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एक बीकर में एक द्रव का ताप समय t पर `theta ` है तथा वातावरण का ताप `theta_0` है तब न्यूटन के शीतलन नियम के अनुसार `log_e(theta-theta_0)` तथा t के बीच सही ग्राफ है:
A. image
B. image
C. image
D. image

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Correct Answer - a
ताप के गिरने की दर ,
`" " (d theta )/(dt) =-k(theta -theta_0)`
`" " int (d theta )/(theta -theta _0)=-kint dt `
` " " log_e(theta -theta_0)=-kt+c`
यह ग्राफ एक सरल रेखा है जिसकी प्रवणता ऋणात्मक है।अतः विकल्प (a ) सही है।

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