A specific gravity bottle contains 51 gm at 20 $^{0}$ C. When it is heated to 100 $^{0}$ C, one gram of the liquid over flows. The co-efficient of apparent expansion of the liquid is:

4 \times 10^{- 5 / 0} C

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25 \times 10^{- 4 / 0} C

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25 \times 10^{- 5 / 0} C

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8 \times 10^{- 5 / 0} C

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Solution

The correct option is D $25 \times 10^{- 5 / 0} C$
$γ_{a} = \frac{m_{2} - m_{1}}{m_{1} * (t_{2} - t_{1})}$ .....(1)
Given,
$m_{1} = 51$
$t_{1} = 20$
$t_{2} = 100$
$m_{2} - m_{1} = 1$
Substituting the values in eqn(1)
$γ_{a} = \frac{1}{51 * (100 - 20)} = 25 \times 10^{- 5}$

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Q. A specific gravity bottle contains liquid of masses 21 gms and 20 gms at 50

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A specific gravity bottle contains 51 gm at 200C. When it is heated to 1000C, one gram of the liquid over flows. The co-efficient of apparent expansion of the liquid is:

The correct option is D 25×10−5/0Cγa=m2−m1m1∗(t2−t1) .....(1)Given,m1=51t1=20t2=100m2−m1=1Substituting the values in eqn(1) γa=151∗(100−20)=25×10−5

A specific gravity bottle contains 51 gm at 20 $^{0}$ C. When it is heated to 100 $^{0}$ C, one gram of the liquid over flows. The co-efficient of apparent expansion of the liquid is:

The correct option is D $25 \times 10^{- 5 / 0} C$
$γ_{a} = \frac{m_{2} - m_{1}}{m_{1} * (t_{2} - t_{1})}$ .....(1)
Given,
$m_{1} = 51$
$t_{1} = 20$
$t_{2} = 100$
$m_{2} - m_{1} = 1$
Substituting the values in eqn(1)
$γ_{a} = \frac{1}{51 * (100 - 20)} = 25 \times 10^{- 5}$