ} C$ it will take

( room temperature is $15^{\circ} C$ )

The correct option is B
6 minutes

$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$

$∴ \frac{100 - 70}{4} = K (\frac{100 + 70}{2} - 15) = 60 K \Rightarrow K = \frac{1}{8}$

Again $\frac{70 - 40}{t} = \frac{1}{8} (\frac{70 + 40}{2} - 15) = 5 \Rightarrow t = 6 m i n .$