A metal rod cools at the rate of 4^ C/ min when its temperature is 90^ C and the rate of 1^ C/ min when its temperature is 30^ C. The temperature of the surrounding is

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Correct answer: C. (c) Let the surrounding temperature is T_0. By Newton's law of cooling, dθ dt=K( - _0) 4=K(90-T_0)….(i) 1=K(30-T_0)….(ii) Divide equation (i) by equation (ii) 4= 90-T_030-T_0 ;120-4T_0=90-T_0 3T_0=30 ;T_0=10^ C

$A metal rod cools at the rate of 4^{\circ} C / min when its temperature is 90^{\circ} C and the rate of 1^{\circ} C / min when its temperature is 30^{\circ} C . The temperature of the surrounding is$

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