A specific gravity bottle contains liquid of masses 21 gms and 20 gms at 50 $^{0}$ C and 100 $^{0}$ C respectively. The coefficient of apparent expansion of liquid is:

10^{- 3 / 0} C

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10^{- 4 / 0} C

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10^{- 5 / 0} C

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10^{- 6 / 0} C

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Solution

The correct option is A $10^{- 3 / 0} C$
$γ_{a} = \frac{m_{2} - m_{1}}{m_{1} * (t_{2} - t_{1})}$ .....(1)
Given,
$m_{1} = 21$
$m_{2} = 20$
$t_{1} = 50$
$t_{2} = 100$
Substituting the values in eqn(1)
$γ_{a} = \frac{21 - 20}{20 * (100 - 50)} = 10 \times 10^{- 4} = 10^{- 3}$

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A specific gravity bottle contains liquid of masses 21 gms and 20 gms at 500C and 1000C respectively. The coefficient of apparent expansion of liquid is:

The correct option is A 10−3/0Cγa=m2−m1m1∗(t2−t1) .....(1)Given,m1=21m2=20t1=50t2=100Substituting the values in eqn(1) γa=21−2020∗(100−50)=10×10−4=10−3

A specific gravity bottle contains liquid of masses 21 gms and 20 gms at 50 $^{0}$ C and 100 $^{0}$ C respectively. The coefficient of apparent expansion of liquid is:

The correct option is A $10^{- 3 / 0} C$
$γ_{a} = \frac{m_{2} - m_{1}}{m_{1} * (t_{2} - t_{1})}$ .....(1)
Given,
$m_{1} = 21$
$m_{2} = 20$
$t_{1} = 50$
$t_{2} = 100$
Substituting the values in eqn(1)
$γ_{a} = \frac{21 - 20}{20 * (100 - 50)} = 10 \times 10^{- 4} = 10^{- 3}$